# "Moon Ben" dilemma

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1. Aug 26, 2014

### Wes Tausend

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The other night my telescope buddy and I were discussing gravities vs time over a campfire and beverages.

I mentioned that a clock on the moon should run faster than an identical clock on earth because the gravity is weaker there. As an example, if we synchronized another rather large Big Ben clock with Big Ben in London, and placed it on the moon, then observed the huge clock-face through a rather powerful telescope from earth's surface, it should initially be observed to read about 1.3 seconds behind Big Ben on earth because of distant electromagnetic delay, that of C.

But since the gravity is weaker on the Moon, a moon-based clock should also run subtly faster there than on earth. So eventually "Moon Big Ben" time should be miraculously observed to "catch-up" to Big Ben on earth... and then surpass it.

My buddy pondered the situation for a moment, and surmised that if this was true, and it had been occurring for millions of years, then everything about the moon should be ahead of earth. Ostensibly, he suggested this would include a substancial gain in the periodic moon orbit. And the Moon should still be orbiting just a little too fast.

If this be true, is it similar to the "fast orbit" of Mercury?

Wes
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2. Aug 26, 2014

### ghwellsjr

Pendulum clocks are useless for your experiment. Try atomic clocks.

On earth, it's already been done. Atomic clocks at Greenwich near sea level run slower than identical ones at Boulder Colorado at about a mile elevation. This, of course, means that time is different for different parts of a planet or moon.

I don't think this has anything to do with the "fast orbit"of Mercury.

3. Aug 26, 2014

### A.T.

Gravitational time dilation depends on the potential, not the field strength. But there is also kinematic time dilation.

Yes, unless gravitational and kinematic time dilation cancel exactly the clocks will go out of synch.

4. Aug 26, 2014

### Simon Bridge

5. Aug 26, 2014

### Staff: Mentor

As A.T. pointed out, the aspect of gravity that affects "rate of time flow" is the potential, not the "force" (i.e., weight), and since the Moon moves relative to the Earth, you have to take that into account as well.

In fact, the gravitational potential can be split up into two pieces: the potential due to the Earth's field at the distance of the Moon, and the potential due to the Moon itself. (In general, you can't just add potentials this way, but the ones involved in this problem are so weak that you can get away with it.) The final formula for time dilation on the Moon, taking all the above effects into account, and using the formula $v = \sqrt{GM / r}$ to express the Moon's orbital velocity in terms of the Earth's mass and the Moon's distance from Earth, is then:

$$J_M = \sqrt{1 - \frac{3 G M_E}{c^2 R_{ME}} - \frac{2 G M_M}{c^2 R_M}}$$

where $M_E$ is the Earth's mass, $R_{ME}$ is the Moon's distance from Earth's center, $M_M$ is the Moon's mass, and $R_M$ is the Moon's radius. This is the "rate of time flow" on the surface of the Moon, as compared to a clock at infinity; but what we actually want to compare to is a clock on Earth's surface, for which we have

$$J_E = \sqrt{1 - \frac{2 G M_E}{c^2 R_E} - \frac{v_E^2}{c^2}}$$

where $R_E$ is the Earth's radius (equatorial radius) and $v_E$ is the velocity of rotation of the Earth (at the equator).

Plugging in numbers, we have $G = 6.67 \times 10^{-11}$, $c = 3.00 \times 10^{8}$, $M_E = 5.97 \times 10^{24}$, $R_E = 6.38 x 10^6$, $R_{ME} = 3.84 \times 10^8$, $M_M = 7.34 \times 10^{22}$, $R_M = 1.74 \times 10^6$, $v_E = 4.65 \times 10^2$, and we obtain

$$J_M = 1 - 4.85 \times 10^{-11}$$

$$J_E = 1 - 6.95 \times 10^{-10}$$

Since $J_M$ is closer to $1$ than $J_E$, clocks on the surface of the Moon do indeed run faster than clocks on the surface of the Earth. Note that we did the calculation for a clock at the Earth's equator, so we included a term for the rotational speed of the Earth at the equator; but we could have done the calculation using the Earth's polar radius instead of equatorial radius, and left out the velocity term, and we would have gotten the same answer (try it and see). This is because the Earth's surface (at least to a good approximation) is an equipotential surface, meaning the "rate of time flow" is the same everywhere on it.

6. Aug 28, 2014

### Wes Tausend

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Thank you all for your replies. I will be back after the holiday.

George, although a pendulum clock, such as Big Ben, may not operate correctly, should the effect really include both mechanical spring/flywheel escapement and atomic clocks?

I guess I was wrong about my assumption that Mercury had a fast orbit. If I now comprehend correctly, I do believe whatever GR is associated with Mercury still has to do with the nearby strong effect of solar gravity, since "corrected space/time, spacetime" is a main elemental difference between Newton and Einstein. So in that respect, Mercury should perhaps logically orbit too slow since it resides in the proximity of a huge solar gravitational field.

My simplistic reasoning for the preceeding would be... other than the size of clock face, what is the difference between our earthern viewing of the periodically rotating hands (tips) of a clock on the immense gravity on the sun, or that of the rotation of a nearby satellite (planet)? Or in our Moon case, an extremely low lunar craft orbit vs a large clock face on the lesser gravity of the lunar field?

Before Simon's links, I previously thought precession was solely related to the orbit speed, but it appears other factors intervene. So actually, when I think about it, Mercury should perhaps appear to orbit (clock) a bit slow which could still possibly affect the odd appearance of a "too fast a related precession". IOW, may the aforementioned time dilation effects be a roundabout way of explaining the more common Mercury precession explanation?
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"a little too fast" for what?" Good question, Simon.

I've sometimes wondered how we can determine the exact mass of another, or any, heavenly body since it appears to a layman that gravitational mass, and respective orbit, is more a calculated ratio and therefore a root of "guessed density" than otherwise measureable density. It seems like bootstrap logic to some degree.

I am supposing we have gone from measuring gravity on earth, to assuming all orbits are directly related timewise. We, and Newton, can therefore assign Mercury an assumed mass based upon it's orbit, and hope it is correct. So Mercury could only orbit too fast if it is less massive than Newton thought, yet suffers a time violation to temper the speed observed from earth. Or on the contrary, more massive than Newton thought, yet orbits too slow based on this proposed hidden mass because of gravitational time dilation... if that would in any way properly affect the observed speed of precession of course.
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PeterDonis,

Thanks for the in-depth math treatise. Qualities are usuallly easy to imagine; quanities not so much. I'm not sure I understand the difference between "potential" and a presumed "force".

Pondering the Moon clock dilema came about because I was trying to imagine if it were possible for time to simply be different in different parts of the universe. In particular I note that both the periodic rotation of spiral galaxies, and distant galaxies traveling in general linear direction, seem to share the common quality of "going too fast". This thought does seem unrelated to my original post however.

Wes
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7. Aug 28, 2014

### ghwellsjr

Yes, the effect includes all independent time pieces.

Wes, you don't seem to be aware that the orbital periods/speeds of objects (planet/moons/asteroids/comets/spacecraft) around our sun is independent of the mass of those objects. If this were not true, then our moon would separate from the earth, wouldn't it?

8. Sep 4, 2014

### Wes Tausend

George,

Whoops. A bit embarrassing, but a valuable lesson, so I thank you for correcting that neophyte misconception of mine. I don't recall seeing the independence of mass vs orbits directly referenced that way before, but I should have known. I could have surmised that from the observation that two very different masses do free-fall mostly at the same rate from the Leaning Tower of Pisa for example. Planets are, of course, also in free-fall.

Setting that error aside for the moment, a curious question remains. Observed from a weaker gravitational field (earth?), and seen from the direction of Polaris, if the hands of a solar-based clock face were large enough to extend just slightly beyond the sun, the outer tips of the hands perhaps reaching out to the orbit of Mercury, would not the obital path of Mercury roughly follow the sweep of such a "gravitationally slowed" clock and Mercury be observed to accumulate an orbit "too slow" for its apparent distance from solar center?

Although my present question may be unrelated, I do now realise that the common "Mercury" reference is to the ellipse of Mercury's orbit which is rotating (precessing) slightly faster than predicted by the traditional theory of Newtonian gravity.

Wes
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9. Sep 4, 2014

### ghwellsjr

A clock under consideration of relativity is always of insignificant dimensions so that its time can be associated with a single point. Any attempt to make a clock so large that its parts are influenced differently by the effects of Time Dilation (due to velocity or gravity) will result in a device that doesn't keep time with itself and cannot be considered a clock. As I said in post #2, try an atomic clock.

I also tried to make it clear that if you had millions of atomic clocks all over the face of the earth, they would all tick at different rates and none of them would tick at the defined time that we use on Earth for the millions of years that you are concerned about. This is because the earth's rotation rate has been slowing down which would effect the tick rate of all the clocks. I believe the moon's gravitational effects on the earth would also influence all the clocks on the earth.

The same thing is true for millions of atomic clocks spread out over the surface of the moon. They would all tick at different rates and not remain stable for millions of years. But all the clocks on the moon would tick faster than all the clocks on the earth.

Now if all the clocks on the surface of the moon tick at different rates and if you consider that atomic clocks embedded throughout the volume of the moon would tick at an even wider difference in rates, then what time are you considering to be the one that controls the moon's orbital period?

10. Sep 4, 2014

### pervect

Staff Emeritus
Something I think should be clarified:

While clocks placed randomly on the surface of the Earth would tick at different rates, it should but noted that there is a surface on the Earth at which clocks would all tick at the same rate. This reference surface is a gravitational equal potential surface at sea level, called the geoid.

See http://relativity.livingreviews.org/open?pubNo=lrr-2003-1&amp;page=articlese3.html [Broken]

The basis of the commonly used atomic coordinate time scale, TAI is to correct a large ensemble of clocks frequencies for their altitude above the geoid, then combine their times to define TAI time. (A synchronization convention is also required).

The coordinate TAI time will be equal to the "proper time" measure by a clock only on the geoid, at other heights the coordinate time will not be the same as the proper time.

It's not particularly strange that a clock on the moon would require detailed (but tiny) relativistic corrections as far as proper time vs coordinate time go. It's a little unclear to me what coordinate time would be best suited to the moon.

Last edited by a moderator: May 6, 2017
11. Sep 4, 2014

### Wes Tausend

That is true, George. Although considering the point size of atomic particles and the immense relative space between them, it seems even atoms must suffer some time discrepancy or be immune by law. A clock as large as a star would suffer severely bent hands I suppose. Sometimes it seems all our test scenarios are near approximations.

Very true. It seems there must exist some sort of average of all points.

Certainly, yes.

If time dilation affected orbit, a further problem would be that an actual point-to-point time mis-match might be supposed to create an accumulated tension in rigidity. The moon might then be seen as a smeared ring from earth until one arrived there, or at least have a stretched volume in the orbital plane. But it is not, and has not. Some other relavistic compensating mechanism may be at play of which I am unaware.

Thank you for your thought provoking input and especially my faulty orbit/mass correction, George.

Wes
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Thank you for the GPS TAI link and your insight, pervect. I am trying to visually understand the Relativities, though lacking a solid advanced mathematical background. It is no mystery my reach often exceeds my grasp.

Wes
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I am quite satisfied we have had an interesting exercise but it may be running out of practical usefulness and I have nothing to add. Thank you all for kindly sharing your comments.

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Last edited by a moderator: May 6, 2017