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Voyager probes and time dilation

  1. Sep 21, 2010 #1
    The Voyager probes are travelling at non-relativistic speeds, at the edge of the solar system where the gravitational field is about 0 so we could consider them a rest frame. If so, from my calculations using the formula for time dilation:

    [tex]\tau(t) = \frac{c}{g} \operatorname {arsinh} \left( \frac{gt}{c} \right) [/tex]

    10 years for the probes would mean about 2.8 years for someone on Earth.

    Is this correct? I'm curious if the gravitational acceleration at the Earth's surface is really so significant.
  2. jcsd
  3. Sep 21, 2010 #2


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    I think I kow what you're doing. You're picking the probes as your inertial frame, and calculate elapsed time on earth following http://en.wikipedia.org/wiki/Time_dilation#Time_dilation_at_constant_acceleration".

    The effect described in the wikipedia article stems from the relative velocity an object would have if - starting from rest - it accelerates for a given time. The acceleration itself contributes exactly nothing to the effect - a circumstance knonw as the "clock hypothesis".

    Even if an observer on earth is accelerating, she's not gaining relative velocity to the probes by this acceleration. So you can't apply this formula. It is not intended for curved spacetime, where acceleration without motion is possible.

    What you should do instead is to combine the effects of http://en.wikipedia.org/wiki/Time_dilation#Time_dilation_due_to_relative_velocity".

    It's a tiny effect, btw.
    Last edited by a moderator: Apr 25, 2017
  4. Sep 21, 2010 #3
    Thank you, this cleared things up for me :)
    Last edited by a moderator: Apr 25, 2017
  5. Sep 21, 2010 #4
    Indeed the dilation at the surface of the Earth is very small compared to a region without any significant gravitation.

    This triggered me to calculate the Pound Rebka experiment using the Schwarzschild solution, and I seem to be stuck.

    We have:

    Mass of the Earth = 0.004435028 meters
    Schwarzschild radius R0 = 0.008870056 meters
    R1 = 6375416 meters
    R2 = 6375438.6 meters (R1 + 22.6)

    Ratio of the two clocks:

    [tex]\sqrt{ {1-R_0/R_1 \over 1-R_0/R_2 } }[/tex]

    This gives me: 1.00000000626084025339746259

    So now how do I go from here to get 2.5×10−15?
    Last edited: Sep 21, 2010
  6. Sep 21, 2010 #5


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    With algebra?

    Maybe you made a sign error.
  7. Sep 21, 2010 #6
    Independently, I get R0 = 0.008871345 which is close enough to your figure because the figures for the mass and radius of the Earth vary a bit depending on the source and whether you are at the equator or not. Using your figures for R0 and R1 I get:

    [tex]\sqrt{ {1-R_0/R_1 \over 1-R_0/R_2 } } = 0.999999999999998 [/tex]


    [tex]1-\sqrt{ {1-R_0/R_1 \over 1-R_0/R_2 } } = 2.44E-15 [/tex]
  8. Sep 21, 2010 #7
    Your figure for R0 has already been divided by c^2, since R0 = 2GM/c^2 = 0.008870056 meters
  9. Sep 21, 2010 #8
    Got it! I missed one single digit somewhere, it's too late here (4:14 AM).

    2.46596 E-15

    With your R0 I get:

    2.46632 E-15

    Yes that's right, I took it already out.
  10. Sep 21, 2010 #9
    If we do not take into account other gravitational fields or the velocity of the Voyager which is obviously totally unrealistic the time differential with respect to the Voyager far away from the Earth should be somewhere like:


    So for 10 years that accumulates to about 0.219 seconds.

    Do we know the current velocity of the Voyager?
    In we have it we can include that in the calculation.
    Last edited: Sep 21, 2010
  11. Sep 22, 2010 #10


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    http://voyager.jpl.nasa.gov/mission/weekly-reports/" [Broken] But Earth is moving, too.

    To make a reasonably accurate guess, you have to factor in the sun's gravitational potential, as well as Earth's motion relative to the sun.
    Last edited by a moderator: May 4, 2017
  12. Sep 22, 2010 #11
    I totally agree.

    In fact the gravitational time dilation from the Sun is much higher than the time dilation of the earth even for a clock on the Earth's surface!

    While gravitational time delay factor due to the Earth is: 0.999999999304 the Sun's gravitational time delay factor, roughly 143 million km removed is: 0.999999989933 two orders of magnitude larger!

    At least according to my calculations.
  13. Sep 22, 2010 #12


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    Yes - MTW has a table of "metric correction factors" - which is just Gm/rc^2 - for various objects. (pg 459)

    Basically larger-scale structures contribute more to time dilation.

    a statue (Venus de Milo) (m=2e5 g, r = 30cm) -> 5e-25
    at the surface of the Earth (m=6e27g, r= 6.4e8cm) -> 6e-10
    at Earth's distance from sun (m=we33g, r=1.5e13mc) ->1e-8
    at Sun's distance from center of galaxy (m=1e44g, r=2.5e23cm) -> 6e-7
    at distance of galaxy from Virgo cluster (m=6e47g, r= 3e25cm) -> 1e-6
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