# A question about time dilation

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## Summary:

SR effect GR effect
OK this is a quote from the physics forum site - it is not mine.
Yes, the GPS clocks 'tick' faster than those on earth due lower gravity [GR effect], which is partially offset by their orbital velocity [SR effect]. The net effect is they would run about 39,000 nanoseconds per day faster than their earthbound counterparts. GPS clocks are precalibrated before launch to cause them to run this much slower, and, voila! Once in orbit they are synchronized with ground clocks. Scientists subsequently tweak the calibration of the GPS clocks to keep them in synch with ground clocks.
So the above quote says that there are two effects which alters the clock ticks, the GR effect and the SR effect for clocks aboard satellites.
We all know that satellite are in free fall and for free fall there must be a steady tangent velocity at a particular orbital distance for the satellite to remain in orbit.
I have read elsewhere that free fall has nothing to do with how a clock ticks (faster or slower)

What am I not understanding?

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## Answers and Replies

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Ibix
Science Advisor
for free fall there must be a steady tangent velocity at a particular orbital distance for the satellite to remain in orbit.
That's not correct - it's true if you want a circular orbit, but that is very far from the only possible free fall path.
I have read elsewhere that free fall has nothing to do with how a clock ticks (faster or slower)

What am I not understanding?
Difficult to tell with a paraphrase of an unreferenced something.

An accurate picture is that time dilation isn't always definable in general relativity. However, it is definable for a near-enough unchanging gravitational field like the Earth's. And in this case you can decompose the time dilation between two clocks into components due to their height difference (gravitational time dilation) and due to their velocity with respect to hovering clocks (kinematic time dilation). These two effects are what's being described (slightly inaccurately) as a GR effect and an SR effect in the post you quoted.

The GPS clocks are higher than us, so the gravitational time dilation effect makes them tick faster. But they are moving in fast orbits, so the kinematic time dilation effect counters that somewhat. Aside from a nitpick about "SR" and "GR" effects, the post you quoted is correct.

A.T.
Science Advisor
I have read elsewhere that free fall has nothing to do with how a clock ticks (faster or slower)
Hard to say what you mean without a direct quote and full context. If you mean the clock hypothesis:

https://en.wikipedia.org/wiki/Time_dilation#Clock_hypothesis

then here is an analogy how proper-acceleration vs. free-fall affects the total elapsed time, without affecting the clock rate:

The analogy I would use is a car at constant speed:

proper acceleration <-> changing the cars direction
proper time <-> traveled distance

Changing direction doesn't directly affect the rate at which distance is traveled, but it can affect the total traveled distance between two points.
But note that in GR it gets even more complicated, because two free falling clocks can have different times elapsed between meetings:

https://www.physicsforums.com/threads/twin-paradox-for-freely-falling-observers.991709/

IBX
true there are elliptical orbits as well

i agree there are more then one condition which change the GPS clock timing etc. I mentioned the condition of free fall because it means the satellite is in a steady gravitational field which must be balanced by the tangent speed to remain in orbit.

I also have read that the gravity effect is real- it actually does change the clock speed aboard the satellite. However the tangent velocity makes the clock APPEAR to run slower- is that right?

https://www.researchgate.net/post/Are_moving_clocks_really_running_at_different_rhythms_just_because_of_velocity said:
Secondly when we consider the speed of a GPS satellite, (~3874 m/s), Einstein's special theory of relativity needs to be applied. The time dilation effect causes the GPS satellite to APPEAR to run slow by about 7 microseconds per day.
If the GPS clock APPEARS to run slow, then a one time offset to the GPS clock must be made because the 7 microseconds per day is not accumulative. Otherwise if the GPS clock actually runs 7 micros slower per day then it is accumulative - so why does the literature use the word APPEAR?

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Ibix
Science Advisor
I mentioned the condition of free fall because it means the satellite is in a steady gravitational field which must be balanced by the tangent speed to remain in orbit.
You seem to be confused about what free fall is. If I throw a ball, the ball is in free fall (neglecting air resistance), as is any unpowered satellite in any orbit. It's not specific to circular orbits, which is what you seem to want to discuss.
I also have read that the gravity effect is real- it actually does change the clock speed aboard the satellite. However the tangent velocity makes the clock APPEAR to run slower- is that right?
"Real" isn't a well-defined term here, and you can make a case for pretty much any answer to this question. However, it is certainly the case that if you record the time shown by a GPS satellite as it passes overhead and try to predict the reading the next time it passes overhead you will get the wrong answer if you neglect either effect. So I would say each one is as "real" as the other.

You may be confusing this with time dilation between inertial frames in special relativity. This is a coordinate effect, and there's no assumption-free way to measure it. That might be what you mean by "not real". Despite a superficial similarity between some of the results, this is quite different from the circular motion case we are discussing here (which is partly why I object to the GR/SR split in the OP - it's not groundless, but I think it's potentially quite confusing).

A satellite is one condition of free fall

The use of the word APPEAR occurs a lot in these types of discussions . If there is rock solid evidence that the tangent speed actually changes the GPS clock then why use the word APPEAR?

If we say that a GPS clock APPEARS to be out of sync with another clock then the information that leads to that conclusion is by measurement which involves the speed at which information can be sent.

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We all know that satellite are in free fall and for free fall there must be a steady tangent velocity at a particular orbital distance for the satellite to remain in orbit.
I have read elsewhere that free fall has nothing to do with how a clock ticks (faster or slower)
According to a person standing on earth there is a gravity field around the earth, and two clock in the international space station may be at different gravitational potential, and if they are, then they must experience different amounts of gravitational time dilation.

According a person inside the ISS all clock inside the ISS tick at the same rate. A person standing on earth must agree with that.

free fall has nothing to do with how a clock ticks (faster or slower)
So when a person standing on earth jumps up, he becomes a freefaller, and according to everyone his clock continues to tick at the same rate as before. It's a very small jump, so that we can ignore the change of gravitational potential and the change of velocity.

(A small correction to the previous paragraph: According to jumper's brain a clock taped on jumper's foot was extra slow when the jumper was standing on the surface of the earth. According to jumper's brain that effect disappeared when the jumper jumped. So it's not quite true that jumper's clocks continued to tick at the same rate as before according to everyone.)

Oh yes, according to the jumping person there is no gravity field around the earth in his neighborhood, so then of course no clock in his neighborhood can experience gravitational time dilation.

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Ibix
Science Advisor
If there is rock solid evidence that the tangent speed actually changes the GPS clock then why use the word APPEAR?
The fact is that "a clock runs slow" is an incomplete concept, since there's no absolute standard of time. So you need to say "a clock runs slow compared to some other clock". When we're talking about GPS clocks we're usually comparing them to earthbound clocks, and compared to earthbound clocks they tick slightly fast. But compared to clocks floating out in deep space, they tick slow.

So I wouldn't say that there is "rock solid evidence that the tangent speed actually changes the GPS clock". That sounds like comparing GPS clocks to a notional absolute reference, which would be nonsense. There is rock solid evidence that a clock orbiting above us ticks faster than our clocks if we compare them, but slower than a clock hovering at their altitude if we compare them. That comparison may be what people mean by "appear", but that applies equally to gravitational time dilation and kinematic time dilation in this context. The reference you cite is simply another forum discussion, and may not be the best example of clear writing.

The maths is absolutely clear, however. Drawing a distinction between "gravitational time dilation" and "kinematic time dilation" in this context is something we choose to do. They come from the same place in the maths, and regarding one as "real" and one as "not real" would be bizarre.

vanhees71
Science Advisor
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2019 Award
I find it a bit confusing to say this or that is a GR and this or that is an SR effect. All there is is the definition of proper time given by the spacetime metric. Since SR is a special case of GR (a situation, where gravity can be neglected) there's no difference here between SR and GR. The proper time of a moving massive body simply is
$$\tau=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \sqrt{g_{\mu \nu} \dot{q}^{\mu} \dot{q}^{\nu}},$$
where ##\lambda## is an arbitrary worldline parameter.

etotheipi, PeroK and Ibix
Ibix
Science Advisor
According a person inside the ISS all clock inside the ISS tick at the same rate. A person standing on earth must agree with that.
If don't think this is true, at least not without some qualification. You can zero both clocks as they pass above some fixed point and stop them the next time they pass over that same point (it's easier to imagine a non-rotating Earth for this, although not necessary). Either the clocks match at the end (to available precision) or they don't, and all observers will agree.

vanhees71
A.T.
Science Advisor
If the GPS clock APPEARS to run slow, then a one time offset to the GPS clock must be made because the 7 microseconds per day is not accumulative. Otherwise if the GPS clock actually runs 7 micros slower per day then it is accumulative - so why does the literature use the word APPEAR?
The net effect is cumulative. If you bring back an uncorrected satelite clock after a long time (so start and landing have negligible effects) the offset to an identical clock that stayed on Earth will depend on the duration in orbit.

Ibix
Science Advisor
I find it a bit confusing to say this or that is a GR and this or that is an SR effect. All there is is the definition of proper time given by the spacetime metric.
Indeed.
Since SR is a special case of GR (a situation, where gravity can be neglected) there's no difference here between SR and GR.
I'd say that's potentially confusingly phrased. By "here" I think you are referring to your previous sentence (about proper time coming from the metric), which I agree with. But it's possible to read "here" as referring to "this scenario", which would be wrong - SR can only be applied to parts of the discussion, and great care is needed if you do because of the issues around clock synchronisation for orbiting bodies.

vanhees71
If don't think this is true, at least not without some qualification. You can zero both clocks as they pass above some fixed point and stop them the next time they pass over that same point (it's easier to imagine a non-rotating Earth for this, although not necessary). Either the clocks match at the end (to available precision) or they don't, and all observers will agree.
Okay, there are some small ticking rate differences. A person inside the ISS observes some small ticking rate differences between clocks inside ISS, and also experiences some small inertial forces, or pseudo gravity fields. When the forces are small, then the ticking rate differences are small too. I assumed zero forces, that's why I assumed zero ticking rate differences.

Ibix
Science Advisor
I assumed zero forces, that's why I assumed zero ticking rate differences.
Yeah, but my point is that then there's zero difference measured by an Earth observer too.

vanhees71
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2019 Award
Indeed.

I'd say that's potentially confusingly phrased. By "here" I think you are referring to your previous sentence (about proper time coming from the metric), which I agree with. But it's possible to read "here" as referring to "this scenario", which would be wrong - SR can only be applied to parts of the discussion, and great care is needed if you do because of the issues around clock synchronisation for orbiting bodies.
Sure, of course, in the context of the GPS you need GR, which is obvious, because gravity is involved.

Ibix
PeroK
Science Advisor
Homework Helper
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The use of the word APPEAR occurs a lot in these types of discussions . If there is rock solid evidence that the tangent speed actually changes the GPS clock then why use the word APPEAR?
This is one of the weaknesses of trying to make the physics accessible to a lay audience that cannot be presented with the mathematical formalism. See post #9. There is no argument about the mathematics. The debate only starts when you try to describe the mathematics by using words like "real" and "appear". If you define an experiment, then the experiment has a measureable outcome. That outcome may be predicted by the theory and mathematics of GR. In terms of experiments you can carry out using GPS satellites, GR makes the correct predictions (whereas classical physics does not).

russ_watters, Ibix and vanhees71
Yeah, but my point is that then there's zero difference measured by an Earth observer too.

I agree completely. I was just saying that the gravitational time dilation is different between clocks at different distances from earth. Do you disagree with that???

There must be some other time dilation effect that cancels out that aforementioned effect. The effect should cause some slowing down of the clocks inside the ISS that are farther away from the earth. (It's the velocity time dilation effect )

Ibix
Science Advisor
Do you disagree with that???
Of course not.
There must be some other time dilation effect that cancels out that aforementioned effect.
Why must there be? The only reason I can think of is the equivalence principle, and that doesn't apply here because the ISS is extended in time, so doesn't count as a "small region of spacetime".

Why must there be? The only reason I can think of is the equivalence principle, and that doesn't apply here because the ISS is extended in time, so doesn't count as a "small region of spacetime".

Does post #13 answer the question why clocks inside ISS must tick at (almost) the same rate?

If post #13 doesn't answer the question, then how about if we consider a huge ring shaped space-station that spins around in space. Crew observes that clock on the floor run slower than clock on the ceiling. They may say that it's because of gravitational potential difference.

Now we bring a large mass into the middle of the ring. Such mass that the observed potential difference inside the ring disappears. IOW such mass that the crew experiences weightlessness. Shouldn't the crew now say that gravity disappeared and so did the gravitational time dilation?

The ring can be million times larger that the orbit of ISS.

Or, you could calculate some ticking rates of some clocks inside ISS. I promise the difference will be very very very small. I mean, if my explanations aren't good, then maybe some simple calculation would make things clear.

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PeroK
Dale
Mentor
The use of the word APPEAR occurs a lot in these types of discussions . If there is rock solid evidence that the tangent speed actually changes the GPS clock then why use the word APPEAR?
There are multiple effects going by the name of "time dilation", some of which are frame invariant and some of which are frame dependent. The word appear usually is used to identify one of those effects which are frame dependent.

On this forum very often the term "time dilation" is used to refer to the frame dependent effects and the term "differential aging" is used to refer to the frame invariant effects. There is rock solid experimental evidence for both types of effects. In other words, there is rock solid evidence that the frame variant effects appear as described by relativity in any given reference frame, and that the frame invariant effects also occur as described by relativity.

The terminology is simply not as rock solid as the evidence.

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russ_watters, Ibix and PeroK
Ibix
Science Advisor
Does post #13 answer the question why clocks inside ISS must tick at (almost) the same rate?
I agree that the differences would be small; but you seemed to be saying that the differences would be larger viewed from an Earth's surface frame. If you weren't saying that then I misunderstood.
Now we bring a large mass into the middle of the ring. Such mass that the observed potential difference inside the ring disappears. IOW such mass that the crew experiences weightlessness. Shouldn't the crew now say that gravity disappeared and so did the gravitational time dilation?
It's possible to pick an altitude where the kinematic and gravitational time dilations cancel, and clocks on satellites match the rate of Earthly clocks, yes. Clocks on the inner and outer edges of the ring would, I think, drift but as you say the effect is small. But still, it wouldn't be larger viewed from any other state of motion and/or potential.

PeterDonis
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2019 Award
When the forces are small, then the ticking rate differences are small too. I assumed zero forces, that's why I assumed zero ticking rate differences.
It might be worth expanding on this some.

First, "ticking rate" is not really what we're talking about here, in the sense of something that can be instantaneously measured. We are talking about some kind of invariant, such as, for example, the elapsed time on clocks in the ISS, as compared to a clock on Earth, over one complete orbit of the ISS--i.e., between two successive passages of the ISS directly overhead of the clock on Earth (if we assume the latter clock to be appropriately located on Earth with respect to the orbit of the ISS). The question then is whether and to what extent clocks in different parts of the ISS will show different elapsed times for one complete orbit, as compared with the reference clock on Earth.

In an Earth-centered frame, at least to a first approximation, there are two things that affect this invariant for a particular orbiting clock: altitude (gravitational potential) and speed (relative to the center of the Earth). (We could include more complicated effects due to things like the Earth's quadrupole moment, but I don't think we need to go to that extreme here.) Different parts of the ISS can be at different altitudes and can have different speeds relative to the center of the Earth, so their ticking rates can in principle be different relative to some standard clock on Earth.

The question then is, will those effects actually cause clocks in different parts of the ISS to have different invariants (elapsed time over one complete orbit)? To even evaluate that, we have to think about how, precisely, the ISS is moving in its orbit and how it is oriented in space. If our idealization is that there should be zero "forces" inside the ISS for the duration of one complete orbit, then in addition to the ISS being in a free-fall orbit and being small enough that tidal gravity effects can be ignored for the duration of one orbit (btw, I have not done the math to see if the actual ISS is indeed that small; my suspicion is that it is not, a suspicion that @Ibix appears to share), the orientation of the ISS must be fixed with respect to an Earth-centered inertial frame. (Actually this is not exactly correct, because of various precession effects, but I don't want to complicate the problem more than it is already.)

Given those requirements for the ISS's orbital motion, we can see that, relative to the Earth, any individual clock on the ISS will have varying altitude and speed relative to the center of the Earth, over one complete orbit. (The only exception would be a clock located precisely at the ISS's center of mass, if the orbit were precisely circular.) I have not done the math, but it seems clear what math would need to be done: compute the altitude and speed of clocks in different parts of the ISS, assuming the orbital motion is as above, as a function of coordinate time in an Earth-centered inertial frame over one complete orbit, and use those functions in an integral to compute the total elapsed time for each clock over one orbit. If your "zero forces = zero ticking rate difference" intuition is correct, that math should give the same result for a clock anywhere on the ISS.

FactChecker
Science Advisor
Gold Member
I have read elsewhere that free fall has nothing to do with how a clock ticks (faster or slower)

What am I not understanding?
The effect of gravity is due to the different gravitational potential at the higher altitude. That is a function of position and free-fall has nothing to do with it. They have tested the effect using clocks at the top of mountains and even at the top of tall buildings. Those clocks are not in free fall but they still show the effect.

H_A_Landman and vanhees71
Ibix
Science Advisor
If our idealization is that there should be zero "forces" inside the ISS for the duration of one complete orbit, then in addition to the ISS being in a free-fall orbit and being small enough that tidal gravity effects can be ignored for the duration of one orbit (btw, I have not done the math to see if the actual ISS is indeed that small; my suspicion is that it is not, a suspicion that @Ibix appears to share), the orientation of the ISS must be fixed with respect to an Earth-centered inertial frame.
I was actually considering the case where the ISS rotates once per orbit, maintaining its orientation as observed by anybody directly below it. The maths for that is fairly straightforward.

Consider two clocks, one in a circular free fall orbit and one in a forced circular orbit directly above the first. Considering the free falling clock, we can use 7.47 and 7.48 from Carroll's GR notes with ##\epsilon=1## for a timelike orbit to write$$\frac{dr}{d\tau}=\sqrt{E^2-\left(1-\frac{R_S}r\right)\left(1+\frac{L^2}{r^2}\right)}$$where ##L## and ##E## are constants of the motion. For a circular orbit at altitude ##r=r_0## we insist ##\frac{dr}{d\tau}=\frac{d^2r}{d\tau^2}=0## and obtain$$\begin{eqnarray*} L^2&=&\frac{R_Sr_0^2}{2r_0-3R_S}\\ E^2&=&\frac{2r_0^2-4R_Sr_0+2R_S^2}{2r_0^2-3R_Sr}\end{eqnarray*}$$Substituting into 7.43 and 7.44 from Carroll we can write$$\begin{eqnarray*} \frac{dt}{d\tau}&=&\sqrt{\frac{2r_0}{2r_0-3R_S}}\\ \frac{d\phi}{d\tau}&=&\sqrt{\frac{R_S}{r_0^2(2r_0-3R_S)}} \end{eqnarray*}$$and hence$$\frac{d\phi}{dt}=\sqrt{\frac{R_S}{2r_0^3}}$$Moving on to the non-free falling clock, we know that its ##\frac{d\phi}{dt}## must be the same in order for it to complete an orbit always directly above the free-falling clock. That means its four velocity must have components ##U^t=A##, ##U^\phi=A\sqrt{\frac{R_S}{2r_0^3}}##, and ##U^r=U^\theta=0##, where ##A## is a constant to be determined. Setting this clock's altitude to ##r=r_1## and normalising the four velocity, we find$$A^2=\frac{2r_1r_0^3}{2(r_1-R_S)r_0^3-r_1^3R_S}$$We can now write down the ratio of the clock rates - this is just the inverse of the ratio of the time components of the clocks' four velocities which, in my rather inconsistent notation, is ##\frac{dt}{d\tau}\frac 1A##, or$$\sqrt{\frac{2(r_1-R_S)r_0^3-r_1^3R_S}{r_0^2r_1(2r_0-3R_S)}}$$Defining ##D## to be the difference in ##r## coordinates of the clocks and writing ##r_1=r_0+D## we get$$\sqrt{1-\frac{3D^2R_S(r_0+R_S)}{2r_0^4+(2D-3R_S)r_0^3-3DR_Sr_0^2}}\approx\sqrt{1-\frac{3D^2R_S}{2r_0^3}}$$where the approximation comes from noting that ##D,R_S\ll r_0##.

Some specific numbers for the ISS are ##r_0## is 6800km, ##R_S## is 4mm, and ##D## is about 10m. Plugging those in, we get a tick rate ratio of about ##1-10^{-21}##. My rule of thumb for atomic clock accuracy is about one part in ##10^{14}##, so that's not detectable with current technology. You'd need about ten kilometres of vertical separation for a detectable rate difference.

Assuming I haven't made any mistakes, anyway. Edit: I did - ##R_S## is about 9mm - also see PAllen's reply beliw.

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Dale and vanhees71
PeterDonis
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2019 Award
I was actually considering the case where the ISS rotates once per orbit, maintaining its orientation as observed by anybody directly below it.
As you note, for this case clocks in the ISS are not, in general, in free fall, except for clocks at the ISS's center of mass. So it doesn't meet the "zero forces" criterion. However, as your calculation shows, for the actual ISS the difference is far too small to detect with our current technology.

As far as the actual ISS is concerned, I believe your scenario (maintaining orientation with respect to Earth as it orbits) is what is done. Note that the thrusters of the ISS would have to be used to accomplish this.

vanhees71