Do you age faster on the moon to your twin on earth?

[Alan McDougall]

Mass/gravity effects time, thus do we age faster on the moon than we do on earth?

 

Alan

[swansont]

Your gravitational potential is smaller on the moon, and this makes time run faster. But the moon is also moving with respect to us (and accelerating) so that makes the clocks run faster. You'd have to run the numbers to see which one dominates.

[NowThatWeKnow]

Using the GPS satellites to compare GR and SR corrections to clocks.

 

"...Special Relativity predicts that the on-board atomic clocks on the satellites should fall behind clocks on the ground by about 7 microseconds per day because of the slower ticking rate due to the time dilation effect of their relative motion ...A calculation using General Relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day..."

 

Not sure of the exact #'s concerning the moon but but Earth's gravity (vs way out in space) will slow clocks by .022 seconds a year.

[Martin]

Using the GPS satellites to compare GR and SR corrections to clocks.

 

"...Special Relativity predicts that the on-board atomic clocks on the satellites should fall behind clocks on the ground by about 7 microseconds per day because of the slower ticking rate due to the time dilation effect of their relative motion ...A calculation using General Relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day..."

...

 

NowThat, thanks for that very useful bit of information!

 

So in the case of GPS satellites, the GR gravitational potential well effect dominates the SR speed and acceleration effect.

 

One can infer that in the case of the moon it would be even more dominant. At least I think one can infer that!

 

Because the moon is moving much slower than the GPS satellite, so the SR kinematic effect of being on the moon would be less.

 

And the difference in gravity between standing on moon and standing on earth is greater (what is it? One sixth gee?) than the difference in potential between earth surface and up there where the GPS are. At least I think so. Swansont please correct me if I'm missing something. This is just a quick reaction.

Edited June 6, 2009 by Martin

[swansont]

NowThat, thanks for that very useful bit of information!

 

So in the case of GPS satellites, the GR gravitational potential well effect dominates the SR speed and acceleration effect.

 

One can infer that in the case of the moon it would be even more dominant. At least I think one can infer that!

 

Because the moon is moving much slower than the GPS satellite, so the SR kinematic effect of being on the moon would be less.

 

And the difference in gravity between standing on moon and standing on earth is greater (what is it? One sixth gee?) than the difference in potential between earth surface and up there where the GPS are. At least I think so. Swansont please correct me if I'm missing something. This is just a quick reaction.

 

One can infer that the gravitational term is small if the body has a negligible gravitational potential, as is true of an artificial satellite. But the moon's gravitational potential can't be ignored.

[Martin]

But the moon's gravitational potential can't be ignored.

 

Thanks for pointing that out. So if might be fun to do the numbers, but in a very back-of-envelope fashion.

 

Not sure of the exact #'s concerning the moon but but Earth's gravity (vs way out in space) will slow clocks by .022 seconds a year.

 

Let me see if I get the same answer you have here, using google calculator.

 

what is the number of seconds in a year? 3.15 x 10^7 I seem to recall.

 

So we have to multiply that by

 

G*mass of earth/(c^2*radius of earth)

 

So what we put into google is

 

3.15*10^7*G*mass of earth/(c^2*radius of earth)

 

Let's see what we get. WOW!!!

Google comes back with

(3.15 * (10^7) * G * mass of Earth) / ((c^2) * radius of Earth) = 0.021906788

 

That is about 0.022, which is what NowThat said!

 

So now we can do the same thing for the moon.

Google comes back with

(3.15 * (10^7) * G * mass of the Moon) / ((c^2) * radius of the moon) = 0.000990759556

That is about 0.001

 

So imagine we have a master clock a long ways out from both earth and moon, moving along with the earth-moon system but effectively "at infinity". And we put two clocks carefully down in the two gravity wells. At the end of a year, the clock on earth will have lost

0.022 seconds just as NowThat says.

The clock on the moon will have lost only 0.001 second.

 

So one can guess that the answer to the question is YES. YOU DO AGE FASTER ON THE MOON.

 

The difference is about 0.021 seconds per year.

 

Now you noticed that I didn't even bother with the special rel effect of the moon's speed, say relative to the center of mass of the earth-moon system. That is because I think of the moon as going too slow to bother with. But I could be wrong. So I'll check it. Yeah it's OK, because the moon is only going a few millionths of the speed of light. So when you do the time dilation beta is like 0.000004, and you square it and it is like 16 trillionths. That is small even in comparison with a small number like

G * mass of Earth) / ((c^2) * radius of Earth)

which is on the order of a billionth. So we can forget the speed correction. You do age faster on the moon, by roughly 0.02 seconds per year.

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Merged post follows:

Consecutive posts merged

I should thank Alan the thread-starter for getting this started. It's fun to see how it works out.

 

there are technical details implicit in the previous like this:

 

sqrt(1 + x)= 1 + (1/2)x - (1/8)x^2 +...

 

and time is slowed by a factor sqrt(1 + 2*potential/c^2) = sqrt(1 - 2*Gm/(c^2 R) ) which by that series expansion is approximately

1 - Gm/(c^2 R), which is what was used in previous post. The potential is already so small that higher power terms in the series are negligible.

 

Here m is the mass of the planet and R is the radius. One could ask why is 2Gm/(c^2 R) always less than one. It is because 2Gm/c^2 is the Schwarzschild black hole radius. R must always be larger (or else the body collapses to hole.) So dividing the Schw radius by R is always less than 1.

Therefore (1 - 2*Gm/(c^2 R)) is always positive and

sqrt(1 - 2*Gm/(c^2 R)) is well defined.

Edited June 7, 2009 by Martin