# Time dilation in a planet-moon system

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• DanMP
In summary: Theory prediction does not have a closed form solution. You would have to calculate it numerically.In summary, the theory of relativity predicts that the time difference between two clocks on a planet and a moon will decrease as the velocity of the moon around the planet decreases. However, this reduction is not always significant and must be determined experimentally.
DanMP
TL;DR Summary
In this thread I want to learn/discuss how time dilation in a planet-moon system is calculated and how it can be tested with accuracy.
In Wikipedia time dilation is considered:
Time dilation is the difference in elapsed time as measured by two clocks, either due to a relative velocity between them (special relativity) or due to a difference in gravitational potential between their locations (general relativity).

As far as I know, in a planet-moon system, the difference in elapsed time between a clock on the planet and a clock on its moon is calculated using GR/proper time, so it's not very obvious if/how the movement/velocity of the moon around the planet would influence the total time difference between the two clocks. Due to the differences in gravitational potential, the clock on the moon should be faster than the one on the planet, but the difference in velocity may reduce a little bit the difference in elapsed time. In this thread I want to learn about the amount of this reduction, if any. One way to do this is to calculate the total difference between the clocks in the situation where the moon is not rotating (is hovering) and compare that result with the normal result, with the moon orbiting the planet. In order to simplify the calculation, we may consider the moon orbit as circular and the clocks on one of the poles. Also, in order to have some numerical values, we can consider the Earth-Moon system, but it's not necessary, because it's nothing special in this particular case.

If we want to experimentally test such a prediction, we may carry a deep space atomic clock in a man mission to the Moon and back, a space Hafele-Keating experiment. Or we may compare the clocks as we do for GPS clocks. Anyway, this experimental part was briefly discussed in another thread, so in this thread I want to learn about what exactly the theory of relativity would predict.

Dale said:
What is the difference between this thread and the one that was already answered and closed
In the other thread I asked about clocks on the Moon and experimental tests already performed. In this one I ask mostly about theory predictions. You have a problem with that?

Please point to the answer, allegedly offered, regarding:
DanMP said:
Due to the differences in gravitational potential, the clock on the moon should be faster than the one on the planet, but the difference in velocity may reduce a little bit the difference in elapsed time. In this thread I want to learn about the amount of this reduction, if any.

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Dale
DanMP said:
In the other thread I asked about clocks on the Moon and experimental tests already performed. In this one I ask mostly about theory predictions. You have a problem with that?
It just seems like you are asking the same question over and over and over hoping for something different. Yes, I have a problem with that. It is exhausting.

In any case, the theory prediction does not have a closed form solution. You would have to calculate it numerically. I don't have the software for that.

russ_watters
DanMP said:
what exactly the theory of relativity would predict.
Because the two clocks are not colocated we have four relevant events: ##T_{A0}## when we record the time on clock A at the start of the measurement interval, ##T_{A1}## when we record the time on clock A at the end of the measurement interval, and likewise ##T_{B0}## and ##T_{B1}## for clock B.

##\Delta A = T_{A1}-T_{A0}## will be equal to the proper time along clock A's worldline and likewise ##\Delta B = T_{B1}-T_{B0}## will be equal to the proper time along clock B's worldline.

You are asking about the relationship between ##\Delta A## and ##\Delta B##; their ratio is the "time dilation" between the two clocks. That value will depend on how we choose our four events (that is, our arbitrary choice of simultaneity convention) and the two proper time calculations (which as @Dale points out above, will have to be done numerically as there is no known closed-form solution to the EFE for this two-body problem). Thus there is no "what exactly" until we've established a simultaneity convention and gone through some very tedious calculations, but once we've done that we will have our exact prediction.

Time dilation isn't actually defined in this situation either since it's a non-stationary spacetime. That's why you have to establish a simultaneity convention before you can answer questions about clocks that don't meet at least twice.

Dale said:
In any case, the theory prediction does not have a closed form solution. You would have to calculate it numerically. I don't have the software for that.
Ok, maybe someone does, or can offer a link regarding the subject.

DanMP said:
Ok, maybe someone does, or can offer a link regarding the subject.
Googling for "General relativity two-body" will get you started. The LIGO observations will be relevant.

Not to be "the engineer" here, but if you had a table of altitude and speed could you get close with a spreadsheet? (Say, 5%?). Or does the fact that you have two massive bodies make for too big of a complication?

That said, I do agree that this feels like just looking at a problem for the problem's sake and doesn't really lead anywhere except to the next problem. E.g., if the answer is 233 microseconds, then what?

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DanMP said:
If we want to experimentally test such a prediction, we may carry a deep space atomic clock in a man mission to the Moon and back, a space Hafele-Keating experiment.
Yes, we could. And GR predicts that the elapsed times on the clocks will be different. To predict the exact difference one would need to know the exact trajectory of the space traveling clock (if we assume that the Earth clock just stays in the same place on the rotating Earth then we know its behavior). But the general approach will be the same as for the analysis of the H-K experiment.

DanMP said:
Or we may compare the clocks as we do for GPS clocks.
We don't compare GPS clocks. We assume that their corrected rates are what they need to be for the signals they emit to carry the correct information for GPS to work. The fact that GPS does work is strong evidence that that assumption is correct, but it's still not the same as actually comparing the clocks directly. For that you would need to bring a GPS satellite back down to Earth and sit it next to a ground GPS clock.

DanMP said:
in this thread I want to learn about what exactly the theory of relativity would predict.
Answered: see above.

Nugatory said:
That value will depend on how we choose our four events (that is, our arbitrary choice of simultaneity convention)
No, the choice of events has nothing to do with a simultaneity convention. The OP has asked about a Hafele-Keating type experiment where one clock stays on Earth and the other clock goes to the Moon and then comes back. So the events are invariant: the event where the clocks separate and the event where they meet again. No comparison ever has to be made between spatially separated clocks, which is what would require a simultaneity convention.

Ibix said:
Time dilation isn't actually defined in this situation either since it's a non-stationary spacetime.
Strictly speaking, it's not, but to get at least a reasonable estimate (though possibly not one accurate enough for atomic clocks) we can use the weak field approximation and simply add the potentials from the Earth and the Moon, and adopt a barycentric inertial frame for the Earth-Moon system. In that approximation the Earth-Moon system can be treated as a stationary system, at least for the duration of the experiment (since we don't have to worry about things like tidal effects on the Earth's spin and the Moon's orbit on that time scale).

Then the time dilation along a trajectory depends only on the potentials and the speed in the barycentric frame. You would still have to integrate the equations numerically since there wouldn't be a closed-form formula for the antiderivative, but Hafele and Keating had to do the same thing in the analysis of their experiment.

Ibix said:
That's why you have to establish a simultaneity convention before you can answer questions about clocks that don't meet at least twice.
These clocks do meet twice. See post #11.

PeterDonis said:
These clocks do meet twice.
Depends if he's planning to do a Hafele-Keating experiment or "compare the clocks as we do for GPS clocks". He says both in the OP.

russ_watters
russ_watters said:
Not to be "the engineer" here, but if you had a table of altitude and speed could you get close with a spreadsheet? (Say, 5%?). Or does the fact that you have two massive bodies make for too big of a complication?

That said, I do agree that this feels like just looking at a problem for the problem's sake and doesn't really lead anywhere except to the next problem. E.g., if the answer is 233 microseconds, then what?
So the OP wants the difference between a clock on the moon with the moon orbiting vs a clock on the moon with the moon suspended. A spreadsheet might get you either of those times with some reasonable precision. But I think you would need a pretty good numerical method to get the small difference the OP is asking about.

I completely agree with your “then what”. The OP is already aware of the direction of the effect. It is small. Does it matter how small? Since this is about a hypothetical non-orbiting moon it cannot be experimentally tested. So of what value is the number?

russ_watters
Ibix said:
Depends if he's planning to do a Hafele-Keating experiment or "compare the clocks as we do for GPS clocks".
I tried to address (and thus eliminate from consideration) the latter in the second part of post #10.

DanMP said:
One way to do this is to calculate the total difference between the clocks in the situation where the moon is not rotating (is hovering) and compare that result with the normal result, with the moon orbiting the planet.
Why would there be any difference? If the moon in both cases is in the same place when the spaceship reaches it and turns around, what does it matter whether the moon is hovering or rotating?

Dale said:
So the OP wants the difference between a clock on the moon with the moon orbiting vs a clock on the moon with the moon suspended. A spreadsheet might get you either of those times with some reasonable precision. But I think you would need a pretty good numerical method to get the small difference the OP is asking about.
Sorry, I was getting confused by multiple scenarios, since he also mentioned a Hafele-Keating experiment. Those scenarios don't seem like a spreadsheet (iterative) calc.

I was thinking if the upcoming Artemis launch had an atomic clock on it on its trip around the moon it would be that type of experiment. A numerical accounting of constantly changing speed and altitude corrections.

Dale
DanMP said:
As far as I know, in a planet-moon system, the difference in elapsed time between a clock on the planet and a clock on its moon is calculated using GR/proper time, so it's not very obvious if/how the movement/velocity of the moon around the planet would influence the total time difference between the two clocks. Due to the differences in gravitational potential, the clock on the moon should be faster than the one on the planet, but the difference in velocity may reduce a little bit the difference in elapsed time.
If the bodies are tidally locked, you could use a rotating frame of reference, where both clocks are at rest. Then you only have the time dilation due to combined potential (gravitational + centrifugal).

PeterDonis said:
The OP has asked about a Hafele-Keating type experiment where one clock stays on Earth and the other clock goes to the Moon and then comes back.
I am not so sure that's an HK test was intended. The next sentence is "Or we may compare the clocks as we do for GPS clocks" as if that would be an equivalent formulation and OP has shown in their previous threads that they do not understand that the two problems are different. But perhaps we should ask OP if they intended to ask about an HK scenario or a comparison of clocks based on a simultaneity convention?

If the former was intended then ##T_{A0}## and ##T_{B0}## are the same event, and likewise ##T_{A1}## and ##T_{B1}## so only two relevant events. The requested prediction is still about the relationship between ##\Delta A## and ##\Delta B##, but at least we know the endpoints.

Klystron, Motore and russ_watters
Nugatory said:
I am not so sure that's an HK test was intended. The next sentence is "Or we may compare the clocks as we do for GPS clocks"
See my post #15, which in turn referenced my post #10.

Nugatory said:
But perhaps we should ask OP if they intended to ask about an HK scenario or a comparison of clocks based on a simultaneity convention?
Since the OP has not responded to hints, I think this is a good idea.

@DanMP, we need to focus the thread discussion on one scenario, and right now we have two: (1) a Moon trip analogue of the Hafele-Keating experiment, which does not require any simultaneity convention since the clocks meet up at the start and end, or (2) some kind of comparison between clocks at the Moon's altitude, perhaps one "hovering" and one orbiting with the Moon, and clocks on Earth, which does require a simultaneity convention since the clocks being compared never meet. Which one do you want to discuss?

The answers are all good, but I suspect that all this will do is cause the OP to add one more complication. It's like the overbalanced wheel - if I add one more thing, I'll get more energy out than in!

Clocks on the moon run slightly faster than clocks on the earth by about ##r_s/r## which is about ##10^{-9}##. This is an approximation because the moon is not an infinity, but is close. The moon is traveling at about 1000 m/s around the earth, so its clocks run slow by an amount too small to matter: ##6 \times10^{-12}##. Then the moon has its own gravitational time dilation which is 6x smaller because of its lower field and 4x bigger since you're that much closer to the center, and that gets you to a net effect of around ##3 \times 10^{-10}## which is 250 microseconds per day.

OK. so now what?

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Klystron, Motore, Ibix and 2 others
Thank you all for your answers.

Vanadium 50 said:
... The moon is traveling at about 1000 m/s around the earth, so its clocks run slow by an amount too small to matter: ##6 \times10^{-12}##. ...
Thank you! That was the main thing I was interested in, as you can see from:
DanMP said:
the difference in velocity may reduce a little bit the difference in elapsed time. In this thread I want to learn about the amount of this reduction, if any.
and also from post #3.You all agree about this estimation (6 x 10-12)? By the way, it is in seconds per day?

The total:
Vanadium 50 said:
net effect of around 3×10−10 which is 250 microseconds per day.
is different from what I found: clocks on the Moon run faster than their terrestrial equivalents – gaining around 56 microseconds or millionths of a second per day. Their exact rate depends on their position on the Moon, ticking differently on the lunar surface than from orbit.

PeterDonis said:
Since the OP has not responded to hints, I think this is a good idea.

@DanMP, we need to focus the thread discussion on one scenario, and right now we have two: (1) a Moon trip analogue of the Hafele-Keating experiment, which does not require any simultaneity convention since the clocks meet up at the start and end, or (2) some kind of comparison between clocks at the Moon's altitude, perhaps one "hovering" and one orbiting with the Moon, and clocks on Earth, which does require a simultaneity convention since the clocks being compared never meet. Which one do you want to discuss?
Scenario number 2 is not of interest for me. Anyway, scenarios for actually testing the GR predictions in the Moon-Earth case were the subject of another, closed, thread, so I'm probably not allowed to talk about experimental issues. For me any scenario is good, as long as the experiment is accurate enough to validate/invalidate the predictions about the total gain per day. For greater accuracy, more days on (or around) the Moon should be considered.

For this thread we can even neglect the travel time between Earth surface and Moon surface, considering it zero (Hafele-Keating thought experiment, if you want). Please just focus on the exact total gain per day and the contribution of the moon speed:
Vanadium 50 said:
The moon is traveling at about 1000 m/s around the earth, so its clocks run slow by an amount ...
##6 \times10^{-12}## is the final answer?

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DanMP said:
You all agree about this estimation (6 x 10-12)? By the way, it is in seconds per day?
I've not been paying a great deal of attention. But I would expect it to be a pure number. ##6 \times 10^{-12}## seconds per second. Or ##6 \times 10^{-12}## days per day.

Let us see if we can figure it out without bothering to ask @Vanadium 50. He had indicated that this number arose from a velocity of 1000 meters per second. Could it be nothing more than the ordinary time dilation factor from Special Relativity for a relative velocity of 1000 meters per second? Let us check that.

As you should know, the time dilation factor, aka the relativistic gamma (##\gamma##) is given by:$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$The given velocity (##v##) is 1000 m/s which is about ##3 \times 10^{-6}## of the speed of light.

The sqare of this ratio (##\frac{v^2}{c^2}##) then is ##1.11 \times 10^{-11}##.

If we subtract that from one and take the square root, we get ##0.99999999999444\dots##

If we take the inverse we get ##1.00000000000555\dots## (you may notice a useful approximation here if you are not wedded to your calculator).

If we count zeroes, this is approximately ##5.5 \times 10^{-12}## which rounds to ##6 \times 10^{-12}##.

So indeed we have our answer without having to attract the eye of Sauron @Vanadium 50.

And no, this is not the final answer. This is just a quick back of the envelope calculation to figure out the size of the "special relativity" effect from relative velocity -- how much does it matter whether we are considering an orbitting moon rather than a hovering moon? Answer: the difference amounts to about 0.5 microseconds per day.

What many of us consider interesting about physics is not specific numbers like "56 microseconds per day" or "250 microseconds per day". We are interested instead in the models, formulas and approximations that can go into calculating those numbers.

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russ_watters and Klystron
DanMP said:
For this thread we can even neglect the travel time between Earth surface and Moon surface, considering it zero (Hafele-Keating thought experiment, if you want).
So your intention is for the clock that goes to the Moon to stay on the Moon for a time that is very long compared to the Earth-Moon travel time? (For example, it stays on the Moon for a year, compared to three days of travel time each way.)

DanMP said:
You all agree about this estimation (6 x 10-12)? By the way, it is in seconds per day?
It seems like the best estimate without doing those heavy duty numerical computations I mentioned. It is unitless.

Say his name three times and he appears.

Just to be clear, my calculation did have some approximations, and there could easily be a factor of 2 dropped or picked up somewhere.

The procedure though is simple:
1. Figure out the gravitational effect on the Earth's clock relative to infinity.
2. Figure out the gravitational effect on the Moon's clock relative to infinity.
3. Combine - and there is about a 2/3 cancellation
4. Figure out the time dilation of the Moon's clocks relative to the Earth. This is smaller.
5. Ignore the Earth and the Moon's rotation - smaller still.
The two scales of the problem are ##r_S/r_\bigoplus## which is about ##10^{-9}## and ##(v_\leftmoon/c)^2## which is about ##10^{-11}## (and here there is definitely a factor of 2 - it's half that).

Vanadium 50 said:
there is about a 2/3 cancellation
What cancellation are you referring to here?

DanMP said:
what exactly the theory of relativity would predict
For this scenario, the weak field, slow motion approximation works, which simplifies the math.

The most convenient frame to use is an Earth-centered frame in which the Earth and Moon are stationary. This frame rotates with a period of one month (more precisely, one sidereal month, or 27.3 days) with respect to an Earth-centered inertial frame. In this frame, we can assume that all motion is along a radial line between the Earth and the Moon, i.e., angular coordinates are constant and the problem reduces to one-dimensional motion only.

With that assumption, the instantaneous time dilation factor for an object at radial coordinate ##r## (where the center of the Earth is ##r = 0##) moving at speed ##v## is

$$\sqrt{1 - \frac{2 G M_E}{c^2 r} - \frac{2 G M_M}{c^2 \left( R_M - r \right)} - \frac{\omega^2 r^2}{c^2} - \frac{v^2}{c^2}}$$

where ##M_E## is the mass of the Earth, ##M_M## is the mass of the Moon, ##R_M## is the ##r## coordinate of the center of the Moon, and ##\omega## is the angular frequency of rotation of the frame (which works out to be ##2.66 \times 10^{-6}##). This factor, btw, is also the ##g_{tt}## coefficient in the (weak field, slow motion approximation to the) metric in this frame, which is the only metric coefficient we will need to deal with.

Since all of the terms other than ##1## under the square root are much smaller than ##1##, we can approximate the above as

$$1 - \frac{G}{c^2} \left( \frac{M_E}{r} + \frac{M_M}{R_M - r} \right) - \frac{1}{2} \left( \frac{\omega^2 r^2 + v^2}{c^2} \right)$$

It should be evident what these terms represent: the Newtonian gravitational potentials due to the Earth and the Moon, the centrifugal potential due to the rotation of the frame, and the SR time dilation due to speed. The weak field, slow motion approximation is what allows us to linearly combine all of these effects to get a total time dilation factor (whereas in general in GR we can't do this because the full Einstein Field Equation is nonlinear).

This time dilation factor is relative to infinity, so comparison between two different locations means taking the ratio of the factors at the two locations. Calculating the total elapsed time for a particular clock traveling on a particular trajectory requires integrating the time dilation factor along the entire trajectory for the desired interval of coordinate time, plugging in the appropriate values of ##r## and ##v## at each instant of coordinate time. This integral will most likely need to be done numerically since you won't be able to find a closed form expression for it.

This is the same kind of analysis that Hafele and Keating did for their experiment; the only difference is that, AFAIK, they used an Earth centered inertial frame, so there was no ##\omega## term, and they did not take into account the gravitational potential of the Moon (which in the vicinity of the Earth is about an order of magitude smaller than the smallest effect Hafele and Keating took into account in their calculations).

jbriggs444 said:
the time dilation factor, aka the relativistic gamma
Actually, the time dilation factor relative to infinity is the reciprocal of gamma. It corresponds to the ##v^2 / c^2## term in the formulas I posted in post #28.

PeterDonis said:
So your intention is for the clock that goes to the Moon to stay on the Moon for a time that is very long compared to the Earth-Moon travel time? (For example, it stays on the Moon for a year, compared to three days of travel time each way.)
Do we need to transport clocks? Couldn't we just measure the frequency shift of signals?

Klystron and Vanadium 50
A.T. said:
Couldn't we just measure the frequency shift of signals?
That can give information about instantaneous relative time dilation factors, yes. (Even there, though, there are complications if the clocks are not at rest, since the time dilation factor of a given clock can change while the signals are traveling.) But translating that into differences in total elapsed time, for clocks that do not meet at the start and end of the experiment, still requires choosing a simultaneity convention.

PeterDonis said:
What cancellation are you referring to here?
If the earth and moon had the same mass and radius, the gravitational effects would cancel. They don't, so the cancellation is only partial: 2/3 or so.

Vanadium 50 said:
If the earth and moon had the same mass and radius, the gravitational effects would cancel.
If by "gravitational effects" you mean the potential, no, those do not cancel. Look at the formulas I gave in post #28. The effects of the potentials of both the Earth and the Moon have the same sign.

The "acceleration due to gravity" due to the Earth and Moon will cancel at a certain point between them, but that has no effect on time dilation, which is the subject of this thread.

PeterDonis said:
That can give information about instantaneous relative time dilation factors, yes. (Even there, though, there are complications if the clocks are not at rest, since the time dilation factor of a given clock can change while the signals are traveling.) But translating that into differences in total elapsed time, for clocks that do not meet at the start and end of the experiment, still requires choosing a simultaneity convention.
I think, for a tidally locked system in perfectly circular orbits you would have a constant frequency shift, which would correspond to difference in elapsed time over long periods of time (where the travel to meetings becomes negligible).

A.T. said:
I think, for a tidally locked system in perfectly circular orbits you would have a constant frequency shift
Not if you're talking about clocks in a Hafele-Keating type experiment where one clock goes to the Moon and then comes back. That clock, at least, will not have constant ##r## and ##v## (in the notation of my post #28), and the frequency shift of signals between it and the other clock (if we assume the other clock is at rest on the rotating Earth) will not be constant.

Even if you only consider the period of time when the Moon clock is at rest on the Moon, the frequency shift between that clock and a clock at rest on the rotating Earth will still not be constant, since the Earth's rotation period is much shorter than the Moon's (the latter is the rotation period of the frame itself). To get a constant frequency shift you would have to have the Earth clock "hovering" above the Earth and rotating about it with a period of one month (which of course is not a free-fall orbit, it would require rocket power) while the Moon clock is at rest on the Moon (and presumably at the point directly facing the Earth). (Even that ignores things like libration, but we are leaving those kinds of complications out here.)

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