# Would Earth clocks tick faster than a clock on a GPS satellite if we disregard gravitational time dilation?

If we disregard the time dilation caused by a GPS satellite being in lower gravity, would earth's clocks tick faster or slower from the satellite's perspective?

Edit. From earth, we understand that a clock on a GPS satellite would tick 38 microseconds per day faster than our clocks. From the satellites perspective, do our clocks tick 38 microseconds per day slower, or 52 microseconds per day slower?

That's 45 microseconds per day due to earth's higher gravitational effecient and 7 microseconds per day due to earth's relative motion.

• The clock on earth runs slower because it is in a higher gravitational field. If the change in the rate of time passage due to gravitational difference is disregarded, then there would be no difference in the rate of passage of time. But that is not a meaningful comparison, since that is to disregard the reason why they are recording a different rate of time passage. Oct 31, 2018 at 0:56
• @ThomasLeeAbshierND I think the OP is interested in time dilation effects from the fact that the Earth clock is moving relative to the frame of the GPS satellite. i.e. Trying to look at effects of SR and not GR Oct 31, 2018 at 1:09
• @AaronStevens, I think you are right. Without that qualification, the question cannot be answered. Good insight! Oct 31, 2018 at 1:28
• @ThomasLeeAbshierND I have been in chat with the OP discussing various aspects of SR (although I wish I could claim excellent insight). I can say with pretty good confidence this is what they mean. But we can wait for the clarification if you would like. Oct 31, 2018 at 1:29
• @AaronStevens, since you have put so much effort into it, go ahead and answer it. Oct 31, 2018 at 1:33

Would earth clocks tick faster than a clock on a GPS satellite if we disregard gravitational time dilation?

Yes.

The fractional difference in clock rates is given by

$$\frac{1}{c^2}\left(\Delta\Phi-\frac{v^2}{2}\right)=5.2\times10^{-10}-0.9\times 10^{-10},$$

where $$\Phi$$ is the gravitational potential. See eq. (53) of Ashby, "Relativity in the Global Positioning System," Living Reviews in Relativity (open access). The gravitational term is bigger than the kinematic term, and they have opposite signs. If the gravitational term was absent, the kinematic term would cause the over-all effect to have the opposite sign.

• Can you explain where the kinematic term comes from? Also, is SR not applicable here due to an accelerating reference frame? Oct 31, 2018 at 2:04
• @AaronStevens: Can you explain where the kinematic term comes from? The kinematic term is the leading term of the Taylor series for $\gamma=(1-v^2/c^2)^{-1/2}$. Also, is SR not applicable here due to an accelerating reference frame? This is the leading-order result from general relativity, not an SR result. (It also is not true that SR cannot handle accelerated frames -- it can.) For a detailed derivation, see N. Ashby, “Relativity in the Global Positioning System,” Living Reviews in Relativity (open access).
– user4552
Oct 31, 2018 at 3:01
• I know SR can handle accelerating frames. I just didn't know if you can apply SR within an accelerated frame. Also, your conclusion is opposite of what the other answer here says. Oct 31, 2018 at 3:08
• @AaronStevens: Also, your conclusion is opposite of what the other answer here says. No, my answer and WillO's agree on the direction of the effect. I know SR can handle accelerating frames. I just didn't know if you can apply SR within an accelerated frame. I'm not clear on what distinction you're trying to make here. In any case, there is no frame of reference being discussed in my answer. This is simply a leading-order approximation to the integral of the line element, $\int ds$.
– user4552
Oct 31, 2018 at 3:10
• You say that according to the GPS, the Earth's clock ticks faster. WillO says it will tick slower. And the frames I am talking about is what reference frame we are choosing to work in. See my comment on WillO's answer. Oct 31, 2018 at 3:19

If you ignore the effects of gravity, then, if the satellite is moving at speed $$v$$, an earthbound observer will say that its clocks are running slow by a factor of $$\sqrt{1-v^2}$$.

Of course an observer on the satellite will say that the earth clocks are running slow by the same factor, but the observer on the satellite keeps moving from one (instantaneous) frame to another, and so keeps revising his opinion of when the earth clock was set to noon.

• is SR still applicable in the accelerating frame of the satellite? I understand that for an observer on Earth (assuming the Earth observer is inertial. If not, just replace the earth observer with one who is at the center of the Earth) we can look at instantaneous inertial frames of reference frames of the satellite. But this is because according to the Earth observer they are in an inertial reference frame, so SR applies between them and an instantaneous satellite frame. But the frame of the satellite is non-inertial to begin with, so does SR apply in that frame? Oct 31, 2018 at 2:25
• The first paragraph makes a correct prediction of the direction of the effect when gravitational time dilation is omitted (i.e., the sign is the opposite of what is actually observed when gravitational time dilation is present). However, the discussion of frames and observers is unnecessary and somewhat misleading conceptually. This is like the twin paradox -- when the twins are reunited, there is no disagreement about which twin is older. No frames of reference are needed in order to explain this, and GR doesn't actually have global frames of reference. The proper time on each clock is [...]
– user4552
Oct 31, 2018 at 3:30
• [...] simply $\int ds$, and this is a mathematical fact that has nothing to do with observers or frames of reference.
– user4552
Oct 31, 2018 at 3:31
• @BenCrowell : Of course I completely agree with you that the time that passes on the satellite's clock between two events on the satellite's worldline is given by the line integral of $ds$, and that if one says it this way, then reference frames are unnecessary. But it's also true that because we're ignoring gravity, we're doing SR, which means that we have global frames of reference, and there's nothing misleading about expressing things in those terms instead. (CTD). Oct 31, 2018 at 3:42
• (CTD) In other words, you're right that the frames of reference don't need to be mentioned if you write down the line integral, but it's equally right that the line integrals don't need to be mentioned if you write down the Lorentz transformations between the frames of reference. Oct 31, 2018 at 3:43