Metric, connection, curvature oh my!

1. Jun 17, 2010

Living_Dog

I just read a sentence in GRAVITATION by MTW (aka the "Princeton Phonebook") that made me realize a confusion wrt the metric, connection, and curvature. In short how are $$g_{\mu\nu}, \Gamma^{\alpha}_{\mu\nu}, and R^{\alpha}_{\beta\mu\nu}$$ distinguished? They all include the description "how space curves."

Here is the little I understand (just so you know where to start):

$$g_{\mu\nu}$$: how the action curves between two events (or two points in spacetime), thus:

$$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$$​

$$\Gamma^{\alpha}_{\mu\nu}$$: how a point travels from one place to another place in the spacetime. (I am only starting to become familiar with modern math terms, alla Wikipedia.org. From what I read, the affine connection is the track laid out by the point as it travels through the manifold.) Thus the equation of geodesic deviation:

$$\f{d^2\xi^{\alpha}}/{d\tau^2} + \Gamma^{\alpha}_{\mu\nu}\f{d\xi^{\mu}}/{d\tau}\f{d\xi^{\nu}}/{d\tau} = 0$$​

$$R^{\alpha}_{\beta\mu\nu}$$: how the global spacetime is curved at every point in the spacetime, thus the EFE:

$$G_{\mu\nu} = R_{\mu\nu} -\f{1}/{2}g_{\mu\nu}R = 8\pi T_{\mu\nu}$$​

Note in fact that $$g_{\mu\nu} \rightarrow \Gamma^{\alpha}_{\mu\nu} \rightarrow R^{\alpha}_{\beta\mu\nu}$$

After writing this out I think I have it correct. If not please correct me. Thanks in advance for all replies.

Last edited: Jun 17, 2010
2. Jun 17, 2010

espen180

Yes.

The metric has to do with the distance between neighbouring points in a manifold.
The Christoffel symbol has to do with straight lines and acceleration. They describe parallel transport of vectors in manifolds.
The Riemann tensor (and Ricci Tensor) describes the curvature of te manifold. Curvature has to do with the angular deviation of a vector after it has been parallel transported through a loop.

The Christoffel symbol is a fuction of the metric and it's derivatives.
The Riemann tensor is a function of the products and derivatives of the Christoffel symbols.

Note that while $$R^{\lambda}_{\mu\sigma\nu}=0$$ implies a flat manifold, $$R_{\mu\nu}=0$$ does not.

3. Jun 17, 2010

Fredrik

Staff Emeritus
You should read "Riemannian manifolds: an introduction to curvature" by John M. Lee.

4. Jun 17, 2010

Living_Dog

...oh, as a reference for the modern math used in GR... Thanks! I was wondering about that very thing.

5. Jun 17, 2010

Living_Dog

Right, the determination of whether the spacetime is flat or not comes from the Riemann - not the Ricci tensor. Thanks for the "head's up!". That could have been an easy avenue for mistakes in the future.

Last edited: Jun 17, 2010
6. Jun 17, 2010

nicksauce

Yes. The point is that a zero Riemann tensor implies a zero Ricci tensor, but a zero Ricci tensor does not imply a zero Riemann tensor (in 4 or higher spacetime dimensions).

7. Jun 17, 2010

Fredrik

Staff Emeritus
Then I should probably have told you about his other books too. The one I mentioned is the best place to read about connections, geodesics, parallel transport and curvature, but it assumes that you already know about manifolds, tensors, etc. All those things and more (Lie groups, differential forms, integration on manifolds,...) are covered in "Introduction to smooth manifolds", also by John M. Lee. Both are excellent. The same guy also wrote a book called "Introduction to topological manifolds". That one is less useful because it won't help you understand the physics, but if you feel that you really want to understand the terms "Hausdorff" and "second countable" that appear in the definition of a manifold, then you need that book too (or some other book that covers general topology, like Munkres or Lipschutz).

8. Jun 17, 2010

AWA

This is important, so if the the Ricci tensor being zero or the Ricci scalar R being zero not necesarily make the manifold flat, my question is if R=0 implies T (Stress-energy tensor) to be zero, absence of matter and therefore no curvature, wouldn't that be enough to imply a flat manifold?

9. Jun 17, 2010

sheaf

Vanishing of the Ricci scalar does not imply vanishing of the stress energy tensor, the stress energy tensor is more closely related to the Ricci tensor not the Ricci scalar.

Absence of matter does not imply a flat manifold - look up vacuum solutions of Einstein's equations, $R_{\mu\nu} = 0$. Some would say all the interesting stuff is in the Weyl tensor anyway !!

10. Jun 17, 2010

JustinLevy

Similar to how Ricci=0 doesn't mean Riemann=0, T (trace of stress energy tensor) =0 doesn't mean the stress energy tensor = 0.

Consider for instance electromagnetic fields. The stress energy tensor is traceless. So T=0 doesn't mean vacuum.

11. Jun 17, 2010

AWA

Thanks. That clears up a few things for me.

So when I see T without the subindices is usually referring to the trace of the tensor, not to the tensor itself?

12. Jun 17, 2010

dulrich

But even if the stress-energy tensor is identically zero, you still may not have flat space-time at that point -- just like you can have a non-zero EM field without any charges (at that point).

$T_{ij} = 0 \implies R_{ij} = 0$ + spherical symmetry is one way to derive the Schwarzschild solution.

13. Jun 18, 2010

Altabeh

You'd better say the spacetime with a vanishing stress-energy tensor must be Ricci-flat but not generally flat. This is because we have spacetimes whose Ricci is zero while the Riemann tensor yet doesn't vanish. Furthermore, spacetimes for which

$$R_{ab}=\frac{1}{2}Rg_{ab}$$

are neither flat nor Ricci-flat with the same matter condition. In fact this completely agrees to nicksauce's post!

AB

14. Jun 18, 2010

Altabeh

I'm quite sure that Fredrik has read the book several times and this book is really charming and in fact it hits the spot but still a very mathematical one that may not be suitable for beginners who want to self-study GR. I think he should try something less technical like Woodhouse's "General Relativity" and then go on consulting other books at Lee's level.

AB

15. Jun 18, 2010

bcrowell

Staff Emeritus
Some physical interpretation that might be helpful: --

$\Gamma$ is essentially the gravitational field. By the equivalence principle, the gravitational field is not a very interesting thing in GR. For instance, you can always say that the field is zero at any given point in spacetime, if you choose the frame of a free-falling observer.

R is the only one of these three quantities that has any meaning that it independent of the choice of coordinates. If R is nonzero in one set of coordinates, it's nonzero in every other set of coordinates.

16. Jun 18, 2010

Fredrik

Staff Emeritus
I have actually only read half of it, quite recently. I learned some of this stuff from Wald and Spivak a long time ago, and forgot most of it. I only used Lee to learn it again, and I was delighted to see how much better it was. I agree that it would be hard for a beginner to start with this one. That's why they also need his "Introduction to smooth manifolds" (to learn and understand the definitions of such things as manifolds and tensor fields). I also agree that it's a good idea to try to pick up the general ideas from a GR book first. (I haven't read Woodhouse, so I can't comment on that).

17. Jun 19, 2010

Altabeh

But definitely this is not indicative of gravitational field not being good in GR because apart from the equations of motion in GR where it may seem to be true that such a thing is not of an abundant interest to be given a precise role, there is a great deal of good discussions that make use of "gravitational fields" as in the "weak approximation" where we raise a frequent gaze at gravitational fields and their now diminished effect on the curvature of spacetime. Of course we can't also neglect their "interesting" and indispensable role as part of components of matter-energy tensor.

To add a little note to your correct conclusion, it is always instructive to say that since R is a (1,3) tensor, then it transforms under a transformation law so R being zero in one frame is preserved in any other frame.

AB