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Twin paradox

  1. Jul 6, 2010 #1
    I was just working on my knowledge of the twin paradox, and had a question that I couldn't find an answer to anywhere:
    If I understand it correctly the paradox is resolved because the two frames are not symmetrical, one is non-inertial, so that frame has it's clock run slower. So if you carry this reasoning back to the beginning of the universe, should there not be some frame that has never undergone any acceleration? The frame of reference would be the 'oldest' frame in that if you had put a clock there since the big bang it would now read more time passed than any other clock in the universe. What is this frame called?
     
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  3. Jul 6, 2010 #2
    That clock would have accelerated in the big bang, so it would be in a non-inertial frame. Acceleration is just the change from reference frame to reference frame, if you describe each reference frame independently then it is easy to see why one clock slows down and another does not. When you take the collection of frames that the clock travels through and call it one frame, it is no longer an inertial reference frame.

    Changing frames is not a relative feature, the observer that changes frames is the only observer effected. If you are in your car driving next to someone and you accelerate, they would never feel the force instead of you.
     
  4. Jul 6, 2010 #3
    Well what I mean is, imagine a scenario with the earth and a spaceship some distance from it. The earth is hit by an asteroid which causes it to accelerate and drift away from the spaceship. Now because the earth is the one that accelerated, a clock there would run slower than on the ship, correct?

    Now apply the twin paradox, one twin accelerates away in a second spaceship and the other remains on earth. The twin that moves away simply accelerates so that he is stationary relative to the first spaceship. Now his clock should be running slower than the clock on earth, but since he is stationary relative to the first ship his clock should be at the same rate as that one, but the original ships clock is moving faster than the earths clock. So how does that work?

    Now since I assume the earth has undergone various accelerations during its lifespan, movement relative to the earth should produce a clock rate of change that could be either slower or faster....shouldn't it?
     
  5. Jul 6, 2010 #4
    When the spaceship decelerates to match the original spaceship's reference frame it is no longer running slower than earth time, it is running faster along with the original spaceship. You must keep track of the positive and negative acceleration, think of it as jumping up and down in an endless ladder of reference frames.

    Edit: Think of the relativistic effects as being related to the relative distance between two ladder rungs, it is the same for both observers. However what is not relative is that one rung is in a different part of the ladder from the other. Einstein did not want the name Relativity Theory, rather wanted to call it Invariance Theory.
     
    Last edited: Jul 6, 2010
  6. Jul 6, 2010 #5
    You keep saying "relative", then you gloss over its implications. LostConjugate really did answer this. There is no one absolute reference frame that supersedes others by being non-inertial ad infinitum. This is about the Relativity of Simultaneity, and the issue of reference frames is the same around earth as it would be around a distant object. It is relative to the observers.
     
  7. Jul 6, 2010 #6

    Fredrik

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    You shouldn't be talking about "frames" here, but about timelike geodesics. There are such curves in FLRW spacetime, and a clock moving as described by one one of them would display the time coordinate of the coordinate system that everyone uses when they're working with FLRW spacetimes. I don't know if there's a standard name for them. FLRW coordinates?

    Not true. One way to see that: What direction would it accelerate in? (Assume that the universe is homogeneous and isotropic).

    It doesn't really make sense to say that an object "is" in a specific frame. I know a lot of people do that, but that doesn't make it right. If you want to say that an object has velocity 0 in a specific frame, just say that. In this case, you should be saying that it's not doing geodesic motion, or that it's accelerating in some specific local inertial frame.

    Coordinate acceleration is the second derivative of the spatial coordinate with respect to the time coordinate. Proper acceleration is a measure of the deviation from non-geodesic motion. Its value is equal to the coordinate acceleration in the comoving inertial frame (at least in SR).

    Expressed in the coordinate system we'd associate with the motion of the ship, yes. But not in the coordinate system we'd associate with the motion of the Earth.

    For standard questions about the twin paradox, please see almost any of the hundreds of other threads about it.

    I also suggest that you stop thinking about clocks as "running faster" or "running slower". They always do what they're supposed to, which is to measure the proper time of the curve in spacetime that represents its motion.
     
  8. Jul 6, 2010 #7
    Which is a long way of saying the same thing right? Since the first derivative of space is just a reference frame itself, velocity.
     
  9. Jul 6, 2010 #8

    Fredrik

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    Not sure which two statements you're referring to. :smile: But I was a bit sloppy, so I'll try again. If you compute dx/dt in any coordinate system, you get the velocity in that coordinate system. The velocity in the comoving inertial frame is =0 by definition of "comoving". If you compute d2x/dt2 in any coordinate system, you get the coordinate acceleration in that coordinate system. Its value in the comoving inertial frame is called the proper acceleration. It can also be defined in a way that's completely independent of all coordinate systems.

    Reference frame = coordinate system = function that assigns a 4-tuple (t,x,y,z) of real numbers to each event.
     
  10. Jul 6, 2010 #9
    Yes, each reference frame is a different coordinate system. If you change your velocity you must transform your coordinate system and therefor are in a new reference frame.
     
  11. Jul 7, 2010 #10

    Fredrik

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    You don't have to use another coordinate system, but if we insist on using the coordinate systems that the standard synchronization procedure associates with the observer's world line ("proper reference frames"), or the coordinate systems that it associates with the tangents of the observer's world line ("comoving inertial frames"), then we do need to change coordinate systems. I really don't like the choice of words "(you) are in a new reference frame", since nothing can just "be" in a function (unless you're using the definition of a function from X into Y as a subset of X×Y with specific properties, and if you are, then the statement still doesn't make sense).
     
  12. Jul 7, 2010 #11
    I'm using this because I'm trying to visualize the problem, and I haven't done much study using geodesics and spacetime curves. I will look into it though.

    Well this is exactly the question I raised in my first post. Regarding the twin paradox, what should have been an acceleration from twin leaving earth is now regarded as a 'deceleration', and so instead of the leaving twin aging slower, the twin that remains behind ages slower. All of this because you put a spaceship up who was not affected by the asteroid hitting and accelerating the earth.

    Now, I don't suppose that you need to actually put a physical spaceship up there. The earth has undergone many accelerations over its lifespan, and we can imagine a spaceship having observed them all. So if we were to do the twin experiment, should we not get different results depending on which direction we shot the twin? Sometimes one would age faster, sometimes the other.

    In which case can you not use this to find in which direction is really a 'deceleration' rather than an acceleration. Should you not then be able to find a frame in which any acceleration produces a slower time. That's what I was asking, I don't see how you've discussed that, unless I'm missing something.

    Thank you.
     
  13. Jul 7, 2010 #12

    Janus

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    You are thinking in terms of an absolute rest frame, and Relativity says that there is none. Deceleration and acceleration are relative terms and are frame dependent. What is acceleration in one frame would be deceleration in another.

    It makes no difference what accelerations the Earth may have experienced in the past, you get the same result from the twin paradox no matter what direction the twin travels in.

    For example, assume you have a large ship that launches a bunch of little ships at the same time in different directions and has them return. Each ship returns with their clock an equal time behind the main ship's clock. The large ship accelerates up to some fraction of the speed of light and repeats the experiment. The results will be exactly as they were before.
     
  14. Jul 7, 2010 #13

    Dale

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    For a flat metric (i.e. little gravity and rectilinear coordinates) a geodesic is a straight line. One twin's path is a straight line, the other twin's path is not. The length of a straight line is different from the length of a bent line.
     
  15. Jul 7, 2010 #14
    Well that's what I assumed was the case, originally. But take the example I gave above:
    LostConjugate seems to be disagreeing with you, I think:
    He seems to be suggesting that the presence of the original spaceship somehow causes the twin's clocks to behave differently. And I don't see how.
     
    Last edited: Jul 7, 2010
  16. Jul 7, 2010 #15

    Janus

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    Again, the problem comes from assuming that one clock running faster or slower than another is something that is true in an absolute sense. If I am in a spaceship with a relative motion with respect to the Earth. (it does not matter how this relative motion occurred), By my determination, the clock on Earth will run slower than my own, and by the determination of someone on the Earth, my clock will be running slower than his.

    You also need to take length contraction ( as measured from each frame) and the Simultaneity of Relativity into account.
     
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