## Changing Length - part 3

### Question 1

Is length contraction always related to a physical object ?

### Question 2

Does length contraction always means that length decreases ?

### Background question 1 and 2

Accordingly to Special Relativity Length contraction(Lorentz Transformation) means that the length of an object becomes shorter, in direction in which the object moves.
The first question tries to answer the question if Length contraction is also applicable for non physical objects. For example: on distances in space.
The second question tries to answer the question if the outcome of any experiment is always that length becomes shorter and never longer.

The same problem was discussed in the newsgroup: news:sci.physics.relativity. See: The Absurd Assertions of the SR Experts

### Description of Experiment 1

In order to answer the second question consider the following experiment which consists of two parts.

Starting point is a grid at rest of equally spaced points pn, a distance d appart.

```   ---->
p0  p1  p2  p3  p4  p5  p6  p7
```
Between the points p0 and p1 there is also a rod with length d, such that the beginning of the rod coincides with point p1 and the end of the rod coincides with point p0.
At each point pn there is also a clock. Before the experiment is done the clocks are synchronised in the rest frame.
At both ends of the rod there is an observer i.e. one at position p0 and one at p1.

The first part of experiment 1 consists of moving the rod or spaceship above the grid towards p7. When each of the observers reaches any of the points they will write down the time on the clock. That means the observer in the back will write down the 6 times at p1 until p6 and the observer at the front the 6 times at p2 until p7.
The obvious question now becomes: will each of the two observers write down the same time sequence (of 6 times)?
Assuming that length contraction takes place the answer is No. When the observer at the end reaches for example point p4 the front end of the rod is not at position p5. The front end will reach p5 later, implying that both observers will not monitor the same time.
It is important to consider that as part of this experiment no time dilation takes place, because all the clocks are at rest in the same frame.

The second part of experiment 1 is almost identical as part one, except that the rod or spaceship initially starts at the points p6 and p7 and moves in the direction of p0. It is important to remark that as part of this experiment the speed of the rod is such that the observer at the back end will write down exactly the same time sequence as in experiment one.
The obvious question is now: will the observer in front also write down the same time sequence as in experiment one ?

1. Accordingly to Special Relativity the answer is:Yes
Both observers in front will write down the same time values and will conclude that length contraction is equal because the speed is the same.
2. IMO the answer is No.
IMO it is not quaranteed when the observers move at a certain speed to the right and detect length contraction that when the observers move to the left with that same speed they will also detect length contraction. It is even possible that they will not detect any change in length and or that length contraction is negative.
See also The Absurd Assertions of the SR experts. Specific comment 2.

### Description of Experiment 2

The starting point of this exepriment is the same as the first part of experiment one i.e. a grid at rest with marker points a distance d apart, clocks at each point and a spaceship of the same length d. The spaceship moves to the right with and length contraction is observed. The spaceship will now have a length d'. With the same speed and direction consider a new second grid with marker points a distance d' apart. Also at each point of this new grid there is a clock. This second grid can also be considerd at rest
We have now two frames:
• Frame one at rest with grid one
• Frame two at rest with grid two. Grid two moves to the right relative from frame 1.
We are now going to perform the same experiment in frame 2. Again the experiment consists of two parts:
• In part one the rod or spaceship moves towards the right.
• In part two the rod or spaceship moves towards the left.
The question is now: are both results be the same ?
1. Accordingly to Special Relativity the answer again is:Yes
2. IMO the answer is: No.
IMO in the second grid when you move to the left (in the spaceship with initial length d') you will definite detect that your spaceship first becomes longer
When you increase the speed more it will first reach a maximum size and then, with still higher speed, become shorter.
IMO when the spaceship reaches its maximum size you can say that its absolute speed is zero. At that speed also the clock in the spaceship runs the fastests

### Description of Experiment 3

The purpose of this experiment is to find an answer on question 1.
The experiment is almost identical as experiment 1. The difference is that we use two rods (space ships) of equal length d: rod1 and rod2
```  rod1 rod2
---->---->
p0  p1  p2  p3  p4  p5  p6  p7
```
The idea behind the experiment that the observers at the back both follow the same flight plan, based on the clocks in the reference frame. For example:
flight plan for rod1: starting point is p0 at t=0, at p1 at t=5, at p2 at t=9 at p3 at t=12 etc.
flight plan for rod2: starting point p1 is t=0, at p2 at t=5, at p3 at t=9 at p4 at t=12 etc.
When you observe the flight plan, the speeds of both rods increase. The question now becomes: does the distance between both rods increase? The answer could be given by an observer in front of rod1 or an observer in the back of rod2. However the best is an observer in the reference (rest) frame. If your answer is yes than length contraction is only related to physical objects.

For a similar discussion See: http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html
The second paragraph of the article starts with the following sentence: To begin, 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 "lab-frame"). Next they write: Since they have the same acceleration, their speeds should be equal at all times and so they should stay a constant distance apart.This raises two comments:

• First if both spaceships have the same acceleration at each instant (in the lab frame) their speeds should be the same, however over a longer period the the speed of each should increase constantly.
• Secondly the distance between the two will only stay constant if the distance between the two is measured between the same locations i.e. the distance from the back of the two.
Next they write: "But after a time they will acquire a large velocity, and so the distance between them should suffer Lorentz contraction". This raises one comment:
• I agree that each space ship will acquire a large speed, but that will result that each space ships will suffer Lorantz contraction, but IMO not the distance between the two, assuming that the above definition of distance is correct
Next they write:"Which is it" Unfortunate the article does not give the answer.
Next they write:I think the best approach is through spacetime diagrams.. I do not agree with this approach. First you should describe what they should observe i.e. how the experiment is performed in detail, what is measured and what the results are. Next you should describe how you explain those results, using mathematics.

### Length Contraction Detection

In the above paragraph Description of Experiment 1 is explained how detect length contraction using a grid (rods) and clocks in two different reference frames.
IMO it is also possible using only grids. What you need (in each reference frame) are two grids:
• One which stays fixed in the reference frame (same as previous used i.e. pn points a fixed distance d apart)
• One which can be moved in the reference frame, just slightly.
The size of the grid of each reference frame is identical. In the initial state of the experiment the fixed and movable grid are at the same position or start position At each point pn of the grid there is one observer.
The experiment goes as follows:
• The moving rod starts between the points p0 (back end) and p1 (front end) and moves towards the right.
• Immediatly when the observer at p0 observes that the back end from the rod leaves p0 he or she moves the movable grid slightly away from the start position.
• The back end moves from p0 to p1 and the front end from p1 to p2.
• When the back end reaches p1 the observer at p1 moves the movable grid back into the start position position.
• At that same instant the observer at p2 observes the same movement.
• Dependent if the front end of the rod has reached point p2, Yes or NO, he or she can now conclude the following:
If the front end of the rod has not yet reached point p2 than length contraction has occured and the length of the rod has decreased.
If the front end has already passed point p2 than the length of the rod has increased.
You can use this concept, in order to detect that length contraction takes place, with both trains in experiment 1. You can also do exactly the same in experiment 2.
In fact what you need are in total 4 identical grids:
• Two grids for experiment 1
• Two grids for experiment 2 which you place both on the train which moves towards the right in experiment 1.

### Clock synchronisation.

The synchronisation of the clocks in a frame if you want to perform measurements and do calculations in a particular frame.
```                           '
A<----->M<----->B<----->M<----->C
d       d       d       d
```
In order to synchronise two clocks A and B a distance 2d apart you proceed as follows.
• The first step is to define the middle M between the two clocks A and B. To do that you need a rod with length d. One end of the rod should touch the point A and the other end should point in the direction of B. That end point defines M.
• Next you place the rod between the points M and B. The rod should touch both points. If it does not your rod has not the correct length. So you continue.
• A different way is to place a mirror at both the points M and B and a light source at A. The reflection towards B and back to A is tB. The reflection time towards M and back to A is tM. M is in the middle whan tM = 0.5 * tB
• Next you place a lamp at point M. You turn the lamp ON. When the observers at both clocks see the light they reset the clock. This synchronises the two clocks A and B.
• If you also want to synchronise clock C with A and B then the easiest way to use a second clock at B. (clock B2 the first one is B1) Again there is a lamp at point M'. When the observers see the light they reset the clocks B2 and C. Now the clocks A with B1 and B2 with C are synchronised. The final step is to change the time of C with the difference of the two clocks B1 and B2. Now all the three clocks A, B,(B1) and C are synchronised.

IMO length contraction is only applicable to a physical object and not of distance in space See also
Description of Experiment 3

IMO, as explained in the paragraph Description of Experiment 1 the answer is No.

### Reflection

The most important experiment, which explains length contraction, is the Michelson and Morley's Experiment. Part of the problem which the experiment is that when you perform it you are not sure if its length is actual getting any shorter and what is more important you will not know when. What you are for sure is that a fixed length is not as fixed as you might think. It can become shorter but it also might become longer. The outcome of the experiment is not decisive.

For a detailed description of the Experiment see: Michelson and Morley's Experiment. There is also a simulation program available from that link.

### Feedback

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Created: 17 December 2000
Modified:9 January 2001, when the Moon was fully covered by the Shadow of the Earth.