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Time Dilation and Differential Aging

  1. Oct 1, 2005 #1
    I know very little about SR -- just the transformation equations I learned in high school.

    I remember reading somewhere that time dilation is a symmetric artifact of the synchronization convention, and that, by itself, it isn't the reason for differential aging. The reason, if I'm remembering accurately, had to do with geometric interpretation and also the idea that 'empty space' isn't really emptly.

    I'm sure you get naive questions all the time from laypersons (like me) who have difficulty getting wrapped around the idea that aging is a function of velocity (or is it a function of acceleration?) Anyway, I'm thinking, velocity wrt what?

    Remembering that the transformations had to do with two observers, each with their own clock, moving relative to each other, I wondered if they would record the same time (for, say, a round trip by one of them from the Earth to the Andromeda galaxy and back) if they were both using the *same* clock (say, revolutions of the Earth around the Sun). Of course, the traveller would record fewer revolutions on his way to Andromeda than on his way back, and the rate at which he recorded them would be different than the rate at which the Earthbound observer recorded them. However, on landing back on Earth, wouldn't the traveller have recorded the same total number of revolutions for his trip as the Earthbound observer?

    If not, why? If so, then what would it mean to say that they aged differently?
     
    Last edited: Oct 1, 2005
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  3. Oct 1, 2005 #2

    jtbell

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    The traveler agrees that the earth has made the same number of orbits around ths sun during his trip, as are counted by the stay-at-home twin. The details of how he comes to that conclusion depend on exactly how he uses the earth's orbits as a clock. If he does it by watching the Earth and Sun through a telescope as he travels out and back, then the situation is exactly like the one analyzed in posting #3 in this thread (click here).

    But in this case, the traveler simply sees the earth revolving at a different rate than it normally does. He experiences the actual passage of time (i.e. grows older) according to clocks that he carries along with him.

    Note that the traveler actually sees the Earth revolving at different rates during the outbound and inbound halves of his trip. What he actually sees with his eyes depends not only on the earth's rate of revolution in his reference frame (via time dilation), but also on the time it takes for light to travel from the solar system to his eyes, which continually changes as he travels (via the Doppler effect).
     
  4. Oct 1, 2005 #3

    JesseM

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    It isn't. If you used the same type of synchronization convention in a Newtonian universe you wouldn't see any time dilation.
    Relative to any inertial frame--although of course, different frames will disagree on the rate that a given person is aging. But if two people separate, move apart for a while and then reunite, all frames will make the same prediction about what ages the people will be when they reunite, and they'll all agree that the one who moved inertially aged more than the one who didn't. You can think of this in analogy with picking two points on a piece of paper, then drawing two paths between them, one straight and one bendy--you could then draw different xy axes on the paper, and in each coordinate system see how x is changing as a function of y along each path and integrate it to find the total length of the path. Even though different coordinate systems would have different functions for how x changes as a function of y along a given path, they'd all get the same answer for the lengths of the two paths between the points, and the straight-line path would always be the shorter one no matter how the bendy path was drawn.
    Sure, otherwise the two observers would be making different predictions about actual physical events, which can't happen. If two clocks cross paths at a single point in spacetime, all frames have to agree about what each clock reads at that moment. And the earth going around the sun can be thought of as a type of clock, you could imagine big numbers displayed on the earth which increment each time it completes an orbit.
     
  5. Oct 1, 2005 #4
    The traveller isn't carrying a clock with him. The time convention that they're using is the earth's revolutions around the sun. They count the same passage of time for the round trip. So, what does it mean to say that the traveller has aged differently?

    Or, we could use the round-trip itself as the unit of time (their common clock), and during this unit of time they count the same number of earth-sun revolutions.

    I'm thinking that maybe SR just doesn't apply here.

    Maybe a better way of framing my question is to have the traveller simply fly around the earth for, say, five years (five earth-sun revolutions). They would both count the same number of revolutions that the traveller has made around the earth during this time.

    I'm thinking of time-keeping as an arbitrary convention. As with all things, what is observed depends on the observational context. The earthbound observer and the traveller are always part of some encompassing motional system. In the last example, wrt the motion of the solar system in its local group, both observers have travelled the same distance in the same amount of time.

    Anyway, the way I'm thinking about it, differential aging shouldn't really have anything to do with time dilation due to Lorentz transformations and Einstein's simultaneity convention, since this is applicable only to a certain, arbitrarily chosen, observational context.
     
    Last edited: Oct 1, 2005
  6. Oct 1, 2005 #5
    Yes, I remember now that it was the *transformations* and the simultaneity convention that produced the time dilation effects.

    If all frames agree that the trip took, say, five years (all observers counted the same number of earth-sun revolutions for the trip), then where's the differential aging?
     
  7. Oct 1, 2005 #6

    JesseM

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    But even that's not quite enough--after all, you could use the same coordinate transformation in a Newtonian universe, but natural physical clocks would not tick at the same rate as the coordinate time in their rest frame. It's the fact that the laws of physics are Lorentz-invariant that produces differential aging, where "Lorentz-invariance" means the laws obey the same equations in all the different coordinate systems produced by the Lorentz transformation.
    Because "differential aging" refers specifically to comparing clocks that move along with both travelers. Anyway, even if the space traveller doesn't carry a clock along with him, all the natural clocks in his body will have elapsed more time (this is why twins are usually used in the thought-experiment, so you can tell that one is visibly older than the other).
     
    Last edited: Oct 1, 2005
  8. Oct 1, 2005 #7

    jtbell

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    It means, for example, that if the trip is long enough, the stay-at-home twin's hair turns grey but the traveller's hair does not.

    This particular effect hasn't been observed yet (differential hair-turning-grey), but a similar phenomenon is routinely observed in high-energy particle physics, where fast-moving particles decay more slowly than slow-moving ones.
     
  9. Oct 1, 2005 #8

    Janus

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    That all depends on how you define a "year". If you define it as 31,558,464 sec, then the spaceship twin will say that the trip took fewer than 5 years while the Earth twin will say that it took 5 years.

    If you define it by "one Earth-Sun revolution, then they will say the trip took the same number of "years', but the Spaceship twin will say that each year lasted for fewer seconds than the Earth will.

    Example:
    Assume that the relative velocity is 0.866c, and each twin has an average heart rate of 60 beats/min. Then, during five Earth-Sun revolutions, the Earth twin will have experienced 157,792,320 heartbeats, and the Spaceship twin will have experienced 78,896,160 heartbeats.
     
  10. Oct 1, 2005 #9
    Effects produced by relative motion are symmetric, isn't this correct? Isn't that one of the reasons why SR was developed -- to deal with these effects symmetrically, rather than asymmetrically as had been done before?

    So, I'm having a difficult time understanding how this applies to differential aging? I remember that the writer of the article that I'm trying to remember (I think it might have been Keating -- he's not some quack or something is he?) said something about time dilation and differential aging being due to different reasons (a different relative history of acceleration or something like that). And, as I was reflecting on this recently it seemed to make sense, sort of.

    That's what I thought. So, if they're using the same clock then there's no differential aging ... I thought. :-) However, the author of the paper said that there is differential aging, but that it was due to empty space having a metric structure, that empty space isn't really empty. If this is true, then does acceleration mean more, or less, interaction between the accelerating object and the structure of space?

    Do you understand my confusion yet? Because I, obviously, don't? :-)

    Isn't the earth-sun system a natural clock?
     
  11. Oct 1, 2005 #10
    Could you give a brief summary of how those experiments work?
     
  12. Oct 1, 2005 #11
    Assuming that's true, then what physical-mechanical interaction is *causing* this differential aging -- or is it an open question?

    For example, I have no doubt that gravitational fields affect the period of anything that one might want to use as a clock. But, with gravitational fields the reason for this is a bit easier for me to conceptualize/visualize than with an object simply accelerating from a previous state of motion.
    Hmmm, is the answer to my question in there somewhere? Is differential aging connected to gravitational fields?
     
  13. Oct 1, 2005 #12

    JesseM

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    Only if the motion is inertial. The laws of physics will look the same to an experimenter in any two windowless boxes that aren't accelerating, but in an accelerating box they'll look different.
    I don't know what you mean by "interaction between the accelerating object and the structure of space". Again, you can think in terms of the analogy I made earlier with the distance between points on an ordinary 2D piece of paper--just as the "structure" of euclidean space is such that the shortest path between two points is a straight line, so it is that the "structure" of spacetime is such that the path between two points in spacetime with the greatest proper time (time as measured by a clock that follows that path) is an inertial one. Just as all non-straight paths between points in 2D space will have a greater length than the straight one, so all non-straight (accelerating) worldlines between two points in spacetime will have a shorter proper time than the straight one. And just as you could use different coordinate systems in a euclidean space which would give different functions y(x) for a given path, and thus different functions s(x)=dy/dx for the slope of the path as a function of x, yet no matter which coordinate system you used you'd get the same answer for the total length of the path when you integrate the appropriate function involving s(x) (I think you'd have to integrate [tex]\int \sqrt{1 + s(x)^2} \, dx[/tex], because if you have a right triangle with horizontal side x and vertical side y then the length of the hypotenuse is [tex]x \sqrt{1 + (y/x)^2}[/tex], but I could have that wrong), so it's analogously true that despite the fact that different coordinate systems have different functions x(t) for a path through spacetime, and thus different functions v(t) for the velocity as a function of time, when you integrate [tex]\int \sqrt{1 - v(t)^2/c^2} \, dv[/tex] in each coordinate system you'll get the same answer for the total amount of time ticked by a clock moving along that path (note that in any reference frame, a clock moving at velocity v will tick at [tex]\sqrt{1 - v^2/c^2}[/tex] the rate of a clock at rest in that reference frame).
    Yes, but it isn't one moving along with the travelling twin. If the travelling twin took an exact copy of the earth-sun system along with him, his earth #2 would make less orbits around sun #2 than the original earth around the original sun. This is because the original sun is moving (approximately) inertially while the sun moving along with the travelling twin is not.
     
    Last edited: Oct 1, 2005
  14. Oct 1, 2005 #13
    Isn't everything accelerating wrt, perhaps, many different gravitational hierarchies. That is, aren't gravitational fields more or less everywhere?

    Neither do I at this point. Just some vague idea having to do with wave interactions.

    In the revised example I gave, the traveller remains, along with the earthbound observer, part of the earth-sun system. Of course, his trip still has him moving differently wrt the gravitational field than the earthbound observer. So, I'm thinking that it's this differential interaction with the gravitational field that has to do with any differential aging that might be observed.
     
  15. Oct 1, 2005 #14
    Isn't that, in effect, what he's doing by observing light from the earth-sun sytem during his trip?
     
  16. Oct 1, 2005 #15

    JesseM

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    Special relativity doesn't deal with gravity, it just deals with flat spacetime, while general relativity treats gravity in terms of curved spacetime.
    What does "part of it" mean? The sole issue here is that the earth twin is moving inertially, or close to it anyway, while the other twin is accelerating significantly.
    No, he is accelerating significantly relative to the sun. If he took a copy of the sun along with him, and he was at rest wrt this copy throughout his trip, and meanwhile the first twin was at rest with respect to the original sun, then the copy sun would undergo fewer orbits.
     
  17. Oct 1, 2005 #16

    Janus

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    No, there would be an age difference even if there were no gravitational fields present. In fact, Special Relativity only deals with situations where there are no gravitational fields. General Relativity is needed to deal with the addtional effects due to the presence of gravitational fields.
     
  18. Oct 1, 2005 #17
    Ok, so my question becomes: what is it about this acceleration that is causing the traveller's natural periods to be altered? Is it in part due to a change in way he is interacting (due to the changes in his motion) with the gravitational fields that he is travelling through. Might it have anything to do with the expansion of the universe?

    The thing is, it doesn't seem to me to be correct to say that the differential aging is caused by time dilation. I mean you can use the transformations to calculate the effect, but that doesn't tell you why it's happening -- does it?
     
  19. Oct 1, 2005 #18

    JesseM

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    I dunno, that's like asking what is it about curved paths between points in euclidean space that make them always be longer than a straight-line path between the same two points.
    No. Both gravity and the expansion of space are part of GR and do not exist in flat spacetime, but the twin paradox certainly works in flat spacetime...SR is solely concerned with what happens in flat spacetime.
    Again, why is it that a curved path between two points in euclidean space is always longer than a straight path between the points? Do you need an explanation of "why" this is besides just noting that it's a consequence of how distance works in euclidean geometry?
     
  20. Oct 1, 2005 #19

    jtbell

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    (about high-energy particle physics experiments where fast-moving particles decay more slowly than slow-moving ones)

    As an example, one of my friends in graduate school was part of a group that studied sigma and xi hyperons. These particles decay with a very short lifetime that had been well-measured at low speeds (small energies). The apparatus produced a high-energy beam of these particles, traveling at a significant fraction of the speed of light. Despite their high speed, if the particles had had the same lifetime as at low speeds, the beam would have extended only a very short distance, maybe a few millimeters or centimeters. (This was 25 years ago, so I don't remember the specific numbers.) However, because of time dilation, these fast-moving particles lived long enough to produce a beam several meters long, which made it a lot easier to study them.

    Note that time dilation was not the point of the experiment; rather, time dilation was simply used as a means of making the experiment feasible. If time dilation didn't work the way relativity predicts, this experiment would not have been possible. I've forgotten what the actual goal of the experiment was... something like measuring the particles' spin magnetic moments.
     
  21. Oct 2, 2005 #20
    How could we know for sure -- since there's always gravitational fields present?
     
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