Twin Paradox


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.

Twins with TV Monitoring

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:
  • 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.
The explanation (part of) is distance. When A correct B's clock with distance he will realize that B's clock runs slower.

What B sees is:

  • 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.
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.

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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