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How would Earth's time slow down?

  1. Dec 2, 2011 #1
    Would it be the rotational speed increasing or the speed at which the Earth revolves around the Sun?

    I take it in both cases Earth's speed being relative to the Sun.

    Now how would time on Earth slow down? Would this happen if the earth rotated faster, revolved faster, or both? Or are they the same thing? Is the Sun what we would usually use as the stationary body in this case?

    Would time just appear to slow down for the FOR of the Sun but we wouldn't feel it?
     
  2. jcsd
  3. Dec 2, 2011 #2

    PAllen

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    No one locally feels anything strange about flow of time. On the other hand, you can locally know if you are inertial or accelerated.

    The answers to your questions depend on who is observing. I think you would be most satisfied with a description from a hypothetical observer far above the plane of earth's orbit, above the sun, in free fall.

    Compared to their clock, they would see clocks run slow, in the following order, slowest to least slow compared to theirs:

    0) Clock on earth. Earth is following inertial path, but this clock is accelerated, it feels force from ground.
    1) Clock in same orbit as earth, but on other side of sun. This path is inertial.
    2) Clock on rocket thrusting so as to remain stationary relative to the sun, in the earth's orbit. This is an accelerated path, in GR. You would feel the force required to hold you stationary.
    3) Clock fired from rocket in (2) to shoot away from the sun and fall back to the rocket. This path is inertial except for firing moment.

    You can try to break these down into speed and gravity effects, but it is not intuitive. Relative to non-inertial (2), you have an inertial path whose clock runs slower (1), and an inertial path whose clock runs faster (3). Non-inertial (0) is basically (1) modified by earth's gravitational time dilation.

    Generally, speeding up the earth's rotation would have little effect because it would make the earth more oblate. The greater distance of the equator from the center would balance greater speed at the equator compared to the poles.

    Speeding up the earth's orbit is more complicated - you have two cases: if you kept the path the same, you would need to continuously accelerate the earth. To get a faster free fall orbit, you would have to bring the earth closer to the sun. Both of these would slow down an earth clock more.
     
    Last edited: Dec 3, 2011
  4. Dec 3, 2011 #3
    I figured you wouldn't actually feel the flow of time differently on Earth.

    Maybe I should have asked the question like this:

    What would it mean for time to slow down on Earth?
     
  5. Dec 3, 2011 #4

    ghwellsjr

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    It means different clocks run at different rates. This is already happening. The atomic clocks at Greenwich England near sea level run at a different rate than identical clocks run at Boulder Colorado at a higher elevation. And all these clocks are running slower than clocks far removed from gravity fields. The GPS clocks run at different rates than any earth bound clocks.

    But all these effects are dominated by gravity, and to a lesser extent by non-inertial speed. That means the predominate effects are not reciprocal and each clock can see the other one running at a different rate.
     
  6. Dec 3, 2011 #5

    PAllen

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    Right.

    One reason I picked a simple observer to describe everything from is that my list of cases did include a potentially confusing reciprocal time dilation as well as most of the rest, which are not reciprocal.

    The reciprocal case in my list was the stationary in earth's orbit clock, and the clock orbiting freely in the earth's orbit (but far from earth). When these pass each other (hopefully, not quite colliding), each sees the others clock running slower. Meanwhile, my chosen distant observer sees the orbiting clock slower. This leads to the confusing, but I believe true scenario:

    Let D be distant observer I described in #2. Let O be orbiting clock. Let S be clock stationary in earth's orbit. Then, when O is passing S:

    - S sees O slow and D fast
    - O sees S slow and D fast
    - D sees S slow and O slower
     
  7. Dec 3, 2011 #6
    If they start at one point with their clocks synchronized then at each orbit when they compare their clocks both will agree on what each clock reads and then obviously one clock must be going slower than the other clock. In fact the stationary clock will go slower than the free falling clock.
     
  8. Dec 3, 2011 #7

    PAllen

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    This is not correct. The orbiting clock will go slower over a complete orbit. This is easy to see from the Schwarzschild metric. Look at the expression for d tau^2 for case of dr = d theta = d phi = 0; then compare to any d theta > 0, and you see the d tau ^2 is smaller. Thus the orbiting clock will go slower.

    This is an exception to the rule of geodesics maximizing proper time. The problem is that between the events of O passing S at some t, and O passing S one orbit later, there are more than one geodesic. One of them maximizes proper time. The others (like this one) do not. The type of geodesic that maximizes proper time is the type that can vary locally - the family of geodesics that can connect events close together on S's world line. These are the ones, so colorfully named by Synge, as the ballistic suicide geodesics - the ones that go radially out and back.

    ---

    On the other issue, the result of a clock comparison at successive passing events is obviously invariant. But the visual appearance of clocks will vary over the orbit for this pair of observers. At the moment of passing, this visual appearance will, indeed, have a reciprocal relationship.

    [Edit: corrected mis-type of O where I meant S]
     
    Last edited: Dec 3, 2011
  9. Dec 3, 2011 #8
    You are right, I meant to say the free falling clock goes slower than the stationary clock.

    If Alice is the one going in orbit and Bob is staying stationary then the ratio of their elapsed times for one orbit is:
    [tex]\Large {\frac {\tau_{{{\it Bob}}}}{\tau_{{{\it Alice}}}}}=\sqrt {{\frac {R-r_
    {{s}}}{R- 1.5\,r_{{s}}}}}[/tex]
    (where R is the r-coordinate of a circular orbit and rs is the Schwarzschild radius)

    Thus Bob is always older than Alice.

    This is different from the case where a clock is tossed (and returns) in a gravitational field in this case the stationary clock goes slower.
     
    Last edited: Dec 3, 2011
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