# Relativity and Aging

*Most recent answer: 10/22/2007*

Q:

In a lecture a confused statement was made and the lecture didn’t answer my question so well so I put it to you:
In relativity if an observer is moving and close to the speed of light (i shall call him observer A) another observer who is in the "still" frame (observer B) will see him slow down he will appear to age slower and the his watch will tick slower according to the "still" frame. similar to this if the observer that is moving at speeds close to c (A), this can be assumed as a "still" frame and the previous "still" observer (B) can be moving relative to (A) at relativistic speed and he can also be seen to slow down and his watch to tick slower, this is fine, the appearene of time distortion, my question is, if the two then enter the same reference from, how will there watches compare and who will be older if both have been seen in this way?
long and complicated i know but im hoping you will be able to respond, faithfully, thank you

- Colin (age 18)

Physics Department uni of manchester

- Colin (age 18)

Physics Department uni of manchester

A:

A:

Hi Colin- This is a form of the ’twin paradox’, which isn’t really a paradox. In fact, it’s a terrific way of seeing how one can start with the idea that the speed of light is the same constantin all frames and come up with truly surprising consequences, such as that twins with different travel itineraries will meet up again having aged differently when they get back together. We already have some answers to questions about the twin paradox on this web site, and so this answer will be a different way to approach the situation.

A consequence of the speed of light being the same in all reference frames is that for every pair of events, (t1,x1) and (t2,x2) (I’ll only worry about one dimension, along the x-axis), the quantity

tau = sqrt( (t1-t2)^2 - ((x1-x2)/c)^2 )

is the same in all reference frames, where c is the speed of light. If an object stays put, that is, x1=x2, then tau is just t1-t2, and corresponds to the amount of time elapsed on a clock on the object between the two events. So if a clock reads zero time at event 1, it will read tau time at event 2. This is true in all reference frames, because specifying a unique location and time at which to read the clock determines what number it will be showing. Note that this tau is for the clock which goes from event 1 to event 2 without accelerating -- if a traveler does not accelerate in one inertial reference frame he accelerates in no other inertial reference frame.

If a traveler is accelerating, then one has to break up the journey into infinitely small pieces and add up the contribution to the total tau sum by integration.

Now there are three important events in the twin paradox situation.

1: Twin B departs A’s location.

2: Twin B turns around (assume this happens quickly)

3: Twin B is reunited with twin A.

For simplicity, I’ll assume twin B travels at a constant velocity except for very short blast-off and turnaround accelerating periods.

Event 2 takes place at a faraway place, say a distance X in twin A’s frame, and at a time T. In Twin A’s frame, these events are:

1: (t=0,x=0)

2: (t=T,x=X)

3: (t=2T,x=0)

Twin B doesn’t have just one frame (he accelerates while turning around -- he spends most of his time in either the outgoing frame or the incoming one). Velocity is relative but acceleration isn’t. You can measure acceleration with an experiment which doesn’t depend on the environment.

The amount of time accumulated on Twin A’s clock from event 1 to event 3 is 2T, and the amount of time accumulated on Twin B’s clock from event 1 to event 3 is 2*(T^2-(X/c)^2), which is less. Since both twins are at events 1 and 3 in all reference frames, this difference between their clock readings can be compared in all reference frames -- twin B will be younger.

In your question you asked "what if twin A and twin B enter the same reference frame?" The answer to this depends on whether twin A and twin B have met up again. If they are not in the same place at the same time, then their relative ages depends on the reference frame in which they are viewed, unfortunately. Either A or B could be younger if they are still apart, and different outside observers would come to different conclusions. One consequence of the idea that the speed of light is the same in all reference frames is that the idea of events being "simultaneous" no longer holds. Events that happen at the same times in different places in one reference frame happen at different times in other reference frames.

Tom J.

Hi Colin- This is a form of the ’twin paradox’, which isn’t really a paradox. In fact, it’s a terrific way of seeing how one can start with the idea that the speed of light is the same constantin all frames and come up with truly surprising consequences, such as that twins with different travel itineraries will meet up again having aged differently when they get back together. We already have some answers to questions about the twin paradox on this web site, and so this answer will be a different way to approach the situation.

A consequence of the speed of light being the same in all reference frames is that for every pair of events, (t1,x1) and (t2,x2) (I’ll only worry about one dimension, along the x-axis), the quantity

tau = sqrt( (t1-t2)^2 - ((x1-x2)/c)^2 )

is the same in all reference frames, where c is the speed of light. If an object stays put, that is, x1=x2, then tau is just t1-t2, and corresponds to the amount of time elapsed on a clock on the object between the two events. So if a clock reads zero time at event 1, it will read tau time at event 2. This is true in all reference frames, because specifying a unique location and time at which to read the clock determines what number it will be showing. Note that this tau is for the clock which goes from event 1 to event 2 without accelerating -- if a traveler does not accelerate in one inertial reference frame he accelerates in no other inertial reference frame.

If a traveler is accelerating, then one has to break up the journey into infinitely small pieces and add up the contribution to the total tau sum by integration.

Now there are three important events in the twin paradox situation.

1: Twin B departs A’s location.

2: Twin B turns around (assume this happens quickly)

3: Twin B is reunited with twin A.

For simplicity, I’ll assume twin B travels at a constant velocity except for very short blast-off and turnaround accelerating periods.

Event 2 takes place at a faraway place, say a distance X in twin A’s frame, and at a time T. In Twin A’s frame, these events are:

1: (t=0,x=0)

2: (t=T,x=X)

3: (t=2T,x=0)

Twin B doesn’t have just one frame (he accelerates while turning around -- he spends most of his time in either the outgoing frame or the incoming one). Velocity is relative but acceleration isn’t. You can measure acceleration with an experiment which doesn’t depend on the environment.

The amount of time accumulated on Twin A’s clock from event 1 to event 3 is 2T, and the amount of time accumulated on Twin B’s clock from event 1 to event 3 is 2*(T^2-(X/c)^2), which is less. Since both twins are at events 1 and 3 in all reference frames, this difference between their clock readings can be compared in all reference frames -- twin B will be younger.

In your question you asked "what if twin A and twin B enter the same reference frame?" The answer to this depends on whether twin A and twin B have met up again. If they are not in the same place at the same time, then their relative ages depends on the reference frame in which they are viewed, unfortunately. Either A or B could be younger if they are still apart, and different outside observers would come to different conclusions. One consequence of the idea that the speed of light is the same in all reference frames is that the idea of events being "simultaneous" no longer holds. Events that happen at the same times in different places in one reference frame happen at different times in other reference frames.

Tom J.

*(published on 10/22/2007)*