# Gravitational time dilation due to one's own mass

1. Jun 21, 2006

### DiamondGeezer

I've been trying to work out something and I've hit a wall of stupid.

Imagine a clock a certain distance r from a large isolated spherically symmetric object of mass M. The rate at which the clock runs compared to the far away time is given by the Schwarzschild relation:

$$d \tau ^2 = \biggl (1- \frac{2M}{r} \biggr ) dt^2 - \frac{dr^2}{ \biggl (1- \frac{2M}{r} \biggr ) } - r^2 d \phi ^2$$

That's fine. But let's suppose that the clock has a finite mass $$m_c$$ (as clocks normally do). In that case, there will be a further time dilation due to that mass.

How do I calculate that extra dilation?

2. Jun 21, 2006

### meemoe_uk

There won't be a further time dilation.

Your title "Gravitational time dilation due to one's own mass", is slightly self inconsistent!
Time dialation is relative. You need at least two objects of reference. If you are interested in how the gravity of a single object affects it's own time perception, you have to at least consider different parts within the mass, model them as a collection of separate entitys, each with their own independant measures of time.
An object always percieves it's own measure of time is 'normal' speed, regardless of itself, or anything.

'Clocks' in relativity are usually assumed to be zero mass.

I'll try and give you the answers you want, but first I'll give you the question I think you wanted to ask!
Q: " Does the mass of an object effect the time dilation it experiences due to other masses?"
The point of rephrasing your question was to get rid of the self inducing time dilation aspect.
A: Yes.

How to work it out...
Consider a zero-mass clock at some point of a massive spherical body. We'll choose the surface, because you can use the fomula you stated. If the clock was at the centre of the body r would be zero in your formula and that would screw it up.
Work out the time dilation the clock experiences due to the MSB, and set this to be 'normal time perception for anything on the surface of the body'.
Natch, any beings on the surface will see the rest of the universe move around quicker than if they were in space.
Note that this will give you the 'extra dilation' you asked for in your post, if thats what you want.

Now work out the time dilation due to another massive body and the first body, and consider the ratio of TD due to both bodys against TD due to only the reference body.

If the reference body is big, and the other body is small, then TD due to extra body will be negliable, and the beings on the surface will not notice any difference to the speed at which the universe moves around.
If the reference body is small, and the other body is big, then TD due to extra body will be relativy large, and the beings will notice the universe seemingly speeding up.

As we see, increasing the mass of the reference body will increase it's own 'time dilation' with respect to the rest of the universe, but it will also reduces the TD effect of other bodys on itself, this is what I meant with my first statement.

Hope this helps.
A nutshell answer to your original question : model the mass of the clock to be a separate mass with a zero-mass clock nearby. Wish I'd just said that straight away.

extra comment:
For a mass to exhibit a substancial gravity force it needs to be quite big, at least the size of a large mountain, and that'll be around a milli-newton. A clock on the other hand is usually so small as to not exhibit a considerable gravity field, so you can model it to have zero mass.

Last edited: Jun 21, 2006
3. Jun 21, 2006

### DiamondGeezer

That's not an answer at all. The mass of the clock itself must alter the geometry of space time. The rate at which time flows is a function of the superposition of individual field contributions, so for example the rate at which time flows on the surface of the Earth relative to "far away time" is a combination of the effects of the Earth, moon and Sun to a first approximation.

Nope. No good for what I'm looking for.

4. Jun 21, 2006

### DiamondGeezer

5. Jun 21, 2006

### masudr

The field equations for gravitation (i.e. general relativity) are non-linear in the sources, unlike, e.g., Maxwell's equations which are. So an initial comment would be that your analysis of the effects at earth being due to the earth, sun and moon, is wrong.

6. Jun 21, 2006

### pervect

Staff Emeritus
If you try to apply the formulas to point masses at r=0, I forsee big problems.

I'm afraid that that's what you might be trying to do - it's not really clear as you haven't given us enough background.

With that warning in mind, let me proceed on.

The remarks about GR being nonlinear are correct in the most general case. However, in the solar system, gravitational fields are low enough that the weak field approximation will give the correct results.

So if you have two bodies, M1 and M2, you can calculate the newtonian potential due to both masses at some point P by the formula

$$\Phi = -U = -G M_1 / r_1 - G M_2 /r_2$$

M_1 is the mass of the first body and r_1 is the distance of the point from M_1, M_2 is the mass of the second body and r_2 is the distance of the point from it.

Note that U is positive.

Then the time dilation will be sqrt(1-2*U/c^2) which will be equal to 1-U/c^2 in the region where the approximation works well. (This will be a number less than 1 reflecting how much slower the clock at the specified point ticks than a clock at infinity).

Note that U/c^2 is dimensionless.

You will note that the formula blows up with a point mass if either r1 or r2=0. Don't use it there.

The formula will work for

point masses if you are not at the point mass
continuous distributions of masses, like a sphere of known size of known density

(There have been many posts about how to calculate the potential phi at the center of a sphere of uniform density. I think you'll find one in the general physics forum right at this moment, for instance.)

Note that the infinite self-energy of point masses isn't really physical. We know that electrons, for instance, have a finite observed mass, rather than the infinite mass that they would have if we took the infinite self-energy term seriously.

7. Jun 22, 2006

### DiamondGeezer

Pervect,

Are you saying that I cannot calculate the gravitational redshift of a signal from a spacecraft with respect to a mass-less receiver in flat spacetime?

Last edited: Jun 22, 2006
8. Jun 22, 2006

### pervect

Staff Emeritus
I'm not saying that at all.

I am saying that gravitational redshfit, to a high degree of accuracy, can be determined from the Newtonian gravitational potential U. (I gave the detailed formulas in my post).

This approach will work for anything you are likely to encounter in the solar system.

I don't know the details for your problem, so I don't know if the usual weak field approximations will be an issue or not an issue.

The Newtonian potential U diverges for point masses at r=0. I don't know if that will be a problem for your potential application or not. As long as your undefined application doesn't involve point masses with r=0 this won't be a problem.

I get the feeling that there's some sort of communication problem here - try reading what I wrote again, and if it still doesn't make sense, asks some more questions and give some more details on the problem.

9. Jun 22, 2006

### meemoe_uk

i.e. with the zero mass clock on the surface of your 'clock' mass.
You could at least give it a go. It wouldn't hurt to try.

Yep, and what better way to take this into account than by my great suggestion?