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Points of reference for time dilation

  1. Sep 1, 2012 #1
    This is something that interested me after reading about time dilation. For the point of making it simpler, ignore relativity relating to gravitational fields

    Lets say for example, that the Earth does not orbit the sun, and the solar system doesnt orbit within the galaxy etc (makes it easier to explain, but the point is the same),... and the Earth is just spinning on its axis in free space, having no resultant movement ( other than rotating) in the universe.
    So if someone were to be in outer space, looking down at the Earth, can we assume that they age "quicker" than someone standing on the Earth, as the person in space would have 0 velocity, and the people on the Earth would be moving with a velocity equal to the rotation of the earth.
    Now, what if someone were to run on the earth in the opposite direction to the earths rotation, with a speed equal to the earths rotation? So within the free space that person would have 0 resulant velocity. The question is that with reference to free space, that running person would be aging "quicker" than someone who is sitting still on the earth, but is this true? Because from the earths point of reference, the running person is travelling very quickly, so time wouldn't apper "slower?" for that person compared to someone sitting still? Does the absolute true value for γ hold only if you take velocities from a point which has no velocity at all ( the idea of free space). And finally, am I correct in saying you would age slower if you are nearer the equator, ( as your speed would be greater?)


    I appreciate anyone who can enlighten me on this. I apologize if I have missed a obvious point here, but I've been thinking about it for a few days and I can't get my head around it! Thanks!
     
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  3. Sep 1, 2012 #2

    PeterDonis

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    Hi, Joshcad55, and welcome to PF!

    So just to be clear, you are asking specifically about time dilation due to relative motion, and only that? I'll assume you are in what follows.

    Ok.

    From the standpoint of the person in outer space, yes, they would "age quicker" and the person at rest on the rotating Earth would age slower. But from the standpoint of the people at rest on the Earth, rotating with it, things would be different. See below.

    From the standpoint of the running person, yes. But not from the standpoint of the person at rest relative to the rotating Earth. See below.

    Yes, from the standpoint of the person standing still (relative to the rotating Earth), the person running away would appear to age slower. So would the person in outer space who is not moving with the Earth's rotation.

    However, eventually the person running around the Earth would come back to meet the person who stood still on the rotating Earth; then things get more interesting. See below.

    There is no such point. That's one of the key lessons of relativity.

    Not necessarily; it depends on whose standpoint you are looking at it from. However, we now have enough pieces in place to make some key observations.

    First, let's label our observers, for convenience. Call the observer in outer space, who is floating free and not rotating with the Earth, O. Call the observer who is standing at rest on the equator of the rotating Earth, E. Call the observer who is running westward along the equator, just fast enough to cancel the Earth's rotation (so he keeps observer O exactly overhead), R. And add one more observer, at the North Pole; call him P.

    Now, if each observer uses special relativity to analyze the rate of time flow for himself vs. the other observers, in the most straightforward and natural way possible, here is what they will conclude:

    E will see everyone else's clocks running slower than his.

    O, P, and R will see E's clock running slower than theirs, and will see the other clocks running at the same rate as theirs.

    However, now let's consider what happens when R runs a complete circuit around the equator and meets up again with E. We'll assume they all set their clocks to zero when R and E last met up. Then each of them can record the time elapsed by their clocks between the two meetups. (Strictly speaking, only E and R can do this directly; the others are spatially separated from them. But since O, P, and R all agree on their clock rates, as above, we can assume that O and P will agree with R's observation of time elapsed.)

    How will E's and R's elapsed times compare? E's elapsed time will be *less* than R's. This is true even though E, above, saw R's (and O's and P's) clocks running slower than his. The comparison of elapsed times is a direct observable, so it has to be the same for both E and R; they can't disagree about it the way they can disagree about the above observations of clock rates when they are not together. So in some sense, E's observation that R's clock was running slower as R ran away from him was an "illusion"; when R comes back to E, E will find that R's clock ticked off more elapsed time between their two meetups.

    And, to answer the question as you actually posed it, since R's clock and P's clock are synchronized, the observer at the equator, rotating with the Earth (E) will indeed age slower, in an absolute sense, than the observer at the North Pole (P). But E will still see P's clock (like R's clock) "ticking slower" in a relative sense.

    I realize all this is counterintuitive, but that's why it's important to draw distinctions between "clock rates" of observers in relative motion, which can appear different to different observers, vs. elapsed time between two common events like the two meetups of E and R, which must be the same for all observers.

    The above is a lot to digest, so I'll stop now. Please feel free to ask more questions if things are not clear after you have digested the above.
     
  4. Sep 1, 2012 #3
    Be careful about making general statements about the situation when one of the observers is accelerating.

    Be careful about statements about the clock rate of the accelerating observer.

    You may want to describe this situation in a little more detail in view of the accelerating observer, avoiding missatements about clock rates for an accelerating frame as reckoned from the vantage point of an inertial frame--or more importantly, about clock rates for the inertial frame as reckoned by from the vantage point of the accelerating observer.

    Although generally you have done a nice job of helping out JoshCAD55 here, this statement is misleading. You might want to work through this a little more carefully. Perhaps you were trying to spare JoshCAD55 the tortuous details of the accelerating example. In that case it's better to first walk Josh through the standard twin paradox for inertial frames, and then graduate to the accelerating frame. The traveling twin accelerates at the turn-around point, but that is not a complicated analysis.
     
    Last edited: Sep 1, 2012
  5. Sep 1, 2012 #4

    PeterDonis

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    It wasn't meant to be a fully general statement about how things look in E's "accelerated reference frame". It was only meant to be a statement about the comparison between the local inertial frames of E and R when they are close enough together to make such a comparison meaningful. We could go into that kind of technical detail, but it might cause more confusion than it solves; that's up to the OP.

    Same comment; I was only talking about the "clock rate" in a local inertial frame, where the issues you are referring to do not arise.

    The comparison of elapsed times between two meetings of R and E does not depend on any statements about "clock rates for an accelerating frame". The entire calculation of elapsed proper time for R and E between two meetings can be done in an inertial frame--R's rest frame--so that none of the issues you are raising arise.

    Not if it only refers to comparison between local inertial frames, as above.

    The difference in elapsed times for E vs. R between two meetups *is* an example of the twin paradox, just with a slightly different type of trajectory for the "traveling twin" (E in this case). You can't have a twin paradox without at least one of the observers being accelerated (at least not in Minkowski spacetime, which is how I'm treating this since the OP said not to consider gravitational time dilation at all).
     
  6. Sep 1, 2012 #5
    Not completely true. You are probably aware of this and just didn't want to complicate the discussion.

    This is fine as long as its understood in what sense we refer to comparisons of clock rates.

    E always observes that R's clock is always running slower? Rethink that.

    Of course this is correct and is the main concept for joshcad55 to take away. But Josh should not be given the impression that the clock rates reckoned by the various observers are illusions.
     
  7. Sep 1, 2012 #6

    PeterDonis

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    Just to be clear, you are referring to the issues you brought up concerning accerated reference frames, correct? If so, then this...

    ...is correct. But as you note, the fact that E's clock shows less elapsed time than R's between two successive meetups is true, period, regardless of any issues about accelerated frames.

    I put "illusion" in scare-quotes precisely because this usage of the term requires considerable unpacking. Perhaps I should have been more consistent about putting "running slower", "aging quicker", etc., etc. in scare-quotes as well. Whether it's worth unpacking all that depends on how much the OP wants to dig into it.
     
  8. Sep 2, 2012 #7

    ghwellsjr

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    I'm glad you want to make it simple. The concept of time dilation is very simple as long as you follow Einstein's procedure presented in his 1905 paper introducing Special Relativity. You start with an inertial Frame of Reference which, for purposes of this discussion, is what you mean by "free space". When you think of "free space", it extends out to infinity in all directions, doesn't it? So can you see how your term "Points of Reference" is at odds with the term "Frame of Reference"? When you're talking about Time Dilation, you don't want to limit it to a Point.

    The first thing you do is determine the speed of each object with respect to that Frame of Reference and you use that speed to calculate γ which will always calculate to be a number greater than one for speeds greater than zero. This is the time dilation factor which means that a second for any moving clock in the Frame of Reference will take longer than one that is not moving. But we usually like to talk about the rate at which a clock ticks so we have to take the reciprocal of γ which will always be a number less than one for any moving clock.

    So this means that moving clocks can only tick slower than non-moving clocks as determined by the Frame of Reference.

    With that little bit of background, which I hope was simple, let's go through the rest of your post.
    Since the person in space is not moving in the Reference Frame of space, he does not experience any time dilation. But the people on the surface of the Earth are moving so they do experience time dilation.
    Yes, it's true but again, we should say that because of his zero speed, his time dilation is zero and we already said the person sitting still on earth is one of the people moving on the surface of earth, so he will be experiencing the same time dilation as everyone else on the surface of the moving earth.
    Now you're deviating from Einstein's simple procedure and it's totally unnecessary. We don't want to think of just a "point of reference". And the surface of the earth is not inertial, so it gets very complicated.
    Again, you want to take velocities from an infinite Frame, not from a point, but, yes, that's how you calculate γ.
    Yes, that is correct, according to the Frame of Reference that you selected.
    Just remember, the time dilation of an object or observer is determined by its speed in an inertial Frame of Reference. If you select a different inertial Frame of Reference moving with respect to the first one, then the speeds of all objects and observers will be different and you have to recalculate all their time dilations.
     
    Last edited: Sep 2, 2012
  9. Sep 2, 2012 #8
    First of all, these are great responses, so thanks all.

    What I meant by here:
    Is how are the people actually aging differently due to time dilation. I mean, I know depending on from where you are observing, γ can be different, but surely you don't age differently depending on where the observer is? Or have I just undermined my own argument there?

    So does that person R age more quickly than E? Because taken from a point where your velocity is 0 (my interpritation of free space?), his velocity is slower.
     
  10. Sep 2, 2012 #9

    ghwellsjr

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    If you want to keep it simple, follow Einstein's procedure. Don't think of "from where you are observing" or "where the observer is". Think in terms of a single inertial Frame of Reference. Your chosen Frame of Reference in this scenario is "free space".
    Again, think of each object's (or observer's) time dilation as independent of everyone else's, they're only dependent on each object's (or observer's) speed according to your chosen Frame of Reference. And time dilation can only result in clocks going slower, never faster just like moving objects can only have speeds greater than zero.

    So, according to your own description of the scenario, the runner has zero speed in your chosen Frame of Reference so he does not experience any time dilation. The people on earth are traveling at a speed depending on their latitude. Someone at one of the poles has zero speed and no time dilation. Someone not running at the equator has the maximum speed in your scenario and therefore the maximum time dilation. The people that are located midway between a pole and the equator are traveling at a speed somewhere in between and have a time dilation that is in between.

    If you think in these terms, it is very simple.

    Now you ask about "actual" aging due to time dilation. That's like asking about "actual" speed. You wanted to remove any motion due to the earth going around the sun or the sun going around our galaxy, which is alright to do, since those motions are not inertial. But you can pick a different Frame of Reference, say one that would approximate the speed of the earth relative to the sun and the speed of the sun relative to the galaxy and transform your scenario into a new inertial Frame of Reference moving at any arbitrary speed in any direction relative to your original Frame of Reference and you will get different speeds for each object/observer and therefore different time dilations. In fact, the people that are on the surface of the earth will be moving at variable speeds which makes your analysis more complicated because you have to integrate or add up all their speeds and time dilations during each orbit of the earth. As long as you do that, you will see that each object/observer ages relative to the others by the same amount. But it will be a more complex analysis, not the simple one you started with.

    So there's no "actual" aging, just like there's no "actual" speed, just relative to a chosen inertial Frame of Reference.
     
    Last edited: Sep 2, 2012
  11. Sep 2, 2012 #10
    Good catch. You were right to question that.

    Here's a link to my previous posting that gives the solution to this problem of circular motion, when the inertial person is located at some fixed position on the circle:

    https://www.physicsforums.com/showpost.php?p=4040633&postcount=286

    The age-correspondence graph shown is for the case where the traveler is moving very fast (v = 0.866c), so that gamma = 2.0. For that speed, from the traveler's "perspective", the inertial person is never aging slower than the traveler. The inertial person concludes that he is always aging exactly twice as fast as the traveler. The traveler concludes that the inertial person is sometimes aging somewhat LESS than twice as fast she herself is (but always at least a little faster), and that sometimes the inertial person is aging somewhat MORE than twice as fast.

    In the case where the traveler's speed is very slow compared to light speed (like in the rotational speed of the earth), gamma is only very slightly greater than 1.0, and there are some segments of the circular motion where the traveler will conclude that the inertial person is aging (very slightly) more slowly than she is, and other segments where the traveler will conclude that the inertial person is aging (very slightly) faster than she is.
     
  12. Sep 2, 2012 #11
    Good job, GrammawSally (are you actually a grandmother?). That's exactly the analysis I had in mind. PeterDonis always seems to be well informed on special relativity, so I thought he just wasn't thinking carefully and would recognize the error in his statement right away and follow up with a correction.

    But your reference was right on the money and shows quite clearly what is going on. Thanks a lot for jumping in. I think joshcad55 will appreciate this.

     
  13. Sep 2, 2012 #12

    PeterDonis

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    I lost track of that thread somewhere in the middle, so I'll have to go back and review the later posts in it; but as I remember, the age-correspondence graphs you made there were done using the CADO equation. That equation gives one way of assigning an "age" or "rate of time flow" to distant objects when an observer is accelerating; but it's not the only possible one. If the OP really wants to dig into that aspect of it, there are plenty of threads we can link to.
     
  14. Sep 2, 2012 #13

    PeterDonis

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    The correction is that, strictly speaking, there is not a unique way to set up an "accelerated reference frame", so what I said above E observing R's clock to "run slow", strictly speaking, requires defining the reference frame I was using. I was using local inertial frames along E's worldline; in any such frame, E is at rest and R is moving, so by the standard rules for how local inertial frames work, R's clock will appear to be running slow compared to E's, in any such frame.

    However, since E is accelerated, his local inertial frame is a *different* inertial frame at each event on his worldline; and there is no way to "patch together" the different local inertial frames into a single global frame that matches up locally with each inertial frame's coordinate assignments. So any "accelerated reference frame" for E that covers E's entire worldline will have to compromise: it will not match up with the coordinate assignments in the local inertial frames at at least some points on E's worldline; and therefore, it is perfectly possible that, when evaluated in such a global "accelerated reference frame", R's clock will appear to be running faster than E's. The CADO equation gives such a result.
     
  15. Sep 2, 2012 #14
    Here's a post from another thread devoted to the twin paradox. MikeLizzi respondes with afformation to GrammawSally's well-put statement about effect of acceleration during the turn-around. joshcad55, you may gain better insight into the effects of acceleration by going back over the standard twin paradox scenario, then coming back to the circular orbit (which is a spiral worldline in the 4-dimensional universe). I think it helps enormously if you also do a google search on "spacetime diagrams." Once with a grasp of the 4-dimensional universe, assisted by use of spacetime diagrams, it makes the understanding of the twin paradox quite clear--you can see graphically what is going on and how, in the sequence of traveling twin frames during turnaround, that the traveling twin's succession of 3-D cross-sections of the 4-dimensional universe present pictures of the stay-at-home twin's clock moving quite fast compared to the traveling twin's clock. This is precisely what GrammawSally and joshcad55 have been communicating to us.

    https://www.physicsforums.com/newreply.php?do=newreply&p=4028043 [Broken]

    Aug9-12, 10:10 PM Re: simple question regarding the twin paradox #64
    MikeLizzi


    Posts: 182 Originally Posted by GrammawSally
    They both use the same laws of physics. But the results they get from those laws depend on whether or not they are accelerating. The person back home never accelerates, but the astronaut does accelerate, at the turnaround.

    Anytime the spacecraft is traveling at a constant velocity with respect to the inertial person back home, the astronaut will say that the person back home is aging more slowly than she is. But while the astronaut is reversing course at the turnaround (which requires her to accelerate), she will say that the person back home is aging much faster than she is.


    Well said, GrammawSally.
     
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  16. Sep 2, 2012 #15

    PeterDonis

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    If you adopt a particular coordinate chart for the traveling twin, this is true. However, as I've noted, there is more than one possible coordinate chart for the traveling twin to adopt.

    The Usenet Physics FAQ on the twin paradox gives a very good discussion, including the multiple ways of interpreting what's going on:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

    GrammawSally's analysis appears to correspond fairly well with the equivalence principle analysis given there.
     
  17. Sep 3, 2012 #16
    joshcad55, I didn't want PeterDonis's comments about different possible coordinate systems to mislead you into thinking that GrammawSally's analysis was not necessarily correct, and that there are other equally good analysis methods that give different results. GrammawSally gave you the actual readings of comparative times that would be obtained from an actual experiment (should you be able to devise such an experiment).

    I've attempted to sketch below a picture representing the worldlines of observers in the 4-dimensional universe. With the possibility of indicating only 3 dimensions in the sketch, the X3 dimension has been suppressed. Here we've depicted a thought experiment in which your runner has enlisted companion runners, all running at the same pace so as to, in effect, be "standing still" (we could go into tortuous detail about what standing still means in this context). So, we have runners at the blue positions, 1,2,3,4,5,6,7,8. These runners share the rest frame shown with the X1, X2, and X4 coordinates. The 4-dimensional worldlines of all eight runners are parallel and extend along the vertical black lines from which we've formed a cylinder. The world line of your E observer who stands at a point on the earth's equator with the earth rotating is shown as the red curve in the sketch.

    All of the blue observers have synchronized their clocks before the start of the experiment so that they all share the same clock time at each of the sequence of instantaneous simultaneous 3-D spaces that they also share (cross-sections of the 4-dimensional universe). We have marked their clock times as blue t1, t2, t3, t4, and t5. The corresponding red worldline times (consistent with GrammaSally) are marked as red t1, t2, t3, t4, and t5. So, in this experiment we have direct comparisons of the blue inertial frame clocks to the red worldline clocks, and these comparisons are exactly those presented by GrammaSally.

    You can choose any other kind of coordinates and and analysis you wish, but they must all yield the same comparisons between the blue and red times (as presented by GrammaSally) or else the analysis is flawed. These events, the intersection points between blue and red worldlines (and their comparative clock readings), are events that would be agreed upon by all other external observers (those other than our blue and red), regardless of their motions.
    Cylindrical_4D_Sphere.jpg
     
    Last edited: Sep 3, 2012
  18. Sep 3, 2012 #17

    PeterDonis

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    I didn't say it wasn't correct; just that it wasn't the only possible analysis.

    All correct methods must give the same results for actual observables, as you note later on (see below).

    This is a good sketch, and it definitely helps in understanding what's going on. And this...

    ...is certainly true.
     
  19. Sep 4, 2012 #18

    ghwellsjr

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    Bobc2, doesn't your analysis/graph show that there is a constant relationship between the blue and red times? Aren't you depicting each "runner" in a common inertial frame where they are all stationary (what GrammaSally calls the inertial person) and that the person at the equator is the non-inertial traveler, the one that GrammaSally calls the accelerating person? And isn't it just the dashed line on GrammaSally's graph that corresponds to your graph?

    (BTW, your link in post #14 is messed up and just before that I think you meant to say MikeLizzi not joshcad55, correct?)
     
  20. Sep 4, 2012 #19
    You already got some answers; I'll try to answer even simpler. :tongue2:
    Yes, if as you assume, that person uses a reference system in which he/she is in rest (that's indeed an obvious choice in this case!).
    "Free space" in your sentence merely has the meaning of "rest frame of that person in space". You seem to not have noticed that import point. In fact the person in space can choose any inertial frame to assign the term "free space" to. Consequently we can't say a thing of who is "actually" aging faster in the sense that you mean, if we compare two inertially moving people. That's in essence what the relativity postulate means. Of course, it's different when one of them is going in circles, as you noticed.
    Ignoring gravitation, then yes indeed people at the equator would age slowest according to everyone over the course of full rotations.

    The reason why I added that last bit is simple. As you remarked yourself: over a short trajectory, according to a with the equator co-moving inertial frame it's just the other way round. Such a disagreement does not exist for full rotations.
     
    Last edited: Sep 4, 2012
  21. Sep 4, 2012 #20

    ghwellsjr

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    You have to be careful here. Just because someone is going around in circles doesn't mean that everyone will agree about how much they have aged. Of course, everyone will agree that their own clocks will have elapsed a specific amount of time after the course of one rotation but that is no different than saying that everyone will agree that an inertial person's heart has beat 60 times during the course of one revolution of the second hand on his clock. What everyone will agree on is that the aging of the rotating person will be a certain amount compared to an inertial person that he periodically meets up with. However, the aging of both of them will be determined by other differently moving inertial observers to be different.
     
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