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I Gravitational time dilation near Earth

  1. Apr 21, 2017 #1
    We know (measured) that a clock on a mountain "ticks" faster than a clock at sea level. At higher altitude, the clock runs even faster.

    Now, if we go much higher, towards the Sun, on the line between the center of the Earth and the center of the Sun, the clock should begin, at some point, to run slower, influenced by Sun's gravitation. Where is that "point" (where the clock would stop increasing its tick rate and then start to decrease again), at what distance from the center of the Earth? It was calculated and/or tested experimentally?

    PS I asked the same question on another forum and made some progress in finding the answer, but it seems that the value I got is not good enough ... Please help.
     
  2. jcsd
  3. Apr 21, 2017 #2

    Dale

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    What assumptions are you making regarding the orbits? I.e. Do you want to ignore earths orbital motion? If not do you want to assume that the clock is in orbit around the sun? Or do you want to assume that the clock is using thrust to maintain some specified location?
     
  4. Apr 21, 2017 #3
    The clock(s) should be (using thrust) on the Sun-Earth line (rotating around the Sun with Earth's angular speed).

    Comparing the difference in ticking/frequency between two clocks on that line (close enough to each other), we should detect the point of inversion.
     
  5. Apr 21, 2017 #4

    Dale

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    I don't know the metric that would describe this system. Maybe someone else here does.
     
  6. Apr 21, 2017 #5

    PeterDonis

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    There isn't a known exact solution for this case. However, we can make a pretty good approximation if we assume weak fields and slow motion, ignore the rotation of the Sun and Earth, and just superpose the Schwarzschild solution for the Sun and the Schwarzschild solution for the Earth (in the weak field slow motion approximation). I'm pretty sure the nonlinear terms involved (i.e., the ones that would not be included in the superposition) are too small to matter.

    For an even simpler first approximation, we could ignore the revolution of the Earth around the Sun and treat the system as static. I'll write down what I get in that case below.

    I don't think the suggestions and calculations in those other discussions are correct.

    The approach I described just above in response to @Dale (where we ignore the Earth's revolution around the Sun and superpose the weak field slow motion approximations for the Earth and the Sun) would result in an equation that looks like this (I am using units in which ##G = c = 1##):

    $$
    \phi(x) = 1 - \frac{M_E}{x} - \frac{M_S}{R - x}
    $$

    where ##M_E## is the mass of the Earth, ##M_S## is the mass of the Sun, ##x## is the distance from the center of the Earth along the Earth-Sun line, and ##R## is the distance between the centers of the Sun and Earth. It should be straightforward to show that ##\phi(x)## has a maximum in the interval ##0 < x < R## and determine what it is.
     
  7. Apr 21, 2017 #6

    Dale

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    That would be my approach too, but I don't know the limitations or subtleties enough to be confident in it. I was not sure if there were other assumptions beyond weak field and slow motion required for a linearized approach.
     
  8. Apr 21, 2017 #7

    PeterDonis

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    The weak field, slow motion limit is sufficient to make linear superposition of solutions a good approximation for this case. Weak field implies ##M / r << 1## for all ##M## and ##r## that are relevant; slow motion implies ##v << 1## for all ##v## that are relevant. Any nonlinear terms in the case under consideration will be products of two or more factors of one of those forms (##M / r## or ##v##), so they are all negligible in this limit.
     
  9. Apr 23, 2017 #8

    David Lewis

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    It only appears to be fast when you observe it from sea level. If you were on top of the mountain, you wouldn't notice anything amiss. Both clocks tick at the same rate (in their own frame) but measure a different amount of time. I believe that's an equivalent way of saying the same thing.
     
  10. Apr 24, 2017 #9
    Your solution/suggestion leads to what I obtained here, also ignoring the rotation (not acceptable, the rotation is present and important).
     
  11. Apr 24, 2017 #10
    If you have 2 accurate synchronized clocks at sea level, take one on a mountain and return after few days, you'll see that the sea level clock is behind the mountain clock, so the difference in ticking is real, although nothing is/was amiss.

    See here:
    I think that with this technique it is possible to experimentally detect the point of inversion/transition. It should be done, because it may be important.
     
  12. Apr 24, 2017 #11

    phinds

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    Which is exactly what David Lewis was pointing out. When you are on the mountain with the clock on the mountain, it is ticking at exactly one second per second. When you are at sea level with the sea level clock it is ticking at exactly one second per second. ALL clocks tick at one second per second in their local frame. Since the two clocks are on different space-time paths, they tick for a different AMOUNT of ticks but that's not because they tick faster or slower. They do not tick faster or slower, they just travel through different space-time paths.
     
  13. Apr 24, 2017 #12
    According to Chou, C. W.; Hume, D. B.; Rosenband, T.; Wineland, D. J. - Science, Volume 329, Issue 5999, pp. 1630- (2010):
    Are they wrong?
     
  14. Apr 24, 2017 #13

    phinds

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    No, you are misunderstanding what is being said vs what I said. What they are saying is that FROM DIFFERENT FRAMES OF REFERENCE you see the clock ticking at a different rate. That is true. What I said is that in the FOR of the clock you do not see it ticking at anything but 1 second per second. You are confusing time dilation with differential aging due to different paths through space time. Time dilation is frame dependent whereas differential aging is not.
     
  15. Apr 24, 2017 #14

    PeterDonis

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    Yes. The force is the gradient of the potential, so at an extremum of the potential (in this case a maximum, as is easy to show by taking the second derivative), the force is zero.

    We are not really "ignoring" the rotation; we are just considering a hypothetical case in which, instead of having a clock that orbits the Sun with the same period as the Earth, we consider a clock that just moves along a radial line between the Sun and some point on the Earth's orbit, starting at the Earth's orbit and moving inwards, and asking at what point along that radial line its clock rate is maximized.

    We could also consider a different, somewhat less hypothetical case, in which we take a clock that is moving around the Sun with the same period as the Earth, and then start it moving inward from the Earth towards the Sun while maintaining the same orbital period around the Sun (another way to think of this would be to imagine a 150 million km long tether between the Earth and the Sun, and imagine the clock moving inward along the tether). We could then ask at what point this clock's rate would be maximized. (See below for more on that.)

    The point of all this is that there is not a single unique answer to the question you asked at the end of your scienceforums.net post: "where is the real point in which a clock going towards the Sun would reverse its ticking rate?" There is no "real point" in any absolute sense; there are different points for clocks with different motions. A more general formula that allows the clock to have an arbitrary velocity with respect to the Sun and the Earth, but still requires weak fields and slow motions (i.e., all velocities much less than ##c##), is (in units where ##G = c = 1##) [Edit--fixed ##v^2## terms]:

    $$
    \phi(r_E, r_S, v_E, v_S) = 1 - \frac{M_E}{r_E} - \frac{M_S}{r_S} - \frac{1}{2} v_E^2 - \frac{1}{2} v_S^2
    $$

    Note that this is a function of four variables, ##r_E## (the distance from the object to the Earth), ##r_S## (the distance from the object to the Sun), ##v_E## (the velocity of the object relative to the Earth), and ##v_S## (the velocity of the object relative to the Sun). In the very simple case I addressed before, I assumed ##v_E = v_S = 0## and used the obvious relationship between ##r_E## and ##r_S## (which I called ##x## and ##R - x##, respectively) for that case.

    For the case where the clock is orbiting the Sun with the same period as the Earth, the relationships are more complicated, but you can still express everything in terms of, say, ##r_E##, since all of the other variables will have a known dependence on that one for this case. That allows you to express ##\phi## as a function of one variable and use the standard technique to find its maximum.
     
    Last edited: Apr 24, 2017
  16. Apr 24, 2017 #15
    The important thing is that we can see/measure disparate clock rates for different heights. Comparing 2 optical clocks connected by optical fiber and "climbing" towards the Sun, one above the other, we should be able to find the point of inversion/transition.
     
  17. Apr 24, 2017 #16

    PeterDonis

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    In principle, yes, this would work. (In practice it would require a very large investment in space propulsion technology in order to keep the "climbing" clock on the proper trajectory, since that trajectory will not be a free fall trajectory except at one point--the inversion point--and will require continuous thrust with very precise control, which we don't currently have the technology to do.) The predicted answer can be derived using the procedure I described in my previous post.
     
  18. Apr 24, 2017 #17

    phinds

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    Agreed, and the optical fiber would not be needed, you could just send a transmitter along with the clock. Appropriate calculations would compensate for the transmission delay just as they would in an optical fiber.
     
  19. Apr 24, 2017 #18
    Thank you, but it seems to lead to the solution I mentioned in OP.

    ##v_E=0## and ##v_S=\omega (R-x)## ...
     
  20. Apr 24, 2017 #19

    PeterDonis

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    I'm not sure ##v_E = 0## is actually correct, because the velocities are supposed to be relative to a non-rotating frame, and we now have the Earth rotating around the Sun, so a clock that stays between the Earth and the Sun is moving in a non-rotating frame centered on the Earth. I think that results in ##v_E = \omega x##.
     
  21. Apr 24, 2017 #20
    You are right.

    The solution is not the same. I'll do the math tomorrow.

    Until then, you are sure that your equation/formula is correct? How you obtained it? What does it mean?
     
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