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Q & A: asymmetrical twins

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Most recent answer: 08/14/2013
Q:
Twin paradox answers always cheat by referring to the acceleration at start and especially at turnaround. Instead, try this experiment: The twins take off together in different ships at high acceleration until they are both at 9/10 light speed, then both cut their engines. Brother A immediately turns around, decelerates, and returns to the starting point. Brother B coasts at 9/10 light speed for ten years, then turns around, decelerates, and returns to join his brother. Each brother spent exactly the same time in acceleration and deceleration, these General Relativity time changes are not an issue. This isolates the ten years spent at near light speed by brother B, a constant velocity indistinguishable to either by Special Relativity. How much difference in age, if any, between the brothers?
- Don Heller (age 74)
Beaumont, CA, USA
A:
Let's look at this from the point of view of someone at rest with respect to the twins before they start accelerating (say triplet C). C has a nice conventional inertial frame. C sees A make a small trip and thus return a little younger than C is. B makes a much bigger trip and returns much younger than C, and thus also younger than A. Anybody, regardless of frame, can look and compare them when they're back together. Thus everyone must agree with C that C is the oldest, then A, then B.

The General Relativity time changes, incidentally, do not cancel. They involve not only the acceleration but also the displacement between the objects. B is accelerating back toward A when the displacement is larger, so B sees a bigger GR speed-up of A's aging, consistent with our conclusion.

Assuming that the acceleration is large enough to take much less than 10 years, the age difference between A and B is approximately 10 yr * (1-sqrt(1-0.92))=
10*(1-sqrt(0.19) )yr = 5.6 yr. To be realistic, the acceleration can't be quite large enough for that approximation to work, but then you'd need to specify the acceleration for us to do the calculation.

Mike W.

(published on 07/21/2009)

Follow-Up #1: twins again

Q:
In your answer, you assume that "everyone must agree" that a longer time at a constant velocity at near light speed causes a permanent lack of aging. You also say that the actual aging of A is larger if observed from a greater distance by B. I don't understand the physics of that. My question is about the aging during the time that neither A nor B is accelerating, and has nothing to do with how much acceleration they can stand. For the purposes of this mind experiment, you can assume that they can stand an acceleration to get them from 0 to 0.9999c in one hour.
- Don Heller (age 74)
Beaumont, Ca, US
A:
Yes, that's the case I described, where the acceleration is unreasonably rapid, because that's the easy case to solve. We described everything from an inertial frame in which we can just use Special Relativity. (There's no gravity involved here, so that makes sense.)

My comments about acceleration were just a reminder that in the frames of the travelers (which cannot use SR, because of the accelerations), the net results are consistent. For that argument we used the General Relativistic effects  of acceleration. Getting the answer, however, did not require doing that calculation.

Mike W.


(published on 08/08/2009)

Follow-Up #2: retwinning

Q:
I’m sorry to be so persistent, but I really want to understand this. First, you did not address my comment on “everyone must agree”. My original question was whether constant velocity caused real aging, and “everyone must agree” does not address that, but begs the question. Second, I was not asking about calculating the aging difference due to GR. Since each twin spent exactly the same amount of proper time and acceleration to and from near-light-speed, those time slow-downs should be the same for each twin. I think. Independently of whatever happens during accelerations, I asked what the SR permanent aging difference would be after the ten years of constant velocity travel for twin B outbound, and 10 more years inbound. At the end of 20 years, plus the 4 or so hours accelerating and decelerating, they are together again. At no time has either twin been near a source of gravity. Third, I truly want to understand your statement that the GR time changes involve not only the acceleration but also the displacement between the objects. You said that B is accelerating back toward A when the displacement is larger, so B sees a bigger GR speed-up of A's aging. We’re talking about permanent aging, not just what B observes, right? And lastly, when the twins re-unite, will B’s ship be shorter than his brother’s ship?
- Don Heller (age 74 1/2)
Beaumont, Ca, US
A:
I'm glad that you're serious about this. I'll go back over some of those points and try to explain more completely.

"I asked what the SR permanent aging difference would be after the ten years of constant velocity travel for twin B outbound, and 10 more years inbound."
That's precisely what we answered, doing the SR calculation using the reference frame at rest with respect to the initial and final traveler positions.

"My original question was whether constant velocity caused real aging, and 'everyone must agree' does not address that, but begs the question.  "
What we have here is a failure to communicate philosophically. The things we consider 'real' are the ones which don't depend on point of view. When we conclude that the differences in age between the travelers don't depend on point of view, we're saying those are real differences.

"Second, I was not asking about calculating the aging difference due to GR." True, and we calculated it using only SR, picking a reference frame in which SR is ok. However, if you instead use the reference frame of either traveler, which are not close to being inertial, you have to use some GR ingredients. We described how, even if you were to do that, your results would be consistent with our simple SR calculation.

"Third, I truly want to understand your statement that the GR time changes involve not only the acceleration but also the displacement between the objects.  You said that B is accelerating back toward A when the displacement is larger, so B sees a bigger GR speed-up of A's aging.  We're talking about permanent aging, not just what B observes, right?"

Again, we know no truth outside the collection of observables from different points of view. When A and B return to the same spot, so that all observers must then agree on which is older, since any can see them simultaneously, the net result from B's point of view agrees with A's and everybody else's. This is a real age difference. Which relativistic effects it's attributed to depends on which reference frame you use.

As for the technical question, a frame accelerating at a with respect to a good inertial frame sees the aging rate of an object at position r change (to lowest order in a) by a factor (1+ar/c2), where "" means vector dot product. If you are on an accelerating rocket, far from gravity sources, a clock at the head will run faster than one at the tail. In gravity, the equivalent real effect is that clocks in attics run faster than ones in basements.

In your example, although each traveler has the same magnitude of a, one has it while r is big and the other while r is small, so they see different effects on the other's aging.

But I think the key point here is the philosophical one, not the technical one.

Mike W.

(published on 08/12/2009)

Follow-Up #3: twins again

Q:
According to special relativity an object that travels fast has its time slowed down. But let us assume that there is an infinitely long road and there are 2 persons A and B on the road. Person 'A' runs at the speed of light(a bit less). 'B' stays fixed with respect to the road.Person 'A' after travelling at the speed of 0.9999c for a distance of 10^100 m he turns back and comes back to the initial point. So according to Special Relativity time would have passed slowly for person 'A' so he would age less than person 'B'. But RELATIVE TO PERSON 'A'-- B would have travelled the distance so person 'B' would have aged less. So who actually is correct? If both of them are correct what will happen when they meet?
- Yogeesh (age 15)
India
A:

see above


(published on 08/14/2013)

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