Twin paradox?

  • Thread starter hawkingfan
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  • #1
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I really don't quite understand the result of time dilation. Let's say that we have two identical twins A and B. Twin A is at rest and twin B is on a high speed rocket ship. Lets say that 10 years has passed in frame A. According to relativity, the twin in frame B is supposed to measure less time. Let's say that only 1 year has passed in frame B. This is what happens if we consider frame A as the rest frame. However, just as an observer in frame A sees the rocket moving with nonzero speed, an observer in frame B (inside the rocket) will think that the rocket is standing still with frame A moving in the opposite direction. Shouldn't time slow down for frame A instead? This seems contradictory. How can you tell which frame will be the one that slows down or speeds up in time?
 

Answers and Replies

  • #2
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The thing is everything is relative... :)

So, the time will appear to slow down in A, with respect to B and vice versa.

So the person in the ship will see the time on earth to run slower and the person on earth will see the time on the ship run slower. It arises from the fact that any non-inertial reference frame can be thought of as 'at rest' to everything else.

With the twin paradox however, when the rocket ship turns around to go back to the earth it must deccelerate (hence becoming an inertial reference frame) and that is why the twin will be younger on the ship then compared to the earth.
 
  • #3
JesseM
Science Advisor
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The thing is everything is relative... :)

So, the time will appear to slow down in A, with respect to B and vice versa.

So the person in the ship will see the time on earth to run slower and the person on earth will see the time on the ship run slower. It arises from the fact that any non-inertial reference frame can be thought of as 'at rest' to everything else.

With the twin paradox however, when the rocket ship turns around to go back to the earth it must deccelerate (hence becoming an inertial reference frame) and that is why the twin will be younger on the ship then compared to the earth.
I think you've got "inertial" and "non-inertial" backwards ;)
 
  • #4
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hawkingfan,
you seem to be confusing 'time-dilation' ( moving clocks appear to run slower) and differential ageing. The first is an instantaneous observartion, the second is the result of adding up the proper-times of the worldline segments, and they can't be compared directly.
 
  • #5
PAllen
Science Advisor
2019 Award
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I know we have FAQs, but I wonder if we should have a sticky thread for the best discussions of the twin 'paradox', since it come up at least weekly, it seems.
 
  • #6
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I think you've got "inertial" and "non-inertial" backwards ;)
Indeed I did, thanks for that :P
 
  • #7
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During the constant-speed legs of the trip, BOTH twins conclude that the other twin is ageing slower. But when the trip is over, they both agree that the stay-at-home twin is older. How is that possible?

It's possible because, during the turnaround, the traveler will conclude that the home twin quickly ages, with very little ageing of the traveler. The home twin concludes that neither of them ages much during the turnaround. When you add up all these segments of ageing, you get the result that the home twin is older (and both twins exactly agree on that).

Years ago, I derived a simple equation (called the "CADO" equation) that explicitly gives the ageing of the home twin during accelerations by the traveler (according to the traveler). The equation is especially easy to use for idealized traveling twin problems with instantaneous speed changes. But it also works for finite accelerations. I've got a detailed example with +-1g accelerations on my webpage:

http://home.comcast.net/~mlfasf [Broken]

And I've published a paper giving the derivation of the CADO equation:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

You might also want to look at my posting in the thread:

https://www.physicsforums.com/showthread.php?t=436131
_______________________________________________________

Here's a brief description of my "CADO" equation:
__________________________________________________ __

Years ago, I derived a simple equation that relates the current ages of the twins, ACCORDING TO EACH TWIN. Over the years, I have found it to be very useful. To save writing, I write "the current age of a distant object", where the "distant object" is the stay-at-home twin, as the "CADO". The CADO has a value for each age t of the traveling twin, written CADO(t). The traveler and the stay-at-home twin come to DIFFERENT conclusions about CADO(t), at any given age t of the traveler. Denote the traveler's conclusion as CADO_T(t), and the stay-at-home twin's conclusion as CADO_H(t). (And in both cases, remember that CADO(t) is the age of the home twin, and t is the age of the traveler).

My simple equation says that

CADO_T(t) = CADO_H(t) - L*v/(c*c),

where

L is their current distance apart, in lightyears,
according to the home twin,

and

v is their current relative speed, in lightyears/year,
according to the home twin. v is positive
when the twins are moving apart.

(Although the dependence is not shown explicitely in the above equation, the quantities L and v are themselves functions of t, the age of the traveler).

The factor (c*c) has value 1 for these units, and is needed only to make the dimensionality correct. For simplicity, you can generally just ignore the c*c factor when using the equation.

The equation explicitly shows how the home twin's age will change abruptly (according to the traveler, not the home twin), whenever the relative speed changes abruptly.

For example, suppose the home twin believes that she is 40 when the traveler is 20, immediately before he turns around. So CADO_H(20-) = 40. (Denote his age immediately before the turnaround as t = 20-, and immediately after the turnaround as t = 20+.)

Suppose they are 30 ly apart (according to the home twin), and that their relative speed is +0.9 ly/y (i.e., 0.9c), when the traveler's age is 20-. Then the traveler will conclude that the home twin is

CADO_T(20-) = 40 - 0.9*30 = 13

years old immediately before his turnaround. Immediately after his turnaround (assumed here to occur in zero time), their relative speed is -0.9 ly/y. The home twin concludes that their distance apart doesn't change during the turnaround: it's still 30 ly. And the home twin concludes that neither of them ages during the turnaround, so that CADO_H(20+) is still 40.

But according to the traveler,

CADO_T(20+) = 40 - (-0.9)*30 = 67,

so he concludes that his twin ages 54 years during his instantaneous turnaround.

The equation works for arbitrary accelerations, not just the idealized instantaneous speed change assumed above. I've got an example with +-1g accelerations on my web page:

http://home.comcast.net/~mlfasf [Broken]

The derivation of the equation is given in my paper

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

Mike Fontenot
 
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  • #8
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OK guys, thank you. I got the answer that I was looking for. The most previous one was the most helpful and detailed.
 

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