# Derivation of formula for time dilation

1. Mar 3, 2014

### wpan

I know how to derive the lorentz time dilation equation. I am wondering how to derive the equation for gravitational time dilation: T=To(1/(sqrt(1-(2GM)/(Rc2)))

2. Mar 3, 2014

### Bill_K

The way you have it written, the formula applies only to the Schwarzschild solution, where R is the Schwarzschild radius. In that case, the way to derive it is to write down the Schwarzschild metric, and look at g00. For a motionless test particle (dr = dθ = dφ = 0),

ds2 = (1 - 2GM/Rc2) dt2,

and from this, ds/dt gives you the time dilation.

3. Mar 3, 2014

### bcrowell

Staff Emeritus
I have a treatment in my SR book, http://www.lightandmatter.com/sr/ , in ch. 5. You don't need GR or the Schwarzschild metric.

4. Mar 3, 2014

### Bill_K

You do if the expression involves the Schwarzschild radius.

5. Mar 3, 2014

### bcrowell

Staff Emeritus
It can be expressed in terms of the gravitational potential.

6. Mar 4, 2014

### BruceW

I don't understand what you mean by "$R$ is the Schwarzchild radius". I would expect that $R$ is just the radial coordinate, in his notation. But I agree with the 'derivation'. Just assume Schwarzschild metric, and then choose a stationary test object outside the event horizon, then you get the time dilation formula. And this is often used as an approximation for the general relativistic equation for the time dilation due to planets, I think. So it's a good metric to use in most of the simple textbook problems on GR time dilation.

pretty cool. I always assumed that anyone that posts on physicsforums wouldn't have time to write a book along with doing their full-time job. hehe. you must have good time-management skills. In the discussion of gravitational potential in chapter 5, I'm not totally sure what you mean though. I understand
$$\frac{\Delta f}{f} \approx -g \Delta x$$
But then I don't understand how that implies the next equation:
$$\frac{\Delta f}{f} \approx - \Delta \Phi$$
Ah wait, duh, I get it now. I thought that this triangle was the Laplacian (and so the right-hand-side would always equal zero in free space). But it is a Delta symbol again. OK that makes sense. haha, sorry, I only realised it was meant to be a Delta symbol when I typed out the equation here, and I realised I wrote a Delta and thought "uh... of course it's a Delta, not a Laplacian".

Anyway, that's kinda interesting. Although I suppose it only works in the Newtonian limit. So in comparison to the Schwarzschild solution, we must choose a radius which is much greater than the event horizon. But in the Schwarzschild solution, we can choose any radius (except the radius of the event horizon, or r=0).

7. Mar 4, 2014

### bcrowell

Staff Emeritus
It's true in general, not just in the Newtonian limit, that $df/f=-d\Phi$. If you integrate both sides, you get $f\propto e^{-\Phi}$, which holds for any static spacetime, including the Schwarzschild spacetime.

8. Mar 4, 2014

### Bill_K

If R is some other radial coordinate and not the Schwarzschild radial coordinate, then the formula is false.

9. Mar 4, 2014

### dauto

Are you sure about that? I'm reading the formula as applying to any radius larger than the Schwarzschild radius. At the Schwarzschild radius the formula produces infinite red-shift.

10. Mar 4, 2014

### BruceW

I think Bill_K does mean "Schwarzschild radial coordinate" as meaning any radius larger than the Schwarzschild radius. I think he is saying "Schwarzschild radial coordinate" to distinguish it from any other choice of radial coordinate that we could make. (I don't know of any other choices though. But I'm sure you could make up some non-useful type of radial coordinate, maybe like the cylindrical radial coordinate).

11. Mar 4, 2014

### BruceW

cool. Is $\Phi$ one of the Hansen potentials? (the mass potential). I just read this on the wikipedia page for stationary spacetime. That's some pretty interesting stuff. So.. it seems like the Newtonian gravitational potential has a (perfect?) analogue in stationary spacetimes in GR. It's not often that a concept carries on so nicely from Newtonian gravity to GR.

12. Mar 4, 2014

### WannabeNewton

As long as you're in a space-time with a time-like Killing field $\xi^{\mu}$ you can always define a gravitational potential by $\Phi = \frac{1}{2}\ln (-\xi_{\mu}\xi^{\mu})$. It's clear why this makes sense physically as observers following orbits of $\xi^{\mu}$ are at rest in the gravitational field, or equivalently the asymptotic Lorentz frame if the space-time is asymptotically flat, and their 4-acceleration is simply $a^{\mu} = \nabla^{\mu}\Phi$ which is (the negative of) the gravitational acceleration of freely falling particles with respect to the stationary ("Copernican") frames defined by $\xi^{\mu}$. If $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} \neq 0$ then we can also write down Newtonian analogues of the centrifugal and Coriolis accelerations of freely falling particles relative to the stationary frames.

Last edited: Mar 4, 2014