# Gravitational time dilation on a non-spherical object

1. Jun 28, 2010

### Dr Chaos

A radius is required for this relativistic formula:
$$t=\frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}$$
However, I notice that some objects are not spherical. Because of this, I am unsure of how to calculate gravitational time dilation for different shapes.

Could somebody help me with this and provide examples for different shapes (like a brick or a heavy stop sign)?

2. Jun 28, 2010

### FunkyDwarf

You would probably need to generate the energy momentum tensor for your object, plug through the field equations to try get a metric and go from there.

3. Jun 28, 2010

### Passionflower

I am not sure what you are trying to do here. Calculating the gravitational time dilation for a body is very difficult and depends on more things than just the spatial dimensions of the body.

4. Jun 29, 2010

### Dr Chaos

What I am looking for is the gravitational time dilation on the surface of a stationary body. Assume the body has uniform density. What other information is required?

5. Jun 29, 2010

### Jonathan Scott

The fractional time dilation is the the same as the Newtonian gravitational potential (potential energy per unit mass) divided by c2.

Provided that the body in question is not as dense as a neutron star, you can just calculate the Newtonian potential by integrating the density divided by the distance from each point of the body in the usual way. For a non-spherical body, the potential normally varies with location on the surface. For a spherical body, it is the usual -Gm/rc2.

6. Jun 29, 2010

### Passionflower

Actually the second part of your posting denies this.

I would rather state:

In general relativity the fractional time dilation is not the same as the Newtonian gravitational potential (potential energy per unit mass) divided by c2. However in the weak field limit it is true because here general relativity becomes equivalent with Newtonian gravity.

I think it is very interesting to study the Weyl form to see the "function" of the Newtonian potential in general relativity. (function clearly between quotes).

7. Jun 29, 2010

### D H

Staff Emeritus
Given your assumption in post #4 ("Assume the body has uniform density"), the weak field limit most definitely applies. The only objects that can have a uniform density must necessarily be small and have negligible mass. A brick, for example. Even planets don't have uniform density, let alone stars or neutron stars. So, for an object of uniform density but of a non-spherical shape, calculating the Newtonian gravitational potential on the surface of the object will work just fine.

8. Jun 29, 2010

### George Jones

Staff Emeritus
There exists a spherically symmetric, constant-density exact solution to Einstein's equation in general relativity that does not have to have negligible mass, and for which the weak-field limit does not necessarily apply.

9. Jun 29, 2010

### D H

Staff Emeritus
You're apparently talking about singularities, George. A black hole will have a fairly simple geometric shape. Is there a brick-shaped solution to Einstein's field equations for black holes?

10. Jun 29, 2010

### George Jones

Staff Emeritus
No, I wasn't talking about black holes or singularities or bricks. Schwarzschild black holes are vacuum solutions to Einstein's equation, and have zero (local) density everywhere, even inside event horizons. I gave an explicit counter-example in general relativity (but not the real world) to the general statements

constant density ==> weak limit applies,

General relativity allows spherical material objects which have non-zero constant-density, and for which the weak field limit does not apply.

Certainly, the weak field limit applies to a real brick.

11. Jun 29, 2010

### TCS

I think that you could model each shell of an atom as a constant density spehereically symmetric solution to Einstein's equation for an atom. although, I guess that electrons would create density variations in the shell and not all atoms spherical.

Last edited: Jun 29, 2010
12. Jun 29, 2010

### Dr Chaos

There seems to be some discrepancy in regard to how this is calculated; could somebody please provide an example for me?

13. Jun 29, 2010

### TCS

You can do it numericly. Create a mesh, assume some linear approximations within each cell of the mesh, solve within the cell, and match the conditions at the boundaries of the cells. If your cells are small enough, you should iterate to a solution.

The book at the following link provides a discusion of the common numerical methods.

This paper is better. the other link isn't the full book.

http://www.tesisenxarxa.net/TESIS_UIB/AVAILABLE/TDX-0923109-130054//tdda1de1.pdf

Last edited: Jun 29, 2010
14. Jun 29, 2010

### Jonathan Scott

As I said before, there's no point in using anything other than a Newtonian approximation for any ordinary object. The Newtonian potential at a reference point due to an object is the volume integral over the object of -G times the mass density divided by the distance from the reference point, but I haven't got the patience to set it in LaTeX right now.

In my earlier post, I intended to be exact, but my terminology may have been non-standard. I meant that "effective Newtonian potential" in a static field (that is, the potential energy per unit mass) divided by c2 is the same thing as the fractional difference in time rate compared with infinity. However, the calculation of that potential only reduces to the usual Newtonian expression of sum(-Gm/r) in the weak approximation.

Last edited: Jun 29, 2010
15. Jul 10, 2010

### Dr Chaos

I'm sorry but I still do not understand. Please forgive my limited physics knowledge but I would still like an example.

16. Jul 10, 2010

### TCS

Why don't you draw the biggest sphere that fits inside, calculate the dialtion for that sphere, then inscribe spheres in each the remaining sections and calculate and correct, and repeat.

17. Jul 11, 2010

### Dr Chaos

That would probably work but it seems very tedious - I would have to keep drawing spheres until I decide that they are small enough, and then calculate each individual sphere's mass and then their time dilation and then multiply it all together. Is there an easier way?

18. Jul 11, 2010

### Jonathan Scott

This is just Newtonian gravity and integration.

If the point where the potential is being measured is the origin, then the potential is given by the following integral over the object:

phi = triple integral of (- G rho/sqrt(x^2+y^2+z^2)) dx dy dz

where rho is the mass per unit volume.

The time dilation relative to infinity is then

(1 + phi/c^2)

I tried entering this using LaTeX, but I couldn't get it to work at all today.

19. Jul 13, 2010

### Dr Chaos

Could you please post the code? I can always use a LaTeX sandbox to view it properly.

$$\phi=\iiint\frac{-G\rho}{\sqrt{x^2+y^2+z^2}}$$ <== is this it?

And what is the full time dilation formula?

20. Jul 13, 2010

### espen180

There is none.

The way the gravitational time dilation is obtained is to solve Einstein's Equation (google it) and obtain a metric, from which it can be obtained. If I were to give a "general formula", it would have to be $$\tau=\frac{\sqrt{g_{tt}}}{c}t$$ where $$\tau$$ is proper time (measured locally) and $$t$$ is coordinate time (measured from infinity).