1 Paul B. Andersen  Re: The Absurd Assertions of the SR Experts  Thursday 11 january 2001 06:17:23 
2 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  Friday 12 january 02:29:44 
3 Paul B. Andersen  Re: The Absurd Assertions of the SR Experts  Friday 12 january 15:55:26 
4 Russell Blackadar  Re: The Absurd Assertions of the SR Experts  Friday 12 january 17:24:02 
5 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  Saterday 13 january 06:38:08 
6 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  Saterday 13 january 2001 06:38:09 
7 Russell Blackadar  Re: The Absurd Assertions of the SR Experts  Sunday 14 january 2001 16:30:10 
8 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  Tuesday 16 january 2001 07:24:03 
9 Russell Blackadar  Re: The Absurd Assertions of the SR Experts  Tuesday 16 january 2001 09:34:06 
10 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  Wensday 17 january 2001 07:39:37 
11 Russell Blackadar  Re: The Absurd Assertions of the SR Experts  Wensday 17 january 2001 13:18:56 
12 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  Thursday 18 january 2001 14:06:00 
13 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  woensdag 24 januari 2001 17:29 
14 Russell Blackadar  Re: The Absurd Assertions of the SR Experts  woensdag 24 januari 2001 20:37 
15 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  vrijdag 26 januari 2001 18:14 
16 Russell Blackadar  Re: The Absurd Assertions of the SR Experts  vrijdag 26 januari 2001 19:18 
17 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  vrijdag 26 januari 2001 23:12 
18 Russell Blackadar  Re: The Absurd Assertions of the SR Experts  zaterdag 27 januari 2001 0:30 
19 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  woensdag 21 februari 2001 14:05 
20 Russell Blackadar  Re: The Absurd Assertions of the SR Experts  woensdag 21 februari 2001 21:45 
21 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  donderdag 22 februari 2001 21:03 
22 Russell Blackadar  Re: The Absurd Assertions of the SR Experts  donderdag 22 februari 2001 22:19 
23 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  maandag 26 februari 2001 11:13 
24 Russell Blackadar  Re: The Absurd Assertions of the SR Experts  dinsdag 27 februari 2001 5:40 
25 Russell Blackadar  Re: The Absurd Assertions of the SR Experts  dinsdag 27 februari 2001 6:55 
26 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  zondag 4 maart 2001 11:51 
27 Russell Blackadar  Re: The Absurd Assertions of the SR Experts  dinsdag 6 maart 2001 4:38 
28 Nicolaas Vroom  Re: The Absurd Assertions of the SR Experts  vrijdag 23 maart 2001 12:17 
Nicolaas Vroom wrote:
> 
Paul B. Andersen 
> >  Nicolaas Vroom wrote: 
> > > 
Paul B. Andersen 
> > 
Simple or not, that's how it is according to SR. 
> > > 
There are two issues: The length of the moving trains. The rate of the moving clocks In the following sketch I have added an extra train which v=0 in track frame v> = v+ and <v=v v+ = v in track frame 
> > > > 
 1. The flash is emitted. Let's call the time t1.   A0 v =0 F0   L0   A' v > F'   L'  *   L  F < v AQuestion: Q1. Am I right when assuming that A' and F are adjacent when the light is emitted? 
> > > 
This means that L' = L and both are equal to L=L0*SQR(1vē/cē) Is that true ? 
> > 
Yes. 
> > > 
Length contraction means that length is contracted. 
> > 
When measured in the track frame, yes. 
> > > 
Now suppose I place the whole setup "track and trains" on the moving train with v> = v+ This setup also has a track frame in rest. Everything on this setup I identify with new. IMO the new train A'F' will also contract However I doubt if the new train FA will contract IMO FA will expand for small values of <v new. IMO the new values for L and L' are not equal. 
> > 
Right. As measured in the "new track frame" where the two trains are moving with different speed, their lenths will be different. I fail to see your point, though. 
> 
IMO when they move there with the same speed measured in this new track frame (How ?  See below in "How do you perform your experiment") the length of both trains will not be identical.   G<B B'>G'   F < A A' > F'  track frame on EarthThe above sketch shows One train FA moving to left in Earth track frame One frain A'F' moving to right Length has decreased of both On train A'F' there are two other trains: Train B'G' moving to the right  This length will decrease always Train BG moving to the left  This length can also increase. At same speed in track frame of A'F' The length of GB and of B'G' are different. 
We still agree on this. So? I still don't see your point.
> > > 

Suppose L and L' are equal. 
> > 
Meaning that we revert to my thought experiment. (Your third train is stationary.) If I get you right, you are talking about what the clocks on trains will show. My thought experiment is not designed to illustrate that point, though. 
> > > 
Above Paul has written: 
> > > >  The flash is emitted. Let's call the time t1. 
> > >  In the above setup there are six observers Each observer has a clock. At least the following equations are true. t1(A0)=t1(F0): t1(A)=t1(F) :t1(A')=t1(F') 
> 
I think that the following three statements are not clear: You can also write them down in a more general form t(A0)=t(F0) (1) t(A)=t(F) (2) : t(A')=t(F') (3)1 means that both observers in track frame in train which stands always measure the same time. 2 means that both observers in train going < always measure the same time 3 means that both observers in train going > always measure the same time Also read below. 
> > > 
The moment of collision/explosion
can be such that all the observers F0, F' and A
are on one line 
> > 
Right. This can be true by definition. 
> > > 
That means that then also t1(A') = t1(F) 
> > 
No. Here is your error. You are forgetting relativity of simultaneity. 
> > > 
I can check this by comparing times when poition of moving observers is identical. 
> > 
Right.
We have two events: 
> 
How do you perform your experiment ? Consider the following: 0 0 1 2 3 4 5 4 3 2 1 0 0 A'F' FALets us see how it works. We are in for a special tread. t0 t=0 in track frame At t0 the train FA is at the right side of the track at the points 00. Length of train is L0 The observer F has a clock with t=0 The observer A has a clock with t=0
At t0 the train F'A' is at the left side of the track
at the points 00. Length of train is L0
t1 t=1 in track frame
Front of the train FA has reached point 1
Length of train is L = L0*SQR(1v^2/c^2)
The observer F has a clock with t=0.9
Front of the train F'A' has reached point 1
Length of train is L = L0*SQR(1v^2/c^2)
The observer F' has a clock with t=0.9
t2 t=2 in track frame
Front of the train FA has reached point 2
The observer F has a clock with t=1.8 Front of the train F'A' has reached point 2 The observer F' has a clock with t=1.8
t5 t=5 in track frame
Front of the train FA has reached point 5
Front of the train F'A' has also reached point 5 t6 t=6 in track frame 6 5 6 FAA'F'Front of the train FA has reached point 6 The observer F has a clock with t=5.5 The observer A has a clock with t=5.5 Observer A has just passed point 5 Observer A' has just passed point 5
Front of the train F'A' has also reached point 5 The train STOPS At t6 t=6 (The ends will touch again  funny) All observers A,F,A',F' have the same time on their clocks t5.5 t=5.5 
So? What is your point? Your two clocks on each train are not synchronous to each other in the train frames while the trains are moving.
> 
(Above this is called t1)
The trains are parallel
A' is at position with F
F' is at position with A
explosion  Ligt flash is emitted 
Where t(A) and t(F') are the reading of the clocks at the event E1: A and F' adjacent.
Still true by definition. That is, the clocks are set so that this is true.
>  IMO: t(A') = t(F) 
Where t(A') and t(F) are the reading of the clocks at the event E2: A' and F adjacent?
You seem to have missed the point completely. What the clocks will read obviously depend on how they are set. And how is that? If both the clocks on each train are to show "train time", then they must be synchronized to each other. That is, the A' clock must simultaneously show the same as F' clock _in the A'F' train frame_, and the F clock must simultaneously show the same as A clock _in the AF train frame_.
If the clocks are synchronized thus, then what I wrote above is true.
> > > > 
 2. The flash reaches A'. Let's call the time t2.   A0 v =0 F0   L0   A' v > F'   L'  *   L  F < v A 
> > > 
At this moment t2(A') = t2(F) t2(A') is time measured by A' in train 
> > > > 
 3. The flash reaches F. Let's call the time t3.   A0 v =0 F0   L0   A' v > F'   L'  *   L  F < v A 
> > > 
At this moment t3(A') = t3(F) How is it possible that t3(F)=t2(A') ? 
> > 
The clocks on the trains run at the same rate as observed in the track frame. But as shown above, the clocks at A' and F do NOT show the same when they are adjacent; the A' clock reads more than the F clock. The extra time _in the track frame_ that the light uses to go from A' to F, is exactly the time needed for the F clock to show the same as the A' clock when they are hit. 
Paul
Paul B. Andersen
>  Nicolaas Vroom wrote: 
> > 
Paul B. Andersen 
> > >  Nicolaas Vroom wrote: 
> > > > 
Paul B. Andersen 
> > > 
Simple or not, that's how it is according to SR. 
> > > > 
There are two issues: The length of the moving trains. The rate of the moving clocks In the following sketch I have added an extra train which v=0 in track frame v> = v+ and <v=v v+ = v in track frame 
> > > > > 
 1. The flash is emitted. Let's call the time t1.   A0 v =0 F0   L0   A' v > F'   L'  *   L  F < v AQuestion: Q1. Am I right when assuming that A' and F are adjacent when the light is emitted? 
> > > > 
This means that L' = L and both are equal to L=L0*SQR(1vē/cē) Is that true ? 
> > > 
Yes. 
> > > > 
Length contraction means that length is contracted. 
> > > 
When measured in the track frame, yes. 
> > > > 
Now suppose I place the whole setup "track and trains" on the moving train with v> = v+ This setup also has a track frame in rest. Everything on this setup I identify with new. IMO the new train A'F' will also contract However I doubt if the new train FA will contract IMO FA will expand for small values of <v new. IMO the new values for L and L' are not equal. 
> > > 
Right. As measured in the "new track frame" where the two trains are moving with different speed, their lenths will be different. I fail to see your point, though. 
> > 
IMO when they move there with the same speed measured in this new track frame (How ?  See below in "How do you perform your experiment") the length of both trains will not be identical.   G<B B'>G'   F < A A' > F'  track frame on EarthThe above sketch shows One train FA moving to left in Earth track frame One frain A'F' moving to right Length has decreased of both On train A'F' there are two other trains: Train B'G' moving to the right  This length will decrease always Train BG moving to the left  This length can also increase. At same speed in track frame of A'F' The length of GB and of B'G' are different. 
> 
We still agree on this. So? I still don't see your point. 
We agreed that the length of the two trains in the old track frame on Earth (in rest) are equal i.e
L = L' = L0 * sqr(1v^2/c^2)
We also agreed that on the "new track frame" of A'F' the length of the two trains is not equal i.e.
L <> L' <> L0 * sqr(1v^2/c^2) (2)
It should be remarked that the values of L0 in both frames are different. This "new track frame" is also at rest. In this "new track frame" the rules are different compared with the "old track frame"
The only way out is that (1) is not true in the "old track frame" i.e. L<>L' in both frames.
IMO the following : L = L' = L0 * sqr(1v^2/c^2) is only true in one frame: A frame in absolute rest. i.e. a frame that itself is not already contracted.
> > > > 

Suppose L and L' are equal. 
> > > 
> > 
How do you perform your experiment ? Consider the following: 0 0 1 2 3 4 5 4 3 2 1 0 0 A'F' FA Lets us see how it works. We are in for a special tread.
t0 t=0 in track frame
At t0 the train FA is at the right side of the track
at the points 00. Length of train is L0
At t0 the train F'A' is at the left side of the track
at the points 00. Length of train is L0
t1 t=1 in track frame
Front of the train FA has reached point 1
Front of the train F'A' has reached point 1
t2 t=2 in track frame
Front of the train FA has reached point 2
Front of the train F'A' has reached point 2
t5 t=5 in track frame
Front of the train FA has reached point 5
Front of the train F'A' has also reached point 5 t6 t=6 in track frame 6 5 6 FAA'F'Front of the train FA has reached point 6 The observer F has a clock with t=5.5 The observer A has a clock with t=5.5 Observer A has just passed point 5 Observer A' has just passed point 5 Front of the train F'A' has also reached point 5 The observer F' has a clock with t=5.5 The observer A' has a clock with t=5.5 The train STOPS At t6 t=6 (The ends will touch again  funny) All observers A,F,A',F' have the same time on their clocks t5.5 t=5.5 
> 
So? What is your point? Your two clocks on each train are not synchronous to each other in the train frames while the trains are moving. 
> > 
(Above this is called t1)
The trains are parallel
A' is at position with F
F' is at position with A
explosion  Ligt flash is emitted 
> 
Where t(A) and t(F') are the reading of the clocks at the event E1: A and F' adjacent. Still true by definition. That is, the clocks are set so that this is true. 
> > 
IMO: t(A') = t(F) 
> 
Where t(A') and t(F) are the reading of the clocks at the event E2: A' and F adjacent? You seem to have missed the point completely. What the clocks will read obviously depend on how they are set. And how is that? If both the clocks on each train are to show "train time", then they must be synchronized to each other. That is, the A' clock must simultaneously show the same as F' clock _in the A'F' train frame_, and the F clock must simultaneously show the same as A clock _in the AF train frame_. If the clocks are synchronized thus, then what I wrote above is true. 
What I have demonstrated above is the following a complete experiment which can be tested. a) In the starting conditions the time of the four observers A F A' F' is identical b) When the train is parallel and when the explosion happens at F' and A the times of t(F') = t(A) and t(A')=t(F) are equal. c) The times when A' and F resp see the light is different. d) In the end conditions the time of the four observers again A F A' F' is identical
You do not agree with c. The times should be identical. You only partly agree with b t(F') = t(A) but t(A')<>t(F)
How do I have to modify the experiment such that what you are saying is correct. I think we have to modify the start conditions however that is tricky. I do not like to change the times on the clock during the trip.
I get the impression from you that there is the following problem Consider a train at rest On this train there are two clocks at A and F both clock are synchronised The train starts to move Accordingly to you: those two clocks are now not any more in sync in the train frame IMO when the train stops both clocks will show the same time.
Why is it important that the two clocks should be in sync in the train frame ? I Expect you have to change the time of the clocks to get them in sync in the train frame. Of course when you do the results of my demonstrations will change and your statement that the time that A' and F see the explosion could become equal
https://www.nicvroom.be/length3.htm
> 
Paul B. Andersen 
> >  Nicolaas Vroom wrote: 
> > > 
Paul B. Andersen 
> > > >  Nicolaas Vroom wrote: 
> > > > > 
Paul B. Andersen 
> > > > 
Simple or not, that's how it is according to SR. 
> > > > > 
There are two issues: The length of the moving trains. The rate of the moving clocks In the following sketch I have added an extra train which v=0 in track frame v> = v+ and <v=v v+ = v in track frame 
> > > > > > 
 1. The flash is emitted. Let's call the time t1.
 A0 v =0 F0   L0   A' v > F'   L'  *   L  F < v AQuestion: Q1. Am I right when assuming that A' and F are adjacent when the light is emitted? 
> > > > > 
This means that L' = L and both are equal to L=L0*SQR(1vē/cē) Is that true ? 
> > > > 
Yes. 
> > > > > 
Length contraction means that length is contracted. 
> > > > 
When measured in the track frame, yes. 
> > > > > 
Now suppose I place the whole setup "track and trains" on the moving train with v> = v+ This setup also has a track frame in rest. Everything on this setup I identify with new. IMO the new train A'F' will also contract However I doubt if the new train FA will contract IMO FA will expand for small values of <v new. IMO the new values for L and L' are not equal. 
> > > > 
Right. As measured in the "new track frame" where the two trains are moving with different speed, their lenths will be different. I fail to see your point, though. 
> > > 
IMO when they move there with the same speed measured in this new track frame (How ?  See below in "How do you perform your experiment") the length of both trains will not be identical.   G<B B'>G'   F < A A' > F'  track frame on EarthThe above sketch shows One train FA moving to left in Earth track frame One frain A'F' moving to right Length has decreased of both On train A'F' there are two other trains: Train B'G' moving to the right  This length will decrease always Train BG moving to the left  This length can also increase. At same speed in track frame of A'F' The length of GB and of B'G' are different. 
> > 
We still agree on this. So? I still don't see your point. 
> 
We agreed that the length of the two trains in the old track frame on Earth (in rest) are equal i.e L = L' = L0 * sqr(1v^2/c^2) 
Right.
> 
We also agreed that on the "new track frame" of A'F'
the length of the two trains is not equal i.e.
L <> L' <> L0 * sqr(1v^2/c^2) (2) 
Right.
>  It should be remarked that the values of L0 in both frames are different. This "new track frame" is also at rest. 
Anything is at rest relative to itself.
>  In this "new track frame" the rules are different compared with the "old track frame" 
No. It is the speeds of the moving trains that are different, not the rules.
>  The only way out is that (1) is not true in the "old track frame" i.e. L<>L' in both frames. 
Can't imagine why you think so.
> 
IMO the following : L = L' = L0 * sqr(1v^2/c^2) is only true in one frame: A frame in absolute rest. 
Of course L = L' is true only in one frame, namely in the frame in which the speeds of the trains are equal. There is nothing absolute about that.
>  i.e. a frame that itself is not already contracted. 
Uh? Nothing is contracted in the frame in which it is stationary. Not even frames.
> > > > > 
Suppose L and L' are equal. 
> > > > 
> > > 
How do you perform your experiment ? Consider the following: 0 0 1 2 3 4 5 4 3 2 1 0 0 A'F' FALets us see how it works. We are in for a special tread.
t0 t=0 in track frame
At t0 the train FA is at the right side of the track
at the points 00. Length of train is L0
At t0 the train F'A' is at the left side of the track
at the points 00. Length of train is L0
t1 t=1 in track frame
Front of the train FA has reached point 1
Front of the train F'A' has reached point 1
t2 t=2 in track frame
Front of the train FA has reached point 2
Front of the train F'A' has reached point 2
t5 t=5 in track frame
Front of the train FA has reached point 5
Front of the train F'A' has also reached point 5 t6 t=6 in track frame 6 5 6 FAA'F'Front of the train FA has reached point 6 The observer F has a clock with t=5.5 The observer A has a clock with t=5.5 Observer A has just passed point 5 Observer A' has just passed point 5
Front of the train F'A' has also reached point 5 The train STOPS At t6 t=6 (The ends will touch again  funny) All observers A,F,A',F' have the same time on their clocks t5.5 t=5.5 
> > 
So? What is your point? Your two clocks on each train are not synchronous to each other in the train frames while the trains are moving. 
> > > 
(Above this is called t1)
The trains are parallel
A' is at position with F
F' is at position with A
explosion  Ligt flash is emitted 
> > 
Where t(A) and t(F') are the reading of the clocks at the event E1: A and F' adjacent. Still true by definition. That is, the clocks are set so that this is true. 
> > > 
IMO: t(A') = t(F) 
> > 
Where t(A') and t(F) are the reading of the clocks at the event E2: A' and F adjacent? You seem to have missed the point completely. What the clocks will read obviously depend on how they are set. And how is that? If both the clocks on each train are to show "train time", then they must be synchronized to each other. That is, the A' clock must simultaneously show the same as F' clock _in the A'F' train frame_, and the F clock must simultaneously show the same as A clock _in the AF train frame_. If the clocks are synchronized thus, then what I wrote above is true. 
> 
What I have demonstrated above is the following a complete experiment which can be tested. 
If you mean a real experiment, hardly. Trains move to slow and are to short for the time differences to be measurable. Possible in principle yes, but the practical difficulties make the experiment unfeasible in the real world.
> 
a) In the starting conditions the time of the
four observers A F A' F' is identical b) When the train is parallel and when the explosion happens at F' and A the times of t(F') = t(A) and t(A')=t(F) are equal. c) The times when A' and F resp see the light is different. d) In the end conditions the time of the four observers again A F A' F' is identical You do not agree with c. The times should be identical. You only partly agree with b t(F') = t(A) but t(A')<>t(F) How do I have to modify the experiment such that what you are saying is correct. 
You must synchronize the clocks on the trains in the train frames.
>  I think we have to modify the start conditions however that is tricky. 
It's simple in a thought experiment. Just state: the clocks on the trains are synchronous. And voila  they are.
>  I do not like to change the times on the clock during the trip. 
You don't have to. Don't start and stop the trains. Just let them move with constant speed. You are free to state: Let one clock on each train show the same as they are adjacent. Let the other clock on each train be synchronized to the first.
>  I get the impression from you that there is the following problem Consider a train at rest On this train there are two clocks at A and F both clock are synchronised The train starts to move Accordingly to you: those two clocks are now not any more in sync in the train frame 
Right.
>  IMO when the train stops both clocks will show the same time. 
In the track frame, right.
>  Why is it important that the two clocks should be in sync in the train frame ? 
Because you wanted to use them to measure the transit time of the light in the train. If you measure a time interval with two clocks, the clocks must be synchronous.
>  I Expect you have to change the time of the clocks to get them in sync in the train frame. 
Yes.
>  Of course when you do the results of my demonstrations will change and your statement that the time that A' and F see the explosion could become equal 
Indeed.
Paul
> 
Nicolaas Vroom wrote: 
> > 
Paul B. Andersen 
> > >  Nicolaas Vroom wrote: 
[snip]
> > > > 
See below in "How do you perform your experiment")
the length of both trains will not be identical.
  G<B B'>G'   F < A A' > F'  track frame on EarthThe above sketch shows One train FA moving to left in Earth track frame One frain A'F' moving to right Length has decreased of both On train A'F' there are two other trains: Train B'G' moving to the right  This length will decrease always Train BG moving to the left  This length can also increase. At same speed in track frame of A'F' The length of GB and of B'G' are different. 
> > > 
We still agree on this. 
Paul, I think he means that GB and B'G' are different in the frame of A'F', even though they have the same speed in that frame. That's wrong, of course!
> > >  So? I still don't see your point. 
> > 
We agreed that the length of the two trains in the old track frame on Earth (in rest) are equal i.e L = L' = L0 * sqr(1v^2/c^2) 
> 
Right. 
Agreed, here I think he means AF and A'F' in the Earth frame.
> 
> > 
We also agreed that on the "new track frame" of A'F' the length of the two trains is not equal i.e. L <> L' <> L0 * sqr(1v^2/c^2) (2) 
> 
Right. 
No, I think here he is talking about BG and B'G' in the A'F' frame, because:
> 
> > 
It should be remarked that the values of L0 in both frames are different. 
This statement could only make sense if he is talking about two different pairs of trains. Of course he has said this in a confusing way because he identifies the train pairs (and their L0's) by frame, which is irrelevant, instead of by their letters.
> >  This "new track frame" is also at rest. 
> 
Anything is at rest relative to itself. 
> > 
In this "new track frame" the rules are different compared with the "old track frame" 
> 
No. It is the speeds of the moving trains that are different, not the rules. 
Or more probably, the answer is No because of the mistake he made above. BG and B'G' have the same speed in that frame, also they have the same length in that frame (contrary to what he said) so we reach the right conclusion: the rules are the same.
Later he writes:
> > > > 
At t0 the train F'A' is at the left side of the track
at the points 00. Length of train is L0 The observer F' has a clock with t=0 The observer A' has a clock with t=0 At t0, t=0 the trains starts to move towards the point 5 
Note in passing, he accelerates the train here, but does not say how the acceleration is to be done along the entire length of the train. This step is therefore not well defined. See:
http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html
Paul B. Andersen
>  Nicolaas Vroom wrote: 
> > 
We agreed that the length of the two trains in the old track frame on Earth (in rest) are equal i.e L = L' = L0 * sqr(1v^2/c^2) 
> 
Right. 
> > 
We also agreed that on the "new track frame" of A'F' the length of the two trains is not equal i.e. L <> L' <> L0 * sqr(1v^2/c^2) (2) 
> 
Right. 
> > 
It should be remarked that the values of L0 in both frames are different. This "new track frame" is also at rest. 
> 
Anything is at rest relative to itself. 
> > 
In this "new track frame" the rules are different compared with the "old track frame" 
> 
No. It is the speeds of the moving trains that are different, not the rules. 
> > 
The only way out is that (1) is not true in the "old track frame" i.e. L<>L' in both frames. 
> 
Can't imagine why you think so. 
> > 
IMO the following : L = L' = L0 * sqr(1v^2/c^2) is only true in one frame: A frame in absolute rest. 
> 
Of course L = L' is true only in one frame, namely in the frame in which the speeds of the trains are equal. There is nothing absolute about that. 
The question is is that the new track frame Or the old track frame Or some other frame
> >  i.e. a frame that itself is not already contracted. 
> 
Uh? Nothing is contracted in the frame in which it is stationary. Not even frames. 
The x coordinate which is something "abstract" is not contracted. The rods and trains (even with v=0) in the new track frame are contracted.
See also my reply to Russell Blackadar
> > > > > > 
Suppose L and L' are equal. 
> > > > > 
> > 
What I have demonstrated above is the following a complete experiment which can be tested. 
> 
If you mean a real experiment, hardly. Trains move to slow and are to short for the time differences to be measurable. Possible in principle yes, but the practical difficulties make the experiment unfeasible in the real world. 
> > 
a) In the starting conditions the time of the four observers A F A' F' is identical b) When the train is parallel and when the explosion happens at F' and A the times of t(F') = t(A) and t(A')=t(F) are equal. c) The times when A' and F resp see the light is different. d) In the end conditions the time of the four observers again A F A' F' is identical You do not agree with c. The times should be identical. You only partly agree with b t(F') = t(A) but t(A')<>t(F) How do I have to modify the experiment such that what you are saying is correct. 
> 
You must synchronize the clocks on the trains in the train frames. 
> > 
I think we have to modify the start conditions however that is tricky. 
> 
It's simple in a thought experiment. Just state: the clocks on the trains are synchronous. And voila  they are. 
I do not like that. In the way I have described the experiment is the following
In the beginning (v=0) I have a set of track frame clocks and a set of clocks which are on the train all those clocks are synchronous At the end of the experiment (again v=0) All the clocks are still synchronous. All the track frame clocks show the same time ta All the train frame clocks show the same time tb ta<>tb
If what you are suggestion is true i.e. you have to synchronise (adjust) SOME of the train frame clocks during the trip when v<>0 then at the end of the experiment (v=0) not all the train clocks show the same time.
https://www.nicvroom.be/length3.htm
Russell Blackadar
>  "Paul B. Andersen" wrote: 
> > 
Nicolaas Vroom wrote: 
> > > 
Paul B. Andersen 
> > > >  Nicolaas Vroom wrote: 
> 
[snip] 
> > > > > 
See below in "How do you perform your experiment") the length of both trains will not be identical.   G<B B'>G'   F < A A' > F'  track frame on EarthThe above sketch shows One train FA moving to left in Earth track frame One frain A'F' moving to right Length has decreased of both On train A'F' there are two other trains: Train B'G' moving to the right  This length will decrease always Train BG moving to the left  This length can also increase. At same speed in track frame of A'F' The length of GB and of B'G' are different. 
> > > > 
We still agree on this. 
> 
Paul, I think he means that GB and B'G' are different in the frame of A'F', even though they have the same speed in that frame. That's wrong, of course! 
> > > > 
So? I still don't see your point. 
> > > 
We agreed that the length of the two trains in the old track frame on Earth (in rest) are equal i.e L = L' = L0 * sqr(1v^2/c^2) 
> > 
Right. 
> 
Agreed, here I think he means AF and A'F' in the Earth frame. 
Exactly
> > 
> > > 
We also agreed that on the "new track frame" of A'F' the length of the two trains is not equal i.e. L <> L' <> L0 * sqr(1v^2/c^2) (2) 
> > 
Right. 
> 
No, I think here he is talking about BG and B'G' in the A'F' frame, because: 
Excalty I was expecting from Paul the Answer No
> > 
> > > 
It should be remarked that the values of L0 in both frames are different. 
> 
This statement could only make sense if he is talking about two different pairs of trains. Of course he has said this in a confusing way because he identifies the train pairs (and their L0's) by frame, which is irrelevant, instead of by their letters. 
What I want to point out is that in both frames I want to perform exactly the same experiment using only ! tools in their own resp. frames.
> > >  This "new track frame" is also at rest. 
> > 
Anything is at rest relative to itself. 
> > > 
In this "new track frame" the rules are different compared with the "old track frame" 
> > 
No. It is the speeds of the moving trains that are different, not the rules. 
> 
Or more probably, the answer is No because of the mistake he made above. 
>  BG and B'G' have the same speed in that frame, 
That is the basis of the experiments (sub experiment) I do that using rods and clocks in the new rest frame Each sub experiment assumes the same speed (absolute) for BG and B'G'
>  also they have the same length in that frame 
That and only that is the question. How do you know that. What (experimental) prove do you have ?
Ofcourse I can easily write down that they should be equal because of: L = L' = L0 * sqr(1v^2/c^2) But that is no proof.
See https://www.nicvroom.be/length3.htm
The reason why I think that the length are not the same is because the train A'F', then new track, L0, and both trains BG and B'G'whenv=0 are all contracted.
When BG moves to the left you wil see that it length will become larger as L0 in the new track frame. It will even become as large as L0 in the old track frame then with more larger v it will become smaller.
> 
(contrary to what he said) so we reach
the right conclusion: the rules are the same.
Later he writes: 
> > > > > 
At t0 the train F'A' is at the left side of the track
at the points 00. Length of train is L0 
> 
Note in passing, he accelerates the train here, but does not say how the acceleration is to be done along the entire length of the train. 
Each train is only accelarated once at t0 During the whole experiment they keep the same speed. At t=6 the train stops immediate.
The observer in the front can make a log book like:
tn,pn track train t0,p0 t = 0 t = 0 t1,p1 t = 1 t = 0.9 t2,p2 t = 2 t = 1.8 t5,p5 t = 5 t = 4,5 t6,p6 t = 6 t = 5,4For example t5,p5 is time position identifion p5=5*L0 from start point t = 5 is track time. t = 4,5 is train time
As you can see during the trip a = 0 v = constant
> 
This step is therefore not well
defined. See:
http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html 
Interested document.
I do not "like" the sentence "Don't tackle the physical situation directly; instead, imagine some pictures."
IMO the most important part is what real experiments tell us i.e. how the physical world behaves.
> 
Russell Blackadar 
> >  "Paul B. Andersen" wrote: 
> > > 
Nicolaas Vroom wrote: 
> > > > 
Paul B. Andersen 
> > > > >  Nicolaas Vroom wrote: 
> > 
[snip] 
> > > > > > 
See below in "How do you perform your experiment") the length of both trains will not be identical.   G<B B'>G'   F < A A' > F'  track frame on EarthThe above sketch shows One train FA moving to left in Earth track frame One frain A'F' moving to right Length has decreased of both On train A'F' there are two other trains: Train B'G' moving to the right  This length will decrease always Train BG moving to the left  This length can also increase. At same speed in track frame of A'F' The length of GB and of B'G' are different. 
> > > > > 
We still agree on this. 
> > 
Paul, I think he means that GB and B'G' are different in the frame of A'F', even though they have the same speed in that frame. That's wrong, of course! 
> > > > > 
So? I still don't see your point. 
> > > > 
We agreed that the length of the two trains in the old track frame on Earth (in rest) are equal i.e L = L' = L0 * sqr(1v^2/c^2) 
> > > 
Right. 
> > 
Agreed, here I think he means AF and A'F' in the Earth frame. 
> 
Exactly 
> > > 
> > > > 
We also agreed that on the "new track frame" of A'F' the length of the two trains is not equal i.e. L <> L' <> L0 * sqr(1v^2/c^2) (2) 
> > > 
Right. 
> > 
No, I think here he is talking about BG and B'G' in the A'F' frame, because: 
> 
Excalty I was expecting from Paul the Answer No 
> > > 
> > > > 
It should be remarked that the values of L0 in both frames are different. 
> > 
This statement could only make sense if he is talking about two different pairs of trains. Of course he has said this in a confusing way because he identifies the train pairs (and their L0's) by frame, which is irrelevant, instead of by their letters. 
> 
What I want to point out is that in both frames I want to perform exactly the same experiment using only ! tools in their own resp. frames. 
> > > > 
This "new track frame" is also at rest. 
> > > 
Anything is at rest relative to itself. 
> > > > 
In this "new track frame" the rules are different compared with the "old track frame" 
> > > 
No. It is the speeds of the moving trains that are different, not the rules. 
> > 
Or more probably, the answer is No because of the mistake he made above. 
>  If I made a mistake than that is the question. If I am not clear, I except. 
> > 
BG and B'G' have the same speed in that frame, 
> 
That is the basis of the experiments (sub experiment) I do that using rods and clocks in the new rest frame Each sub experiment assumes the same speed (absolute) for BG and B'G' 
> > 
also they have the same length in that frame 
> 
That and only that is the question. How do you know that. What (experimental) prove do you have ? 
Experiments don't prove. But, every experiment that confirms Lorentz invariance is to some degree a confirmation of my statement. The number of such experiments is very large. On the other hand, I doubt that any sufficiently precise experiments have been performed using trains running on tracks mounted on a train.
> 
Ofcourse I can easily write down that they should be equal because of: L = L' = L0 * sqr(1v^2/c^2) But that is no proof. See https://www.nicvroom.be/length3.htm The reason why I think that the length are not the same is because the train A'F', then new track, L0, and both trains BG and B'G'whenv=0 are all contracted. 
No, L0 is not contracted; it's defined as the rest length of the train and is therefore invariant. You are misusing this notation if you say L0 can change depending on frame. I didn't remark on that last time, but it's important. We have to agree on terminology.
> 
When BG moves to the left you wil see that it length will become larger as L0 in the new track frame. 
No, its length is *never* larger than L0. Not in any frame. In the new track frame, it is definitely less than L0. In the Earth frame, it can be less than or equal to L0 (it is equal if the right v and the left v cancel out to zero).
Maybe you are confused about the phrase "in a frame", which has different colloquial meanings depending on context. For an object (e.g. a train) we sometimes say it is "in a frame" if it is motionless in that frame. But if we are talking about a measurement, such as a length, "in a frame" just means that we are using the measuring tools of that frame; it makes no claim that the object we are measuring is motionless in that frame. That's why I can talk about the length of BG as it now is in the new frame, *or* as it now is in the old frame, and of course the two are different in general.
>  It will even become as large as L0 in the old track frame then with more larger v it will become smaller. 
The increase in length which you describe is for a measurement of BG in the *old* track frame, where "in" is used as I describe in my previous paragraph above. Again, I'll remark that L0 itself does not depend on frame.
> 
> > 
(contrary to what he said) so we reach the right conclusion: the rules are the same. Later he writes: 
> > > > > > 
At t0 the train F'A' is at the left side of the track
at the points 00. Length of train is L0 
> > 
Note in passing, he accelerates the train here, but does not say how the acceleration is to be done along the entire length of the train. 
Sorry, I should have said trains (plural). Bell's paradox applies to each train individually.
> 
Each train is only accelarated once at t0 During the whole experiment they keep the same speed. At t=6 the train stops immediate. 
Yes, I understood that you were saying this; my point was that you apparently do not appreciate how complex those simple statements really are in SR. You simply can't do what you said and still get the numbers you calculated.
> 
The observer in the front can make a log book like: tn,pn track train t0,p0 t = 0 t = 0 t1,p1 t = 1 t = 0.9 t2,p2 t = 2 t = 1.8 t5,p5 t = 5 t = 4,5 t6,p6 t = 6 t = 5,4 For example t5,p5 is time position identifion p5=5*L0 from start point t = 5 is track time. t = 4,5 is train time 
Right, those numbers are OK for the fronts of the trains. But if we assume they are correct, your numbers for the tails were wrong.
For example, when you started the front at t=0 in the track frame, the train was uncontracted in that frame; but immediately after, you said it was contracted. How did the tail jump from its uncontracted position to its contracted position at t=0? It would have had to move faster than light, which is impossible.
There are a couple ways it *could* contract: (1) the tail could start accelerating *before* t=0 in the track frame, so that by t=0 it has reached its correct contracted position and the front can start accelerating at t=0; or (2) the train could stretch out mechanically at t=0 so that it begins its motion uncontracted, i.e. with a proper length that is too long; then the tail could move forward at a faster rate (like a spring) until it catches up to its proper position some time later.
The problem with either of these alternatives is that the clocks in the tails are moving differently from the clocks in the fronts, for at least some period of time, and this will put them out of sync. In other words the tail clocks do *not* show the same time as the clocks at the front, contrary to what you posted.
> 
> > 
This step is therefore not well defined. See: http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html 
> 
Interested document. I do not "like" the sentence "Don't tackle the physical situation directly; instead, imagine some pictures." IMO the most important part is what real experiments tell us i.e. how the physical world behaves. 
I agree, but please understand that the FAQ is meant to be pedagogical in nature: the authors are trying to teach a concept that, presumably, is new to the reader. Their advice is (they think) the best way to understand this issue for the first time. Probably their hope is that later, you *can* tackle the physical situation. OTOH in physics it almost always helps to draw pictures first, at least in your imagination.
Russell Blackadar
>  Nicolaas Vroom wrote: 
> > 
> > > > 
> > > 
[snip] No, I think here he is talking about BG and B'G' in the A'F' frame, because: 
> > 
Exactly I was expecting from Paul the Answer No 
> > > > 
> > > > > 
It should be remarked that the values of L0 in both frames are different. 
> > > 
This statement could only make sense if he is talking about two different pairs of trains. Of course he has said this in a confusing way because he identifies the train pairs (and their L0's) by frame, which is irrelevant, instead of by their letters. 
> > 
What I want to point out is that in both frames I want to perform exactly the same experiment using only ! tools in their own resp. frames. 
> > > > > 
This "new track frame" is also at rest. 
> > > > 
Anything is at rest relative to itself. 
> > > > > 
In this "new track frame" the rules are different compared with the "old track frame" 
> > > > 
No. It is the speeds of the moving trains that are different, not the rules. 
> > > 
Or more probably, the answer is No because of the mistake he made above. 
> >  If I made a mistake than that is the question. If I am not clear, I except. 
> > > 
BG and B'G' have the same speed in that frame, 
> > 
That is the basis of the experiments (sub experiment) I do that using rods and clocks in the new rest frame Each sub experiment assumes the same speed (absolute) for BG and B'G' 
> > > 
also they have the same length in that frame 
> > 
That and only that is the question. How do you know that. What (experimental) prove do you have ? 
> 
Experiments don't prove. But, every experiment that confirms Lorentz invariance is to some degree a confirmation of my statement. The number of such experiments is very large. On the other hand, I doubt that any sufficiently precise experiments have been performed using trains running on tracks mounted on a train. 
> > 
Ofcourse I can easily write down that they should be equal because of: L = L' = L0 * sqr(1v^2/c^2) But that is no proof. https://www.nicvroom.be/length3.htm The reason why I think that the length are not the same is because the train A'F', then new track, L0, and both trains BG and B'G'whenv=0 are all contracted. 
> 
No, L0 is not contracted 
L0 in the new rest frame (L0new) is physical contracted. So are the trains AF and A'F' in the old rest frame. How do we demonstrate (proof) that ? Is that in principle possible by performing an experiment. ?
I have described one at:
https://www.nicvroom.be/length3.htm#lengthcd The above experiment only shows if length contraction takes place, yes or no, but not which how much.
If you to know how much you need at least clocks in the old reference frame.
>  it's defined as the rest length of the train and is therefore invariant. You are misusing this notation if you say L0 can change depending on frame. 
L0new is the length of the train in rest (i.e. with v=0) in the new reference frame. However L0new is exactly the same as L' in the old reference frame. L'old is physical contracted and so is L0new.
In fact when you decrease the speed of A'F' L'old becomes larger and equal to L0old
>  I didn't remark on that last time, but it's important. We have to agree on terminology. 
First we have to agree how we demonstrate that both L and L'old are contracted.
> > 
When BG moves to the left you wil see that it length will become larger as L0 in the new track frame. 
> 
No, its length is *never* larger than L0. 
How do you know that ? It is very easy for me to agree with you with what you say, but that will not truely convince me.
Have a look at https://www.nicvroom.be/sketch.htm the second sketch explains this issue
>  Not in any frame. In the new track frame, it is definitely less than L0. 
For you that is the case with any frame In the old frame we have two trains with +v and v both trains are smaller then L0 (L0old)
In the new frame (which has a speed of +v rel to the old frame) we also have two trains with +v and v both trains are smaller then L0 (L0new)
The problem is that this train with v has a speed of zero(1) in the old frame and its length is small This small length is difficult for me to understand. (In fact its length should be roughly L0old)
(1) the value of zero is not correct because you cannot add those two speeds The value of zero is an approximation.
The second train with speed of +v has a speed of roughly +2v in the old frame Its length is small but that is not difficult to understand and is in agreement with SR
>  In the Earth frame, it can be less than or equal to L0 (it is equal if the right v and the left v cancel out to zero). Maybe you are confused about the phrase "in a frame", which has different colloquial meanings depending on context. For an object (e.g. a train) we sometimes say it is "in a frame" if it is motionless in that frame. But if we are talking about a measurement, such as a length, "in a frame" just means that we are using the measuring tools of that frame; it makes no claim that the object we are measuring is motionless in that frame. That's why I can talk about the length of BG as it now is in the new frame, *or* as it now is in the old frame, and of course the two are different in general. 
> > 
It will even become as large as L0 in the old track frame then with more larger v it will become smaller. 
> 
The increase in length which you describe is for a measurement of BG in the *old* track frame, where "in" is used as I describe in my previous paragraph above. Again, I'll remark that L0 itself does not depend on frame. 
> > 
Each train is only accelarated once at t0 During the whole experiment they keep the same speed. At t=6 the train stops immediate. 
> 
Yes, I understood that you were saying this; my point was that you apparently do not appreciate how complex those simple statements really are in SR. You simply can't do what you said and still get the numbers you calculated. 
Have a look at https://www.nicvroom.be/sketch.htm The first sketch explains the example raised by Paul.
> > 
The observer in the front can make a log book like: tn,pn track train t0,p0 t = 0 t = 0 t1,p1 t = 1 t = 0.9 t2,p2 t = 2 t = 1.8 t5,p5 t = 5 t = 4,5 t6,p6 t = 6 t = 5,4For example t5,p5 is time position identifion p5=5*L0 from start point t = 5 is track time. t = 4,5 is train time As you can see during the trip a = 0 v = constant 
> 
Right, those numbers are OK for the fronts of the trains. But if we assume they are correct, your numbers for the tails were wrong. For example, when you started the front at t=0 in the track frame, the train was uncontracted in that frame; but immediately after, you said it was contracted. How did the tail jump from its uncontracted position to its contracted position at t=0? It would have had to move faster than light, which is impossible. 
I 100% agree with you
Assume that train slowly increases from 0 to 1 and reach it constant speed there after Assume also that the train slowly deaccelerates after point 6. You could also assume that both ends contracts towards the moving center of the train.
>  The problem with either of these alternatives is that the clocks in the tails are moving differently from the clocks in the fronts, for at least some period of time, and this will put them out of sync. In other words the tail clocks do *not* show the same time as the clocks at the front, contrary to what you posted. 
I agree with you in principle however this part can be neglected for v < 1000 km/sec.
The true issue is synchronisation in the train frame.
During my exercise I keep the clocks in the train synchronised with the track frame. The result is that when you stop both trains all the clocks in the both trains show the same time. As a result are the moments t3(A') <> t3(F) t3(A') is the time that A' sees the explosion. t3(F) is the time that F sees the explosion.
What Paul does he synchronises the two clocks A' and F' train A'F' when the speed is +v (in train frame) and the two clocks A and F in the train AF when the speed is v What that means is that he has to modify in each train at least one clock. He has to take care that at the moment of explosion t2 t2(F') = t2(A) If you do that then also t3(A') = t3(F)
If you want to perform the experiment again with different v then you have to synchronise again in the train frames.
This explains the results obtained by Paul
> > 
> > > 
This step is therefore not well defined. See: http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html 
> > 
Interested document. I do not "like" the sentence "Don't tackle the physical situation directly; instead, imagine some pictures." IMO the most important part is what real experiments tell us i.e. how the physical world behaves. 
> 
I agree, but please understand that the FAQ is meant to be pedagogical in nature: the authors are trying to teach a concept that, presumably, is new to the reader. Their advice is (they think) the best way to understand this issue for the first time. Probably their hope is that later, you *can* tackle the physical situation. OTOH in physics it almost always helps to draw pictures first, at least in your imagination. 
> 
Russell Blackadar 
[snip]
> >  No, L0 is not contracted 
> 
L0 in the new rest frame (L0new) is physical contracted. 
No. If you are going to use the formulas from SR (or from LET) then you need to define your terms correctly.
L0 is the length *as measured in the rest frame of the train*. If you don't agree that L0 has this meaning, then *none* of your equations have any validity in your arguments for or against SR. They have no relationship at all to the equations of SR.
> 
So are the trains AF and A'F' in the old rest frame.
How do we demonstrate (proof) that ?
Is that in principle possible by performing an experiment. ?
I have described one at: https://www.nicvroom.be/length3.htm#lengthcd The above experiment only shows if length contraction takes place, yes or no, but not which how much. If you to know how much you need at least clocks in the old reference frame. 
> > 
it's defined as the rest length of the train and is therefore invariant. You are misusing this notation if you say L0 can change depending on frame. 
> 
L0new is the length of the train in rest (i.e. with v=0) in the new reference frame. 
I take it that this is your *definition* of L0new? If so, SR says that L0new = L0.
>  However L0new is exactly the same as L' in the old reference frame. 
No! Not in SR. L0new is the length of the train at rest, and L' is the measurement of that *same* train according to an observer moving relative to the train. SR says these are *different* numbers.
Now, you could argue that SR is wrong. But in that case it would make no sense to use the SR formulas (wrongly!) to predict what you will see in your experiment. See my point?
>  L'old is physical contracted and so is L0new. 
What in heaven do you mean by L'old? That's the first I've seen this notation in any of your postings.
I'm sorry to say this, but it seems that you are too confused on these matters to be able to define your terms and notation, and to use them in equations, in anything like a consistent fashion so that others can understand what you are trying to say. In light of that, I really don't have time to respond to you further. I wish you well.
[rest snipped]
Russell Blackadar
>  I'm sorry to say this, but it seems that you are too confused on these matters to be able to define your terms and notation, and to use them in equations, in anything like a consistent fashion so that others can understand what you are trying to say. In light of that, I really don't have time to respond to you further. I wish you well. 
Let me start with a fresh new experiement Hopefully it is clear.
Starting point is the example at page 23 of the book: "Introducing Einstein's Relativity" by Ray d'Inverno, in a slightly different way.
We start with a train AF at rest in track reference frame which is placed (i.e. the track and the train) on a MOVING platform relative to the Earth frame. On the train there two observers: one at Front (F) and one at back (A). Each Observer has a clock. Near each observer there is a light. A light will flash when the firing device hits a contact near the track. There are two contacts near the track. (When the train is positioned above those contacts both lamps are on) Currently the firing devices are just in front of those contacts and both lamps are off.
The first part of the experiment consists of clock synchronization. This is done placing a lamp in the middle of the train or contacts or observers. (2) The lamp is turned on and when an observer receives the flash he resets his clock. (1)
The clocks are now synchronous in the track frame (which moves rel to the Earth)
The second part of the experiment consists of moving the train a little bit. both lamps will flash and Each observers monitors the time of the flash at his lamp i.e. the lamp close to him.
Question will they monitor both the same time.?
IMO the answer is NO.
(1) The clocks were earlier synchronised in the Earth frame. When they performed the synchronisation (moving platform) they monitored also the time on each clock just before the clock was reseted. The difference between both was "5 secs".
(2) In this experiment the position of an observer, his lamp, the firing device on the train and the contact near the track are the same.
> 
Russell Blackadar 
> > 
I'm sorry to say this, but it seems that you are too confused on these matters to be able to define your terms and notation, and to use them in equations, in anything like a consistent fashion so that others can understand what you are trying to say. In light of that, I really don't have time to respond to you further. I wish you well. 
> 
Let me start with a fresh new experiement 
Ok, I'll have a look. Just this once.
>  Hopefully it is clear. 
It wasn't completely clear, to me. See below.
> 
Starting point is the example at page 23 of the book: "Introducing Einstein's Relativity" by Ray d'Inverno, in a slightly different way. We start with a train AF at rest in track reference frame which is placed (i.e. the track and the train) on a MOVING platform relative to the Earth frame. On the train there two observers: one at Front (F) and one at back (A). Each Observer has a clock. Near each observer there is a light. A light will flash when the firing device hits a contact near the track. There are two contacts near the track. (When the train is positioned above those contacts both lamps are on) Currently the firing devices are just in front of those contacts and both lamps are off. The first part of the experiment consists of clock synchronization. This is done placing a lamp in the middle of the train or contacts or observers. (2) The lamp is turned on and when an observer receives the flash he resets his clock. (1) The clocks are now synchronous in the track frame (which moves rel to the Earth) The second part of the experiment consists of moving the train a little bit. 
Exactly *how* do you move it? You don't seem to understand that this has a crucial effect on the outcome.
> 
both lamps will flash and Each observers
monitors the time of the flash at his lamp
i.e. the lamp close to him.
Question will they monitor both the same time.? 
They *won't* monitor the same time if you move the train just by pulling on the front, because no train is perfectly rigid. This result is uninteresting because it would be silly to base our notion of time on the elastic properties of a particular train.
Alternatively, if you apply the same force to both front and back "at the same time", as measured by the clocks, then of course the clocks *will* show the same time. (Duh!) This is even less interesting than the previous result.
Or, if you use some *other* clocks to determine what the "same time" is, and those other clocks are moving wrt your observers, then the prediction of SR is that your observers will *not* monitor the same time on their clocks. Only this last prediction is different from what Galilean Relativity would predict, and to me that makes it interesting.
So, do we assume you are doing the interesting experiment, or a totally uninteresting one?
> 
IMO the answer is NO. 
Your opinion is correct according to SR, assuming you did the interesting thing. So, what's your point?
> 
(1) The clocks were earlier synchronised in the Earth frame. When they performed the synchronisation (moving platform) they monitored also the time on each clock just before the clock was reseted. The difference between both was "5 secs". 
I don't see why you bother mentioning this detail; it's made irrelevant by your resetting the clocks. (You *do* mean that after being reset, they read the same time, don't you? Otherwise why bother.)
> 
(2) In this experiment the position of an observer, his lamp, the firing device on the train and the contact near the track are the same. 
Except, the contact is not exactly in the same place as the observer *before* you start the train moving. Your omission of this crucial detail is what causes all the ambiguity.
Russell Blackadar
>  Nicolaas Vroom wrote: 
> > 
Let me start with a fresh new experiement 
> 
Ok, I'll have a look. Just this once. 
> > 
Hopefully it is clear. 
> 
It wasn't completely clear, to me. See below. 
> > 
Starting point is the example at page 23 of the book: "Introducing Einstein's Relativity" by Ray d'Inverno, in a slightly different way. We start with a train AF at rest in track reference frame which is placed (i.e. the track and the train) on a MOVING platform relative to the Earth frame. On the train there two observers: one at Front (F) and one at back (A). Each Observer has a clock. Near each observer there is a light. A light will flash when the firing device hits a contact near the track. There are two contacts near the track. (When the train is positioned above those contacts both lamps are on) Currently the firing devices are just in front of those contacts and both lamps are off. The first part of the experiment consists of clock synchronization. This is done placing a lamp in the middle of the train or contacts or observers. (2) The lamp is turned on and when an observer receives the flash he resets his clock. (1) The clocks are now synchronous in the track frame (which moves rel to the Earth) The second part of the experiment consists of moving the train a little bit. 
> 
Exactly *how* do you move it? You don't seem to understand that this has a crucial effect on the outcome. 
I think I understand it.
> > 
both lamps will flash and Each observers
monitors the time of the flash at his lamp
i.e. the lamp close to him.
Question will they monitor both the same time.? 
> 
They *won't* monitor the same time if you move the train just by pulling on the front, because no train is perfectly rigid. This result is uninteresting because it would be silly to base our notion of time on the elastic properties of a particular train. 
But they should accordingly chapter 2.10 "The relativity of simultanity" In that chapter they use the same train configuration as I have described. In that chapter they have placed an observer A in the center between the contacts resp firing devices. The text is. "From the configuration it is clear that A will judge that the two events, when the light sources first switch on, occur simultaneously" My interpretation why A makes this judgement is because A actual sees light from the two light sources simultaneous. Next the text states (modified) "B on the train sees light from light source 2 before light from light source 1" If this judgement from A is correct than the two observers near the light sources should also conclude (based on the clocks they have) that the lights flashed at the same time.
Now you write above that when you perform the experiment that this is "not true". (3)
>  Alternatively, if you apply the same force to both front and back "at the same time", as measured by the clocks, then of course the clocks *will* show the same time. (Duh!) This is even less interesting than the previous result. 
If you do that observer A in the center of the track will actual see the two lights simultaneous. Why don't they write such a scenario in the text of the book ?
> 
Or, if you use some *other* clocks to determine what
the "same time" is, and those other clocks are moving
wrt your observers, then the prediction of SR is that
your observers will *not* monitor the same time on their
clocks. Only this last prediction is different from
what Galilean Relativity would predict, and to me that
makes it interesting.
So, do we assume you are doing the interesting experiment, or a totally uninteresting one? 
See above
> > 
IMO the answer is NO. 
> 
Your opinion is correct according to SR, assuming you did the interesting thing. So, what's your point? 
Suppose that both observers see the light simultaneous as decided by reading their clocks.
Next I change the speed of the whole platform including the track and train Next I repeat the whole experiment starting first with clock synchronization. The outcome should be identical: both observers see the light simultaneous.
Next I change the speed of the whole platform again, and again both observers see the light simultaneous.
In fact this gives me an easy way to synchronize the clocks of the two observers: I only have to move the trains slightly.
I do not trust this final sentence.
> > 
(1) The clocks were earlier synchronised in the Earth frame. When they performed the synchronisation (moving platform) they monitored also the time on each clock just before the clock was reseted. The difference between both was "5 secs". 
> 
I don't see why you bother mentioning this detail; it's made irrelevant by your resetting the clocks. (You *do* mean that after being reset, they read the same time, don't you? Otherwise why bother.) 
In some way I want to quantify the results. (3) maybe also here some quantumfication is required  how much "not true"
> > 
(2) In this experiment the position of an observer, his lamp, the firing device on the train and the contact near the track are the same. 
> 
Except, the contact is not exactly in the same place as the observer *before* you start the train moving. Your omission of this crucial detail is what causes all the ambiguity. 
In the book they use a moving train. Moving trains contract, (which cause that the lights do not flash simultaneous) That's why I use a train which is very near the contacts and which you only have to give a minuscule push. The contacts should be something like:
V V train contact (firing device) ^ ^ track contact
Russell Blackadar
> 
See:
http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html 
I have a problem with that article.
Consider two spaceships (rod 1 and rod 2) one after an other and touching each other:
<><> > direction of movement rod 1 rod 2The two spaceship start to move with constant acceleration 1 After some time t acceleration is zero and speed of boths rods (spaceships) is the same. Everything is measured in lab frame. (track frame)
When the speed is constant (acceleration = 0) there will be a distance delta d between the two space ships This distance delta d will stay constant as long as there is no change in speed.
The above picture becomes:
<> <> rod1 rod 2The length of both rods is identical and both are contracted.
Now let us see what happens in the moving frame of rod 1
In that frame the speed of rod 1 is zero.
Q1 In that frame is there a distance between the rods ?
Q2 Is that distance constant ?
Q3 Is the length of both rods identical ?
Q4 Is the speed of rod 2 also zero ?
> 
Russell Blackadar 
> > 
See:
http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html 
> 
I have a problem with that article. Consider two spaceships (rod 1 and rod 2) one after an other and touching each other: <><> > direction of movement rod 1 rod 2The two spaceship start to move with constant acceleration 1 
If you think that's a wellspecified statement, you didn't understand the FAQ article! Anyhow, I'll add the assumption that they accelerate Bornrigidly, and equally.
> 
After some time t acceleration is zero and speed of boths rods (spaceships) is the same.
Everything is measured in lab frame. (track frame)
When the speed is constant (acceleration = 0) there will be a distance delta d between the two space ships This distance delta d will stay constant as long as there is no change in speed. The above picture becomes: <> <> rod1 rod 2The length of both rods is identical and both are contracted. Now let us see what happens in the moving frame of rod 1 
I assume you mean the final frame, which is inertial.
> 
In that frame the speed of rod 1 is zero. Q1 In that frame is there a distance between the rods ? 
Yes. And the distance is nonzero, if the acceleration was Bornrigid.
>  Q2 Is that distance constant ? 
Yes.
>  Q3 Is the length of both rods identical ? 
Yes.
>  Q4 Is the speed of rod 2 also zero ? 
Yes.
Russell Blackadar
>  Nicolaas Vroom wrote: 
> > 
Russell Blackadar 
> > > 
See:
http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html 
> > 
I have a problem with that article. Consider two spaceships (rod 1 and rod 2) one after an other and touching each other: <><> > direction of movement rod 1 rod 2The two spaceship start to move with constant acceleration 1 
>  If you think that's a wellspecified statement, you didn't understand the FAQ article! Anyhow, I'll add the assumption that they accelerate Bornrigidly, and equally. 
What do you mean with Bornrigid ?
The Idea is that both spaceships starts at the same time
at that they both accelerate during the same period (lab frame)
with the same amount (lab frame)
IMO there speed will than all the time be equal. (and the distance to their origin) All lab frame.
> 
> > 
After some time t acceleration is zero and speed of boths rods (spaceships) is the same.
Everything is measured in lab frame. (track frame) The above picture becomes: <> <> rod1 rod 2The length of both rods is identical and both are contracted. Now let us see what happens in the moving frame of rod 1 
> 
I assume you mean the final frame, which is inertial. 
I mean the same as in the article.
They make a distinction between a "lab frame" and
the frame of the "comoving observer"
IMO both are inertial.
I mean: the frame of the comoving observer of rod 1
> 
> > 
In that frame the speed of rod 1 is zero. 
>  Yes. And the distance is nonzero, if the acceleration was Bornrigid. 
Why do you write with Bornrigid ?
> >  Q2 Is that distance constant ? 
>  Yes. 
> >  Q3 Is the length of both rods identical ? 
>  Yes. 
> >  Q4 Is the speed of rod 2 also zero ? 
>  Yes. 
If I understand you correct than you agree with me that in both frames (lab frame and frame of comoving observer) there is a distance between both rods.
IMO the explanation for both frames is: length contraction.
Now let us continue with this experiment in the
frame of the moving observer in rod 1
(I call this state 0, with V = 0)
1. We will increase the speed of both rods
at the same time
in the same direction as before.
Q: will the distance between both rods increase ?
2. We go back to state 0 and instead of increasing the speed of the rods in the same direction as before we do it in the opposite direction (i.e. towards where we came from)
Q: will the distance between both rods increase ?
https://www.nicvroom.be/sketch.htm
> 
Russell Blackadar 
> >  Nicolaas Vroom wrote: 
> > > 
Russell Blackadar 
> > > > 
See:
http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html 
> > > 
I have a problem with that article. Consider two spaceships (rod 1 and rod 2) one after an other and touching each other: <><> > direction of movement rod 1 rod 2 The two spaceship start to move with constant acceleration 1 
> > 
If you think that's a wellspecified statement, you didn't understand the FAQ article! Anyhow, I'll add the assumption that they accelerate Bornrigidly, and equally. 
>  What do you mean with Bornrigid ? 
As I said, it's evident you didn't understand the FAQ.
>  The Idea is that both spaceships starts at the same time at that they both accelerate during the same period (lab frame) > with the same amount (lab frame) 
Yes, I know that's what you said. If each rocket were a single point (or even approximately a point) then I would have no objection, no reason to talk about Bornrigidity.
But that's not the case here. Here, you are very much concerned with the length of each rod; your whole line of reasoning depends on it. Your rods are not points. And so, you *must* tell us whether they accelerate Bornrigidly, or not. Otherwise, the problem is not well defined.
>  IMO there speed will than all the time be equal. (and the distance to their origin) All lab frame. 
Ok  so if what you are saying is, that *all parts* of both rockets accelerate equally in the lab frame, then there will be no contraction; the proper length of each rocket will increase and there will be no gap. (This is the opposite of Bornrigidity.)
> > > 
After some time t acceleration is zero and speed of boths rods (spaceships) is the same.
Everything is measured in lab frame. (track frame) When the speed is constant (acceleration = 0) there will be a distance delta d between the two space ships 
Delta d will be zero, because, to achieve what you say you have achieved, you had to apply a force sufficient to stretch the rockets in their rest frame. (Assuming they are stretchable. If not, you have torn them apart.) Read the FAQ article again, more carefully.
> > >  This distance delta d will stay constant as long as there is no change in speed. 
> 
The above picture becomes:
<> <> rod1 rod 2The length of both rods is identical and both are contracted. 
> > >  Now let us see what happens in the moving frame of rod 1 
> >  I assume you mean the final frame, which is inertial. 
> 
I mean the same as in the article. They make a distinction between a "lab frame" and the frame of the
"comoving observer IMO both are inertial. I mean: the frame of the comoving observer of rod 1 
> > 
> > > 
In that frame the speed of rod 1 is zero. 
> > 
Yes. And the distance is nonzero, if the acceleration was Bornrigid. 
> 
Why do you write with Bornrigid ? 
Because it's important. Btw my answer to your question is now "no", because you have stated your assumptions more clearly, and they differ with the assumptions I made when I said yes.
> 
> > 
> > > 
Q2 Is that distance constant ? 
> > 
Yes. 
> > > 
Q3 Is the length of both rods identical ? 
> > 
Yes. 
> > > 
Q4 Is the speed of rod 2 also zero ? 
> > 
Yes. 
> 
If I understand you correct than you agree with me that in both frames (lab frame and frame of comoving observer) there is a distance between both rods. 
*If* the acceleration is Bornrigid, or even partially so. But *not* if all parts of each rocket accelerate equally. Reread the FAQ.
> 
IMO the explanation for both frames is: length contraction.
Now let us continue with this experiment in the frame of the moving observer in rod 1 Q: will the distance between both rods increase ? 2. We go back to state 0 and instead of increasing the speed of the rods in the same direction as before we do it in the opposite direction (i.e. towards where we came from) Q: will the distance between both rods increase ? 
Why do you ask these things? The answers are the same of course. And they have the same dependence on *how* you do the acceleration to each rocket.
Russell Blackadar
>  Nicolaas Vroom wrote: 
> > 
Russell Blackadar 
> > >  Nicolaas Vroom wrote: 
> > > > 
Russell Blackadar 
> > > > > 
See: http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html 
> > > > 
I have a problem with that article. Consider two spaceships (rod 1 and rod 2) one after an other and touching each other: <><> > direction of movement rod 1 rod 2The two spaceship start to move with constant acceleration 1 
> > > 
If you think that's a wellspecified statement, you didn't understand the FAQ article! Anyhow, I'll add the assumption that they accelerate Bornrigidly, and equally. 
> > 
What do you mean with Bornrigid ? 
> 
As I said, it's evident you didn't understand the FAQ. 
I think the article you want me to read is: The Rigid Rotating Disk in Relativity. http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html That article mentions born rigid.
What I have in mind is the following sentence from that article: "If we accelerate a rod rigidly in the longitudinal direction, then the rod suffers the usual Lorentz contraction."
When I wrote down above the following: "The two spaceship start to move with constant acceleration 1" I should have added: based from two equivalent points (either from both the front, or the back or the middle)
(Q: If you take the middle are then not both front and back accelerating equally ? )
> 
> > 
The Idea is that both spaceships starts at the same time and that they both accelerate during the same period (lab frame) with the same amount (lab frame) 
> 
Yes, I know that's what you said. If each rocket were a single point (or even approximately a point) then I would have no objection, no reason to talk about Bornrigidity. But that's not the case here. Here, you are very much concerned with the length of each rod; your whole line of reasoning depends on it. Your rods are not points. And so, you *must* tell us whether they accelerate Bornrigidly, or not. Otherwise, the problem is not well defined. 
It is interesting to note that in the book Ray d'Inverno (who uses a train) that he does not mention this problem.
> 
> > 
IMO there speed will than all the time be equal. (and the distance to their origin) All lab frame. 
> 
Ok  so if what you are saying is, that *all parts* of both rockets accelerate equally in the lab frame, then there will be no contraction; the proper length of each rocket will increase and there will be no gap. (This is the opposite of Bornrigidity.) 
I do not mean that all parts of both rockets accelerate equally. I mean one point.
> 
> > 
> > > 
> > > > 
After some time t acceleration is zero and speed of boths rods (spaceships) is the same. Everything is measured in lab frame. (track frame) When the speed is constant (acceleration = 0) there will be a distance delta d between the two space ships 
> 
Delta d will be zero, because, to achieve what you say you have achieved, you had to apply a force sufficient to stretch the rockets in their rest frame. (Assuming they are stretchable. If not, you have torn them apart.) Read the FAQ article again, more carefully. 
> > > > 
This distance delta d will stay constant as long as there is no change in speed. The above picture becomes: <> <> rod1 rod 2The length of both rods is identical and both are contracted. Now let us see what happens in the moving frame of rod 1 
> > > 
I assume you mean the final frame, which is inertial. 
> > 
I mean the same as in the article. They make a distinction between a "lab frame" and the frame of the "comoving observer" IMO both are inertial. I mean: the frame of the comoving observer of rod 1 
> > > 
> > > > 
In that frame the speed of rod 1 is zero. 
> > > 
Yes. And the distance is nonzero, if the acceleration was Bornrigid. 
> > 
Why do you write with Bornrigid ? 
> 
Because it's important. Btw my answer to your question is now "no", because you have stated your assumptions more clearly, and they differ with the assumptions I made when I said yes. 
It is interesting to note that in the spaceship puzzel problem the concept born rigid is not mentioned. Except in the link to the "rigid disk" URL
> > 
> > > 
> > > > 
Q2 Is that distance constant ? 
> > > 
Yes. 
> > > > 
Q3 Is the length of both rods identical ? 
> > > 
Yes. 
> > > > 
Q4 Is the speed of rod 2 also zero ? 
> > > 
Yes. 
> > 
If I understand you correct than you agree with me that in both frames (lab frame and frame of comoving observer) there is a distance between both rods. 
> 
*If* the acceleration is Bornrigid, or even partially so. But *not* if all parts of each rocket accelerate equally. Reread the FAQ. 
Done
> 
> > 
IMO the explanation for both frames is: length contraction.
Now let us continue with this experiment in the frame of the moving observer in rod 1
(I call this state 0, with V = 0) Q: will the distance between both rods increase ? 2. We go back to state 0 and instead of increasing the speed of the rods in the same direction as before we do it in the opposite direction (i.e. towards where we came from) Q: will the distance between both rods increase ? 
> 
Why do you ask these things? The answers are the same of course. And they have the same dependence on *how* you do the acceleration to each rocket. 
In the case of born rigid acceleration do the length of both rods in both directions decrease ?
I hope you reply on the above again.
> 
Russell Blackadar 
> >  Nicolaas Vroom wrote: 
> > > 
Russell Blackadar 
> > > >  Nicolaas Vroom wrote: 
> > > > > 
Russell Blackadar 
> > > > > > 
See: http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html 
> > > > > 
I have a problem with that article. Consider two spaceships (rod 1 and rod 2) one after an other and touching each other: <><> > direction of movement rod 1 rod 2The two spaceship start to move with constant acceleration 1 
> > > > 
If you think that's a wellspecified statement, you didn't understand the FAQ article! Anyhow, I'll add the assumption that they accelerate Bornrigidly, and equally. 
> > > 
What do you mean with Bornrigid ? 
> > 
As I said, it's evident you didn't understand the FAQ. 
I meant that you didn't understand its implications; see below how I apply the twospaceship example to a single rod.
>  I think the article you want me to read is: The Rigid Rotating Disk in Relativity. http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html That article mentions born rigid. 
Right, sorry for the confusion. I forgot that I'd followed a link to it. Actually, the discussion on this page is a bit complicated by its need to talk about rotatory motion; for linear acceleration you might prefer the article "Re: Three questions involving an accelerated frame of reference" <3A6DE22E.3F07E594@lucent.com> which Tom Roberts posted here recently.
> 
What I have in mind is the following sentence from that article: "If we accelerate a rod rigidly in the longitudinal direction, then the rod suffers the usual Lorentz contraction." When I wrote down above the following: "The two spaceship start to move with constant acceleration 1" I should have added: based from two equivalent points (either from both the front, or the back or the middle) (Q: If you take the middle are then not both front and back accelerating equally ? ) 
Not necessarily, and that's exactly my point.
E.g. if (say) you grabbed each rod by its middle and gently accelerated it so that it were free to Lorentzcontract in the original frame, how can it contract unless the back accelerates very very slightly more than the front? What happens is that as the tail starts to fall behind, tension in the crystal lattice increases so that the force of tension on the tail is very slightly greater than the force of compression on the front, causing the tail to accelerate more. Of course this is too small to measure in practical experiments done with metal rods, at least for the foreseeable future. But this is what we mean by Bornrigid motion. And for this kind of acceleration, you get a gap.
OTOH if you put many very strong grabbers all along the full length of each rod, at every point, and forced each one of the grabbers to accelerate at the same rate everywhere, what you would find is that the rods would physically stretch exactly enough to compensate for the Lorentz contraction that also occurs, with the result that there is no gap between the rods.
> 
> > 
> > > 
The Idea is that both spaceships starts at the same time at that they both accelerate during the same period (lab frame) with the same amount (lab frame) 
> > 
Yes, I know that's what you said. If each rocket were a single point (or even approximately a point) then I would have no objection, no reason to talk about Bornrigidity. But that's not the case here. Here, you are very much concerned with the length of each rod; your whole line of reasoning depends on it. Your rods are not points. And so, you *must* tell us whether they accelerate Bornrigidly, or not. Otherwise, the problem is not well defined. 
>  It is interesting to note that in the book Ray d'Inverno (who uses a train) that he does not mention this problem. 
Does he *accelerate* the train, in the passage you are referring to? I bet not. At any rate I am sure he is careful not to get into Bell's paradox trouble by asking illdefined questions, such as what is the distance between the tail of one accelerating train and the head of another that starts out touching. Not without first clarifying the full nature of the acceleration.
> 
> > 
> > > 
IMO there speed will than all the time be equal. (and the distance to their origin) All lab frame. 
> > 
Ok  so if what you are saying is, that *all parts* of both rockets accelerate equally in the lab frame, then there will be no contraction; the proper length of each rocket will increase and there will be no gap. (This is the opposite of Bornrigidity.) 
> 
I do not mean that all parts of both rockets accelerate equally. I mean one point. 
Then you *do* mean Bornrigid acceleration? Ok.
> 
> > 
> > > 
> > > > 
> > > > > 
After some time t acceleration is zero and speed of boths rods (spaceships) is the same. Everything is measured in lab frame. (track frame) When the speed is constant (acceleration = 0) there will be a distance delta d between the two space ships 
> > 
Delta d will be zero, because, to achieve what you say you have achieved, you had to apply a force sufficient to
stretch the rockets in their rest frame. (Assuming they are stretchable. If not, you have torn them apart.) 
> > > > > 
This distance delta d will stay constant as long as there is no change in speed. The above picture becomes: <> <> rod1 rod 2The length of both rods is identical and both are contracted. Now let us see what happens in the moving frame of rod 1 
> > > > 
I assume you mean the final frame, which is inertial. 
> > >  I mean the same as in the article. They make a distinction between a "lab frame" and the frame of the "comoving observer" IMO both are inertial. I mean: the frame of the comoving observer of rod 1 
> > > > 
> > > > > 
In that frame the speed of rod 1 is zero. 
> > > > 
Yes. And the distance is nonzero, if the acceleration was Bornrigid. 
> > >  Why do you write with Bornrigid ? 
> >  Because it's important. Btw my answer to your question is now "no", because you have stated your assumptions more clearly, and they differ with the assumptions I made when I said yes. 
>  It is interesting to note that in the spaceship puzzel problem the concept born rigid is not mentioned. Except in the link to the "rigid disk" URL 
Right; in the spaceship article the ships are assumed to be pointlike, i.e. not extended in space. (IOW their lengths are irrelevant to the problem.) The relevance of this page comes when you consider the front of one rod as the first spaceship, and the back of the *same* rod as the second spaceship. That is, the length of your rod 1 is analogous to the distance between the two "spaceships" in the article. And same for rod 2. Now do you see my point?
> 
> > > 
> > > > 
> > > > > 
Q2 Is that distance constant ? 
> > > > 
Yes. 
> > > > > 
Q3 Is the length of both rods identical ? 
> > > > 
Yes. 
> > > > > 
Q4 Is the speed of rod 2 also zero ? 
> > > > 
Yes. 
> > > 
If I understand you correct than you agree with me that in both frames (lab frame and frame of comoving observer) there is a distance between both rods. 
> > 
*If* the acceleration is Bornrigid, or even partially so. But *not* if all parts of each rocket accelerate equally. Reread the FAQ. 
>  Done 
> > 
> > > 
IMO the explanation for both frames is: length contraction.
Now let us continue with this experiment in the
frame of the moving observer in rod 1
(I call this state 0, with V = 0) Q: will the distance between both rods increase ? 2. We go back to state 0 and instead of increasing the speed of the rods in the same direction as before we do it in the opposite direction (i.e. towards where we came from) Q: will the distance between both rods increase ? 
> > 
Why do you ask these things? The answers are the same of course. And they have the same dependence on *how* you do the acceleration to each rocket. 
> 
In the case of born rigid acceleration do the length of both rods in both directions decrease ? 
In the track frame, the measured lengths decrease if the speed increases, and increase if the speed decreases. You know this is true for the case of Bornrigid motion because the intrinsic length of the rod (in its own rest frame) is *preserved*, hence the rod's length as measured in a different frame is given directly by the LTE.
Russell Blackadar
>  Nicolaas Vroom wrote: 
> > 
What I have in mind is the following sentence from that article: "If we accelerate a rod rigidly in the longitudinal direction, then the rod suffers the usual Lorentz contraction." When I wrote down above the following: "The two spaceship start to move with constant acceleration 1" I should have added: based from two equivalent points (either from both the front, or the back or the middle) (Q: If you take the middle are then not both front and back accelerating equally ? ) 
> 
Not necessarily, and that's exactly my point. E.g. if (say) you grabbed each rod by its middle and gently accelerated it so that it were free to Lorentzcontract in the original frame, how can it contract unless the back accelerates very very slightly more than the front? What happens is that as the tail starts to fall behind, tension in the crystal lattice increases so that the force of tension on the tail is very slightly greater than the force of compression on the front, causing the tail to accelerate more. Of course this is too small to measure in practical experiments done with metal rods, at least for the foreseeable future. But this is what we mean by Bornrigid motion. And for this kind of acceleration you get a gap. OTOH if you put many very strong grabbers all along the full length of each rod, at every point, and forced each one of the grabbers to accelerate at the same rate everywhere, what you would find is that the rods would physically stretch exactly enough to compensate for the Lorentz contraction that also occurs, with the result that there is no gap between the rods. 
> > 
> > > 
> > > > 
The Idea is that both spaceships starts at the same time at that they both accelerate during the same period (lab frame) with the same amount (lab frame) 
> > > 
Yes, I know that's what you said. If each rocket were a single point (or even approximately a point) then I would have no objection, no reason to talk about Bornrigidity. But that's not the case here. Here, you are very much concerned with the length of each rod; your whole line of reasoning depends on it. Your rods are not points. And so, you *must* tell us whether they accelerate Bornrigidly, or not. Otherwise, the problem is not well defined. 
> > 
It is interesting to note that in the book Ray d'Inverno (who uses a train) that he does not mention this problem. 
>  Does he *accelerate* the train, in the passage you are referring to? I bet not. 
During the experiment acceleration is zero. Speed of the train v = constant.
Under the same conditions I perform the following
experiment:
Length of the rod in rest frame (track frame, lab frame) = L0
The rod has a constant speed v in the rest frame.
(Measured from the back (but front should have the same))
The observer A is at position A at rest in the rest frame.
When the back B of the rod is at A the observer sents
a light signal.
This signal is reflected back at the front end F (at tF)
The signal is back at Observer A at tA.
What does tA measure:
1) tA = 2 * L0 / (cv)
2) tA <> 2 * L0 / (cv)
3) tA = 2 * L0 * SQR(1  v^2/c^2) /(cv)
>  At any rate I am sure he is careful not to get into Bell's paradox trouble by asking illdefined questions, such as what is the distance between the tail of one accelerating train and the head of another that starts out touching. Not without first clarifying the full nature of the acceleration. 
> > 
> > > > > 
Yes. And the distance is nonzero, if the acceleration was Bornrigid. 
> > > > 
Why do you write with Bornrigid ? 
> > > 
Because it's important. Btw my answer to your question is now "no", because you have stated your assumptions more clearly, and they differ with the assumptions I made when I said yes. 
> > 
It is interesting to note that in the spaceship puzzel problem the concept born rigid is not mentioned. Except in the link to the "rigid disk" URL 
> 
Right; in the spaceship article the ships are assumed to be pointlike, i.e. not extended in space.
(IOW their lengths are irrelevant to the problem.) 
Almost the same problem as above. Consider two space ships A and B. At "rest" they are a distance L apart. This is the initial position at v = 0
During a certain period they both accelerate
exactly in the same manner.
(Measured from the rest frame)
Then they both make their a = 0
and both will have the same speed.
A is behind and B is in the front.
A sents a light signal to B (From the back of A to the back of B) this signal is reflected back to B
The signal is back at Observer A at tA.
What does tA measure:
(Using a clock in the rest frame)
1) tA = L / (cv) + L / (c+v)
2) tA <> L / (cv) + L / (c+v)
3) tA = (L / (cv) + L / (c+v)) * SQR(1  v^2/c^2)
> > > >  Q: will the distance between both rods increase ? 
> > > 
Why do you ask these things? The answers are the same of course. And they have the same dependence on *how* you do the acceleration to each rocket. 
> > 
In the case of born rigid acceleration do the length of both rods in both directions decrease ? 
> 
In the track frame, the measured lengths decrease if the speed increases, and increase if the speed decreases. You know this is true for the case of Bornrigid motion because the intrinsic length of the rod (in its own rest frame) is *preserved*, hence the rod's length as measured in a different frame is given directly by the LTE. 
> 
Russell Blackadar 
Hi, Nicolaas. Nice to see you again, it's been a while.
(regarding Born rigidity) Nicolaas writes:
> > >  It is interesting to note that in the book Ray d'Inverno (who uses a train) that he does not mention this problem. 
> > 
Does he *accelerate* the train, in the passage you are referring to? I bet not. 
> 
During the experiment acceleration is zero. Speed of the train v = constant. Under the same conditions I perform the following experiment: Length of the rod in rest frame (track frame, lab frame) = L0 The rod has a constant speed v in the rest frame. 
In other words, you mean its length *now* in the lab frame is L0? If so, be aware, your notation is very confusing. According to the usual convention, if you write an L with a subscript 0, you mean *proper* length. But here that is apparently not what you mean.
> 
(Measured from the back (but front should have the same))
The observer A is at position A at rest in the rest frame.
When the back B of the rod is at A the observer sents a light signal.
This signal is reflected back at the front end F (at tF)
The signal is back at Observer A at tA.
What does tA measure: 1) tA = 2 * L0 / (cv) 
The above equation is correct, using L0 as you've defined it.
> 
2) tA <> 2 * L0 / (cv) 3) tA = 2 * L0 * SQR(1  v^2/c^2) /(cv) 
Equation 3 *would* have been correct if you had defined L0 in the usual way. (And of course in that case, inequality 2 would also have been correct for nonzero v.)
> 
Almost the same problem as above. Consider two space ships A and B. At "rest" they are a distance L apart. This is the initial position at v = 0 During a certain period they both accelerate exactly in the same manner. (Measured from the rest frame) 
I assume you mean the *original* rest frame (i.e. the frame analogous to your track frame above). Right?
>  Then they both make their a = 0 and both will have the same speed. A is behind and B is in the front. 
When you say "then", I assume you are still talking about the *original* rest frame (or track frame), right?
This is important to specify, because the way I stated it, B actually accelerates for a greater *proper* time than does A.
After the accelerations, the separation in the original rest frame is still L. But in the frame of A and B, it is gamma * L.
>  A sents a light signal to B (From the back of A to the back of B) this signal is reflected back to B 
I assume you mean reflected *from* B back toward A.
>  The signal is back at Observer A at tA. 
Does observer A move with ship A, or is he motionless
in the original frame? I will assume the latter,
since you said this problem is almost the same as the
previous one. In that case the answer is
tA = 2 * L / (cv)
> 
What does tA measure:
(Using a clock in the rest frame) 1) tA = L / (cv) + L / (c+v) 2) tA <> L / (cv) + L / (c+v) 3) tA = (L / (cv) + L / (c+v)) * SQR(1  v^2/c^2) 
Only your inequality 2 is correct, under the assumptions I stated.
Equation 3 *would* have been correct if you meant the
time measured by an observer riding on ship A, using
his own clock. (That is, the righthand side is just
an algebraic rearrangement of 2*gamma*L/c.)
Also in
that same case, Equation 1 would have expressed the
elapsed time in the original restframe coordinates
(it is a rearrangement of gamma*2*gamma*L/c).
In other words, the correct answer depends on what question you are really asking. I tried to read your words carefully, and based on that reading, my answer is #2.
Do you have a point, or are you just posting problems for us to solve for our pleasure?
Russell Blackadar
>  Nicolaas Vroom wrote: 
> > 
Russell Blackadar 
> 
Hi, Nicolaas. Nice to see you again, it's been a while.
(regarding Born rigidity) Nicolaas writes: 
> > > >  It is interesting to note that in the book Ray d'Inverno (who uses a train) that he does not mention this problem. 
> > >  Does he *accelerate* the train, in the passage you are referring to? I bet not. 
> > 
During the experiment acceleration is zero. Speed of the train v = constant.
Under the same conditions I perform the following experiment: 
> 
In other words, you mean its length *now* in the lab frame is L0?
If so, be aware, your notation is very confusing. According to the usual convention, if you write an L with a subscript 0, you mean *proper* length. But here that is apparently not what you mean. 
With L0 I mean the length when v=0. With L0 I mean the length that you calculate based on the time that you measure when you sent a signal to the front of the rod and back (in the rest frame) i.e. t = 2*L0/c
It is very important to note that the Observer stays at rest at position A.
> > 
(Measured from the back (but front should have the same))
The observer A is at position A at rest in the rest frame.
When the back B of the rod is at A the observer sents a light signal.
This signal is reflected back at the front end F (at tF)
The signal is back at Observer A at tA.
What does tA measure: 1) tA = 2 * L0 / (cv) 
>  The above equation is correct, using L0 as you've defined it. 
> > 
2) tA <> 2 * L0 / (cv) 3) tA = 2 * L0 * SQR(1  v^2/c^2) /(cv) 
> 
Equation 3 *would* have been correct if you had defined L0
in the usual way. (And of course in that case, inequality 2
would also have been correct for nonzero v.)
"more snippage" 
> > 
Almost the same problem as above.
Consider two space ships A and B.
At "rest" they are a distance L apart.
This is the initial position at v = 0
During a certain period they both accelerate exactly in the same manner. (Measured from the rest frame) 
>  I assume you mean the *original* rest frame (i.e. the frame analogous to your track frame above). Right? 
> >  Then they both make their a = 0 and both will have the same speed. A is behind and B is in the front. 
> 
When you say "then", I assume you are still talking
about the *original* rest frame (or track frame), right?
This is important to specify, because the way I stated it, B actually accelerates for a greater *proper* time than does A. 
The test is performed using a string of tests points X The distance between each two points is L
X X X A B X X X X >At each point there is a clock synchronised in rest frame. Both space ships start at t=0
>  After the accelerations, the separation in the original rest frame is still L. But in the frame of A and B, it is gamma * L. 
> > 
A sents a light signal to B (From the back of A to the back of B) this signal is reflected back to B 
> 
I assume you mean reflected *from* B back toward A. 
Yes I see, I made a mistake.
> > 
The signal is back at Observer A at tA. 
> 
Does observer A move with ship A, or is he motionless in the original frame? I will assume the latter, since you said this problem is almost the same as the previous one. In that case the answer is tA = 2 * L / (cv) 
I agree with you but I meant the next. That is A moves with space ship A but uses clocks in the original restframe.
> > 
What does tA measure:
(Using a clock in the rest frame)
1) tA = L / (cv) + L / (c+v) 2) tA <> L / (cv) + L / (c+v) 3) tA = (L / (cv) + L / (c+v)) * SQR(1  v^2/c^2) 
> 
Only your inequality 2 is correct, under the assumptions I stated.
Equation 3 *would* have been correct if you meant the time measured by an observer riding on ship A,
using his own clock. (That is, the righthand side is just an algebraic rearrangement of 2*gamma*L/c.) 
I agree with you.
Specific with equation 1
If I perform the experiment and the result is accordingly
to equation 1 then I know that the distance between
the two space ships is still L.
i.e. the distance is not contracted
(as a function of v.)
>  In other words, the correct answer depends on what question you are really asking. I tried to read your words carefully, and based on that reading, my answer is #2. 
I groszly appreciate your effort !
>  Do you have a point, or are you just posting problems for us to solve for our pleasure? 
I am not absolute sure what your answer is in the first example. I want to have the solution in the original restframe coordinates. No moving clocks are involved.
My first example adresses the moving train problem and I want to see how you establish if the length of the train actually is contracted as a function of v.
The second example adresses the distance issue of two objects with equal speed. No "distance" contraction is involved.
> 

"Asking question is not easy....."
https://www.nicvroom.be/length3.htm
https://www.nicvroom.be/length4.htm
> 
Russell Blackadar 
> >  Nicolaas Vroom wrote: 
> > > 
Russell Blackadar 
> > 
Hi, Nicolaas. Nice to see you again, it's been a while. (regarding Born rigidity) Nicolaas writes: 
> > > > >  It is interesting to note that in the book Ray d'Inverno (who uses a train) that he does not mention this problem. 
> > > > 
Does he *accelerate* the train, in the passage you are referring to? I bet not. 
> > > 
During the experiment acceleration is zero. Speed of the train v = constant. Under the same conditions I perform the following experiment: Length of the rod in rest frame (track frame, lab frame) = L0 The rod has a constant speed v in the rest frame. 
> > 
In other words, you mean its length *now* in the lab frame is L0? If so, be aware, your notation is very confusing. According to the usual convention, if you write an L with a subscript 0, you mean *proper* length. But here that is apparently not what you mean. 
> 
With L0 I mean the length when v=0. 
Ok, thanks for the clarification. (You see how important it is to be clear.) Hopefully I also made it clear, below, that this would change my answer.
>  With L0 I mean the length that you calculate based on the time that you measure when you sent a signal to the front of the rod and back (in the rest frame) i.e. t = 2*L0/c 
Ok. Or you could just measure it with meter sticks.
> 
It is very important to note that the Observer stays at rest at position A. 
> > > 
(Measured from the back (but front should have the same))
The observer A is at position A at rest in the rest frame.
When the back B of the rod is at A the observer sents a light signal.
This signal is reflected back at the front end F (at tF)
The signal is back at Observer A at tA.
What does tA measure: 
> > 
The above equation is correct, using L0 as you've defined it. 
> > > 
2) tA <> 2 * L0 / (cv) 
> > 
Equation 3 *would* have been correct if you had defined L0 in the usual way. (And of course in that case, inequality 2 would also have been correct for nonzero v.) 
Equation 3 is the correct answer.
> > 

> > > 
Almost the same problem as above. Consider two space ships A and B. At "rest" they are a distance L apart. This is the initial position at v = 0 During a certain period they both accelerate exactly in the same manner. (Measured from the rest frame) 
> > 
I assume you mean the *original* rest frame (i.e. the frame analogous to your track frame above). Right? 
> > > 
Then they both make their a = 0 and both will have the same speed. A is behind and B is in the front. 
> > 
When you say "then", I assume you are still talking about the *original* rest frame (or track frame), right? This is important to specify, because the way I stated it, B actually accelerates for a greater *proper* time than does A. 
> 
The test is performed using a string of tests points X The distance between each two points is L X X X A B X X X X >At each point there is a clock synchronised in rest frame. Both space ships start at t=0 
It's best not to include the acceleration in your timing, otherwise the problem becomes complicated as I've said in previous posts. Let's say t=0 is the time when they both *stop* accelerating, in the original rest frame, i.e. the time when they begin their inertial motion at the new final speed v.
> 
> > 
After the accelerations, the separation in the original rest frame is still L. But in the frame of A and B, it is gamma * L. 
> > > 
A sents a light signal to B (From the back of A to the back of B) this signal is reflected back to B 
> > 
I assume you mean reflected *from* B back toward A. 
> 
Yes I see I made a mistake. 
> > > 
The signal is back at Observer A at tA. 
> > 
Does observer A move with ship A, or is he motionless in the original frame? I will assume the latter,
since you said this problem is almost the same as the previous one. tA = 2 * L / (cv) 
> 
I agree with you but I meant the next. That is A moves with space ship A but uses clocks in the original restframe. 
I assume you mean, he looks out at the X point he is passing when he sends the signal, and takes a reading off the clock he sees there, then later when the signal returns, he reads the clock at a *different* X point that he happens to be passing at that later moment. (Of course you will need fractional X points for that to work precisely, not just the few you have drawn at a discrete interval of L.) Right?
In that case, the correct answer is equation 1.
> 
> > 
> > > 
What does tA measure:
(Using a clock in the rest frame) 
> > 
Only your inequality 2 is correct, under the assumptions I stated. Equation 3 *would* have been correct if you meant the time measured by an observer riding on ship A, using his own clock. (That is, the righthand side is just an algebraic rearrangement of 2*gamma*L/c.) Also in that same case, Equation 1 would have expressed the elapsed time in the original restframe coordinates (it is a rearrangement of gamma*2*gamma*L/c). 
> 
I agree with you. Specific with equation 1 If I perform the experiment and the result is accordingly to equation 1 then I know that the distance between the two space ships is still L. 
In the original frame, yes.
>  i.e. the distance is not contracted (as a function of v.) 
Actually it *is* contracted. The proper separation between the two ships (i.e. in their own frame) has increased from L to gamma*L, due to the unequal accelerations of the two ships in their comoving frames. The contraction in the original frame brings the distance in that original frame back down to L.
> 
> > 
In other words, the correct answer depends on what question you are really asking. I tried to read your words carefully, and based on that reading, my answer is #2. 
> 
I groszly appreciate your effort ! 
> > 
Do you have a point, or are you just posting problems for us to solve for our pleasure? 
> 
I am not absolute sure what your answer is in the first example. 
I gave you the wrong answer in my previous post due to my misunderstanding of your question. The correct answer is Equation 3.
>  I want to have the solution in the original restframe coordinates. No moving clocks are involved. 
Right.
>  My first example adresses the moving train problem and I want to see how you establish if the length of the train actually is contracted as a function of v. 
And my answer says, it is contracted. Note, I've implicitly assumed you haven't physically stretched or compressed the train somehow during the acceleration, or at least if you have, it has settled back to its original strainfree state in the new frame. In other words, its proper length stays at L0.
>  The second example adresses the distance issue of two objects with equal speed. No "distance" contraction is involved. 
But you need to understand, the final distance apart depends on the details of the acceleration. I made assumptions about the acceleration which might not be the ones you had in mind; and if they were not, my answer is not correct. Furthermore I am not happy to see you making, apparently, a general conclusion based on this answer of mine, which I assure you is definitely *not* general.
Russell Blackadar
>  Nicolaas Vroom wrote: 
> > 
Russell Blackadar 
> > > > 
2) tA <> 2 * L0 / (cv) 3) tA = 2 * L0 * SQR(1  v^2/c^2) /(cv) 
> > > 
Equation 3 *would* have been correct if you had defined L0 in the usual way. (And of course in that case, inequality 2 would also have been correct for nonzero v.) 
> 
Equation 3 is the correct answer. 
Now we both agree we can go back to the moving train problem.
Starting point is a track AB and a train A'B'
The following sketch shows this
M' A'B' v=0 AB M Figure 1The length of both in rest/track frame is identical i.e. L0 At the points A' and B' there are lamps on the train At the points A and B there are firing devices near the track. In the drawn position (A=A') , (B=B') both lamps are on. M is an observer equidistant from A and B.
At A and B there are also clocks in the rest/track frame. The clocks are synchronised by sending a light signal from M
After synchronisation we can perform our first experiment: Two observers at A and B will turn on/off the lamps (Using a separate switch) at the same clock reading. Result : The Observer at M will see those signals simultaneous.
Next we perform this second experiment.
We place the train a distance away to the left
We move the train at a constant speed v to the right.
At a certain moment the situation becomes the following:
M' A'B' v<>0 > AB M Figure 2At that moment A will make contact with A' and the lamp at A will flash. But not at B. IMO M will not see the lights simultaneous.
In the book"Introducing Einstein's Relativity"
by Ray D'Inverno at page 23 the same problem
is discussed. There is "written":
"Imagine a carriage travelling with velocity v realitive
to an Observer M on the bank of the track...
We assume that there are two electrical devices on
the track which are the length of the carriage apart
and equidistant from M.
When the carriage goes over thes devices, they fire
and activate two light sources situated at each end of
the carriage (Fig 2.13)
From the configuration, it is clear that M will judge that
the two events, when the light sources first switch on,
occur simultaneously."
Fig 2.13 is like figure 1 i.e. the line A'B' (carriage) has the same length as AB.
The question now becomes: How do they do that. IMO it is clear that the carriage of fig 2.13 when at rest (V=0) is longer as the distance AB i.e. L0
It is clear that when you draw AB equal to A'B' that the light at M will switch on simultaneous but that same is not obvious when you do the actual experiment
IMO the real issue is the value of the two clocks
at A and B in figure 1 i.e. Fig 2.13
When of the moving Cariage in Fig 2.13
point A touches A' at tA
and point B touches B' at tB
is then also tA = tB ? (Measured track frame)
IMO the last sentence should be modified as follows:
"From the configuration, it is clear that when M sees
light from the two light sources simultaneous
that M will judge that the length of the moving carriage
is equal to the difference between the length of the
electrical devices on the track."
> >  I agree with you but I meant the next. That is A moves with space ship A but uses clocks in the original restframe. 
> 
I assume you mean, he looks out at the X point he is passing when he sends the signal, and takes a reading off the clock he sees there, then later when the signal returns, he reads the clock at a *different* X point that he happens to be passing at that later moment. (Of course you will need fractional X points for that to work precisely, not just the few you have drawn at a discrete interval of L.) Right? 
Exactly.
I prefer to study only in the rest frame.
> 
In that case, the correct answer is equation 1. 
> > 
> > > 
> > > > 
What does tA measure:
(Using a clock in the rest frame)
1) tA = L / (cv) + L / (c+v) 
> > > 
Only your inequality 2 is correct, under the assumptions I stated. Equation 3 *would* have been correct if you meant the time measured by an observer riding on ship A, using his own clock. (That is, the righthand side is just an algebraic rearrangement of 2*gamma*L/c.) Also in that same case, Equation 1 would have expressed the elapsed time in the original restframe coordinates (it is a rearrangement of gamma*2*gamma*L/c). 
> > 
I agree with you. Specific with equation 1 If I perform the experiment and the result is accordingly to equation 1 then I know that the distance between the two space ships is still L. 
> 
In the original frame, yes. 
> > 
i.e. the distance is not contracted (as a function of v.) 
> 
Actually it *is* contracted. 
I do not understand this. What is contracted is the length of each space ship.
If you perform the same experiment as above with Observer A in the back of space ship A and reflection via the front of the space ship of A (not back of B) and back towards Observer A then you will get equation 3 with L replaced by L0 That means (1) that space ship A is contracted. (For space ship B the same is true)
On the other hand (2) the distance between the back of space ship B and the front of space ship A is enlarged. (all measured in rest frame and no comoving clocks)
What is contracted in your opinion ?
>  The proper separation between the two ships (i.e. in their own frame) has increased from L to gamma*L, due to the unequal accelerations of the two ships in their comoving frames. 
This is a rather tricky sentence specific because you are speaking of comoving frames (i.e one of space ship A and one of B) and most probably you are right. I think (but not sure) that you should consider one comoving frame A and explain what the speed of space ship B is in that frame. (Both can be zero) However the reason IMO is first that a clock in this comoving plane runs different as a clock in the rest frame. Secondly different clocks in this comoving frame require synchronisation.
>  The contraction in the original frame brings the distance in that original frame back down to L. 
I do not understand this sentence.
What equation 1 tells me is that
1)A (in the back of A) sents out a signal at t=0
2)B (in the back of B) receive and reflects this
signal back to A at t = L/(cv)
3)A receives the signal at t = L/(cv)+L/(c+v)
All t, L and v are measured in the rest frame.
If that is true than there is no length contraction involved.
> 
> > 
> > > 
In other words, the correct answer depends on what question you are really asking. I tried to read your words carefully, and based on that reading, my answer is #2. 
> > 
I groszly appreciate your effort ! I want to have the solution in the original restframe coordinates. No moving clocks are involved. 
> 
Right. 
> > 
My first example adresses the moving train problem and I want to see how you establish if the length of the train actually is contracted as a function of v. 
> 
And my answer says, it is contracted. Note, I've implicitly assumed you haven't physically stretched or compressed the train somehow during the acceleration, or at least if you have, it has settled back to its original strainfree state in the new frame. In other words, its proper length stays at L0. 
I have not physical stretched nor compressed the train. On the other hand I want to come back to this issue later.
> 
> > 
The second example adresses the distance issue of two objects with equal speed. No "distance" contraction is involved. 
> 
But you need to understand, the final distance apart depends on the details of the acceleration. 
The accelaration I used is measured from the rest
frame.
I assume that both spaceships start at the same time
and have the same acceleration for the same duration.
Then each their acceleration becomes zero.
IMO during this whole experiment the speed of both
spaceships is identical.
>  I made assumptions about the acceleration which might not be the ones you had in mind; and if they were not, my answer is not correct. Furthermore I am not happy to see you making, apparently, a general conclusion based on this answer of mine, which I assure you is definitely *not* general. 
Then what is the general conclusion.
Bell' spaceship Paradox is as follows:
"To begin, a statement of the paradox,Bell asks us
to consider two rocket ships, each accelerating
at the same constant rate, one chasing the other.
The ships start out at rest in some coordinate system
(the "labframe").
Since they have the same acceleration, their speeds
should be equal at all times (relative to the labframe)
and so they should stay a constant distance apart
(in the labframe).
But after a time they will acquire a large velocity,
and so the distance between them should suffer
Lorentz contraction. Which is it? "
IMO at least the last sentence should be modified
as follows:
"At the same time their speed is constantly increasing
and so the distance between them should suffer
more and more Lorentz contraction."
IMO the question should be: "How can both be true" or...
In order to find that answer what you have is the following: In spaceship A observer A is in the back. In spaceship B observer B is in the back. Both observers each have two clocks. All the clocks are synchronised in the rest frame. Assume a constant acceleration During this period both A and B will see that their clocks runs slower as the clocks in the rest frame. (Along the whole length of the trip their are clocks positioned which run synchronised in the rest frame) The two clocks of B are not synchronised with the clocks in the comoving frame of A. B will synchronise one of his clocks with A's clock. Because the speed of A,B is constantly increasing B has to do that constantly.
Using this synchronised clock of B in order to calculate B' s Acceleration and you compare that with A's Accelaration using A's clock, ofcourse the two are not the same.
The reason is synchronisation (in the comoving frame)
If you make acceleration equal to zero than you can state the following:
Using A's clock to measure A's length
you will get:
tA = (L0 / (cv) + L0 / (c+v)) * SQR(1  v^2/c^2) * SQR(1v^2/c^2)
tA = 2 * L0 / c
The reason is time dilation and length contraction
If you want to measure the distance between the
two space ships (back to back) using A's clock you will
get:
tA = (L/ (cv) + L / (c+v)) * SQR(1  v^2/c^2)
Reason is time dilation.
The difference is the distance between the front of A and the back of B This value is increasing.
This will be my last post in this thread; I simply do not have the time to answer all your questions, Nicolaas.
> 
Russell Blackadar 
> >  Nicolaas Vroom wrote: 
> > > 
Russell Blackadar 
> > > > > 
2) tA <> 2 * L0 / (cv) 3) tA = 2 * L0 * SQR(1  v^2/c^2) /(cv) 
> > > > 
Equation 3 *would* have been correct if you had defined L0 in the usual way. (And of course in that case, inequality 2 would also have been correct for nonzero v.) 
> > 
Equation 3 is the correct answer. 
> 
Now we both agree we can go back to the
moving train problem.
Starting point is a track AB and a train A'B' The following sketch shows this M' A'B' v=0 AB M Figure 1 
Are we to assume that the device at A fires when A' passes it, and the device at B fires when B' passes it? Ok.
> 
In the drawn position (A=A') , (B=B') both lamps are on.
M is an observer equidistant from A and B.
At A and B there are also clocks in the rest/track frame. The clocks are synchronised by sending a light signal from M After synchronisation we can perform our first experiment: Two observers at A and B will turn on/off the lamps (Using a separate switch) at the same clock reading. Result : The Observer at M will see those signals simultaneous. 
Yes.
> 
Next we perform this second experiment.
We place the train a distance away to the left
We move the train at a constant speed v to the right.
At a certain moment the situation becomes the following: M' A'B' v<>0 > AB M Figure 2At that moment A will make contact with A' and the lamp at A will flash. But not at B. IMO M will not see the lights simultaneous. 
Correct. Note, this does not conflict with your previous experiment.
>  In the book"Introducing Einstein's Relativity" by Ray D'Inverno at page 23 the same problem is discussed. 
I don't have that book, and I'm not really interested in critiquing it simply on the basis of quotes by you. However, as far as I can tell, it is *not* the same problem.
> 
There is "written": "Imagine a carriage travelling with velocity v realitive to an Observer M on the bank of the track... We assume that there are two electrical devices on the track which are the length of the carriage apart and equidistant from M. When the carriage goes over thes devices, they fire and activate two light sources situated at each end of the carriage (Fig 2.13) From the configuration, it is clear that M will judge that the two events, when the light sources first switch on, occur simultaneously." 
The only way this would be the same as your problem, would be if d'Inverno had meant "proper length" when he said "length". But since he didn't say proper length, and since we can easily tell by context that he didn't mean proper length, there is no doubt that he is talking about a different problem.
He is talking about a case where the proper length of the train is *greater* than the length between the marks on the track, so that the length measured in the track frame (hence contracted, because v > 0 in that frame) is equal to the length between the marks on the track.
>  Fig 2.13 is like figure 1 i.e. the line A'B' (carriage) has the same length as AB. 
It's like your Figure 1 with one important exception: namely, your notation "v=0". In d'Inverno's case, v is not 0. Also he has probably said nothing about how the length A'B' (in the train frame) compares to AB in the track frame.
>  The question now becomes: How do they do that. IMO it is clear that the carriage of fig 2.13 when at rest (V=0) is longer as the distance AB i.e. L0 
Longer *than* the distance AB. Yes.
>  It is clear that when you draw AB equal to A'B' that the light at M will switch on simultaneous but that same is not obvious when you do the actual experiment 
I don't understand why you think there is a problem. As you say, the train is longer in its own frame, but in the track frame (where the length is contracted) it is equal; that makes the switches simultaneous in the track frame (but not in the train frame).
I agree this is not what our naive intuitions would tell us, but still, it's not hard to see that it is mathematically consistent.
>  IMO the real issue is the value of the two clocks at A and B in figure 1 i.e. Fig 2.13 When of the moving Cariage in Fig 2.13 point A touches A' at tA and point B touches B' at tB is then also tA = tB ? (Measured track frame) 
Yes, in this example.
> 
IMO the last sentence should be modified as follows: "From the configuration, it is clear that when M sees light from the two light sources simultaneous that M will judge that the length of the moving carriage is equal to the difference between the length of the electrical devices on the track." 
Which last sentence? It doesn't seem to me that you have contradicted anything in the quoted d'Inverno passage.
Yes, that will be the train's *measured length* in the track frame. The train's *proper length* (i.e. its length in a frame where its v=0) will be different.
> 
> > > 
I agree with you but I meant the next. That is A moves with space ship A but uses clocks in the original restframe. 
> > 
I assume you mean, he looks out at the X point he is passing when he sends the signal, and takes a reading off the clock he sees there, then later when the signal returns, he reads the clock at a *different* X point that he happens to be passing at that later moment. (Of course you will need fractional X points for that to work precisely, not just the few you have drawn at a discrete interval of L.) Right? 
> 
Exactly. I prefer to study only in the rest frame. 
Careful, here. In relativity, *any* frame can be the "rest frame"; that term is ambiguous unless you tell us what object or objects (or observer) you consider to be at rest. So, to avoid ambiguity, you should always state the frame clearly, e.g. *original* rest frame, in this problem.
> > 
In that case, the correct answer is equation 1. 
> > > 
> > > > 
> > > > > 
What does tA measure:
(Using a clock in the rest frame) 
> > > > 
Only your inequality 2 is correct, under the assumptions I stated. Equation 3 *would* have been correct if you meant the time measured by an observer riding on ship A, using his own clock. (That is, the righthand side is just an algebraic rearrangement of 2*gamma*L/c.) Also in that same case, Equation 1 would have expressed the elapsed time in the original restframe coordinates (it is a rearrangement of gamma*2*gamma*L/c). 
> > > 
I agree with you. Specific with equation 1 If I perform the experiment and the result is accordingly to equation 1 then I know that the distance between the two space ships is still L. 
> > 
In the original frame, yes. 
> > > 
i.e. the distance is not contracted (as a function of v.) 
> > 
Actually it *is* contracted. 
> 
I do not understand this. 
Sorry, I didn't mean to confuse you. The contraction I was talking about was the difference between the final "proper" distance (i.e. the distance as measured in the rockets' final frame, gamma*L) vs. the final distance as measured in the original frame (namely, gamma*L/gamma = L). It's an effect of perspective (i.e. transforming between frames), not a physical effect. You can choose not to use that terminology if you prefer, but you should at least understand it so you can follow what other people say  because my usage is quite common.
The important point here is that I don't want you to think that contraction (in *my* sense, i.e. due to the transform between frames) does not occur. It *does* occur. But in *this particular example*, you have specified the acceleration in such a way that a physical expansion exactly counteracts the contraction, with the result that the L stays the same in the original frame.
>  What is contracted is the length of each space ship. 
Yes, those are contracted too, but we weren't talking about them.
> 
If you perform the same experiment as above with Observer A in the back of space ship A and reflection via the front of the space ship of A (not back of B) and back towards Observer A then you will get equation 3 with L replaced by L0 That means (1) that space ship A is contracted. (For space ship B the same is true) On the other hand (2) the distance between the back of space ship B and the front of space ship A is enlarged. (all measured in rest frame and no comoving clocks) 
Yes, it is slightly enlarged, but we weren't talking about that either.
>  What is contracted in your opinion ? 
See above. The contraction is not a physical effect. The expansion (which is specific to your acceleration spec) *is* a physical effect. Put the two together, and you get no change in the distance as measured in the original frame, for this particular case.
> >  The proper separation between the two ships (i.e. in their own frame) has increased from L to gamma*L, due to the unequal accelerations of the two ships in their comoving frames. 
> 
This is a rather tricky sentence specific because you are speaking of comoving frames (i.e one of space ship A and one of B) and most probably you are right. I think (but not sure) that you should consider one comoving frame A and explain what the speed of space ship B is in that frame. (Both can be zero) 
Both *are* zero, in the final state.
>  However the reason IMO is first that a clock in this comoving plane runs different as a clock in the rest frame. Secondly different clocks in this comoving frame require synchronisation. 
Right, accelerations make everything quite tricky; I'm glad you appreciate this.
> 
> > 
The contraction in the original frame brings the distance in that original frame back down to L. 
> 
I do not understand this sentence.
What equation 1 tells me is that If that is true than there is no length contraction involved. 
Yes, it's the same distance in that frame as earlier. That's because there was a Lorentz contraction to counteract the physical expansion of the distance between the rockets (i.e. their separation in their own frame).
> 
> > 
> > > 
> > > > 
In other words, the correct answer depends on what question you are really asking. I tried to read your words carefully, and based on that reading, my answer is #2. 
> > > 
I groszly appreciate your effort ! I want to have the solution in the original restframe coordinates. No moving clocks are involved. 
> > 
Right. 
> > > 
My first example adresses the moving train problem and I want to see how you establish if the length of the train actually is contracted as a function of v. 
> > 
And my answer says, it is contracted. Note, I've implicitly assumed you haven't physically stretched or compressed the train somehow during the acceleration, or at least if you have, it has settled back to its original strainfree state in the new frame. In other words, its proper length stays at L0. 
> 
I have not physical stretched nor compressed the train. On the other hand I want to come back to this issue later. 
> > 
> > > 
The second example adresses the distance issue of two objects with equal speed. No "distance" contraction is involved. 
> > 
But you need to understand, the final distance apart depends on the details of the acceleration. 
> 
The accelaration I used is measured from the rest frame. I assume that both spaceships start at the same time and have the same acceleration for the same duration. Then each their acceleration becomes zero. IMO during this whole experiment the speed of both spaceships is identical. 
Right. But you could have done it differently, and if you had, the outcome would be different.
> 
> > 
I made assumptions about the acceleration which might not be the ones you had in mind; and if they were not, my answer is not correct. Furthermore I am not happy to see you making, apparently, a general conclusion based on this answer of mine, which I assure you is definitely *not* general. 
> 
Then what is the general conclusion. 
That when you transform from one frame to another, distances change by a factor of gamma.
> 
Bell' spaceship Paradox is as follows: "To begin, a statement of the paradox,Bell asks us to consider two rocket ships, each accelerating at the same constant rate, one chasing the other. The ships start out at rest in some coordinate system (the "labframe"). Since they have the same acceleration, their speeds should be equal at all times (relative to the labframe) and so they should stay a constant distance apart (in the labframe). But after a time they will acquire a large velocity, and so the distance between them should suffer Lorentz contraction. Which is it? "
IMO at least the last sentence should be modified
as follows: 
That's a good way to put it, but IMHO the original is not really wrong either. Remember, this is a statement of a seeming (but actually illusory) paradox, so it's not meant to be precise.
>  IMO the question should be: "How can both be true" or... 
Yes, that might be better. So, here's the answer to *your* question:
Both *can* be true if the physical distance increases by a factor of gamma, because the measured distance (in the original frame) is just a Lorentzcontracted measurement of the physical distance. In other words, the measured L can stay the same if the gammas cancel. Whether or not it *will* stay the same, in a particular case, depends on how you do the acceleration.
> 
In order to find that answer what you have is the following: In spaceship A observer A is in the back. In spaceship B observer B is in the back. Both observers each have two clocks. All the clocks are synchronised in the rest frame. Assume a constant acceleration. During this period both A and B will see that their clocks runs slower as the clocks in the rest frame. 
No, A and B will say the "rest frame" clocks are running slower, due to an effect of the relativity of simultaneity. I realize this is backwards from what you might naively think, but that is what SR says.
> 
(Along the whole length of the trip their are clocks
positioned which run synchronised in the rest frame) The two clocks of B are not synchronised with the clocks in the comoving frame of A. B will synchronise one of his clocks with A's clock. Because the speed of A,B is constantly increasing B has to do that constantly. 
Note, if he does this, B's clock will not indicate his proper time. I don't see much reason why B should bother to do this, since (as you've specified the acceleration) A and B won't be in the same inertial frame at any time between the original and final states.
>  Using this synchronised clock of B in order to calculate B' s Acceleration and you compare that with A's Accelaration using A's clock, ofcourse the two are not the same. 
Right, and even if you use proper times, you will find that A's proper acceleration differs slightly from B's proper acceleration, in this example as you've posed it. B's will be greater but will last for a shorter proper time (I think I said that wrong in my previous post). The result is that the final v's will be the same, but the final proper separation will be greater.
> 
The reason is synchronisation (in the comoving frame) If you make acceleration equal to zero than you can state the following: Using A's clock to measure A's length you will get: tA = (L0 / (cv) + L0 / (c+v)) * SQR(1  v^2/c^2) * SQR(1v^2/c^2) tA = 2 * L0 / c The reason is time dilation and length contraction If you want to measure the distance between the two space ships (back to back) using A's clock you will get: tA = (L/ (cv) + L / (c+v)) * SQR(1  v^2/c^2) Reason is time dilation. 
No, the reason is the physical increase in the distance between A and B, due to their unequal proper acceleration as you've specified it in this problem.
Time dilation is a nonphysical effect of transforming between frames. Since A is measuring a distance using his own clock, there is no transform in this case, so time dilation is not relevant.
>  The difference is the distance between the front of A and the back of B This value is increasing. 
Yes it's increasing, but this also is irrelevant to what we were talking about.
I am sorry to end our dialog on this note, but I really cannot spend any more time here. Best of luck in your continuing studies.
>  This will be my last post in this thread 
Woops, except for (at least) one correction...
...
>  Note, if he does this, B's clock will not indicate his proper time. I don't see much reason why B should bother to do this, 
[i.e. resynchronize his clock with A]
>  since (as you've specified the acceleration) A and B won't be in the same inertial frame at any time between the original and final states. 
Clearly wrong. If the acceleration is the same in the original frame, then there will always be a comoving inertial frame in which both rockets are motionless.
Accounting for the difference in proper times (essentially a twinparadox sort of thing) requires a more subtle explanation, one I'm not prepared to give at this time. Ah well.
Russell Blackadar
>  This will be my last post in this thread; I simply do not have the time to answer all your questions, Nicolaas. 
I really appreciate all your efforts.
I hope that other readers of this newsgroup will
try to answer my questions and remarks.
>  Nicolaas Vroom wrote: 
> > 
In the book"Introducing Einstein's Relativity" by Ray D'Inverno at page 23 the same problem is discussed. 
> 
I don't have that book, and I'm not really interested in critiquing it simply on the basis of quotes by you. However, as far as I can tell, it is *not* the same problem. 
> > 
There is "written": 
> 
The only way this would be the same as your problem, would be if d'Inverno had meant "proper length" when he said "length". But since he didn't say proper length, and since we can easily tell by context that he didn't mean proper length, there is no doubt that he is talking about a different problem. He is talking about a case where the proper length of the train is *greater* than the length between the marks on the track, so that the length measured in the track frame (hence contracted, because v > 0 in that frame) is equal to the length between the marks on the track. 
I agree with you. As such we agree that the length L of the moving train A'B' is contracted i.e. is shorter as its rest length L0 However that does not make everything clear
What we have are two observers: M in the middle between AB near the track. M' in the middle between A'B' on the moving train. M sees the flashes simultaneous. M' sees the flashes not simultaneous. The train is length contracted.
How do we explain all of this?
I have doubts if M sees the flashes simultaneous. IMO it is even possible that M' sees the flashes simultaneous.
What you should do is consider the whole experiment in a more *abstract* way. in a more symetrical fashion.
A'M'B' v>  Track AMB v=0
We have two trains AB and A'B' We have one (Multi purpose) track. AB is at rest. A'B' moves to the right. M is in the middle between AB and sees the flashes simultaneous. M' is in the middle between A'B' and does not sees the flashes simultaneous. At each train there is also a rod with the same length as the train. When train AB meets A'B' there one spark at both ends.
For M' to see the flashes simultaneous M' has to move. But than M' is not any more in the middle between A'B' That means When M' sees the two sparks simultaneous At the position of M' M' breaks the (moving) rod A'B' in two parts: A'M' and M'B' and places them side by side they are not of equal length. Even when the train /rod stops.
You can repeat this same experiment with a different train A''B'' with a speed v'' The two rods A''M'' and M''B'' are not equal.
(Any number of trains n will give AnMn <> MnBn)
You can also do that with the rod AB of M when M
(at rest) sees the two flashes simultaneous.
Is AM equal to MB ?
IMO NO.
Is the train AB with M special in some way ?
NO.
Is it possible that AxMx = MxBx voor some train x ? Yes but than v<>0 .......
*snip*
> 
> > 
Exactly. I prefer to study only in the rest frame. 
> 
Careful, here. In relativity, *any* frame can be the "rest frame"; that term is ambiguous unless you tell us what object or objects (or observer) you consider to be at rest. So, to avoid ambiguity, you should always state the frame clearly, e.g. *original* rest frame, in this problem. 
> > > 
In that case, the correct answer is equation 1. 
> > > > 
> > > > > 
> > > > > > 
What does tA measure:
(Using a clock in the rest frame) 
> > > > > 
Only your inequality 2 is correct, under the assumptions I stated. Equation 3 *would* have been correct if you meant the time measured by an observer riding on ship A, using his own clock. (That is, the righthand side is just an algebraic rearrangement of 2*gamma*L/c.) Also in that same case, Equation 1 would have expressed the elapsed time in the original restframe coordinates (it is a rearrangement of gamma*2*gamma*L/c). 
> > > > 
I agree with you. Specific with equation 1 If I perform the experiment and the result is accordingly to equation 1 then I know that the distance between the two space ships is still L. 
> > > 
In the original frame, yes. 
> > > > 
i.e. the distance is not contracted (as a function of v.) 
> > > 
Actually it *is* contracted. 
> >  I do not understand this. 
> 
Sorry, I didn't mean to confuse you. The contraction I was talking about was the difference between the final "proper" distance (i.e. the distance as measured in the rockets' final frame, gamma*L) vs. the final distance as measured in the original frame (namely, gamma*L/gamma = L). It's an effect of perspective (i.e. transforming between frames), not a physical effect. 
Yes you are confusing me.
When the length of the shape ship is equal to the
distance during this experiment,
then considered from the Rest Frame:
1. The distance (back to back) stays constant
and is accordingly to equation 1
2. The length of the space ship physical decreases with dl
and is accordingly to equation 3
3. The distance (front to back) increases with dl
i.e. EQ1  EQ 3
In the rockets final frame the measured values are different.
1. The distance (back to back) is equal to:
tA = L / (cv) + L / (c+v)* SQR(1v^2/c^2) Eq 42. The length of the space ship physical decreases
tA = (L / (cv) + L / (c+v)) * SQR(1  v^2/c^2) * SQR(1  v^2/c^2) tA = 2*L/c Eq53. The distance (front to back) increases with EQ4  Eq 5
IMO it is confusion to say that Eq4 represents length expansion. Nothing is physical expanded. It *seems* that the distance increases but that comes (not completely) because we measure a distance with a clock which runs slower.
> 
You can choose not to use that terminology if
you prefer, but you should at least understand it so
you can follow what other people say  because my usage
is quite common.
The important point here is that I don't want you to think that contraction (in *my* sense, i.e. due to the transform between frames) does not occur. It *does* occur. But in *this particular example*, you have specified the acceleration in such a way that a physical expansion exactly counteracts the contraction, with the result that the L stays the same in the original frame. 
> > 
What is contracted is the length of each space ship. 
> 
Yes, those are contracted too, but we weren't talking about them. 
It is important to include inorder to "see the total picture"
> > 
If you perform the same experiment as above with Observer A in the back of space ship A and reflection via the front of the space ship of A (not back of B) and back towards Observer A then you will get equation 3 with L replaced by L0 That means (1) that space ship A is contracted. (For space ship B the same is true) On the other hand (2) the distance between the back of space ship B and the front of space ship A is enlarged. (all measured in rest frame and no comoving clocks) 
> 
Yes, it is slightly enlarged, but we weren't talking about that either. 
> > 
What is contracted in your opinion ? 
> 
See above. The contraction is not a physical effect. The expansion (which is specific to your acceleration spec) *is* a physical effect. Put the two together, and you get no change in the distance as measured in the original frame, for this particular case. 
I do not fully grasp this. IMO in the rest frame nothing expans nor contracts.
Equation 1 is what we actual measure. Equation 1 describes the situation (a model) of two points a distance L apart (in rest frame) moving both with speed v (in rest frame). Why do we need such a "difficult" explanation.
Equation 4 is the same situation but now measured in the moving frame of the space ships. The difference is the factor SQR(1v^2/c^2) which represents the physical effect that moving clocks are running slower.
> > 
> > > 
The proper separation between the two ships (i.e. in their own frame) has increased from L to gamma*L, due to the unequal accelerations of the two ships in their comoving frames. 
> > 
This is a rather tricky sentence specific because you are speaking of comoving frames (i.e one of space ship A and one of B) and most probably you are right. I think (but not sure) that you should consider one comoving frame A and explain what the speed of space ship B is in that frame. (Both can be zero) 
> 
Both *are* zero, in the final state. 
> > 
However the reason IMO is first that a clock in this comoving plane runs different as a clock in the rest frame. Secondly different clocks in this comoving frame require synchronisation. 
> 
Right, accelerations make everything quite tricky; I'm glad you appreciate this. 
> > 
> > > 
The contraction in the original frame brings the distance in that original frame back down to L. 
> > 
I do not understand this sentence.
What equation 1 tells me is that If that is true than there is no length contraction involved. 
> 
Yes, it's the same distance in that frame as earlier. That's because there was a Lorentz contraction to counteract the physical expansion of the distance between the rockets (i.e. their separation in their own frame). 
> > 
Bell' spaceship Paradox is as follows:
IMO at least the last sentence should be modified
as follows: 
> 
That's a good way to put it, but IMHO the original is not really wrong either. Remember, this is a statement of a seeming (but actually illusory) paradox, so it's not meant to be precise. 
What is it actual illusory in this experiment ? Why is it not precise ? Why does not it have to be precise ?
IMO it has to be precise in order to establish what is right (or wrong) and why.
> > 
IMO the question should be: "How can both be true" or... 
> 
Yes, that might be better. So, here's the answer to *your* question: Both *can* be true if the physical distance increases by a factor of gamma, because the measured distance (in the original frame) is just a Lorentzcontracted measurement of the physical distance. In other words, the measured L can stay the same if the gammas cancel. Whether or not it *will* stay the same, in a particular case, depends on how you do the acceleration. 
> >  In order to find that answer what you have is the following: In spaceship A observer A is in the back. In spaceship B observer B is in the back. Both observers each have two clocks. All the clocks are synchronised in the rest frame. Assume a constant acceleration During this period both A and B will see that their clocks runs slower as the clocks in the rest frame. 
>  No, A and B will say the "rest frame" clocks are running slower, due to an effect of the relativity of simultaneity. I realize this is backwards from what you might naively think, but that is what SR says. 
This is only true for the clock at the rest which is at the position of t=0 for A. A will recieve time pulses from that clock at a slower rate because the distance with that clock increases. However if you place clocks in the rest frame all along the path that A follows than A will realize that his moving clock runs slower (All the time)
> >  (Along the whole length of the trip their are clocks positioned which run synchronised in the rest frame) The two clocks of B are not synchronised with the clocks in the comoving frame of A. B will synchronise one of his clocks with A's clock. Because the speed of A,B is constantly increasing B has to do that constantly. 
>  Note, if he does this, B's clock will not indicate his proper time. I don't see much reason why B should bother to do this, since (as you've specified the acceleration) A and B won't be in the same inertial frame at any time between the original and final states. 
> >  Using this synchronised clock of B in order to calculate B' s Acceleration and you compare that with A's Accelaration using A's clock, ofcourse the two are not the same. 
>  Right, and even if you use proper times, you will find that A's proper acceleration differs slightly from B's proper acceleration, in this example as you've posed it. B's will be greater but will last for a shorter proper time (I think I said that wrong in my previous post). The result is that the final v's will be the same, but the final proper separation will be greater. 
> > 
The reason is synchronisation (in the comoving frame)
If you make acceleration equal to zero than you can state the following:
Using A's clock to measure A's length
you will get:
If you want to measure the distance between the
two space ships (back to back) using A's clock you will
get: 
>  No, the reason is the physical increase in the distance between A and B, due to their unequal proper acceleration as you've specified it in this problem. 
How do you measure this acceleration ?
>  Time dilation is a nonphysical effect of transforming between frames. Since A is measuring a distance using his own clock, there is no transform in this case, so time dilation is not relevant. 
> >  The difference is the distance between the front of A and the back of B This value is increasing. 
> 
Yes it's increasing, but this also is irrelevant to what
we were talking about.
I am sorry to end our dialog on this note, but I really cannot spend any more time here. Best of luck in your continuing studies. 
> 
Russell Blackadar 
> >  This will be my last post in this thread; I simply do not have the time to answer all your questions, Nicolaas. 
Ok, here's one more. But no guarantees for the future. Pay particular attention to my advice at the beginning (next couple of paragraphs) please.
> 
I really appreciate all your efforts. I hope that other readers of this newsgroup will try to answer my questions and remarks. 
Perhaps they will, but it's difficult and it never seems to lead anywhere. You just keep repeating the same questions, or slight variations of them.
Usenet, even at its best, is a very limited medium.
I think the best option for you is to take a course. You seem interested enough in the subject, and there simply is no substitute for getting the actual instruction from a living person who can explain things by drawing pictures on a chalkboard, etc. Also, getting feedback on homework assignments is crucial.
> 
> > 
Nicolaas Vroom wrote: 
> > > 
In the book"Introducing Einstein's Relativity" by Ray D'Inverno at page 23 the same problem is discussed. 
> > 
I don't have that book, and I'm not really interested in critiquing it simply on the basis of quotes by you. However, as far as I can tell, it is *not* the same problem. 
> > > 
There is "written": "Imagine a carriage travelling with velocity v realitive to an Observer M on the bank of the track... We assume that there are two electrical devices on the track which are the length of the carriage apart and equidistant from M. When the carriage goes over thes devices, they fire and activate two light sources situated at each end of the carriage (Fig 2.13) From the configuration, it is clear that M will judge that the two events, when the light sources first switch on, occur simultaneously." 
> > 
The only way this would be the same as your problem, would be if d'Inverno had meant "proper length" when he said "length". But since he didn't say proper length, and since we can easily tell by context that he didn't mean proper length, there is no doubt that he is talking about a different problem. He is talking about a case where the proper length of the train is *greater* than the length between the marks on the track, so that the length measured in the track frame (hence contracted, because v > 0 in that frame) is equal to the length between the marks on the track. 
> 
I agree with you. As such we agree that the length L of the moving train A'B' is contracted i.e. is shorter as its rest length L0 
You are not careful enough with your language. When you say it is contracted, you should say in what frame the measurement is made. (Yes I remember, last time you said that you prefer the track frame always, but you still should say that explicitly, every time, to avoid impling things that are untrue. A train is never "contracted" independent of frame.)
> 
However that does not make everything clear
What we have are two observers: M in the middle between AB near the track. M' in the middle between A'B' on the moving train. M sees the flashes simultaneous. M' sees the flashes not simultaneous. 
Note, this is because in his frame, the flashes literally *are not* simultaneous.
>  The train is length contracted. 
In the track frame.
> 
How do we explain all of this? I have doubts if M sees the flashes simultaneous. 
Why? You *set up* the problem so that he will. The flashes literally *are* simultaneous in this frame, and M is equidistant from the flash points (and remains so, since his v=0). And we assume light speed is c in both directions. So what is there to doubt?
>  IMO it is even possible that M' sees the flashes simultaneous. 
Your opinion does not matter; what matters is what the calculations (and of course experiments) show.
> 
What you should do is consider the whole experiment in a more *abstract* way. in a more symetrical fashion. A'M'B' v>  Track AMB v=0 
Sorry, but this is not symmetrical at all. Symmetry would require the lower train to move *left* at speed v. As it is, your A'B' train has a greater proper length than AB. But let's move on...
>  We have two trains AB and A'B' We have one (Multi purpose) track. AB is at rest. A'B' moves to the right. M is in the middle between AB and sees the flashes simultaneous. M' is in the middle between A'B' and does not sees the flashes simultaneous. At each train there is also a rod with the same length as the train. 
Why do you bother with rods? A train is as good as a rod (or at least we can pretend it is, in a thought experiment). Let's keep things as simple as possible.
>  When train AB meets A'B' there one spark at both ends. 
You say "when" without mentioning frame. Don't always assume we understand "track frame" without saying that *explicitly*. It makes your writing unclear at best.
> 
For M' to see the flashes simultaneous M' has to move. 
That's just another way of saying he sees them *not* simultaneously.
>  But than M' is not any more in the middle between A'B' That means When M' sees the two sparks simultaneous At the position of M' 
You are getting confused. Is M' a position, or is M' a person? If some person M' at position M' moves, then I would say his position is no longer M'. Your terminology frequently suffers from such confusion IMHO.
>  M' breaks the (moving) rod A'B' in two parts: A'M' and M'B' and places them side by side they are not of equal length. Even when the train /rod stops. 
Sure, but this line of thinking gets you nowhere. If the train stops and M' is not in the middle, he simply won't see the flashes simultaneously. He only sees them simultaneously (in the new M' position no longer at the midpoint) if he is moving rightward at speed v. And btw only then because you have adjusted this position precisely to make this be so.
> 
You can repeat this same experiment with a different train A''B'' with a speed v'' The two rods A''M'' and M''B'' are not equal. (Any number of trains n will give AnMn <> MnBn) You can also do that with the rod AB of M when M (at rest) sees the two flashes simultaneous. Is AM equal to MB ? IMO NO. 
Wrong.
>  Is the train AB with M special in some way ? NO. 
Wrong. It is special in *this* problem because of all the frames in which we could have specified that the flashes are set off simultaneously, we did so in *this* frame, namely, the frame in which train AB is motionless.
Yes there is nothing *intrinsically* special about train AB. Do you see the distinction?
> 
Is it possible that AxMx = MxBx voor some train x ? Yes but than v<>0 ....... 
You have confused yourself. AB is *the* train that fulfills this condition; this is how we set up the flashes, specifically so that it would be true. It is a given. (AB is the only train whose v=0 in the frame in which the flashes are simultaneous.)
> 
*snip* 
> > 
> > > 
Exactly. I prefer to study only in the rest frame. 
> > 
Careful, here. In relativity, *any* frame can be the "rest frame"; that term is ambiguous unless you tell us what object or objects (or observer) you consider to be at rest. So, to avoid ambiguity, you should always state the frame clearly, e.g. *original* rest frame, in this problem. 
> > > > 
In that case, the correct answer is equation 1. 
> > > > > 
> > > > > > 
> > > > > > > 
What does tA measure:
(Using a clock in the rest frame) 
> > > > > > 
Only your inequality 2 is correct, under the assumptions I stated. Equation 3 *would* have been correct if you meant the time measured by an observer riding on ship A, using his own clock. (That is, the righthand side is just an algebraic rearrangement of 2*gamma*L/c.) Also in that same case, Equation 1 would have expressed the elapsed time in the original restframe coordinates (it is a rearrangement of gamma*2*gamma*L/c). 
> > > > > 
I agree with you. Specific with equation 1 If I perform the experiment and the result is accordingly to equation 1 then I know that the distance between the two space ships is still L. 
> > > > 
In the original frame, yes. 
> > > > > 
i.e. the distance is not contracted (as a function of v.) 
> > > > 
Actually it *is* contracted. 
> > >  I do not understand this. 
> > 
Sorry, I didn't mean to confuse you. The contraction I was talking about was the difference between the final "proper" distance (i.e. the distance as measured in the rockets' final frame, gamma*L) vs. the final distance as measured in the original frame (namely, gamma*L/gamma = L). It's an effect of perspective (i.e. transforming between frames), not a physical effect. 
> 
Yes you are confusing me. 
Sorry, I don't know how to say it much better.
The problem is, you seem to think that contraction is something that happens only to *things* (e.g. spaceships etc.). But that's wrong. Contraction is first of all only a measurement effect (not a physical effect) and secondly, it happens to *any* measurement of a distance, even a distance *between* two things.
IOW, when you make a measurement of (say) L between two objects moving at the same speed in your frame, that L is a *contracted* length. The intrinsic distance between them (i.e. the distance in their own frame) is gamma*L.
In your spaceship scenario, the backtoback *intrinsic* distance between the spaceships increased from L to gamma*L during their acceleration, so that now, as measured in the original frame (hence contracted) the separation is still L.
>  When the length of the shape ship is equal to the distance during this experiment, 
Nicolaas, you are asking for trouble if you do this. I thought you had learned your lesson and were considering only tiny pointlike spaceships. But ok, let's consider what you are saying now. (I think you will get confused.)
>  then considered from the Rest Frame: 
Again, you should say *original* rest frame to avoid confusion.
>  1. The distance (back to back) stays constant and is accordingly to equation 1 
Hmm. I guess you could say that Equation 1 could be used to get the light path length, which in turn could be used to deduce the backtoback distance in that frame, which turns out to be L. So, OK.
>  2. The length of the space ship physical decreases with dl and is accordingly to equation 3 
Wrong in two ways: first, the contraction is not physical. Second, it is not given by equation 3, rather it is L/gamma.
Note, this means the front of the trailing spaceship is no longer touching the back of the leading spaceship. A gap has developed between the ships, as I think you are saying below:
> 
3. The distance (front to back) increases with dl i.e. EQ1  EQ 3 
Wrong formula. The front to back distance in the original rest frame is L  L/gamma.
> 
In the rockets final frame the measured values are different.
1. The distance (back to back) is equal to: 
Hmm, well of course that isn't a distance, but I assume you mean to divide by 2/c or some such, and if you do, the above eq'n is nonsense AFAICT. (Did you forget to parenthesize something?)
The correct answer should be gamma*L, which is the same as your Eq 3 divided by 2/c.
> 
2. The length of the space ship physical decreases tA = (L / (cv) + L / (c+v)) * SQR(1  v^2/c^2) * SQR(1  v^2/c^2) tA = 2*L/c Eq 5 
No it doesn't physically decrease, it stays the same: L. This is a given of the problem.
>  3. The distance (front to back) increases with EQ4  Eq 5 
Wrong formula.
> 
IMO it is confusion to say that Eq4 represents length expansion. Nothing is physical expanded. 
Wrong, the backtoback distance between the rockets is physically expanded i.e. they are further apart in their final frame. The reason for this is that the lead rocket started accelerating earlier in that final frame and so (in that frame) was traveling in a negative direction for less time than the "trailing" rocket. Remember, simultaneity is relative.
>  It *seems* that the distance increases but that comes (not completely) because we measure a distance with a clock which runs slower. 
First of all, in principle you don't measure distances with clocks  you use standard meter sticks. Yes, a measured flight time of a light beam can be used in place of a meter stick but that is a derivative measurement which makes assumptions about light speed. (I.e. that it is always c.)
Second, clocks don't run slower in SR. They are measured (by observers in other frames) to run slower, but that is a different thing.
Third, the intrinsic backtoback distance in this case *does really* increase. It increases because that's how you said the rockets were to accelerate. It is a given of the problem. You could have done it differently, and then the answer would have been different.
> 
> > 
You can choose not to use that terminology if you prefer, but you should at least understand it so you can follow what other people say  because my usage is quite common. The important point here is that I don't want you to think that contraction (in *my* sense, i.e. due to the transform between frames) does not occur. It *does* occur. But in *this particular example*, you have specified the acceleration in such a way that a physical expansion exactly counteracts the contraction, with the result that the L stays the same in the original frame. 
> > > 
What is contracted is the length of each space ship. 
> > 
Yes, those are contracted too, but we weren't talking about them. 
> 
It is important to include inorder to "see the total picture" 
IMHO it's good to pick and choose your details sometimes. Too much detail can lead to confusion. But I agree, you should be the judge of that yourself; try to find what works for you.
> 
> > > 
If you perform the same experiment as above with Observer A in the back of space ship A and reflection via the front of the space ship of A (not back of B) and back towards Observer A then you will get equation 3 with L replaced by L0 That means (1) that space ship A is contracted. (For space ship B the same is true) On the other hand (2) the distance between the back of space ship B and the front of space ship A is enlarged. (all measured in rest frame and no comoving clocks) 
> > 
Yes, it is slightly enlarged, but we weren't talking about that either. 
> > > 
What is contracted in your opinion ? 
> > 
See above. The contraction is not a physical effect. The expansion (which is specific to your acceleration spec) *is* a physical effect. Put the two together, and you get no change in the distance as measured in the original frame, for this particular case. 
> 
I do not fully grasp this. IMO in the rest frame nothing expans nor contracts. 
No, the ships (as measured) contract. Surely even you agree with this.
Yes the backtofront distance (as measured) stays the same, but that was an accident of how you specified the acceleration, using simultaneity conditions of the original rest frame.
> 
Equation 1 is what we actual measure. Equation 1 describes the situation (a model) of two points a distance L apart (in rest frame) moving both with speed v (in rest frame). Why do we need such a "difficult" explanation. 
It might seem overly difficult in this particular case, but we need it to apply to *all* cases, not just ones where the final separation happens (by accident of the problem statement) to equal the original.
> 
Equation 4 is the same situation but now measured in the moving frame of the space ships. The difference is the factor SQR(1v^2/c^2) which represents the physical effect that moving clocks are running slower. 
I don't know how you got Eq 4 but it's wrong, AFAICT.
The clocks are *not* running slower; however, simultaneity conditions *are* different and this accounts for the differences in measured lengths etc, as SR explains it.
> 
> > > 
> > > > 
The proper separation between the two ships (i.e. in their own frame) has increased from L to gamma*L, due to the unequal accelerations of the two ships in their comoving frames. 
> > > 
This is a rather tricky sentence specific because you are speaking of comoving frames (i.e one of space ship A and one of B) and most probably you are right. I think (but not sure) that you should consider one comoving frame A and explain what the speed of space ship B is in that frame. (Both can be zero) 
> > 
Both *are* zero, in the final state. 
> > > 
However the reason IMO is first that a clock in this comoving plane runs different as a clock in the rest frame. Secondly different clocks in this comoving frame require synchronisation. 
> > 
Right, accelerations make everything quite tricky; I'm glad you appreciate this. 
> > > 
> > > > 
The contraction in the original frame brings the distance in that original frame back down to L. 
> > > 
I do not understand this sentence.
What equation 1 tells me is that If that is true than there is no length contraction involved. 
> > 
Yes, it's the same distance in that frame as earlier. That's because there was a Lorentz contraction to counteract the physical expansion of the distance between the rockets (i.e. their separation in their own frame). 
> > > 
Bell' spaceship Paradox is as follows: "To begin, a statement of the paradox,Bell asks us to consider two rocket ships, each accelerating at the same constant rate, one chasing the other. The ships start out at rest in some coordinate system (the "labframe"). Since they have the same acceleration, their speeds should be equal at all times (relative to the labframe) and so they should stay a constant distance apart (in the labframe). But after a time they will acquire a large velocity, and so the distance between them should suffer Lorentz contraction. Which is it? " IMO at least the last sentence should be modified as follows: "At the same time their speed is constantly increasing and so the distance between them should suffer more and more Lorentz contraction." 
> > 
That's a good way to put it, but IMHO the original is not really wrong either. Remember, this is a statement of a seeming (but actually illusory) paradox, so it's not meant to be precise. 
> 
What is it actual illusory in this experiment ? 
It is the description as "paradox" that is illusory. Correctly explained, there is no real paradox in this experiment.
> 
Why is it not precise ?
Why does not it have to be precise ?
IMO it has to be precise in order to establish what is right (or wrong) and why. 
Look, I was wrong to imply it was imprecise. To my reading, the sentence in question was perfectly fine. IMHO your version of it didn't say anything more of substance.
IMHO if you are worried about precise wordsmithing, you should direct your attention more at your *own* writing, especially to the places where you are not careful to identify frame.
> 
> > > 
IMO the question should be: "How can both be true" or... 
> > 
Yes, that might be better. So, here's the answer to *your* question: Both *can* be true if the physical distance increases by a factor of gamma, because the measured distance (in the original frame) is just a Lorentzcontracted measurement of the physical distance. In other words, the measured L can stay the same if the gammas cancel. Whether or not it *will* stay the same, in a particular case, depends on how you do the acceleration. 
> > > 
In order to find that answer what you have is the following: In spaceship A observer A is in the back. In spaceship B observer B is in the back. Both observers each have two clocks. All the clocks are synchronised in the rest frame. Assume a constant acceleration During this period both A and B will see that their clocks runs slower as the clocks in the rest frame. 
> > 
No, A and B will say the "rest frame" clocks are running slower, due to an effect of the relativity of simultaneity. I realize this is backwards from what you might naively think, but that is what SR says. 
> 
This is only true for the clock at the rest which is at the position of t=0 for A. A will recieve time pulses from that clock at a slower rate because the distance with that clock increases. 
You are talking about the Doppler effect. Yes that occurs, but it is not the only thing that occurs; the clock is measured to run slower even *after* you correct for Doppler. Note, it is also measured to run slower if you have collaborators with synchronized clocks measure the traveling clock locally, where there is *no* Doppler to consider.
When we talk about "measurements" in SR, by convention we mean that Doppler (if applicable) has already been corrected for.
Anyway you are wrong if you think it only happens for the clock in the original rest frame. The observers in the ships see clocks in the *original* frame to be dilated. They must, because the original frame is moving at speed v (leftward) from their point of view. And physical laws do not depend on frame. (Nor on right vs. left.)
>  However if you place clocks in the rest frame all along the path that A follows than A will realize that his moving clock runs slower (All the time) 
You seem to have misunderstood the whole point of relativity. Yes, if the traveler uses the original frame (i.e. the wayside clocks) to define "time", then he will see his own clock falling more and more behind each of clock that passes by. But don't you see, that is a statement about a measurement in the *original* frame, not in the traveler's frame. We already know his clock runs slow as measured in that original frame. I was talking about something different: what he would say if he let his *own* clocks define time.
You might naively think in that case he'd get the same answer, but you'd be wrong. Time dilation is symmetrical, not invariant.
> 
> > > 
(Along the whole length of the trip their are clocks positioned which run synchronised in the rest frame) 
I wasn't talking about that. The point of relativity is to be able to use *any* frame as the "rest frame".
(And don't tell me you always prefer just to use *your* rest frame. For one thing, since you are riding on the Earth, your rest frame is not the same all the time anyway. If you ever need to convert from one frame to another, you need relativity.)
> > >  The two clocks of B are not synchronised with the clocks in the comoving frame of A. B will synchronise one of his clocks with A's clock. Because the speed of A,B is constantly increasing B has to do that constantly. 
> > 
Note, if he does this, B's clock will not indicate his proper time. I don't see much reason why B should bother to do this, since (as you've specified the acceleration) A and B won't be in the same inertial frame at any time between the original and final states. 
Aside: the above is correct, if you take my words "at any time" to refer to the simultaneity of the hypothetical (in fact nonexistent) inertial frame in question. I unwisely retracted this correct statement in a later post.
> > 
> > > 
Using this synchronised clock of B in order to calculate B' s Acceleration and you compare that with A's Accelaration using A's clock, ofcourse the two are not the same. 
> > 
OTOH I bit off way more than I could chew with the following:
> >  Right, and even if you use proper times, you will find that A's proper acceleration differs slightly from B's 
Oof, no. If A and B each record their accelerometer readings on a strip chart (vs. proper time) the charts should come out identical. The asymmetry is more subtle than this.
> >  proper acceleration, in this example as you've posed it. B's will be greater but will last for a shorter proper time (I think I said that wrong in my previous post). 
No, I said it wrong here, too. The *proper time* is not shorter, rather the *time* for B to accelerate, in either A's accelerated coords or B's accelerated coords, is shorter. This is because B is at a higher potential in whoseever accelerating coords you choose.
(Best to skim over this section, Nicolaas  I am definitely not the best person to tutor you in GR stuff. I'm just correcting the record, I hope. ;)
> >  The result is that the final v's will be the same, but the final proper separation will be greater. 
My conclusion was right. B spends more time at the higher velocity, from the standpoint of either A or B.
> > 
> > > 
The reason is synchronisation (in the comoving frame) If you make acceleration equal to zero than you can state the following:
Using A's clock to measure A's length
you will get:
If you want to measure the distance between the
two space ships (back to back) using A's clock you will
get: 
> > 
No, the reason is the physical increase in the distance between A and B, due to their unequal proper acceleration as you've specified it in this problem. 
My statement was partly incorrect (see above). Although the reason was indeed because of how you specified the acceleration, I got the details wrong. It would have been better to say, it was because you specified that the accelerations were to *begin* simultaneously in the initial comoving frame, rather than *end* simultaneously in the final comoving frame. They can't do both and be equal.
> 
How do you measure this acceleration ? 
A spring scale with 1 kg mass on it would be one way.
> 
> > 
Time dilation is a nonphysical effect of transforming between frames. Since A is measuring a distance using his own clock, there is no transform in this case, so time dilation is not relevant. 
> > > 
The difference is the distance between the front of A and the back of B This value is increasing. 
> > 
Yes it's increasing, but this also is irrelevant to what we were talking about. I am sorry to end our dialog on this note, but I really cannot spend any more time here. Best of luck in your continuing studies. 
Good luck again. Do take that course please, and do the homework too.
Russell Blackadar
>  Nicolaas Vroom wrote: 
> > 
Russell Blackadar 
> > >  Nicolaas Vroom wrote: 
> > > > 
In the book"Introducing Einstein's Relativity" by Ray D'Inverno at page 23 the same problem is discussed. There is "written": "Imagine a carriage travelling with velocity v realitive to an Observer M on the bank of the track... We assume that there are two electrical devices on the track which are the length of the carriage apart and equidistant from M. When the carriage goes over thes devices, they fire and activate two light sources situated at each end of the carriage (Fig 2.13) From the configuration, it is clear that M will judge that the two events, when the light sources first switch on, occur simultaneously." 
> > > 
The only way this would be the same as your problem, would be if d'Inverno had meant "proper length" when he said "length". But since he didn't say proper length, and since we can easily tell by context that he didn't mean proper length, there is no doubt that he is talking about a different problem. He is talking about a case where the proper length of the train is *greater* than the length between the marks on the track, so that the length measured in the track frame (hence contracted, because v > 0 in that frame) is equal to the length between the marks on the track. 
> > 
I agree with you. As such we agree that the length L of the moving train A'B' is contracted i.e. is shorter as its rest length L0 
> 
You are not careful enough with your language. When you say it is contracted, you should say in what frame the measurement is made. (Yes I remember, last time you said that you prefer the track frame always, but you still should say that explicitly, every time, to avoid impling things that are untrue. A train is never "contracted" independent of frame.) 
Yes in the track frame
> > 
However that does not make everything clear
What we have are two observers: 
> 
Note, this is because in his frame, the flashes literally *are not* simultaneous. 
The word literally is not necessary. I have a slight problem with the word: are In your terminology "seeing" and "are" are synonym
> >  The train is length contracted. 
> 
In the track frame. 
> > 
How do we explain all of this? I have doubts if M sees the flashes simultaneous. 
> 
Why? You *set up* the problem so that he will. The flashes literally *are* simultaneous in this frame, and M is equidistant from the flash points (and remains so, since his v=0). And we assume light speed is c in both directions. So what is there to doubt? 
Why, because I have certain doubts. Mainly because of v=0. I will explain later.
> >  IMO it is even possible that M' sees the flashes simultaneous. 
> 
Your opinion does not matter; what matters is what the calculations (and of course experiments) show. 
Most important are experiments I will come back to this subject later.
> > 
What you should do is consider the whole experiment in a more *abstract* way. in a more symmetrical fashion. A'M'B' v>  Track AMB v=0 
> 
Sorry, but this is not symmetrical at all. Symmetry would require the lower train to move *left* at speed v. As it is, your A'B' train has a greater proper length than AB. But let's move on... 
I agree that this is not truelly symmetrical. That is why is used the word fashion.
> >  We have two trains AB and A'B' We have one (Multi purpose) track. AB is at rest. A'B' moves to the right. M is in the middle between AB and sees the flashes simultaneous. M' is in the middle between A'B' and does not sees the flashes simultaneous. At each train there is also a rod with the same length as the train. 
> 
Why do you bother with rods? A train is as good as a rod (or at least we can pretend it is, in a thought experiment). Let's keep things as simple as possible. 
> > 
When train AB meets A'B' there one spark at both ends. 
> 
You say "when" without mentioning frame. 
There are sparks in the frame of AB and in the frame of A'B' M in train AB (frame AB) sees the sparks simultaneous. M' in train A'B' (frame A'B') does not see the sparks simultaneous. Using your terminology In the frame AB the sparks are simultaneous. In the frame A'B' the sparks are not simultaneous.
See figure 2:
https://www.nicvroom.be/vabsolute.htm#fig2
>  Don't always assume we understand "track frame" without saying that *explicitly*. It makes your writing unclear at best. 
> > 
For M' to see the flashes simultaneous M' has to move. 
> 
That's just another way of saying he sees them *not* simultaneously. 
> > 
But than M' is not any more in the middle between A'B' That means When M' sees the two sparks simultaneous At the position of M' 
This paragraph should have been as follows: For M' to see the flashes simultaneous M' has to move to a new position. At that new position M' is not any more in the middle between A'B' That means When M' sees the two sparks simultaneous, at that new position, M' breaks the (moving) rod A'B' in two parts: A'M' and M'B'
This is the point x in:
https://www.nicvroom.be/vabsolute.htm#fig2
>  You are getting confused. Is M' a position, or is M' a person? If some person M' at position M' moves, then I would say his position is no longer M'. Your terminology frequently suffers from such confusion IMHO. 
> > 
You can repeat this same experiment with a different
train A''B'' with a speed v''
The two rods A''M'' and M''B'' are not equal.
(Any number of trains n will give AnMn <> MnBn) You can also do that with the rod AB of M when M (at rest) sees the two flashes simultaneous. Is AM equal to MB ? IMO NO. 
Here I completely messed up.
I should have written: "You can also do that with the rod AB of M when M' sees the two flashes simultaneous."
The idea is that in the frame of AB with v=0 of of train AB and with v>0 for train A'B' that you adjust the length of A'B' such that M' sees the two flashes simultaneous.
See figure 3:
https://www.nicvroom.be/vabsolute.htm#fig3
The next step is to draw that same situation but now from the frame of train A'B' In frame of train A'B' v of train A'B' = 0 v of train AB <> 0
See figure 4:
https://www.nicvroom.be/vabsolute.htm#fig4
In fact figure 2 and figure 4 are (almost) identical.
Figure 2 describes the situation where M sees the two sparks
simultaneous in his frame
Figure 4 describes the situation where M' sees the two sparks
simultaneous in his frame.
Are they both correct ? I do not think so. Both figures give the impression that each observer (resp M and M') see the sparks with equal strength (intensity). I think that that is contrary to observation / experiment.
> >  *snip* 
> > > 
> > > > 
Exactly. I prefer to study only in the rest frame. 
> > > 
Careful, here. In relativity, *any* frame can be the "rest frame"; that term is ambiguous unless you tell us what object or objects (or observer) you consider to be at rest. So, to avoid ambiguity, you should always state the frame clearly, e.g. *original* rest frame, in this problem. 
> > > > > 
In that case, the correct answer is equation 1. 
> > > > > > 
> > > > > > > 
> > > > > > > > 
What does tA measure:
(Using a clock in the rest frame) 
> > > > > > > 
Only your inequality 2 is correct, under the assumptions I stated. Equation 3 *would* have been correct if you meant the time measured by an observer riding on ship A, using his own clock. (That is, the righthand side is just an algebraic rearrangement of 2*gamma*L/c.) Also in that same case, Equation 1 would have expressed the elapsed time in the original restframe coordinates (it is a rearrangement of gamma*2*gamma*L/c). 
> > > > > > 
I agree with you. Specific with equation 1 If I perform the experiment and the result is accordingly to equation 1 then I know that the distance between the two space ships is still L. 
> > > > > 
In the original frame, yes. 
> > > > > > 
i.e. the distance is not contracted (as a function of v.) 
> > > > > 
Actually it *is* contracted. 
> > > >  I do not understand this. 
> > > 
Sorry, I didn't mean to confuse you. The contraction I was talking about was the difference between the final "proper" distance (i.e. the distance as measured in the rockets' final frame, gamma*L) vs. the final distance as measured in the original frame (namely, gamma*L/gamma = L). It's an effect of perspective (i.e. transforming between frames), not a physical effect. 
> > 
Yes you are confusing me. 
> 
Sorry, I don't know how to say it much better. The problem is, you seem to think that contraction is something that happens only to *things* (e.g. spaceships etc.). But that's wrong. Contraction is first of all only a measurement effect (not a physical effect) and secondly, it happens to *any* measurement of a distance, even a distance *between* two things. 
For me it is very important to understand when an effect is real or physical versus illusionary
The moving train A'B', described above (figure2), its contraction is real when v of A'B' >0
Because:
1) When v of both trains = 0 (Initial position)
length of A'B' > length of AB
2) M only sees the flashes simultaneous when
length of A'B' = length of AB
If this is accordingly to experiment than length of A'B'
is physically contracted.
In the Twin experiment the clock of the moving twin runs slower as the clock of the staying at home twin. This is a physical effect.
You can use both clocks (one with v=0 and one with v<>0)
in an experiment to measure the distance between the
staying at home clock and an object which is at rest with
the staying at home clock (t = 2*L/c)
The result will be different
a) because the distance measured by the moving clock
and the staying home are different.
b) because the clock rate of both clocks is different.
> 
IOW, when you make a measurement of (say) L between two
objects moving at the same speed in your frame, that L
is a *contracted* length. The intrinsic distance between
them (i.e. the distance in their own frame) is gamma*L.
In your spaceship scenario, the backtoback *intrinsic* distance between the spaceships increased from L to gamma*L during their acceleration, so that now, as measured in the original frame (hence contracted) the separation is still L. 
> > 
When the length of the shape ship is equal to the distance during this experiment, 
> 
Nicolaas, you are asking for trouble if you do this. I thought you had learned your lesson and were considering only tiny pointlike spaceships. But ok, let's consider what you are saying now. (I think you will get confused.) 
> > 
then considered from the Rest Frame: 
> 
Again, you should say *original* rest frame to avoid confusion. 
> > 
1. The distance (back to back) stays constant and is accordingly to equation 1 
> 
Hmm. I guess you could say that Equation 1 could be used to get the light path length, which in turn could be used to deduce the backtoback distance in that frame, which turns out to be L. So, OK. 
> >  2. The length of the space ship physical decreases with dl and is accordingly to equation 3 
> 
Wrong in two ways: first, the contraction is not physical. Second, it is not given by equation 3, rather it is L/gamma. 
This item is important for rest of discussion.
>  Note, this means the front of the trailing spaceship is no longer touching the back of the leading spaceship. A gap has developed between the ships, as I think you are saying below: 
> > 
3. The distance (front to back) increases with dl i.e. EQ1  EQ 3 
> 
Wrong formula. The front to back distance in the original rest frame is L  L/gamma. 
> > 
In the rockets final frame the measured values are different.
1. The distance (back to back) is equal to: 
> 
Hmm, well of course that isn't a distance, but I assume you mean to divide by 2/c or some such, and if you do, the above eq'n is nonsense AFAICT. (Did you forget to parenthesize something?) 
>  The correct answer should be gamma*L, which is the same as your Eq 3 divided by 2/c. 
> > 
2. The length of the space ship physical decreases 
> 
No it doesn't physically decrease, it stays the same: L. This is a given of the problem. 
> > 
3. The distance (front to back) increases with EQ4  Eq 5 
> 
Wrong formula. 
> > 
IMO it is confusion to say that Eq4 represents length expansion. Nothing is physical expanded. 
> 
Wrong, the backtoback distance between the rockets is physically expanded i.e. they are further apart in their final frame. The reason for this is that the lead rocket started accelerating earlier in that final frame and so (in that frame) was traveling in a negative direction for less time than the "trailing" rocket. Remember, simultaneity is relative. 
The explanation of above is:
1) First there is physical length contraction of the moving rods.
2) The clocks in the moving rods run slower.
3) Clock synchronisation of moving clocks in moving rods.
This has to be done each time when you change the speed of
the rods.
The result of 3 is that is the accelaration the two space ships
measured in the moving frame is different.
> >  It *seems* that the distance increases but that comes (not completely) because we measure a distance with a clock which runs slower. 
> 
First of all, in principle you don't measure distances with clocks  you use standard meter sticks. Yes, a measured flight time of a light beam can be used in place of a meter stick but that is a derivative measurement which makes assumptions about light speed. (I.e. that it is always c.) 
In the book Introducing Einsteins Relativity by Ray d'Inverno
at page 19 we read:
"We previously defined an observer in the Newtonian theory
as someone equipped with a clock and a ruler
with which to map the events of the universe.
However, the approach of the kcalculs is to dispense with the
rigid ruler and use radar methods for measuring distances.
( What is rigidity anyway? If a moving frame appears nonrigid
in another frame, which, if either, is the rigid one?)
Thus..
The distance is then simply defined as half the time
difference between emission and reception."
If a distance in a rest frame is for example 10 standerd meter sticks than you will find, that If you use moving standard meter sticks to measure that same distance, that you need more. Implying IMO physical length contraction.
>  Second, clocks don't run slower in SR. They are measured (by observers in other frames) to run slower, but that is a different thing. 
>  Third, the intrinsic backtoback distance in this case *does really* increase. It increases because that's how you said the rockets were to accelerate. It is a given of the problem. You could have done it differently, and then the answer would have been different. 
> > 
Equation 1 is what we actual measure. Equation 1 describes the situation (a model) of two points a distance L apart (in rest frame) moving both with speed v (in rest frame). Why do we need such a "difficult" explanation. 
> 
It might seem overly difficult in this particular case, but we need it to apply to *all* cases, not just ones where the final separation happens (by accident of the problem statement) to equal the original. 
SNIP
> > > >  Bell' spaceship Paradox is as follows: "To begin, a statement of the paradox,Bell asks us to consider two rocket ships, each accelerating at the same constant rate, one chasing the other. The ships start out at rest in some coordinate system (the "labframe"). Since they have the same acceleration, their speeds should be equal at all times (relative to the labframe) and so they should stay a constant distance apart (in the labframe). But after a time they will acquire a large velocity, and so the distance between them should suffer Lorentz contraction. Which is it? " 
The next step of this discussion is: Start in the moving frame with two space ships back to back. Both space ships are initially at rest in this moving frame. Both space ship start to move in the moving frame at the same time, with the same speed in the moving frame.
If you measure the distance between the space ships Does equation 1 apply? If you measure the length of the trailing space ships Does equation 3 apply? All in the moving frame.
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