Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relative motion and time dilation

  1. Nov 16, 2009 #1
    I've been reading the threads in this forum about relative motion and time dilation and I would like to know if I am understanding things correctly.

    The atomic clocks aboard the GPS satellites are programmed with a "factory offset" (according to Wikipedia) of +7us/day for relative motion and -45us/day for gravitational time dilation for a total of -38us/day. So, if I were in the same earth orbit as the GPS satellite with two atomic clocks, and I want to send one of them to the earth's surface, how would I need to program it's clock so that it would be in sync with the one that remains with me on the satellite?

    My answer is +52us/day. Is this correct?
  2. jcsd
  3. Nov 16, 2009 #2
    My initial guess is +38us/day.
  4. Nov 16, 2009 #3
    That was my initial guess also. But then I changed my mind after reading some threads here. So I guess I'm still confused. :)
  5. Nov 16, 2009 #4
    Now I'm thinking that +38us/day is the correct answer. My reason: The clock hypothesis, which has been experimentally tested to be correct to a high degree of precision, states that the tick rate of a clock when measured in an inertial frame depends only upon its velocity relative to that frame, and is independent of its acceleration or higher derivatives. The key words here being "inertial frame" and "relative to that frame". So if I consider the earth and the satellite as being in the same frame, then in order for the frame to be inertial I must consider the frame to be referenced to the center of mass. Viewing it this way it becomes obvious that the velocity of the clock on the satellite is higher than the velocity of the clock on the earths surface. Therefore, time adjustments to a clock sent from the satellite to the earths surface must be adjusted -7us/day (for relative motion) because of the decrease in velocity.

    Is this correct, or am I still confused?
  6. Nov 17, 2009 #5
    Does anyone here know if +38us/day is the correct answer? Regardless of whether my reason is correct or not.
  7. Nov 17, 2009 #6


    User Avatar
    Science Advisor

    No frame covering a non-infinitesimal region of curved spacetime can qualify as "inertial", inertial frames only exist in flat spacetime, or "locally inertial frames" in infinitesimal regions of curved spacetime (see the http://www.aei.mpg.de/einsteinOnline/en/spotlights/equivalence_principle/index.html [Broken]).

    Aside from that, can you explain you're getting these numbers? Pervect's post #35 here suggests you can't actually break down the time dilation experienced by satellite clocks in an Earth-centered coordinate system into a simple sum of velocity-based time dilation and gravitational time dilation, as you seem to be suggesting.
    Last edited by a moderator: May 4, 2017
  8. Nov 17, 2009 #7
    Thanks for the reply JesseM. So my reason is incorrect but not my answer?
    I guess I'm not understanding this at all because it seems contradictory to me. Isn't all "real" space time curved?

    According to Wikipedia a COM frame can be an inertial frame.
    I got the numbers from Wikipedia. And it appears that the programming of the GPS clocks for time dilation are a simple sum of relative motion based time dilation and gravitational time dilation. However, I admit that I may be misinterpreting this.
  9. Nov 17, 2009 #8
    Starting with this equation given by pervect here

    [tex]c^2 \left( \frac{d\tau}{dt} \right)^2 = c^2 \, \left(1-\frac{r_s}{r}\right)^2\ - r^2 \left(\frac{d\phi}{dt}\right)^2[/tex]

    and given that instantaneous tangential coordinate velocity u is related to the angular velocity by:

    [tex]u = r \left(\frac{d\phi}{dt}\right)[/tex]

    then pervect's equation can be rewritten as:

    [tex]c^2 \left( \frac{d\tau}{dt} \right)^2 = c^2 \, \left(1-\frac{r_s}{r}\right)\ - u^2[/tex]


    [tex] \left( \frac{d\tau}{dt} \right)^2 = \, \left(1-\frac{r_s}{r}\right)\ - \frac{u^2}{c^2}[/tex]

    Now u in the above equation is the instantaneous tangential velocity as measured by the Schwarzschild observer at infinity.
    The local velocity v is related to the coordinate velocity by:

    [tex] u = v \left(1-\frac{r_s}{r}\right)[/tex]


    [tex] \left( \frac{d\tau}{dt} \right)^2 = \, \left(1-\frac{r_s}{r}\right)\ - \frac{v^2}{c^2} \left(1-\frac{r_s}{r}\right) [/tex]


    [tex] \left( \frac{d\tau}{dt} \right)^2 = \, \left(1-\frac{r_s}{r}\right)\ \left(1 - \frac{v^2}{c^2} \right) [/tex]

    Taking the square root of both sides:

    [tex] \frac{d\tau}{dt} = \, \sqrt{1-\frac{r_s}{r}}\ \sqrt{1 - \frac{v^2}{c^2} } [/tex]

    so there appears (if I have done the calculations right) to be a sense in which the total time dilation is the product of the gravitational time dilation and the velocity based time dilation, when the local instantaneous velocity is used.
    Last edited: Nov 18, 2009
  10. Nov 17, 2009 #9


    User Avatar
    Science Advisor

    Didn't pervect show that it was approximately but not exactly the product of the two at the end of his post? He derived this equation:
    Since the object is assumed to be orbiting at constant angular velocity, its tangential velocity should be equal to [tex] r \frac{d\theta}{dt}[/tex], so if [tex]g_{tt}[/tex] is close to one the right side looks almost like the SR equation [tex]\frac{d\tau}{dt}= \sqrt{1 - \frac{v^2}{c^2}}[/tex], but it's not exact because of that extra factor of [tex]\frac{1}{g_{tt}}[/tex].
  11. Nov 17, 2009 #10
    We both started with the same equation. The reason pervect did not get exactly the product of the two is that he was using coordinate (angular) velocity and I am using local velocity.
  12. Nov 17, 2009 #11


    User Avatar
    Science Advisor

    Ah, I see now the steps where you noted that the coordinate velocity is [tex]u = r \left(\frac{d\phi}{dt}\right)[/tex], and then later said "The local velocity v is related to the coordinate velocity by: [tex]u = v \left(1-\frac{r_s}{r}\right)[/tex]". Does "local velocity" in this case mean velocity in the locally inertial frame of a freefalling observer passing next to the orbiting body? If so what is being assumed about the freefalling observer's own path through spacetime--presumably freefalling observers passing the object in different directions would get different values for the object's velocity in their own local inertial frames?
  13. Nov 17, 2009 #12
    By local velocity (in Schwarzschild coordinates) I had in mind the velocity as measured by an (accelerating) observer located at radius r with zero angular coordinate velocity and zero radial coordinate velocity. For example an observer located on a tower of height r on a non rotating planet.
  14. Nov 17, 2009 #13


    User Avatar
    Science Advisor

    But how is this observer defining "velocity", if not in terms of change in Schwarzschild coordinate position over change in Schwarzschild coordinate time (which would just give us the velocity u)? Are you defining v in terms of change in Schwarzschild coordinate position over change in clock time for a set of clocks hovering at that radius (and synchronized in Schwarzschild coordinates)?
  15. Nov 17, 2009 #14
    The local observer could use a horizontal ruler. The ruler is calibrated by timing light signals. Let's say it takes one second for a light signal to travel the length of this ruler, then if this horizontal ruler is transported to any radial height it will always take a light signal one second to traverse the length of the ruler. The clocks at a given radius always run at the same rate so there is no problem with using more than one clock and of course a mirror could always be placed at one end of the ruler and an average of the two way light travel time taken using a single clock. So the local velocity is measured using a local ruler and local clocks. How to tell if the rulers and clocks have zero angular velocity? Simply place two objects in orbit going in opposite directions. If they both have equal velocitities by your local measurement and maintain an orbital radius of r, then you have zero angular velocity.
  16. Nov 17, 2009 #15
    Here are a couple of references for my previous posts:

    Last edited by a moderator: May 4, 2017
  17. Nov 17, 2009 #16


    User Avatar
    Science Advisor

    Yes, SR is an idealization. But the practical upshot of the equivalence principle is that if you pick a region of spacetime that's small enough that tidal forces (which measure spacetime curvature) are negligible according to your equipment, then you can construct a coordinate system in this region in which the laws of physics are close enough to those of ideal inertial frames in SR that your equipment won't notice the difference (though with more sensitive equipment you'd be able to).
    COM = center of mass? They might have been talking about classical mechanics where there's no spacetime curvature, or they might have been talking in the sort of approximate sense I discussed above, where the curvature due to the object's mass is too small to make any measurable difference.
    Which wikipedia page? From pervect and kev's calculations it seems like it'd be more accurate to treat total time dilation as a product of gravitational and SR time dilation.

    edit: I see you posted a link to a site in your last post, unfortunately the author doesn't give any details as to how the numbers were arrived at...
  18. Nov 17, 2009 #17
    Yes COM is center of mass. You must not have followed the link I provided for "Earth-Centered Inertial Frame" (ECI).
    The ECI coordinate system is the one used for GPS.
    What other details are needed to show that the author's method is simple addition of SR time dilation and gravitational time dilation?
  19. Nov 17, 2009 #18


    User Avatar
    Science Advisor

    Which link? I didn't see a link to a wikipedia article in your posts.
    None, but I was saying it was unfortunate that the author doesn't provide details so we could see how this squares with the derivation of total time dilation given by pervect and kev.
  20. Nov 17, 2009 #19


    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, it should be a product. However, the product should be able to be approximated by a sum.

    (1+alpha)(1+beta) = 1 + alpha + beta + alpha*beta. But the last term is small, of second order when alpha << 1 and beta << 1.

    BTW, as kev points out, with the right definition of velocity, the multiplicative relationship is exact.
  21. Nov 17, 2009 #20
    The link is in the underlined "Earth-Centered Inertial or ECI frame". Just click on it. Sorry, I should have made it a separate link for clarity.
  22. Nov 18, 2009 #21
    Last edited by a moderator: Apr 24, 2017
  23. Nov 19, 2009 #22

    Gravitational constant (G) = 6.67488E-11 m3/kg/s2
    Mass of Earth (M) = 5.98E24 kgs
    Orbital radius of GPS satellite ([itex]R_S[/itex]) = 26578000 m
    Equatorial radius of Earth ([itex]R_E[/itex]) = 6378000 m
    Sidereal day on the surface of the Earth (D) = 86164 s
    Speed of light (c) = 299792458 m/s

    The orbital period of the satellite is given by:

    [tex]P = 2\pi \sqrt{\frac{R_S^3}{GM}} = 43092.7 s[/tex]

    The tangential velocity of the satellite is:

    [tex]V_S = 2\pi R_S /P = 3875.2 m/s [/tex]

    The tangential velocity on the surface of the Earth is:

    [tex]V_E = 2\pi R_E /D = 465.1 m/s [/tex]

    The ratio of the velocity time dilation factor for the Earth surface and the satellite is:

    [tex] \frac{\sqrt{1-V_E^2/c^2}}{\sqrt{1-V_S^2/c^2}} = 1.000000000082342569791133268 [/tex]

    During a sidereal day on the surface of the Earth, the orbiting clock falls behind by:

    D - D*1.000000000082342569791133268 = -0.000007094965183483206886590666862 s

    due to velocity time dilation, which is about 7 microseconds.

    The ratio of the gravity time dilation factor for the Earth surface and the satellite is:

    [tex] \frac{\sqrt{1-2GM/(R_E\ c^2)}}{\sqrt{1-2GM/(R_S\ c^2)}} = 0.9999999994707970061739772883 [/tex]

    During a sidereal day on the surface of the Earth, the orbiting clock gets ahead by:

    D - D*0.9999999994707970061739772883 = 0.00004559824676002542093384086709 s

    due to gravitational time dilation, which is about 45 microseconds.

    The total additive time offset of the satellite per sidereal day is then:

    0.00003850328157654221404725020023 s or about 38 microseconds.

    Multiplying the velocity time dilation ratio by the gravitational time dilation ratio gives a total multiplicative time dilation ratio of:


    During a sidereal day on the surface of the Earth, the orbiting clock gets ahead by:

    D - D*0.9999999995531395759215346216 = 0.00003850328158029689086344090986 s

    due to the total multiplicative time dilation, which is also about 38 microseconds and only differs from the additive time dilation after 14 decimal places! However in the strongly curved spacetime near a black hole the additive time dilation approximation will differ greatly from the multiplicative time dilation formula.

    The multiplicative ratio given above is the preset factory offset applied to the gps satellite clock frequency before it is launched. There are many other effects that may have to be taken into account (e.g. for military gps receivers) such as Sagnac effect, ionosphere delays, Shapiro delay, orbit eccentricity, etc but these effects are generally much smaller than the main two relativistic effects.
  24. Nov 19, 2009 #23
    Thank you for the informative post Kev. And thanks for taking the time to present it in a way that is easy to understand (for a layman). It is greatly appreciated. I understand completely. :)
    Last edited: Nov 19, 2009
  25. Nov 19, 2009 #24
    Thanks :smile: It is nice to have some positive feedback now and then. Cheers :wink:
  26. Jun 4, 2010 #25
    You start with the wrong metric, therefore you get the wrong result.

    The correct metric is:


    There is no [tex](1-\frac{r_s}{r})^2[/tex]

    This was a rare goof by pevect , perpetrated by you.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook