# Time dilation as seen by different observers

1. Dec 9, 2014

### dougy

Let's say the gravity around a planet of mass M is accurately modeled by the Schwarzschild metric. Then an observer (A) at infinity sees a stationary clock (B) at distance R from the center of the planet being gravitationally time dilated by the factor $\sqrt{1-\frac{2GM}{Rc^2}}$, i.e. if B sends a light pulse every second in its own time towards A, A will receive them every $\frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}$ second in its own time.

More formally, if tA indicates A's time and tB indicates B time, and their clocks are initially synchronized, then
$$t_A = t_B\frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}$$
If I'm not mistaken we arrive at this result by solving the Schwarzschild metric when dr, dθ and dφ = 0:
$$c^2dt_B^2 = \left(1-\frac{2GM}{Rc^2}\right)c^2dt_A^2 - \frac{dr^2}{\left(1-\frac{2GM}{Rc^2}\right)} - r^2 \left(d\theta^2+\sin^2(\theta)d\phi^2\right)$$
And if I'm not mistaken there is a reciprocity in what B observes, i.e. if A sends a light pulse every second in its own time towards B, B will receive them every $\sqrt{1-\frac{2GM}{Rc^2}}$ second in its own time. And to see that all we have to do is go from $t_A = t_B\frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}$ to $t_B = t_A\sqrt{1-\frac{2GM}{Rc^2}}$, am I right?

Now assuming the above is true, let's say we have another stationary clock (C) at distance RC from the center of the planet. In a similar way we can say that if A sends a light pulse every second towards C, then C will receive them every $\sqrt{1-\frac{2GM}{R_Cc^2}}$ in its own time, or $t_C = t_A\sqrt{1-\frac{2GM}{R_Cc^2}}$.

And now the crux of my question: from the point of view of C, how much would B appear to be time dilated? And from the point of view of B, how much would C appear to be time dilated? I guess a naive answer would be to say that since $t_B = t_A\sqrt{1-\frac{2GM}{Rc^2}}$ and $t_C = t_A\sqrt{1-\frac{2GM}{R_Cc^2}}$, then: $$\frac{t_B}{t_C} = \frac{\sqrt{1-\frac{2GM}{Rc^2}}}{\sqrt{1-\frac{2GM}{R_Cc^2}}}$$
But aren't these time dilations only valid in the reference frame of A? We have arrived at the relations between tA and tB and tA and tC by solving the Schwartzschild metric when the time coordinate is tA at infinity. And so I would guess that it is not correct to simply replace tA in the metric by tB or tC, because in the Schwarzschild metric the time coordinate on the right hand side of the equation is always implicitly the time experienced at infinity by an observer ticking faster than all other observers? Is this right or is this just my misunderstanding and we can use the metric from whatever point of view we want?

I have many more questions about how time dilation would be seen by other observers in motion, for instance I've heard that from the reference frame of an observer orbiting the planet, the planet is the one orbiting the observer and so that the Schwarzschild metric is not the right one to model the situation and calculate the time dilation experienced by other people as seen by that observer.

Also I think it is useful to mention that the above is not homework, I haven't been enrolled in an educational institution for a while now, this is pure curiosity. Ultimately the reason I am asking is that I wonder how a clock on the surface of the Earth would appear to be time dilated from the point of view of a GPS satellite in orbit. From the point of view of an observer at infinity a clock on the surface of the Earth is more time dilated than a GPS satellite by about 0.447ns every second, and the clocks on the satellites are slowed down accordingly, but since a GPS satellite is not in the same frame as an observer at infinity then a clock on Earth shouldn't appear to be dilated by the same amount, and I just do not understand why this isn't taken into account.

2. Dec 9, 2014

### Jonathan Scott

That looks right. In general, for static situations, the time dilation factors simply multiply and divide in the obvious way.

3. Dec 9, 2014

### Staff: Mentor

Yes, for both cases, A vs. B and B vs. C.

No. The "time dilation" you are referring to here is a direct observable; it can be measured by exchanging light signals between the observers, exactly as you describe. Direct observables can't depend on the frame of reference you use.

No, you arrived at them by relating the change in proper time (the time elapsed on A's or B's or C's clock) along the worldlines of A or B or C, for a given interval of coordinate time. You didn't define $t_A$ as coordinate time; you defined it as "A's time", meaning the time elapsed on A's clock between two particular events (in this case, the arrival of two successive light signals from B). It just so happens that, since A is at infinity, the change in his proper time, $t_A$, is the same as the change in coordinate time, $t$. But the two are still conceptually distinct. Strictly speaking, what your calculation says is that the same interval of coordinate time corresponds to different intervals of proper time for A, B, and C, according to the formulas you gave.

You are correct that the replacement you describe would be wrong. But it's easier to avoid the temptation to do it if you enforce a clear distinction between coordinate time and proper time. The usual way of doing that is to use $\tau$ for proper time; so the general equation for time dilation becomes

$$\tau = \sqrt{1 - \frac{2 G M}{c^2 R}} t$$

where $t$ is the interval of coordinate time and $\tau$ is the corresponding interval of proper time. Then you would just plug in the appropriate value of $R$ for each observer. For A, $R = \infty$, so the factor in the square root becomes 1, and you have $\tau_A = t$, as I noted above; but now it's clear that this is equating two conceptually different things, the interval of coordinate time $t$ and the time elapsed on A's clock, $\tau_A$.

For B and C, you would plug $R_B$ and $R_C$ into the formula above to get $\tau_B$ and $\tau_C$. You could then simply take the ratio of the two to get the equation you wrote down; note that $t$ cancels out in the ratio, so you don't even have to use the fact that $t = \tau_A$. In fact, strictly speaking, you always have to take ratios of $\tau$ quantities to get the direct observables you describe (how much time elapses by one observer's clock between two successive light signals from another observer); it's just that, since $\tau_A = t$, any observable that includes $\tau_A$ has a simpler formula because one of the square root factors is just 1.

4. Dec 9, 2014

### Staff: Mentor

This case is more complicated because both the satellite and the Earth's surface are moving with respect to an idealized observer at infinity, so if you want to use the Schwarzschild metric you have to account for these motions as well as for the difference in altitude. In the metric, this means that, in the simplest case (circular motion), $d\phi$ is nonzero as well as $dt$. So you end up with a time dilation formula that includes a term for velocity as well as a term for altitude. But the general rule I gave in my previous post, that the relative time dilation between two observers is the ratio of their $\tau$ values, still holds.

5. Dec 9, 2014

### dougy

But in the case where two observers are moving away from each other, they also can measure their respective time dilation by exchanging light signals, and yet these time dilations depend on the reference frame and so they shouldn't be direct observables?

Very clear thank you. So it was conceptually wrong for me to write $dt_A^2$ on the right-hand side in the Schwarzschild metric, the one to write is always the coordinate time $dt$, it just happened to be equal to A's proper time.

More generally is the coordinate time always the one ticking the fastest and corresponding to a region where the spacetime is flat, or is it possible to pick whatever coordinate time we want and change the expression of the line element of the metric accordingly? For instance let's say I want $t_B$ to be the coordinate time (and thus to appear on the right-hand side in the expression of the metric), would that be possible by some appropriate change? If so how would it look like to express the Schwarzschild metric when $t_B$ is the coordinate time?

Is this rule always true in any situation and in any metric, that the relative time dilation between two observers is the ratio of their proper times?

Where I'm getting confused in the example of the satellite and the clock on Earth's surface, is that on top of their difference in gravitational potentials the satellite and the clock are moving relative to each other. Due to their different gravitational potentials the satellite's clock runs faster by about 0.53 ns every second compared to the clock on the Earth ( $\frac{\sqrt{1-\frac{2GM}{R_Sc^2}}}{\sqrt{1-\frac{2GM}{R_Ec^2}}}$ where RS = 26540 km and RE = 6357 km), and if I understand well both observers can objectively agree on that value when they are not in relative motion. However since they are in relative motion, the satellite's clock would see Earth's clock running slower by a little more than 0.53ns every second, while Earth's clock would see the satellite's clock running faster by a little less than 0.53ns every second. If $\tau_S$ is the proper time of the satellite and $\tau_E$ the proper time of Earth's clock, then this would mean we can no longer write $\tau_S$ = $K\tau_E$ where K is the time dilation factor, because depending on who is measuring the other's light signals, the time dilation factor would not be the same, and this is what I can't wrap my mind around.

I'm gonna try to explain my question more formally. If we neglect the rotation of the Earth, then in the frame of an observer at infinity whose proper time $\tau_A$ equals the coordinate time $t$, Earth's clock is stationary while the satellite's clock is in motion. So the proper time of the clock on Earth is exactly $\tau_E = \sqrt{1-\frac{2GM}{R_Ec^2}}t$, while the motion of the satellite causes the proper time of the satellite's clock to be a little bit less than $\sqrt{1-\frac{2GM}{R_Sc^2}}t$, the correct value being I believe $\tau_S = \sqrt{1-\frac{3GM}{R_Sc^2}}t$ on a circular orbit. Then $\tau_E = (1-6.98*10^{-10})t$ and $\tau_S = (1-2.50*10^{-10})t$

So assuming the relative time dilation between two observers is the ratio of their proper times, if the clock on Earth sends a light pulse every second in its own time towards the satellite, then the satellite will receive them every $\frac{\tau_S}{\tau_E} = 1+4.5*10^{-10}$ second in its own time (the satellite's clock runs faster by 0.45ns), and conversely if the satellite sends a light pulse every second in its own time towards the clock on Earth, then that clock will receive them every $\frac{\tau_E}{\tau_S} = 1-4.5*10^{-10}$ second in its own time (Earth's clock runs slower by 0.45ns).

Now the heart of the problem. If the satellite was instead stationary, we would have $\frac{\tau_S}{\tau_E} = 1+5.3*10^{-10}$ (the satellite's clock runs faster by 0.53ns) and $\frac{\tau_E}{\tau_S} = 1-5.3*10^{-10}$ (Earth's clock runs slower by 0.53ns). When the satellite is in motion, Earth's clock should see the satellite's clock run a little less faster, which is indeed what we calculated (0.45ns faster instead of 0.53ns). However from the reference frame of the satellite, Earth's clock is in motion too, and so the satellite's clock should see Earth's clock run a little more slower as well, which is not what we calculated! Instead of seeing Earth's clock run slower by more than 0.53ns, the satellite sees Earth's clock run slower by less than 0.53ns (0.45ns). How to reconcile this contradiction?

There are two assumptions that lead to this apparent contradiction here. The first is that the relative time dilation between two observers is the ratio of their proper times, and the second that two observers in relative motion should see the clock of the other run slower due to that motion. I thought the second assumption was universally true in relativity but I may be mistaken, does the presence of gravity somehow break this symmetry and then we can say that the satellite is the one objectively in motion? Either that or the first assumption is wrong, which is what I thought when I started this thread, but you stated the opposite so now I'm still confused. Why doesn't the satellite see the clock on the Earth run a bit slower when the satellite is in motion compared to when it isn't?

6. Dec 9, 2014

### Staff: Mentor

I would use the term "redshift" for the direct observable and use "time dilation" for the ratio of coordinate time to proper time.

7. Dec 9, 2014

### Staff: Mentor

You can pick whatever coordinate time you want and change the expression accordingly. For an easy example, you could do the coordinate transformation t'=kt. You could use that so that the coordinate time matched proper time for a static observer on earth's surface instead of at infinity.

8. Dec 9, 2014

### Staff: Mentor

No, they can't; what they actually measure is redshift (or blueshift), as DaleSpam pointed out. That's the direct observable, so wherever I was referring to the direct observable I should have said "redshift/blueshift" instead of "time dilation". Sorry for the sloppy terminology on my part.

As for how redshift/blueshift and time dilation are related, see below.

Since you meant to use $dt_A$ to denote A's proper time, yes, that was wrong.

The latter.

You would just rescale the time coordinate as DaleSpam described. That would add an extra factor in the $dt^2$ term in the metric, but wouldn't change anything else.

It depends on what you are trying to denote by the term "relative time dilation". The key question is this: is there some invariant way of picking out "corresponding" events on the two observers' worldlines? If there is, then the ratio of their respective proper times between corresponding events can be interpreted as their "relative time dilation" in an invariant sense. But only in that case.

For example, the observers A and B in your OP are at rest relative to each other, and the spacetime they are in is static, so we can say that "corresponding" events on A's and B's worldlines are events that have the same coordinate time, and this can be given an invariant meaning. So the ratio we calculated earlier is indeed the "relative time dilation" of A and B in an invariant sense.

Now consider another observer, D, who is in a circular orbit about the gravitating body at the same altitude as B. Once per orbit, D will pass by B, and each such event is obviously a "corresponding" event on both their worldlines (since the worldlines cross at each such event). So the ratio of elapsed proper times over one orbit of D again can be interpreted as the "relative time dilation" of D and B. This can also be used transitively to give the "relative time dilation" of D and A, or D and C, or D and any other observer who can be given "corresponding" events to B. (And note that this method can also be used to handle the satellite case--the only further complication there is that the second observer, E, is also moving; but you can still use the periodicity of both motions to pick out "corresponding" events on both worldlines--the events where each one would pass a static observer.)

But in the case of two inertial observers in relative motion in flat spacetime, if both remain inertial forever, there is no invariant way to pick out more than one "corresponding" event on their worldlines (and you can only even pick out one if the two worldlines cross). So there is no invariant way to define their "relative time dilation". Each observer individually can define the other observer's "time dilation" relative to him, but the two definitions are not compatible.

9. Dec 9, 2014

### Staff: Mentor

Neither of these reference frames are the right one for calculating the numbers you're quoting using the formulas you're using. Those formulas are only valid in an "inertial frame" (see below for why that's in quotes) in which the center of the Earth is at rest. Changing frames to one in which either the Earth's (rotating) surface or the satellite is at rest doesn't just invert the ratios; it changes the whole calculation, because you're changing from an inertial frame to a non-inertial frame.

There is one respect in which the Earth-centered frame being used is not quite a standard global inertial frame: the nonzero gravitational potential. However, the formula you are using for that is also only valid in that particular frame; you can't just transfer it to a frame in which the Earth's rotating surface or the satellite is at rest. In those frames, the notion of "gravitational potential" is not even well-defined; you have to do a more complicated calculation to derive the numbers you are quoting.

(Note that the change of frames you are trying to make here is not the same as just recaling the time coordinate from infinity to some finite altitude, as was discussed in an earlier post. That kind of change still requires the frame to be static--the Earth's rotating surface and the satellite would still be moving in such a frame.)

10. Dec 10, 2014

### dougy

Thank you for the replies DaleSpam and PeterDonis. However I still do not understand why exactly the satellite doesn't see the clock on Earth run slower when the satellite is in orbit compared to when it isn't.

Let me clarify the situation. For simplicity let's say the Earth doesn't rotate on itself (we could pick an hypothetical planet with the same mass as that of the Earth that doesn't rotate on itself). In the inertial frame in which the center of the Earth is at rest, observer A at infinity is static, observer B in altitude (at distance $R_B$ from the center of the Earth) is static, observer C on the surface of the Earth (at distance $R_C$ from the center) is static, observer D in circular orbit at the same altitude as B (at distance $R_D$ = $R_B$ from the center of the Earth) is in motion. In this frame the gravitational effects of the Earth can be accurately modeled by the Schwarzschild metric.

Noting $\tau_A$, $\tau_B$, $\tau_C$ and $\tau_D$ the proper times of A, B, C and D we have:

$$\tau_B = \tau_A \sqrt{1-\frac{2GM}{R_Bc^2}}\\ \tau_C = \tau_A \sqrt{1-\frac{2GM}{R_Cc^2}}\\ \tau_D = \tau_A \sqrt{1-\frac{3GM}{R_Bc^2}}$$

To be clear, are these relations only valid in the inertial frame where the center of the Earth is at rest, or valid in all inertial frames, or valid in all frames? From what you explained I would think they are only valid in the inertial frame where the center of the Earth is at rest, and so only from the point of view of A, B and C.

All observers begin to exchange light signals between each other, sending light pulses every second of their own time. From my understanding:
A would receive B's pulses every $\frac{\tau_A}{\tau_B}$ second in its own time, C's pulses every $\frac{\tau_A}{\tau_C}$ second in its own time, and D's pulses every $\frac{\tau_A}{\tau_D}$ second in its own time.
B would receive A's pulses every $\frac{\tau_B}{\tau_A}$ second in its own time, C's pulses every $\frac{\tau_B}{\tau_C}$ second in its own time, and D's pulses every $\frac{\tau_B}{\tau_D}$ second in its own time.
C would receive A's pulses every $\frac{\tau_C}{\tau_A}$ second in its own time, B's pulses every $\frac{\tau_C}{\tau_B}$ second in its own time, and D's pulses every $\frac{\tau_B}{\tau_D}$ second in its own time.
However D is not at rest in Earth's inertial frame, which means D does not receive A's pulses every $\frac{\tau_D}{\tau_A}$ second in its own time, B's pulses every $\frac{\tau_D}{\tau_B}$ second in its own time, and C's pulses every $\frac{\tau_D}{\tau_C}$ second in its own time.

Are the above 4 assertions true? I hope so otherwise I'm really misunderstanding something fundamental.

Then my question would be, at which intervals of time does D receive A's pulses, B's pulses and C's pulses?

Thank you

edit: Actually I have realized just now that since the distance between D and the other observers is not constant, D will not receive A's, B's and C's pulses at constant intervals, and the observers A, B, C will not receive D's pulses at constant intervals either.

So a more relevant question would be, what are the proper times of A, B and C in the reference frame centered on D where D is at rest?

Last edited: Dec 10, 2014
11. Dec 10, 2014

### Staff: Mentor

It's not a question of what frames they're valid in; it's a question of what, physically, these numbers represent. Assume that A, B, and C all lie along a single radial line. Then the numbers represent the proper time elapsed on each clock between two successive passages of D across that radial line--since D is at the same altitude as B, this also means two successive instants where D passes B. These successive passages are invariants; they are spacetime events, independent of any frame. So the ratios of the proper times are also invariants, independent of any frame.

As far as which frame the relations are valid in, technically they are only valid in the Schwarzschild coordinate chart centered on the Earth, since that's how you derived them. Note, by the way, that this is not an "inertial frame where the center of the Earth is at rest"; there is no such thing, because spacetime around the Earth is curved, not flat. However, if you look at the formulas, you will see that all the quantities in them are invariants, except possibly the $R$ values; but even they can be given an invariant meaning (in terms of the circumference of circular orbits at the corresponding altitudes). So actually, if you are careful to write the formulas the way you have and define the $R$ values appropriately, the formulas are frame-independent, because they only involve invariants; no frame-dependent quantities appear anywhere.

For the ratios among A, B, and C, yes. For the ratios of any of them with D, not completely. D's motion not only changes the time intervals at which D receives pulses from the others; it changes the time intervals at which the others receive pulses from D. I haven't done the calculation for what these time intervals would actually be, and I don't have time to do it now; but I know the general statement I just made holds.

Proper times between what events? If you mean, proper times between successive instants where D passes B, those are invariants, independent of any frame, as I said above; when you calculate them in any frame, you must get the same answer (if not, you've done something wrong).

In other words, proper time intervals such as those you have written down are not properties of observers per se; they are properties of particular finite segments of the observers' worldlines. Once you have defined which finite segments you are interested in, it doesn't matter what frame you use; they must have the same length, because that's an invariant geometric property of the segments.

12. Dec 13, 2014

### dougy

Okay I'm starting to see things clearer, but I still have several issues.

Is it right to say that A receives D's pulses every $\frac{\tau_A}{\tau_D} + \frac{v_{DA}}{c}$ second in its own time, where $v_{DA}$ is the component of D's velocity in the direction of the line joining A and D (approximating $v_{DA}$ to be constant over one second)? And similarly that B receives D's pulses every $\frac{\tau_B}{\tau_D} + \frac{v_{DB}}{c}$ second in its own time, and C receives D's pulses every $\frac{\tau_C}{\tau_D} + \frac{v_{DC}}{c}$ every second in its own time?

Then does D receive A's, B's and C's pulses respectively every $\frac{\tau_D}{\tau_A} + \frac{v_{AD}}{c}$, $\frac{\tau_D}{\tau_B} + \frac{v_{BD}}{c}$ and $\frac{\tau_D}{\tau_C} + \frac{v_{CD}}{c}$ in its own time, or are those values different from the point of view of D?

Besides, I understand (thanks to your explanations) that after one orbit (when they meet again) both B and D will agree that $\frac{\tau_B}{\tau_D}$ times more time has elapsed on B's clock than on D's clock, what I don't understand is why D does not see B's clock run slower than his when they are in relative motion in orbit. In special relativity when two observers are in relative motion it is said that both see the clock of the other run slower than their own, and I don't understand why we cannot say the same here. From the point of view of D, B is the one in motion.

In an earlier post you said:

What would change exactly in the calculations in the frame where the satellite (D) is at rest (assuming that Earth does not rotate on itself)? We couldn't use the Schwarzschild metric anymore and we would have to use another one (which one would it be?), or are there other changes?

13. Dec 13, 2014

### Staff: Mentor

No. The times in all the cases involving D are not even constant; they vary as D moves.

He does, if he uses a momentarily comoving inertial frame (MCIF) at a given event in his orbit. This gives a different definition of relative clock rates than counting clock ticks between successive meetings of D and B. And the MCIF definition has a key disadvantage compared to the other one: it isn't invariant, because there is no set of common events for D and B corresponding to the successive meetings.

You would have to use a different metric. I don't know of a simple expression for that metric. There is an analogue to it in flat spacetime called Born coordinates, which might give some of the flavor, but since these coordinates are in flat spacetime, they don't include the effects of the gravitational potential due to the central body.

14. Dec 15, 2014

### dougy

Yes I know that the intervals of time between each pulse received are not constant, in the way I defined it $v_{DA}$ is not constant, it is a function of time (just like $v_{DB}$ and $v_{DC}$). $v_{DA}$ is the component of D's velocity in A's frame in the direction of the line joining A and D. I only approximated $v_{DA}$ to be constant over one second, but its value wouldn't be the same from one second to the other. But actually I just realized I had made a mistake anyway, the ratio $\frac{v_{DA}}{c}$ should also be multiplied by the factor $\frac{\tau_A}{\tau_D}$.

So correct question now, without the approximation and the mistake: is it right to say that A receives D's pulses every $\frac{\tau_A}{\tau_D}\left(1 + \frac{1}{c}\int v_{DA}(t)dt\right)$ second in its own time where $\int v_{DA}(t)dt$ is how much the distance between A and D increases over one second? And similarly for B and C? Then from the point of view of D, does D receive A's pulses every $\frac{\tau_D}{\tau_A}\left(1 + \frac{1}{c}\int v_{AD}(t)dt\right)$ second in its own time and similarly for B and C?

All I'm trying to understand here is whether, when two observers are in relative motion in a gravitational field (both staying in a constant potential), the intervals of time at which they receive the other's light pulses only depends on their relative time dilation as calculated from the Schwarzschild metric and on the distance between them (as it was the case for observers A, B and C stationary relative to the metric), or if D's motion relative to the metric changes something else in the calculations (besides the fact the distance between D and the other observers is not constant)?

When you say that D does see B's clock run slower than his "if he uses a momentarily comoving inertial frame at a given event in his orbit", does this mean the answer to the above question is yes (it only depends on their relative time dilation as calculated in the Schwarzschild metric and on the distance between them) or am I misinterpreting something?

Thanks again for the help

Also D is in free fall around the Earth so isn't he in an inertial frame all along? If so what would be the time dilations of A, B and C in that inertial frame?

Last edited: Dec 15, 2014
15. Dec 15, 2014

### Staff: Mentor

D's motion obviously changes something else in the calculation; your formula for $\tau_D$ has a different quantity under the square root sign (the $3GM/c^2R$ instead of $2GM/c^2R$).

If you're asking if that plus the changing distance (and therefore the changing "light speed time delay") is all that changes, the answer is "sort of" ;). There is another effect involved, to do with the fact that the spacetime curvature due to the Earth affects how long it takes light to travel between the various observers; you can't calculate the light travel time in terms the radial and other coordinates as you can in flat spacetime. This effect doesn't matter for A, B, and C, because they are all at rest relative to each other; but it does matter for D because of D's motion. For the Earth's field, the effect is small enough that we can ignore it here; but it's there in principle, and it has been measured for the Sun's field (Google "Shapiro time delay").

You're misinterpreting something. Any comparison of clock rates requires a common simultaneity convention among all the clocks being compared. All of the formulas you wrote down assume the simultaneity convention that is "natural" for A, B, and C, i.e., the simultaneity convention that's built in to the standard Schwarzschild coordinates. Since D's motion is periodic (he passes B once per orbit), this simultaneity convention also makes sense for comparing D's clock rate to those of A, B, and C. But you still need to specify that convention in order for the clock rate comparison in terms of your formulas to be valid.

If we talk about D seeing B's clock running slower in D's MCIF, we mean in terms of the "natural" simultaneity convention of that MCIF, which, since D is moving relative to A, B, and C, will be different from the "natural" simultaneity convention of A, B, and C, the one built in to all your formulas. (In fact, since D's MCIF changes as he moves around his orbit, the "natural" simultaneity convention for him using his MCIF changes too, so D doesn't really have a single "natural" simultaneity convention all the time, whereas A, B, and C do. This is another reason why it makes sense to use the simultaneity convention of A, B, and C in this scenario.)

No. A free-falling observer is only at rest in a single inertial frame all the time in flat spacetime. In curved spacetime (i.e., in the presence of gravity), even a free-falling observer changes inertial frames as he moves. More precisely, he changes MCIFs, since there is no such thing as a global inertial frame in curved spacetime.

16. Dec 15, 2014

### dougy

Does this effect really matter when D remains in a constant gravitational potential? Even though the distance between D and the other observers changes the light signal still has to cross the same difference in potentials. Intuitively I'm thinking that if a light ray travels twice the distance across the same difference of gravitational potentials it will take twice the time, but my intuition may be wrong here.

What are the simultaneity conventions used by A, B, C and used by D's MCIF? Ultimately what I want to do is compare the clock rates between A and D at every instant, not just once per orbit. I thought calculating the intervals of time at which A receives the light signals from D (and vice versa) was the way to do it, but we've got quite far in the discussion and now with all that talk of MCIF, simultaneity convention and so on I'm not sure I'm any closer to the answer. I do appreciate your replies which have allowed me to understand things that I didn't understand before so I thank you for that, but the way this discussion is going I feel I am going to have new questions after every of your replies again and again for weeks, so in order to waste as little as possible of your time I think that I should get straight to the point: How can A determine the clock rate of D, and how can D determine the clock rate of A, at every instant?

In the end what this thread is about is I don't understand how the clocks on GPS satellites can be synchronized with the clocks back on Earth. I don't understand how clocks in relative motion can be synchronized. In flat spacetime two observers in relative motion see the clock of the other run slower than his own, so any attempt to synchronize by slowing down or speeding up the rate of one of the clocks will synchronize in one way but offset in the other way even more. With the GPS there is the difference in gravitational potentials to take into account but the satellites and the clocks on Earth are still in relative motion so I don't understand how synchronization can be achieved both ways.

I don't understand how D can measure A's clock rate according to him and how A can measure D's clock rate according to him. What does it mean to measure the clock rate of another observer in the first place, how is that done in practice? By exchanging light signals and measuring their times of arrival as I described? How does D go from measuring the intervals of time between the signals he receives from A to determining the clock rate of A?

17. Dec 15, 2014

### Staff: Mentor

Yes.

Yes, but twice what time? The point is that the time when spacetime is curved is different than what it would be if spacetime were flat.

A, B, and C's simultaneity convention is the "natural" one for the global coordinate chart in which they are all at rest; physically, it can be realized by sending round-trip light signals. For example, if A sends a round-trip light signal to B, the event on A's worldline simultaneous with the event of B receiving (and reflecting) the signal is exactly halfway between the events of A emitting the signal and A receiving the return signal.

D's MCIF simultaneity convention is the natural one for the MCIF, considered as an inertial frame (and, as I said before, it's a different MCIF for each point on D's worldline)

And how you do that depends on what simultaneity convention you choose. Again, since D's motion is periodic, and since D's MCIF changes at each point while A, B, and C all remain at rest relative to each other, the "natural" one to choose for the comparison is the simultaneity convention of A, B, and C. But you could choose some other convention if you insisted, and your choice would affect the answer you obtained.

There's no way to do it by direct observation, because of the light travel time delay. However, since both A's and D's motion are known, A can do it by calculation, once he has picked a simultaneity convention. (So can D, for that matter.) See further comments below.

They can only be synchronized if the relative motion is periodic, and the synchronization requires one of the clocks to adjust its rate. To eliminate gravitational potential from the discussion, let's first consider a simple scenario in flat spacetime: A is floating in empty space at some point. B is floating at rest relative to A and some distance away. D is moving in a circle with A at its center, and with a radius such that D passes by B once per revolution.

A and B can obviously synchronize their clocks, since they are both at rest in the same inertial frame. So let's assume they've done that. Then D can simply set his clock by B's clock when he passes B (or, in a more sophisticated scenario, A could emit light signals in all directions, and D could continuously set his clock by the signals, applying the appropriate corrections for the distance from A). If D does this, he will find that the rate of his clock will not be the same as its "natural" rate--or, to make it easier to see what's going on, suppose D carries two clocks, one which just ticks naturally, while the other is set by B's clock or by light signals from A as above. Then D will see that the second clock runs faster than the first; this is the invariant sense in which D's clock "runs slower" than A's and B's clock.

Of course D is not forced to use the second clock at all; his only reason for doing so is to maintain clock synchronization with A and B. But you should be able to see that if D wants to maintain clock synchronization with A and B, he has to do it the way I described above. The reason for that is, as I said, that D's motion is periodic--he passes B once per revolution--and the periodicity of the motion defines a set of events, each equally separated in time, that give a common reference for judging "clock rate" (because they lie on both D's and B's worldlines). There is no other common reference to use if D wants to maintain clock synchronization with A and B.

In the case of GPS, the same sort of thing is going on, but there is an additional adjustment due to gravitational time dilation (and another small variation from the simpler case above because the Earth is rotating). An observer on Earth's surface plays the role of A above, but with the difference that he is not at the center of D's orbit (and also that he is moving with the Earth's rotation). The GPS satellite plays the role of D. There is no actual object in the GPS system that plays the role of B, but we can imagine that there is someone in a rocket "hovering" above Earth (and not rotating with the Earth) at the same altitude as the GPS satellite (D), so that D passes B once per orbit, just as in the flat spacetime case above.

We can then apply the same sort of logic to GPS that we applied in flat spacetime: D passes B once per orbit, so there is an invariant way to synchronize D's and B's clocks, by D setting his clock by B's clock every time he passes B. D's clock then runs slower than B's, just as in the above case. However, because of gravitational time dilation, B's clock rate is not the same as A's; it is faster, because B is at a higher altitude in the gravitational field. (And it is a bit faster still because A is moving with the Earth's rotation whereas B is at rest.) It turns out that, when the correction from B's clock rate to A's clock rate is applied, A's clock turns out to be slower than D's clock (because the slowdown due to the decrease in altitude from B to A is larger than the slowdown due to motion from B to D). So D's clock rate actually has to be adjusted downward in order to maintain clock synchronization with A. (The actual GPS satellites actually have something like the two-clock arrangement I described above; they have a base clock that runs at its "natural" rate, and then they have a separate frequency generator that corrects that clock's rate to be the same as the rate of a clock at rest on the rotating Earth at the altitude of the geoid, meaning roughly mean sea level, and outputs a clock signal accordingly. The latter clock signal is what is sent out with the GPS signals that each satellite broadcasts.)

(Also, since A is not at the center of D's orbit, D has to do more complicated calculations to continuously adjust his clock to A's based on light signals from A. In the real GPS system, there are multiple ground stations that send signals back and forth from the satellites to keep all the clocks in sync, applying sophisticated corrections based not only on relativistic effects, but also on things like signal delays in the Earth's ionosphere. But the basic idea is still the same: to keep D's clock synchronized with A's clock, using the simultaneity convention of, in this case, B, since B is the only one at rest in a non-rotating frame centered on the Earth.)