# Gamma in GR

1. Apr 20, 2008

### snoopies622

Does the Lorentz factor show up much (or at all) in general relativity? For instance, is it appropriate to use it as part of the four-momentum vector when finding the components of the stress-energy tensor even in a non-flat spacetime? I was wondering because the derivations of it that I've seen always assume either a Euclidean or Minkowskian metric.

2. Apr 20, 2008

### lbrits

$$\gamma$$ is the ratio of (Lorentz) coordinate time to proper time, and so it only shows up when you are using local Lorentz coordinates. In other words, it almost never shows up in GR.

Although you are right that it shows up in the derivation of the SE tensor for a fluid, since there you want to convert the rest frame picture into a general frame picture.

3. Apr 20, 2008

### snoopies622

I'm a little confused (what else is new?): if there is fluid, the spacetime isn't flat. Why then is it valid to use the Lorentz factor at all?

4. Apr 20, 2008

### lbrits

Well, there is a local inertial frame in which the metric is flat around one point. Alternatively, you can always look at the components of the tensor in an orthonormal frame.

MTW has a good discussion.

5. Apr 20, 2008

### snoopies622

So, are you saying that since spacetime is a manifold, if we take a small enough section of it, we can treat it as flat, even in the presence of mass-energy? If so, then I think I understand.

6. Apr 20, 2008

### lbrits

Yes. That is correct. A very useful thing to realize in GR.

Here's a resource you might find useful.
http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec12.html [Broken]

Last edited by a moderator: May 3, 2017
7. Apr 20, 2008

### snoopies622

Thanks, Ibrits (or is it lbrits?). I'll check out that 'Gravitation' book, too.

8. Apr 21, 2008

### off-diagonal

I think its called "local cartesian coordinate"

9. Apr 21, 2008

### shalayka

From what I understand, gamma does play a direct role in general relativity.

Both gravitational and kinematic time dilation contribute toward the relativistic precession of an orbiting mass (ex: Mercury around the Sun), where kinematic time dilation is what arises from special relativity.

In terms of "orbits per orbit" of relativistic precession, where r is the average orbit radius (semi-major axis), a common formula is:

$$\delta = \frac{3GM}{rc^2}$$

Two-thirds of this is from gravitational time dilation (due to the gravitational field of the source mass), and one-third from kinematic time dilation (due to the current velocity of the orbiting mass in relation to the source mass).

The value $$\delta 2\pi$$ gives the relativistic precession in terms of radians per orbit.

Last edited: Apr 21, 2008
10. Apr 21, 2008

### snoopies622

So in that case the kinematic time dilation would involve the Lorentz factor, even though the metric is non-Minkowskian. Hmm.

11. Apr 21, 2008

### shalayka

Yep. I believe that's a fair summary.

One other point I left out was that when a mass is radially in-falling (no transverse acceleration), kinematic and gravitational time dilation are actually the same thing.

Last edited: Apr 21, 2008
12. Apr 22, 2008

### snoopies622

Thanks, shalayka. I will look further into this. That 3GM/(rc^2) term looks familiar.

13. Apr 22, 2008

### shalayka

It might also help to see it from the perspective of:

$$3 \frac{GM}{rc^2} \frac{1}{1 - e^2} = 3 \frac{v^2}{c^2} \frac{1}{1 - e^2},$$

$$v = \sqrt{\frac{GM}{r}}.$$

... where e is eccentricity, and r is the average orbit radius (semi-major axis).

From what we know about special relativity, kinematic time dilation, in relation to an orbiting body, is given by:

$$\tau = t \sqrt{1 - \frac{v^2}{c^2}} = t \sqrt{1 - \frac{GM}{rc^2}},$$

From general relativity, gravitational time dilation is given by:

$$\tau = t \sqrt{1 - \frac{2GM}{rc^2}}.$$

As you can see, both forms of time dilation are quite similar when considering it from the perspective of an orbiting body.

For fun, we can obtain the Newtonian acceleration in a backwards sort of way, by differentiating the gravitational time dilation formula with respect to r:

$$\tau \approx t \approx 1,$$

$$\frac{\partial \tau}{\partial r} \approx \frac{GM}{r^2 c^2},$$

$$a = \frac{\partial \tau}{\partial r} {c^2} = \frac{GM}{r^2},$$

$$v = \sqrt{a r} = \sqrt{\frac{GM}{r}}.$$

Last edited: Apr 22, 2008
14. Apr 22, 2008

### snoopies622

Fascinating. This is from a general geodesic equation using the Schwarzschild metric?

15. Apr 22, 2008

### shalayka

Yep. The gravitational time dilation formula appears in the standard Schwarzschild solution's line element / metric, in the form of:

$${\rm d}s^2 = {\rm d}t^2 \left(\sqrt{1 - \frac{2GM}{rc^2}}\right)^2 - \frac{{\rm d}r^2}{\left(\sqrt{1 - \frac{2GM}{rc^2}}\right)^2} - ...$$

The fact that Einstein managed to work backwards from Newtonian velocity equation to get the gravitational time dilation equation just blows my mind!

Last edited: Apr 22, 2008
16. Apr 22, 2008

### pmb_phy

yes. It does show up in gr. In gr, gamma (ratio of coordinate tmie to proper time) is a function of both speed and the gravitational potential (especially when g_0t =0).

Pete

17. Apr 22, 2008

### snoopies622

Thanks. What is "g_0t"? Aren't zero and t as metric tensor subscripts used to mean the same thing?

18. Apr 23, 2008

### pmb_phy

Oops! You're right! Sorry about that. I meant to write g_0k = 0, k = 1,2,3. I can send you a PM which shows the calculation if you'd like? Its far too difficult for me to do that in PM since all I have to do is send you a URL (which we're not allowed to post in open forum).

Pete

19. Apr 23, 2008

### shalayka

That's wierd that you can't post URLs. I am always interested in reading new material related to GR, so I would make as much use of it as snoopies.

20. Apr 23, 2008

### pmb_phy

Correction: People are not allowed to post URLe to pages under their personal websites. The powers that be never made sense to me on this point so if you want to know the real reason you'd be better off asking a moderator why this is policy.

Pete