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Twin paradox and the nature of time?

  1. Nov 11, 2008 #1
    I'm trying to make sense of the way time dimension is related to the other 3, the example is twin paradox:

    Observer 1 is moving away at 0.866c from observer 2, who is standing still, then turns around at a defined point and goes back at the same speed, arriving back to observer 2 position, and see him 0.5t older than classically expected.

    Now, the problem with that visualization is why are the points D and C are the ones that are in the same "now"?

    Why does 1 not arrive at point B instead and see the 2 at 0.5t also?
    Are the points B and D not at the same space&time coordinates?

    Alternatively, why, if C and D are the same, can they see each other?
    Because it implies that things could be seen across a distance in time dimension, which contradicts the fact that we don't see our past and future trailing around us, and things can't be seen in two places at once.

    Or, is there something fundamentally wrong with that visualization attempt?

  2. jcsd
  3. Nov 11, 2008 #2


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    I can't calculate this off the top of my head, but when the twins meet, they do have the same space and time coordinates. The key is that ageing is the "length" of the path they travelled in spacetime. The geometry of spacetime is such that the straight worldline between two spacetime points has the longest "length" or "proper time".
  4. Nov 11, 2008 #3


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    Check out #3 and #142 (page 9) in this thread.
  5. Nov 11, 2008 #4


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    An alternative visualization is to use the proper time of the twins, instead of the coordinate time as dimension. Then you see that the twins don't meet at the same proper time coordinate. Here are both diagram types for comparison:
    The main problem of this visualization is that you cannot draw one simple space time diagram for the twin that turns around, because space time is curved in non inertial frames. But you can split it in two inertial parts for leaving & returning as done here.
  6. Nov 11, 2008 #5
    You are not drawing the spacetime diagram correctly.

    First and least important, by convention, we draw them with time going up the page, not down. Not important mathematically, but if you stick to the conventions it will be easier for others to understand.

    Second and really important, someone somewhere has told you "all bodies travel through spacetime at a speed of c," and you have taken this waay too seriously. You are trying to draw line segments of equal length for equal periods of coordinate time because you think that since everything travels at c, they all travel the same "distance" for the same "time," but in different directions. This is incorrect.

    For instance, take a light ray, which must make a 45 degree angle in a spacetime diagram. However, lines of constant coordinate time are horizontal lines corresponding to t=0, t=1, t=2, etc. If you draw a spacetime diagram for a stationary observer and a light ray, then the observer starts at the origin and ends up at the point (0,1). The light ray starts at the origin and ends up at (1,1). One second of coordinate time has passed, and the length of the observer's worldline is 1 on the diagram, and the length of the light ray's worldline is [itex]\sqrt 2[/itex]. Faster bodies have longer worldlines on a spacetime diagram, because in 1 unit of coordinate time, they move up the diagram by one unit but also horizontally by an amount [itex]\beta[/itex].
  7. Nov 11, 2008 #6


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    It is correct for the space-propertime diagram (Epstein diagram), in which you also directly see the age difference between the twins. It is the bottom one here:
    Last edited: Nov 11, 2008
  8. Nov 11, 2008 #7
    Artlav you do not need to use “Space Time Diagrams” to do this, IMO at you level you are better off not using that as it can become a crutch without understanding the problem. Just use what you have already laid out, BUT you must be much more complete.

    You are missing some definable events and measuring points. And, have worked out the solution in only one POV or reference frame. You cannot expect to understand the issue unless you work out all the elements of the problem including calculating them from the POV of all frames of reference – you only used one.

    You define events A B C for observer 1 which stays stationary in ref frame 1.
    But you only give D for observer 2.
    And you only work out the solution in ref frame 2.

    First the missing events:
    The start for observer 1 needs to be defined: Call it A1
    The turnaround needs to be defined: call it T
    (B is not T; B is when and where observer 2 is at the same time that the turn T is made)
    And you did define the ending point for observer 2 as D.

    But you also need to define the start point: call it SP
    And the turn point: call it TP
    And the end point call it EP
    Note these are locations that do not move, not events.

    Now SP, TP & EP are easy if as you did only workout the problem based on ref frame 1. These locations are based on ref frame 2 where SP & EP are the same location. But they are not at the same location in the other reference frames!
    You need to show this more clearly for you own benefit.
    Do not assume the arbitrary frame choice of ref frame 2 sufficient in any way to treat it as “preferred”.

    You have two addition frames not well defined here;
    call them rf1East & rf1West.

    Now with a more complete discription of all the elements involved you are ready to go back to work.
    From all these locations, movements, events and frames of reference what can you say you know with absolute certainty no matter what else might appear to be true?
    Only two things:
    1) events A1 & A happened simultaneously at SP
    2) events D & C happened simultaneously at EP

    All Frames using their POV must agree that these events are two things are true in a realistic reality.

    You need to work out a complete solution giving the time and location for every point at all event times.
    And redo the same solution from the POV of the other two reference frames as well.
    Sorry that means you need to actually do the math! But other than just taking our word for it, that is the best way to actually understand this stuff. and two or three pages of work for each ref frame POV is small price to pay.

    IMO, That is the best way to make clear that the classical cannot reconcile this paradox,
    but SR does by giving results that show all frames agree on the times and locations given by the other frames for events A1 & A and events D & C.
    But other than those two facts none of the frames agree on any other element of the problem. Including where and when SP and EP are physically located at any other event times and the never agree on where and when T, B, or TP are.
    That leads directly to how Einstein defined the Simultaneity Rule, meaning you cannot trust the synchronization of any frame as a correct measure of Simultaneous events.
  9. Nov 11, 2008 #8


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    I usually agree with RandallB, but we have opposite opinions about this particular thing. I think everything is easier with spacetime diagrams. When I first learned about them I was really upset that we didn't use them from the beginning. I would teach them to students very early, even before the Lorentz transformation.
  10. Nov 12, 2008 #9
    Re: Twin space time diagrams

    There is a elementary question here I dont understand:

    If we assume an acceleration rate equal to earth gravity, wouldn't the pictures of the two twins be identical. The gravitational time dilation would be equivelant.
    Wouldnt the space/time diagrams drawn from either frame be identical????

    Wouldn't each see the other accelerate away , travel , stop ,reverse motion and return?

    Does simply our knowledge of the mechanics of the space craft or the experience of the vibration of acceleration cause a change of perspective or give more relevance to the earth frame so we expect the other twin to age less?
    I have read some of the other threads and am still perplexed.
  11. Nov 12, 2008 #10


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    Re: Twin space time diagrams

    The twin paradox is usually assumed to have no gravity. So the twin on the "earth" is not really on the earth with gravity and everything. Without gravity (and with gravity too), the ageing of the twins is due to the different spacetime "lengths" of their worldlines. In his special relativity paper, Einstein actually predicted that a clock on the equator would go more slowly than a clock at the north pole, but that didn't take gravitry into account. Later, after Einstein discovered general relativity, it was realised that the equatorial and polar clocks would both have the same rate because the gravitational and special relativistic time dilations sort of "cancel" each other, as you expected.
  12. Nov 12, 2008 #11


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    Re: Twin space time diagrams

    You can define a coordinate system such that a spacetime diagram drawn using those coordinates shows the accelerating twin as a straight line and the other one as two shorter straight lines, but this wouldn't be an inertial frame.

    The time dilation formula [itex]t'=\gamma t[/itex] tells you the relationship between the time t between two events on the time axis of one inertial frame and the time t' between the same two events in another inertial frame. Note the word "inertial" here. You won't be able to use the time dilation formula to prove that from the accelerating twin's point of view, it's the other twin who's younger.
  13. Nov 12, 2008 #12


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    Hi Austin0,

    For an example of the kind of non-inertial coordinate system with the travelling twin "at rest" that Fredrik is talking about please see figures 8 and 9 of http://arxiv.org/abs/gr-qc/0104077" [Broken].
    Last edited by a moderator: May 3, 2017
  14. Nov 13, 2008 #13
    Hi Dale and Fredrik--

    I was not trying to prove the earthbound twin would age less ,I expect she would age more. I guess my confusion centers on my understanding of what constitutes a valid inertial frame.
    For instance ; What if you put the space twin in an isolated cabin on the ship and then accelerate away at the earth rate. At some point in the journey you carefully reduce forward acceration while reorienting the ship so the centrifical quasi gravity effect is equal to the forward reduction with a comparable slight increase after, so you end up now accelerating back toward earth even though the forward motion is still away from earth. At some point much later, in the earth frame, the ship has slowed and stops and begins accelerating back towards earth but on the ship this is a non-event as far as the internal inertial conditions go ,this is an uninterrupted constant acceleration begun earlier. At the appropriate point you repeat the re-orientation maneuver and arrive back.

    So my questions are:
    Are there any possible tests that the space twin could perform along the way to ascertain that she was in motion or had even left earth?
    If not ,wouldn't this then meet the criteria for an inertial frame.?
    A state of motion wherein the rules of physics apply as usual.?
    Is a state of constant acceelration a valid frame and if not, what is the conceptual basis for making this distinction?
    Couldn't the Lorentz factor be applied differentially in this situation?

    The question of acceleration vs uniform motion seems integral to many situations involving time dilation or synchronicity. I know this has created some confusion in my mind and from what I have read, there seems to be some general uncertainty in this regard.
    From the paper you suggested above: Radar Twins, I gathered that they at least, felt that the acceleration effect was negligeble , but is this the concensus view?
    Sorry if this is a lot of questions.

    Thanks Dale for your helpful input on the frame idea. You zeroed in on the real problem in both the concept and my understanding.
  15. Nov 13, 2008 #14


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    There is no objective truth about who is "in motion" in SR--velocity is relative, so for any two objects moving apart at constant velocity, you can consider either one to be at rest. However, acceleration is not relative--if you're accelerating, you feel G-forces, which can be measured by an accelerometer or weight on a scale. Inertial observers always feel weightless in SR.
  16. Nov 13, 2008 #15
    Originally Posted by Austin0
    So my questions are:
    Are there any possible tests that the space twin could perform along the way to ascertain that she was in motion or had even left earth?
    There is no objective truth about who is "in motion" in SR--velocity is relative, so for any two objects moving apart at constant velocity, you can consider either one to be at rest.

    Sorry if I was unclear I was not suggesting that her motion was in any way true.
    I was refering to it in the purely Galilean Cabin sense where it is relative motion being considered and the point is the inability to detect it without going on deck and looking.
    In the space twins case it would mean leaving her cabin which she is prohibited from doing.

    However, acceleration is not relative--if you're accelerating, you feel G-forces, which can be measured by an accelerometer or weight on a scale.

    Isn't the foundation of the Equivalence Principle the recognition that, in fact, you cannot
    determine a constant acceleration by those very means you suggest?
    That within the elevator physics plays out just like it does in every other valid inertial frame.

    Inertial observers always feel weightless in SR.

    This is really the core of my question. Is this then the basic requirement for a valid frame?
    If this is so, does this mean there is a fundamental disjunction between SR and GR ?

    The effects of both realms are inextricably intertwined in the real world so it is confusing to me if the basic principles of both disciplines aren't in agreement.

    Thanks for your answer to my question.
  17. Nov 13, 2008 #16


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    The acceleration isn't constant in the scenario you're describing in #13. It can be directed "towards the floor" the whole time, but only if the rocket is rotating, and if it is the rotation is detectable.

    The equivalence principle says that there are no local experiments you can perform to distinguish between constant proper acceleration and a constant gravitational field, but the word "local" refers to the fact that the equivalence is only exact in the limit where the "diameter" of the region of spacetime where the experiment is performed goes to zero. The final ages of the two twins depend on the curves in spacetime that represent their motion. There's nothing "local" about that.
  18. Nov 13, 2008 #17
    I had thought that the local consideration was simply because at a large range the lack of divergence in the direction of the force would be revealed. And is the scale in x limited??
    The basic conditions of the experiment would seem to indicate that it isnt.

    In any case I now begin to see the objections. I still cant help but think this is a severe limitation for a theory that deals specifically with systems and objects in motion. And you cannot achieve motion without acceleration. Not to mention that by this definition the earth itself is not a valid frame, being both gravitational and accelerated. Yet that is where the actual application of the theory is enacted.
    Well thanks

  19. Nov 13, 2008 #18
    When Galileo developed the special principle of relativity he was talking about dropping a couple things from the top of a ships mast. One landing on deck the other falling with it far enough out to go over the side and land on a the dock. But when released as they pass by the end of the dock the second would miss the dock as the dock “would be gone” by the time it fell due to the relative speed. Being “in the cabin” did not require not looking out the window to see that the dock was “moving”.

    Nothing in the Twins prevents you from looking out the window to see the relative speed between Earth and ship. That is not the same as “leaving the cabin”.
  20. Nov 13, 2008 #19


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    Here you're talking about motion in the curved spacetime of GR rather than the flat spacetime of SR. In fact, physicists say that a freefalling observer is moving in an "inertial" manner in GR, you wouldn't say the falling object is being accelerated by gravity like you would in Newtonian physics where gravity is assumed to be a force and the surface of the Earth can be treated as an inertial frame (an observer on the surface of a gravitating body like Earth would be considered to be moving non-inertially in GR).
    No, see my comment above--it's still true that observers moving "inertially" in GR are weightless, because moving inertially in GR means freely-falling, moving along a geodesic path in curved spacetime.
    Last edited: Nov 13, 2008
  21. Nov 13, 2008 #20


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    I'm not sure what limitation you're talking about.
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