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Pure twin paradox

  1. May 9, 2008 #1
    We know what is twin paradox is, and also why it is not a paradox, but my question is, "if we can not define absolute motion, how can we decide that the twin which went to space will age less, and not the home one?"

    That is to say that, when peter at home sees paul's clock moving slow (in space ship), similarly, paul should see peter's clock to be slowed down. So for paul, peter should be the younger! How is it that when paul comes, he is the one who is younger?

    for example, let's say (arbitrarily) that, the absolute speed of earth (which is of course not measurable) is 70,000 km/s. the speed of paul's spaceship is 30,000 km/s. and they are moving in opposite direction. So, if we talk about the absolute speed, peter should be younger, and if we talk about relative speed, both should be younger for each other. What is it that lets us decide, that each time, paul will be the one who aged less?
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  3. May 9, 2008 #2

    Doc Al

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    One twin accelerates, the other doesn't; that breaks the symmetry. The accelerating twin experiences less proper time.
  4. May 9, 2008 #3
    Oh yes, but if the accelerating twin looks back, he will think, the earth is accelerating with respect to him. Or acceleration is not relative, while velocity is?
  5. May 9, 2008 #4

    Doc Al

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    Acceleration is not relative.
  6. May 9, 2008 #5
    How come acceleration is not relative?
    Does that mean that when paul will look back, he will find that earth is in uniform relative motion with some velocity v with respect to him (and not accelerated with respect to him)? This is highly improbable.
  7. May 9, 2008 #6
    The twin on the spaceship is the one who is accelerating away from the inertial frame of reference (Earth) and will undergo such effects associated with the act of acceleration. Barring the effects of gravity and the motion of Earth around the Sun, the twin on Earth maintains the original inertial frame of reference. That's how I understood it.
  8. May 9, 2008 #7
    I understood more or less the same way.
    But my question is, removing everything else, other than the earth and paul's spaceship (i.e. keeping only two frames of reference, earth and the spaceship), If paul is accelerated, will not he believe that the earth is accelerated as well with respect to him. accepted that he can measure his own acceleration without the reference to the earth (with an accelerometer), yet when he would look-back, he would notice the earth to be accelerating at the same rate, and hence, the time dilation for paul due to his own acceleration as observed by peter, will be applicable to peter as well when paul observes him.
  9. May 9, 2008 #8


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    Since he knows he is accelerating (after all, he's firing his engines), he will consider the earth stationary wrt him.
  10. May 10, 2008 #9


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    And the twin on the earth doesn't "feel" anything happening to himself when his twin fires the spaceship's engines.
  11. May 10, 2008 #10
    "He's firing his engines" is not an answer! In such a case, If two space ships are going in the same direction (no relative velocity), and if they don't consider any reference to the other spaceship, or the earth, because they know that their engines are working, they will believe their clocks are slower? But I think that was just a friendly comment and not explanation.

    In relativity, we are concerned about what one observer feels (or observes) about other, and not what one feels about himself! The time for paul in his frame is going at a uniform speed for him. It's what peter's observation that paul's clock is slow. And thus, if it is the acceleration of paul, which let's him slow down his time, absolutely irrespective of peter, then that doesn't seem relativity to me.

    Consider paul is in the ship, but he doesn't drive the ship, and just observes the relative motion, not visually, but by some radio signal strength or sort of thing. Now, before the engine is fired, some heavenly body strikes earth and accelerates to same amount, as would have been case if the engine would have been fired. Will he not feel that the engine has been fired? Now in such a situation, whose time will be slow? Paul's or Peter's?
    Last edited: May 10, 2008
  12. May 10, 2008 #11


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    Force equals mass times acceleration. The twin who is accelerating will fill a force during acceleration. If it were the earth that were accelerating, the twin on earth would feel an additional force.

    This is all in "special relativity". In "general relativity", an "external" force can be interpreted as additonal gravity rather than acceleration. However, since a higher gravitational force results in slowing clocks, the result is the same.
  13. May 10, 2008 #12


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    (This comment accentuates those of russ_watters and jtbell.)

    Consider a block (an ice cube) on a frictionless (icy) surface.
    This setup is in the ship of each observer.

    For the inertial observer, that block remains in the same position on the table throughout the experiment.

    For the observer that accelerates (or is accelerated by the "driver"), that block is not in the same position.
  14. May 10, 2008 #13
    You are not listening AntigenX, people here are trying to help you with your problems in understanding relativity, but with comments like the ones above I fear not for long.
  15. May 10, 2008 #14
    You are right, pardon me for being so arrogant and what not...

    I have read comments of everybody, but I think we are missing the point here.

    Simply put, If acceleration, which is absolute, and doesn't require any external frame of reference for its measurement (i.e. is self evident in the observer's own frame), then how come paul's time slow for peter, when it isn't relative to peter at all? I consider the gravity point of view, but how would we give explanation of this so called paradox in SR?
  16. May 10, 2008 #15


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    you appear to mixing two different things here. The twin case applies to a situation where the the twins are together in one place, then one goes off on a round trip, and they compare clocks when the traveller returns. In this case the twin who did the most accelerating will have fewer ticks on his clock.

    The other situation is what they will get if they measure each others clock rates while the travelling twin is in progress. In this case, each one will measure the others clock as being slower.

    This contrasts with what they will see if they observe one another with telescopes, which is another case.
  17. May 10, 2008 #16
    May be I'm not putting the situation correctly. I am not mixing two situations though. As referred in my last post, If acceleration, which is absolute (and not relative) dilates paul's time, without any reference to peter, how should it matter to peter at all. As is our argument formulated, during the whole journey, paul is bound to accelerate at least 4 times, during which, his clock runs slower. My point is, slower with respect to which clock? Because, as noted above, acceleration is self evident in paul's frame for paul, and is not relative to peter.
    If we say, with respect to peter's clock, then the time dilation in a situation of paul's acceleration is relative to peter (i.e. is not absolute), and hence, the same time dilation should be evident for peter as well, as observed from paul's point of view.
  18. May 10, 2008 #17


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    My recommendation is to forget the acceleration. It only serves to break the symmetry and does not cause time dilation by itself. Instead, use the spacetime interval which allows you to extend the analysis to arbitrarily accelerating twins.

    Your scenario here is fine, but you didn't actually do the math, so your conclusion is wrong.

    In frame A (for "absolute") ve is .00023 c and vp is initially -.00010 c. The two start together at spacetime coordinates (0,0). If the outbound portion of the trip takes 1 year in frame A then at the end of the year the earth will be at spacetime coordinates (1,.00023) and paul will be at spacetime coordinates (1,-.00010) where both are measured in lightyears. At this point paul turns around and heads back to the earth where they re-unite at spacetime coordinates (2,.00046).

    The spacetime interval for the earth's clock is
    |(1,.00023)-(0,0)| + |(2,.00046)-(1,.00023)| = 1.9999999471 years.

    The spacetime interval for paul's clock is
    |(1,-.0001)-(0,0)| + |(2,.00046)-(1,-.0001)| = 1.9999998382 years.

    So paul's clock reads less than the earth's clock by 3.4 seconds.

    The point is that you can pick any arbitrary frame and reach the same conclusion that paul aged less. We don't have to "decide", we just have to follow the same rules in any inertial frame.
    Last edited: May 10, 2008
  19. May 10, 2008 #18


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    Paul's acceleration does not cause his time dilation, rather it effects what he measures is happening to Peter's clock. You have to use different rules when dealing with the measurements of an accelerating observer. What Paul determines is happening to Peter's while he (Paul) is accelerating is determined by three things: The magnitude of the acceleration, The direction of the acceleration, and the distance between Peter and Paul as measured by Paul along a line parallel to the direction of the acceleration.

    When Paul is accelerating away from Peter he measures Peter's clock as running slow, The greater the distance between the two, the more pronounced the slowing of Peter's clock. This is in addition to any slowing due to the relative velocity between them.

    When Paul is accelerating towards Peter, he measures Peter's clock as running fast, minus any slowing due to relative velocity.Again, this becomes more pronounced as the distance between them is increased. Note that accelerating "towards" doesn't always means "moving towards". Braking to a stop while moving away counts as acceleration towards.

    Thus Paul measures the following as happening to Peter and his clock.

    As Paul accelerates away from Peter he sees Peter's clock run slow both due to the increasing velocity difference and his acceleration away. Since at this time the distance between the two is not large, the acceleration does not add much.

    Once Paul reaches the desired cruising speed, he quits accelerating and coasts. During this time Peter's clock will run slow at a fixed rate.

    Paul reaches the turn around point and begins to brake and then accelerate back towards Peter. During this period, he determines that Peter's clock speeds up due to his acceleration towards Peter. Since at this point, the distance between the two is its greatest, this speeding up of Peter's clock is enough to not only overcome any slowing due to relative velocity but actually causes Peter's clock to jump ahead of Paul's clock.

    Paul begins the coasting leg of the return trip. He again sees Peter's clock as running slow. However Peter's clock starts out as ahead.

    Peter and Paul reunite. The slowing of Peter's clock during the return trip is not enough to completely offset the time it gained during the turn around phase, and Paul finds that he and his clock has accumulated less time than Peter's.
  20. May 10, 2008 #19
    For anyone having trouble understanding why the acceleration is crucial, take a look at this link. If you don't understand it (I didn't the first time through), then slow down and read it very carefully. I've found skimming through these sorts of things is just a waste of time.

    Last edited: May 10, 2008
  21. May 10, 2008 #20


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    Drawing a space-time diagram makes it very clear. Curvy world-lines represent accelerated frames, and they are longer than straight lines ( inertial frames ) joining any two points. The relative clock speeds are related to the lengths of the worldlines between the points of coincidence. It is immediately obvious why (how?) accelerations slow clocks.
  22. May 10, 2008 #21
    In spacetime the curved lines are shorter not longer than straight lines between two events. The length of the lines represent the total elapsed time. It is not correct to say that the clock on the curved line went slower, it went with the same rate as the clock on the straight line, it is just that the path was shorter thus less time was elapsed.
  23. May 11, 2008 #22
    Note that while velocity is relative in special relativity acceleration is relative under general relativity.
  24. May 12, 2008 #23
    I have said it before, but it may not hurt to say it again.

    There is really no twin paradox. It is a false paradox due to an imperfect framing of the scenario.

    Acceleration is not really the answer, although the acceleration does give you an indication of symmetry breaking.

    Think of it this way: the two twins are knocked out, and then they are revived just as they pass each other such that neither knows which of them could be considered to be stationary (it is actually neither of them in absolute terms). They both have identical clocks which are set in motion as they pass each other. When they reach a certain separation, they are both knocked out again and their clocks are stopped. They are allowed to keep moving relative to each other, then they undergo relative deceleration until their relative velocity is negative - so that they now approach each other at the same speed as they separated before.

    When they reach the same separation as they had when they were knocked out, they are revived again and their clocks restarted. They are knocked out for a third time when they reach each other and their clocks are stopped for a second time. The experiment stops when they are collocated and stationary with respect to each other. They can now compare clocks, which only functioned during inertial portions of their voyages.

    Which clock will read less? In the scenario as stated, both clocks should read the same. There is nothing to distinguish one twin from the other and, most importantly, the relative distance travelled is expressed as a separation from each other.

    This is where there is a difference - in the standard expression of the "twin paradox", one twin travels to a location which is stationary with respect to the other twin then turns around and comes home again.

    Say that you modify my scenario slightly:

    Now you can introduce Lorentz transformations to see that if the relative distance travelled according to the twin "at rest" is 2L then, according to the other twin the relative distance travelled is:

    2L' = ((L-vt)+(L+vt)) / sqrt(1-v^2/c^2) = 2L / sqrt(1-v^2/c^2)

    This requires some interpretation, since the twin "at rest" doesn't go anywhere relative to the turning point. However, according to the other twin, the twin "at rest" does move, along with the turning point. 2L' is this perceived travelling distance, a greater distance than "actually" travelled. The time needed to travel this greater distance is greater given that the relative velocity is common to both, which is why the twin "at rest" will have a clock reading which is greater than the "travelling" twin's.

    The acceleration itself has absolutely no effect. What has an effect is choosing a turning point which is at rest relative to one of the twins and it this which causes the symmetry break - not the acceleration.


  25. May 12, 2008 #24
    There is a problem here DocZaius.

    Consider this, Joe and Jane do what I discussed in the previous post with a slight twist.

    Jane travels to a spot which is a distance L from Joe. She travels inertially from Joe to this spot, starting her clock as she passes Joe (telling him to start his clock) and stopping it as she passes the reference spot. At that time she sends a message to Joe to stop his clock.

    Then she turns around, travels back again, past the reference spot again, past Joe again and turns around a second time, passes Joe a third time, restarts her clock and tells Joe to restart his clock. She stops her clock again when she passes the reference spot for the third time and sends a message back to Joe to stop his clock.

    When they join up again, they compare clocks.

    Which will read the most?

    We can work it out using relative distance travelled. According to Joe, Jane travelled from Joe to distant spot twice, a distance of 2L. According to Jane, the whole length Joe-distant spot moved past her twice. Because that length was in relative motion, it was contracted - so 2L'=2L . sqrt (1 - v^2/c^2).

    According to Joe, the time taken for Jane to travel Joe -> distant spot was 2L/v.

    According to Jane, the time take for the length Joe-distant spot to move past here was

    Therefore Joe's clock will read more.

    No need for deceleration or acceleration. No need for comparison of world lines.

    Again, the deceleration/acceleration is a useful indicator and the world lines are interesting for explaining what goes on ... but these are symptoms, not causes.

    Here's a little mind bender:

    Let's say Jane travels until Joe sends a message to stop and turn around.

    Jane will take longer to get back to Joe because the message takes time to catch up to her. So if Joe sends the message at time t, the message has to travel a distance of vt plus whatever distance Jane travels in that time (and so on).

    Then instead say that Jane travels till she gets sick of it (or lets the universe pass her by, if you prefer), then sends a message to Joe that she is turning around (or telling the universe to change relative direction).

    Jane will get back to Joe quicker than expected, because the message takes time to get to him, time during which Jane is already on her way home.

    But what will their clocks read, relative to each other? Will Jane be younger in both instances?

    (To make it easier, try thinking of it from another perspective, what if Joe was in charge? If he decides that he wants Jane back, he can change the direction that the universe is moving relative to Jane (including himself) immediately. If Jane wants to return to Joe, she has to send him a message asking him to be so kind as to change the direction the universe is moving.)

    The answer highlights that acceleration is not the culprit here.


  26. May 12, 2008 #25

    Doc Al

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    I fail to see how this example has much to do with the standard "twin paradox" scenario.

    Also, don't forget to include the time it takes for Jane's signal to Joe to reach Joe.
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