# Actual clock rate difference between surface of Earth and surface of Moon

1. Jun 9, 2012

### arindamsinha

Is there any way in relativity to mathematically compute what the rate difference would be between two identical clocks, one on the surface of the Earth and one on the surface of the Moon? The points on the surfaces to be considered are on the line joining the individual centers of gravity (i.e. closest points between the two bodies), and ignoring the Earth's small rotational velocity.

(Assuming both bodies to be perfect solid spheres of uniform density, and in an isolated two-body gravitational system where external gravitational influences are negligible, e.g. as done for the computation of GPS time dilation.)

Q1) How much is the relative velocity time dilation? (This is the key question, as there seems to be no clear answer in the relativity equations) It will be very small compared to the gravitational time dilation, but it must still exist. How much is it for either clock?

Q2) The gravitational time dilation from the individual bodies can be worked out at each point under consideration, but
- Do we add or subtract these individual time dilations to get the net effect on an actual clock at either point? Add probably?

2. Jun 9, 2012

### tiny-tim

hi arindamsinha!
the velocity and gravitational time dilation are completely separate

use the standard formulas

the gravitational time dilation is a factor (√(-goo)) measured relative to "a point at infinity" …

divide by one factor to get from earth to infinity, multiply by the other factor to get from infinity to the moon

3. Jun 9, 2012

4. Jun 9, 2012

### arindamsinha

Thanks for the responses tiny-tim and Simon Bridge.

I was thinking more in terms of the Schwarzschild solution, where both gravitation and velocity time dilation are accounted for. I believe both the gravitational and velocity time dilations are captured in that (SR being an integral part of GR), looking at the instantaneous distance and velocity.

Does the factor (√(-goo)) account for both? I mean to say, as the moon is accelerating, the velocity time dilation factor may be modified, but doesn't disappear, right? In fact, the instantaneous velocity is a fair approximation to inertial motion - providing the v2 factor in 2GM/Rc2+v2/c2 formulation of the Schwarzschild solution. In GR, the time dilation factor of an orbiting body comes out as 1+3GM/Rc2 (or equivalently 1+2GM/Rc2+v2/c2)?

Or am I mistaking something?

5. Jun 9, 2012

### Staff: Mentor

Not in the sense of proper acceleration. The Moon is weightless; it feels no acceleration.

6. Jun 9, 2012

### Staff: Mentor

No. The formula you are talking about only includes gravitational time dilation due to altitude; it does not include the effects of velocity.

The Moon is not "accelerating" in any invariant sense; see my reply to Simon Bridge's post. You could calculate sqrt(-g_00) for the Earth's field at the altitude of the Moon, but that would not include the effects of the Moon's velocity on "rate of time flow". It would also not include the effects of the Moon's gravity; see below.

There should be minus signs instead of plus signs there, but yes, that formula is correct for a body in a free-fall orbit. However, the quantity 1 - 3GM/Rc^2 is *not* -g_00; it is not read directly off the Schwarzschild metric.

To fully answer the question in the OP, a number of things need to be taken into account:

(1) There is no exact solution for two gravitating bodies orbiting each other, such as the Earth and the Moon, in GR. So we have to approximate.

(2) Gravity is weak enough for both the Earth and the Moon that we can approximate the field at a given point by just adding together the Earth's effect and the Moon's effect. "Adding together" in this case means adding potentials, which means adding terms under the square root sign in the formulas we will give. (In general this doesn't work because gravity is nonlinear; two solutions do not add together to form another solution. But the nonlinear terms in this scenario are too small to matter. Also, we are actually multiplying time dilation factors, where each factor is sqrt(1 + 2phi/c^2), where phi is a potential; but the potentials are so small that we can just add the phi terms under the square root sign, which again amounts to ignoring nonlinear terms. Also each phi is negative, since the potential is normalized to be zero at infinity.)

(3) Since the Earth has a much greater mass, it is easier to use its field as the "primary" field and to ignore the effects of the Moon's field on objects close to the Earth.

(4) Since the Moon is rotating very slowly, compared to the Earth, we can ignore any effects of the Moon's rotational velocity on rate of time flow. But I would not recommend ignoring the effects of the Earth's rotation; its rotational velocity is large enough to have an appreciable (though not a large) effect on the answer.

(5) Since the Moon is orbiting the Earth, we can use the formula for an orbiting object to get Earth's field at the Moon. But an object on the surface of the Earth is not in orbit, so we have to include the velocity term separately since it's much less than the orbital velocity.

Putting all of the above together, we get the following:

$$\frac{d\tau_{Earth}}{dt} = \sqrt{1 - \frac{2 G M_{E}}{R_{E} c^{2}} - \frac{v_{E}^{2}}{c^{2}}}$$

$$\frac{d\tau_{Moon}}{dt} = \sqrt{1 - \frac{3 G M_{E}}{R_{EM} c^{2}} - \frac{2 G M_{M}}{R_{M} c^{2}}}$$

where $M_{E}$, $M_{M}$ are the masses of the Earth and Moon; $R_{E}$, $R_{M}$ are the radii of same; $R_{EM}$ is the distance from the Earth to the Moon; and $v_{E}$ is the velocity of an object at the equator on Earth's surface, due to the Earth's rotation.

Both of these numbers are time dilation "relative to infinity", so the ratio of the two gives the relative time dilation between the two points.

Last edited: Jun 9, 2012
7. Jun 9, 2012

### arindamsinha

I was using the + sign as a first order approximation, but you are right, in that case I should have divided the other terms after 1 by a factor of 2.

I was intrigued by your statement "There is no exact solution for two gravitating bodies orbiting ... in GR". Is this a limitation of GR theory itself, or is the mathematics too complicated to solve for this situation exactly?

8. Jun 9, 2012

### arindamsinha

PeterDonis - a follow up question. The equation you have given for the Moon would be exactly that of a GPS satellite, putting MM ≈ 0.

However, I think we cannot extend this further to a case where the Moon has a mass similar to Earth (i.e. they become binary planets). Based on the equations given, we would still have an actual time dilation between Earth and Moon in such a case, while looking at the symmetry there should be no relative time dilation (in case they are equal masses)?

So, is there a missing component somewhere, which would affect the maths as we scale up a satellite from the size of a GPS satellite to the size of the Moon and then to something the same size as Earth?

I realize we can make the equations exactly symmetric when the satellite is exactly the size of Earth, but what if say the satellite is 9/10th the size of Earth, and therefore its gravity cannot be ignored?

9. Jun 9, 2012

### Staff: Mentor

And changing the radius appropriately, yes.

Sure we can, at least to a good enough approximation for this case. The added term in $M_{M}$ is what accounts for the "mass of the satellite". The Moon is being treated here as just a large satellite with enough mass to produce an appreciable time dilation for nearby objects on its own, in addition to the time dilation resulting from the Earth's field. This is an approximation, but it works fine in this case. See further comments below, though.

No *relative* time dilation; but the equations I gave, by themselves, aren't equations for relative time dilation. They're equations for time dilation compared to "infinity". You have to take the ratio of the two numbers to get the relative time dilation between two points; if the parameters are such that both equations give the same number, then the ratio will be 1 and there will be no relative time dilation between those two points (though both will still see time dilation compared to an observer at infinity).

If the Moon's mass were large enough compared to the Earth's mass, we would have to include the effects of its field on the "Earth" object just as we included the effects of the Earth's field on the "Moon" object. That would change the formulas some, yes; the key changes I can see would be:

(1) We would have to separate out velocity terms in both formulas, since we would not be able to consider either massive body as a "small" object orbiting in the other's field, so we could not just use the shortcut formula that we used for the Moon (the 3GM/Rc^2 term instead of the 2GM/Rc^2 term). We would have to calculate each massive body's velocity due to its orbital motion about the common center of mass, and include it explicitly.

(2) We would have to include 2GM/Rc^2 terms for both massive bodies in both formulas. In the formulas I posted we did not include a term for the Moon's field in the "Earth" object's time dilation because it was too small; we would have to include that if the Moon's mass were large enough compared to the Earth's.

So we would have, for any object in the region around both bodies, something like this:

$$\frac{d\tau}{dt} = \sqrt{1 - \frac{2 G M_{1}}{R_{1} c^{2}} - \frac{2 G M_{2}}{R_{2} c^{2}} - \frac{v_{total}^{2}}{c^{2}}}$$

where M1 is the mass of body 1, M2 is the mass of body 2, R1 is the distance of the object from the center of body 1, R2 is the distance of the object from the center of body 2, and v_total is the total velocity of the object, including its velocity due to the rotation of the body it's on (if it's sitting at rest on the body's surface) or due to its orbit about that body (if it's orbiting it like, say, a GPS satellite), plus its velocity due to the orbital motion of that body around the common center of mass of both bodies (the two would add vectorially).

In fact, the formula I gave for the Moon could be restated in exactly this form, by separating out the orbital velocity as a v^2/c^2 term instead of including it in the 3GM/Rc^2 term. We were ignoring the (too small to matter) velocity due to the Moon's rotation, so that orbital velocity would just be v_total.

10. Jun 9, 2012

### arindamsinha

Great, thanks a lot.