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Twin paradox: who decided who is the younger one

  1. Oct 30, 2011 #1
    Hello,

    Still puzzled about the twin paradox (one guy stays on the ground, the other goes travelling). If we see the two twins as points in 3D space, the only thing that changes, is the distance between two. That is, the two guys are in totally symmetrical positions, and there should be no reason one is ageing slower than the other. If we take into account the Earth, the 1st twins distance from the Earth is fixed, and the others — varies. Is it the presence of Earth (ie, energy of the planet matter that bends the spacetime) that causes the difference in ageing for each twin?

    This confuses me, because the special relativity doesn't speak about the way how matter bends the spacetime, and the twin paradox is typically used for explaining the special relativity.

    Passiday
     
  2. jcsd
  3. Oct 30, 2011 #2

    ghwellsjr

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    No, we ignore any real attributes of earth when discussing the Special Relativity aspects of the Twin Paradox.

    You are correct, if the twins remain symmetrical, then each thinks the other one is aging more slowly. But if one of them accelerates to join up with his twin, when they get back together, that is the one who will be younger.
     
  4. Oct 30, 2011 #3

    Matterwave

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    One of the twins must change direction, and therefore go into a different inertial frame than the one he started from, this is the a-symmetrical part of the twin paradox. There is no twin paradox if the situation remained perfectly symmetrical (e.g. if both twins moved away from Earth, traveling at the same speed, turning back at the same time etc).
     
  5. Oct 30, 2011 #4

    phinds

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    I'm not an expert on this but I think I have an issue w/ that ghwellsjr ... that is the statement "if one of them accelerates to join up ... "

    Tell me if I have something wrong with this thought experiment. Twin A travels in a magical thought-experiment spaceship in orbit around the earth and accelerates up to .99c and then decelerates down to non-relativistic speeds and lands. I think that is a situation in which twin A did NOT accelerate back to meet his brother but he's still younger when he gets there.

    Is my analysis of this situation perhaps wrong because I am failing to take into account the fact that twin A spends half his time getting farther away from his brother and half getting closer?
     
  6. Oct 30, 2011 #5

    phinds

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    But they would both be younger than those they left behind so if one of them had a son, there would be the "father/son paradox". :smile:
     
  7. Oct 30, 2011 #6

    DrGreg

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    While orbiting the Earth at 0.99c, A is undergoing a very high radial acceleration towards the Earth.
     
  8. Oct 30, 2011 #7

    tom.stoer

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    Let's do some math.

    Assume we have two twins located at (t,x) = (0,0) in one specific coordinate system. They will meet again at a later time T but at the same location x=0, i.e. at (T,0). The question now is "what are T and T' prime in which coordinate system?".

    Now let's avoid coordinates.

    Assume one twin is traveling along a curve C from point A to point B in spacetime. The second twin is traveling along a different curve C' from point A to point B in spacetime. Of course we could introduce the coordinates for A and B, but that is not necessary.

    Now you have to believe me that the proper time tau of a twin along his curve between A and B is given by the "length" of the curve through spacetime.

    [tex]\tau = \int_C d\tau[/tex]

    Here the "length" and therefore the proper time is calculated according to the strange 4-dim. relativistic Pythagoras t² - x².

    As the two curves C and C' through spacetime are different for the two twins their proper times will differ.

    [tex]\Delta\tau_{A\to B} = \int_{C_{A\to B}} d\tau - \int_{C^\prime_{A\to B}} d\tau[/tex]
     
    Last edited: Oct 31, 2011
  9. Oct 30, 2011 #8

    phinds

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    Fair enought. Thanks.
     
  10. Oct 30, 2011 #9
    I think yours is the clearest explanation. Here is a space-time diagram illustrating your point. The red guy in the diagram stays home and sees his twin return after 13 red years. But it's only ten years on the blue guy's calendar. The blue guy who made the trip took a short cut through space-time. Each observer moves along his X4 at the speed of light (and T = X4/c), but the blue guy in the diagram took the shorter path as you showed with your integral.

    It's the "strange 4-dimensional relativistic Pythagoras..." -- that's the point. Excellent post. Thanks.
    Twin_Paradox_Proper2B.jpg
     
    Last edited: Oct 30, 2011
  11. Oct 30, 2011 #10

    tom.stoer

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    Thanks for the drawing - and one caveat: it's not possible to measure the length directly in such a coordinate system b/c one must not use cartesian coordinates in Minkowski space.
     
  12. Oct 30, 2011 #11
    Yes. You are right about that. But look closely and you will see that I included that hyberbolic calibration curves so that we could make comparisons between the two coordinate systems. I used the 5 year calibration curves for each of the two inertial start events for the traveling twin.
     
  13. Oct 30, 2011 #12

    tom.stoer

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    Of course the caveat was not meant for you :-)
     
  14. Oct 31, 2011 #13

    ghwellsjr

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    What do the diagonal red lines represent?
     
  15. Oct 31, 2011 #14
    Thanks for the explanations, however, I am talking about a point of confusion before we get to the Minkowski space. Ok, one guy takes a shortcut in spacetime, thus ages a bit slower. But my question is, what is he doing to use that shortcut? At this moment, educated layman would shout to me, you stupid, he is achieving speed close to c, while the other guy is not. But, I am asking, against what entity his superior speed is measured? If we ignore the presence of the field of gravity of the Earth/Sun/Galaxy, then all we are left with, is two points on a line, whose distance changes in time. Is it the extraordinary acceleration experience that makes the guy age slower? Then that would be involving the general rather special relativity in the explanation (which is fine with me, but perhaps not fine for all those books explaining the special relativity using this paradox).
     
  16. Oct 31, 2011 #15

    ghwellsjr

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    Special Relativity is all about choosing an arbitrary inertial (non-accelerating) Frame of Reference from which all speeds are referenced. Any object/observer moving in that frame will experience time dilation. The faster they go, the more time dilation or the slower their clocks run. Special Relativity has no problem with an object/observer accelerating but the FoR must remain inertial (non-accelerating).

    You don't have to have any other material objects besides the two twins in your chosen Frame of Reference. If you pick your FoR so that one of the twins remains stationary in it throughout the entire scenario so that neither he nor the FoR are accelerated, then he will never experience time dilation. The other twin travels away and back and experiences time dilation for the entire trip. He has to accelerate in order to do this. It's not his acceleration that causes him to experience time dilation, it's his acceleration that causes him to travel at a high speed in the FoR and it's his high speed in the FoR that causes his time dilation. When he gets back to the first twin, he will have aged less than the one that never experienced acceleration or speed according to the FoR.

    As I said in my first post, if both twins are moving with respect to each other, then you could choose a FoR in which either of them was at rest and the other one would be experiencing time dilation. In this case, the jury is out, the issue of which one is really aging less is meaningless. You could also pick a FoR in which they were both traveling at the same speed but in opposite directions and although they would both be experiencing time dilation, it would be the same for both of them so they would age at the same rate. But unless at least one of them accelerates so that he can rejoin his brother they won't be able to get back together and it is only when they are together that they can compare accumulated aging.

    Now you might ask why you can't pick a FoR in which the traveling twin is at rest and the answer is that you could for one half of the trip but then on the way back, he will be traveling at an even higher speed and experiencing even more time dilation than his brother.

    The salient point when dealing with issues like this is that unless you can get the same answer to a problem with any arbitrary FoR that you select, then the problem can't be solved. But if it is a solvable problem, and we know the Twin Paradox is, then you can pick the simplest FoR to analyze the problem and you'll get the correct answer.
     
  17. Oct 31, 2011 #16

    tom.stoer

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    My mathematical explanation does not use any reference frame, but only two curves C and C' through spacetime with different 'length'. Along C and C' speed and acceleration may differ which leads to different curves and therefore different proper time along these curves.

    There is no shortcut, but simply two different curves. It's like going from A to B on a sheet of paper; you can do that along different curves with different length. In SR (or GR) you have to take into account that spacelike and timelike directions will differ and that you are therefore not able to 'measure' proper time on a sheet of paper using a ruler.

    Regarding the 'entity' against which this difference of proper time is measured: there is no such entity, proper time along C is measured against proper time along C' - nothing else.
     
  18. Oct 31, 2011 #17
    I think this is what I have problem with understanding. Since we don't take the presence of Earth into account, I imagine starting the thought experiment far away in space, where there's no any heavenly body to mess with the speeds and accelerations. Pretty much the environment with no matter other than the two twins and spacecraft with loads of fuel. If one guy gets into the rocket and travels, while the other stays hanging in void, the only thing that differentiates the travelling guy is that he's using reactive force to accelerate. He feels the acceleration and thus knows that he is changing the speed. If he has gyroscope on the board then he can even know when ship turns 180 degrees to travel back. I guess it's the mystery of why there should be any acceleration whatsoever in totally empty space what escapes my understanding.
     
  19. Oct 31, 2011 #18

    pervect

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    1) You can tell when you're accelerating by the seat of your pants - literally. If you accelerate, you, or an accelerometer, can feel it or measure the forces. It doesn't matter if you are in empty space or not. This is in contrast to velocity, which you can't feel.

    2) The first twin does not have to turn around for the two twins to meet. This was mentioned, though you got distracted by the other issue, I gather.

    For the two twins to meet without twin #1 turning around and going back, all that has to happen is that twin #2 waits a bit, then accelerates even harder to catch up. In this scenario, it's twin #1 that ages the most.
     
  20. Oct 31, 2011 #19
    According to SR, acceleration has "absolute" effects: it matters who accelerates. Even in deep space. One explanation is that "totally empty space" isn't truly totally empty. The first full presentation of the twin paradox scenario was even given in support of that view - see from p.47 of http://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time
     
  21. Oct 31, 2011 #20

    phinds

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    Passiday, as ghwellsjr pointed out, it is not the acceleration that causes time dilation ... the acceleration is a necessary side effect of the process because it's needed to get the moving twin up to a high speed.

    Time dilation is ONLY due to high speed of one FoR relative to another (non-accelerating) FoR , and the two FoR have to meet up again in order for it to have any meaning (again, this is as ghwellsjr pointed out).

    So consider this scenario: Twin B sits on the ground of a non-accelerating planet and according to his FoR, his Twin A goes accelerating off, travels at a very high speed for a while, decellerates, and at the end of the process ends up back at Twin A's side.

    WHILE he was traveling at high speed, here's what happens:

    http://www.phinds.com/time%20dilation/ [Broken]

    It's just this simple.
     
    Last edited by a moderator: May 5, 2017
  22. Oct 31, 2011 #21
    Your explanation used the class of inertial reference frames, just as all the other explanations have done, and this is precisely what the OP is challenging. To define the 'lengths' of the curves you said "Here the "length" and therefore the proper time is calculated according to the strange 4-dim. relativistic Pythagoras t² - x²." The variables "t" and "x" in that expression are not arbitrary, they must be inertial coordinates, so you are referring to an inertial frame. Otherwise the expression doesn't give the proper time. Of course, as others have pointed out, it doesn't matter which system of inertial coordinates you use, but they must be inertial coordinates. Your explanation did not circumvent this requirement.

    Basically the OP is asking about the origin of inertia. Ernst Mach (for example) argued that ultimately inertia must be defined by the relations of an object to every other object in the universe. It happens to be a fact that the inertial coordinate systems are those at rest or in uniform motion relative to the frame in which all the matter of the universe (that we can see) is isotropic, i.e., the same Doppler shift spectrum in all directions. Whether this is cause or effect is debatable. Einstein originally thought general relativity fulfilled Mach's prediction, but later he realized that it (probably) doesn't. The origin of inertia remains mysterious, even in general relativity and quantum field theory. (Note that the Higgs has not been found at the LHC.)
     
  23. Oct 31, 2011 #22
    You should consider what is not symmetrical. When one of the twins returns there must be acceleration which breaks the symmetry. Also consider the proper time. The twin that is accelerated back should have a spacial displacement in all reference frames or you simply cannot find a inertial reference frame that sees the twin travelling at rest. Therefore, I think that dτ2=dt2-dx2 would make him younger.
     
  24. Oct 31, 2011 #23

    ghwellsjr

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    I did point out what you said in the first paragraph but I did not say anything like what you are saying in the second paragraph. This is totally false.

    Here is the truth: Time dilation is only due to the speed of each object/observer relative to an inertial (non-accelerating) Frame of Reference. There's only one FoR considered at a time. It has nothing to do with the relationship between two FoR's or two FoR's meeting up again, whatever that means. Every FoR extends out in all directions, here, there and everywhere, and includes all time, past, present and future, and every object/observer that you want to consider is in any FoR you want to consider. So every FoR already "meets" every other FoR at all locations and at all times.

    But each object/observer can experience a different time dilation in each different FoR. You have to pick a FoR, do your analysis to get an answer and then pick another one and see if you get the same answer. If you do, then the answer is legitimate, if you don't, then the question is meaningless.

    So if you have two twins traveling away from each other and you ask how each one ages in the rest frame of one of them, you will determine that only the other one is experiencing time dilation. If you then do the same thing for the rest frame of the other one, you will determine that only the first one is experiencing time dilation. If you then pick a FoR in which they are both traveling at the same speed in opposite directions, you will determine that they both age at the same time dilated rate but neither is getting older than the other. You get three different answers to the question of which one is really experiencing time dilation (or which one is younger) which means the question itself is meaningless. And you can get an infinite number of other answers by merely picking an infinite number of FoR's in which both twins are moving at different speeds.

    But if the two twins get back together, then the question can be answered using any FoR. They all yield the same answer. But you should use just one FoR at a time. Work it out, see what the two clocks say after they come back together. Pick another FoR, work it out again, you get the same answer.
     
    Last edited: Oct 31, 2011
  25. Nov 2, 2011 #24

    Interesting you should ask. The slanted lines are there to show the relative progress of each observer as they move along their respective 4th dimensions at the speed of light. When the traveling twin meets up with the black stay-at-home twin, the home twin is 13 years while the twin guy is 10 years. But wait--the home twin really couldn't be there could he -- if he has so far moved at the speed of light along his X4 to just his own proper time 10 yr point on X4???

    If you think of observers moving at light speed along their 4th dimension, the home twin hasn't really arrived yet. But this is one reason why many physicists have concluded that objects in 4-dimensional space are 4-dimensional themselves. Thus the travel twin is meeting up with a different cross-section view of home twins 4-dimensional body. And the physical bodies really aren't doing any traveling anyway, because they are four-dimensional objects fixed --frozen in a 4-D universe.

    It's ugly, I know. My subjective self does not want to accept that picture at all, but as I've said before, there is no alternative external objective picture out there that contradicts the 4-dimensional universe populated by 4-dimensional objects. I think many just abandon the idea of an external objective reality.
    Twin_Paradox_Proper2B.jpg
     
    Last edited: Nov 2, 2011
  26. Nov 2, 2011 #25
    Here's another example. The hyperbolic calibration curves were computed in MatLab. Notice that when the home twin is 20 yrs old his view of the universe includes the travel twin with his (red's) clock showing 10 yrs. Red (travel twin) is about 17.5 light-years away from home (along the black X1 axis, going about 0.875c) at this point. That's the usual time dilation.
    Hyperbolic_Matlab1.jpg
     
    Last edited: Nov 2, 2011
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