question: Which is the law that describes each of the two twins.
The players in the Twin Paradox are two twins called A and B. Each has a clock.
Twin A stays at home.
Twin B travels with a uniform velocity to a point X and returns than back home.
When they meet, both clocks are not equal, A's clock has advanced the most i.e. A is the oldest.
In order to explain the behaviour of the clocks, Special Relativity is used and the formula: T2 = T1 * SQR(1-vē/cē)
Using that formula A can explain the Behaviour of B's clock, which has a relative velocity of v, but also B can explain the behaviour of A's clock, which has the same relative velocity v.
The result will be that both A will claim that B's clock will run slower and B will claim that A's clock will run slower (with equal amounts).
This is a contradiction i.e. a paradox.
In order to solve the Twin Paradox two TV Monitors are used i.e. A's clock is monitored and B's clock is monitored. The pictures are transmitted to the other party. i.e. A will get an image of B's clock and vice versa.
However we will do slighty more: At A's side not only A's clock is monitored but also B's clock.
We get then the following set up:
At A's side At B's side
A B' A" B A' B"
A is A's clock.
B' is B's clock transmitted.
A" is A's clock 2 times transmitted (via B)
What A sees is:
The explanation (part of) is distance. When A correct B's clock with distance he will realize that B's clock runs slower.
- His own clock A
- An image of B's clock, delayed and backwards in time
- An image of his own clock, also delayed and more backwards in time. When his speed is more than 0.5*c the time will be before he even started.
What B sees is:
The explanation (part of) is distance. However, when B correct A's clock with distance he will realize that A's clock runs faster, during the whole trip.
- His own clock B
- An image of A's clock, delayed and backwards in time
- An image of his own clock, also delayed and more backwards in time.
In order to make this visible, a simulation in the form of a program in Quick Basic, is supplied.
To get a copy select:TWIN.BAS
Reflection part 1
The common explanation for the Twin Paradox is that A is at rest in an inertial frame and B is in an accelerated frame (which requires General Relativity).
The problem with this explanation is that when B is at the turn around point X (Just before and just after, when B's speed changes from +v to -v), that the speed of A's clock does not change. (No so called "jumping" is observed)
Reflection part 2
The speed of B in the Twin Paradox is not constant. There is a change at least at the Start and at the End.
How do we know that A's speed is constant (and not subject to any force) ?
Last modified: 17 March 1998
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