Time dilation in a planet-moon system

[PeterDonis]

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.

 

[Vanadium 50]

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.

 

[PeterDonis]

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.

 

[A.T.]

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).

 

[PeterDonis]

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.)

 

[A.T.]

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.

I'm talking about a tidally locked two body system with perfectly circular orbits, where you continuously send signals from a fixed point on one surface to a fixed point on the other surface. The frequency shift would be constant here.

 

[PeterDonis]

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.

Actually, I need to correct this. The $g_{tt}$ coefficient will not include the $v^2 / c^2$ term, since that is not a property of the geometry or the choice of coordinates. That term comes from the $\sqrt{1 - v^2 / c^2}$ SR time dilation. The fact that all the corrections are small compared to $1$ is what allows the $v^2 / c^2$ term to come under the square root in the first formula.

 

[Vanadium 50]

no, those do not cancel

If you have a planet of radius R and mass M orbiting another planet of radius R and mass M (technically, the berycenter) the clocks run at the same rate. The cancellation is perfect. If the bodies have different masses and/or radii, the cancellation is only approximate. In the Earth-Moon case, it's abour 2/3.

 

[PeterDonis]

If you have a planet of radius R and mass M orbiting another planet of radius R and mass M (technically, the berycenter) the clocks run at the same rate.

Relative to each other, yes.

Relative to infinity, their clock rate, assuming we are still within the domain of validity of the weak field, slow motion approximation, will be the combined effect of both of their potentials, plus the effect of their orbital speed (which will appear either as a $v^2 / c^2$ term in a barycentric inertial frame, or an $\omega^2 r^2 / c^2$ term in a barycentric rotating frame). In other words, their common time dilation factor relative to infinity will look like the formula I posted in post #28.

 

[PeterDonis]

I'm talking about a tidally locked two body system with perfectly circular orbits, where you continuously send signals from a fixed point on one surface to a fixed point on the other surface. The frequency shift would be constant here.

As you describe it here, it wouldn't in the Earth-Moon case at present (see the second paragraph of the post of mine that you quoted). If the Earth were fully tidally locked to the Moon (i.e., its rotation had slowed to the same period as the Moon's orbit), then yes.

 

[A.T.]

If the Earth were fully tidally locked to the Moon

Yes, that's what I meant. Or just sent the signal from the pole.

 

[Gleb1964]

The total:

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.

May I ask about bold selected quote.
Why would be clock's rate depending on their position on the Moon? Isn't the Moon surface equipotential where all the clocks beat at the same rate?

 

[jbriggs444]

May I ask about bold selected quote.
Why would be clock's rate depending on their position on the Moon? Isn't the Moon surface equipotential where all the clocks beat at the same rate?

The "ticking differently on the lunar surface than from orbit" seems to be key. We are not talking about clocks at rest on an equipotential surface.

 

[DanMP]

Relative to infinity, their clock rate, assuming we are still within the domain of validity of the weak field, slow motion approximation, will be the combined effect of both of their potentials, plus the effect of their orbital speed (which will appear either as a v2/c2 term in a barycentric inertial frame, or an ω2r2/c2 term in a barycentric rotating frame). In other words, their common time dilation factor relative to infinity will look like the formula I posted in post #28.

Thank you for your answers!

So, we can choose, for the Earth-Moon system, or any planet-moon system, the barycentric inertial frame. In this frame, the planet and its moon would have, usually, different orbital speeds, meaning that their clock rate relative to infinity would be affected by speed differently. That's what I wrote:

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.

Now,

in the situation where the moon is not rotating (is hovering)

, relative to infinity, their clock rate, assuming we are still within the domain of validity of the weak field, slow motion approximation, will be only the effect of their potentials.

So, according to GR, the moon orbital speed would influence the total difference per day between planet/moon clocks, and this influence can be calculated.Regarding the experimental test:

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.)

Yes, something like that. But I also suggest to send signals between clocks, all this time, in order to monitor the progress. I would send one signal from the Moon clock at every 12 hours and record the exact time of arrival at the Earth clock/clocks. If, from the Earth clock, a reply signal is sent instantly, we may approximate the signal travel time and make a prety good progress chart.

 

[jbriggs444]

Yes, something like that. But I also suggest to send signals between clocks, all this time, in order to monitor the progress. I would send one signal from the Moon clock at every 12 hours and record the exact time of arrival at the Earth clock/clocks. If, from the Earth clock, a reply signal is sent instantly, we may approximate the signal travel time and make a prety good progress chart.

So in your mind's eye, you have a Earth tidally locked to the moon and vice versa. You assume approximate long term stability, ignoring any gravitational radiation that may be slowing the assembly down over the very long term. You exchange time-stamped radio signals periodically and observe a drift between the local clock and the received remote clock readings. Your peer at the remote clock sees an inverse drift. This, even before the drift reaches a point where the clocks are out of synchronization by more than one round trip time.

i.e. you've invented NTP.

 

[PeterDonis]

I also suggest to send signals between clocks, all this time, in order to monitor the progress

Doing this requires adopting a simultaneity convention. But there is no common simultaneity convention that is "natural" for both clocks, since they are in relative motion. And the obvious simplest simultaneity convention to adopt, that of the barycentric inertial frame, is not "natural" for either clock, since both clocks are moving in this frame.

 

[DanMP]

Doing this requires adopting a simultaneity convention. But ...

Ok, it's not easy, nor accurate, but this signals exchange may still be useful to monitor the divergence of the clocks. The final, more accurate, result would be available when the moon clock returns.

With Lunar Laser Ranging experiment we were able to get a lot of information by bouncing laser light pulses on retroreflectors installed on the Moon, so I wont dismiss completely the idea to send signals between clocks, as described in post #44.

Anyway, in 10 years, when lunar GPS clocks will be on the Moon, we'll probably see if/how the different passage of time between Earth and Moon can be measured.

 

[Vanadium 50]

What does seeing whether "passage of time differs between the Earth and the moon" is more interesting than the same measurement between Singapore and Barrow, Alaska? Or the Earth and (near) the Sun with Cassini?

The moon is just not going that fast - between the speed of a commercial and military jet.

 

[PeterDonis]

when lunar GPS clocks will be on the Moon

They won't be using the same simultaneity convention as the Earth-based GPS system does. And their rates will be adjusted from their "natural" rates just as Earth-based GPS satellite clock rates are--but adjusted to a different standard.

 

[DanMP]

What does seeing whether "passage of time differs between the Earth and the moon" is more interesting than the same measurement between Singapore and Barrow, Alaska? Or the Earth and (near) the Sun with Cassini?

I'm interested in much more than this particular subject (as you can see from my threads/posts), even in other time measurement scenarios, but now/here I'm interested in this planet-moon scenario.

If you want to talk about

the same measurement between Singapore and Barrow, Alaska or the Earth and (near) the Sun with Cassini

you can open dedicated threads. I don't mind.

In fact, "the Earth and (near) the Sun with Cassini" scenario reminded me about my older thread Time dilation for the Earth's orbit around the Sun. Also, "Barrow, Alaska" reminds me about another interesting scenario (for me), with clocks on Earth's poles. It was done? Anyway, any test may be interesting.

The moon is just not going that fast - between the speed of a commercial and military jet.

It is enough, as Hafele-Keating experiment proved.

their rates will be adjusted from their "natural" rates just as Earth-based GPS satellite clock rates are--but adjusted to a different standard.

Probably, but even so, if there is a discrepancy between theory and reality, the clocks on the Moon would diverge from the clocks on the Earth more than expected. Over time this divergence would increase/accumulate and we would notice it, eventually.

 

[Dale]

if there is a discrepancy between theory and reality, the clocks on the Moon would diverge from the clocks on the Earth more than expected

Why would the discrepancy between theory and reality be more noticeable with clocks on the moon than with different pairs of clocks on the earth? That makes no sense whatsoever.

One way to see that this makes no sense is to try to figure out which post-Newtonian parameter such an experiment would be sensitive to. Then you can look for the existing experiments that already constrain that parameter.

 

[PeterDonis]

if there is a discrepancy between theory and reality, the clocks on the Moon would diverge from the clocks on the Earth more than expected

If there were such a divergence we would already have seen it, in errors in the GPS system, for example. As has already been pointed out, the Moon's orbital speed is just not that fast, comparatively speaking.

 

[PeterDonis]

It is enough, as Hafele-Keating experiment proved.

Yes, which means that experiments we have already done are sufficient to test for any kind of divergence between theory and reality that the experiments you are proposing would test for. So the experiments you are proposing do not push the boundaries of our testing of theory against reality at all.

 

[Vanadium 50]

(as you can see from my threads/posts

Yes.

I would encourage you to drop this anti-relativity spin until you understand it better. I would also encourage you to work through a textbook (e.g. Taylor and Wheeler, 1st edition) as it will help your understanding more than a parade of increasingly complex scenarios. And, most importantly, I would encourage you to look at which experiments measure the same thing and how large the possible deviations from SR can be - otherwise it's just "yes, but has it ever been measured in Fresno on a Tuesday?"

You will discover that either test bodies are large and the effects are small or the other way around. That's just another way of saying we don't have starships.

 

[DanMP]

Why would the discrepancy between theory and reality be ...

What discrepancy? I wrote "if" bolded (if):

Probably, but even so, if there is a discrepancy between theory and reality, the clocks on the Moon would diverge from the clocks on the Earth more than expected. Over time this divergence would increase/accumulate and we would notice it, eventually.

and I wrote the above in order to point that even when the moon-based clock

rates will be adjusted from their "natural" rates just as Earth-based GPS satellite clock rates are--but adjusted to a different standard.

we still can use them to prove/disprove the theory.

I should have added that, as for the Earth-based GPS satellite clocks (with their adjusted rates), we still can use future adjusted Moon GPS clocks to proudly say that our theory of relativity was, yet again, confirmed.

It is better now?

 

[jbriggs444]

What discrepancy? I wrote "if" bolded (if):

You seem to be missing the point by a considerable margin.

If one is going to perform a test of relativity and has a test in mind, it would be efficient to spend some time thinking about what aspect of relativity that test measures. It would then be efficient to consider the possibility of other less expensive or more accurate tests.

we still can use them to prove/disprove the theory.

In terms of proving the theory, that ship has sailed. The theory is well accepted. The tick rate of clocks in the neighborhood of the moon is expected to follow the predictions of general relativity.

The information content of a confirming experiment would be essentially nil. The expected information content of an experiment (before knowing whether it is confirming or not) would also be essentially nil.

If you want to maximize the expected information yield of your experiments per unit cost, you want experiments where the outcomes are more like 50:50 rather than 1:bazillion.

 

[PeterDonis]

we still can use them to prove/disprove the theory.

Prove in the sense of providing yet more confirming data to add to the mountain of confirming data we already have, yes. But not in the sense of pushing the boundaries of the domain in which we've confirmed the theory; as I've already said, your proposed experiments do not do that at all.

Disprove? No. Not going to happen, because if any of the experiments you propose could produce data that conflicted with the theory, we would already have seen the same conflict in experiments we have already run. That's why I said your proposed experiments do not push the boundaries at all: because we already know what results your proposed experiments will have to give. In a "pushing the boundaries" experiment, we wouldn't, because we would be experimenting in a domain we had not experimentally probed before.

 

[Vanadium 50]

"None of our tests disprove SR"
"So hey - let;s run a less sensitive test!"

 

[jedishrfu]

It seems this is a good time to close this thread. The OP's question has been debated and now we are just rehashing our words.

Thank you all for contributing here.

Jedi