Space-time Curvature

[RichIsnang]

How do you work out the space-time curvature according to general relativity? Iv seen the special relativity equations, but never the general

[timo]

The equation relating the structure of spacetime to its content (matter) is called Einstein equations.

[elfmotat]

The stress-energy tensor contains all of the information about the energy and momentum distribution in spacetime. The SET is denoted with the symbol [math]\bold{T}[/math] (bold symbols mean it's a tensor). The Einstein tensor, [math]\bold{G}[/math], contains the information about the curvature of spacetime. The Einstein field equation tell us that the SET is related to the Einstein tensor by the following:

 

[math]\bold{G}=\frac{8\pi G}{c^4} \bold{T}[/math]

 

where G is Newton's gravitational constant (even though it uses the same symbol as the Einstein tensor, they aren't related) and c is the speed of light. So if you know the energy and momentum distribution in spacetime, the Einstein field equation will tell you what the curvature of spacetime is.

[Aethelwulf]

How do you work out the space-time curvature according to general relativity? Iv seen the special relativity equations, but never the general

 

You work out the spacetime curvature using what is called the Curvature Tensor. I will write a bit up later on it to help explain it to you.

 

I will write something up in about half an hour, I am just getting something to eat first.

 

I have going to have to explain this as quick as I can because my life just got very busy lol

 

You're gonna have to know some math to completely understand this, even some knowledge on tensors. I can't explain everything today. The thing which calculates curvature in General Relativity is the Riemann Tensor and its given as

 

[math]R^{\rho}_{\sigma \mu \nu} = \partial_{\mu} \Gamma^{\rho}_{\mu \sigma} - \partial^{\rho}_{\nu \sigma} + \Gamma^{\rho}_{\mu \lambda} \Gamma^{\lambda}_{\nu \sigma} - \Gamma^{\rho}_{\nu \lambda} \Gamma^{\lambda}_{\mu \sigma}[/math]

 

The part [math]\Gamma_{\mu}\Gamma_{\nu}[/math] is what you call the commutator of two matrices.

 

You can rewrite it more compactly when you bracket expressions and realize that these are the derivatives of the connection ''Gamma''

 

[math]\frac{\partial \Gamma_{\mu}}{\partial x^{\nu}} - \frac{\partial \Gamma_{\nu}}{\partial x^{\mu}} + \Gamma_{\nu}\Gamma_{\mu} - \Gamma_{\mu}\Gamma_{\nu}[/math]

 

You can only get the Riemann tensor by contracting the ''Ricci Tensor''. Notice that one alpha is on the upper indices and one is on the lower indices:

 

[math]R_{\mu \nu} = R^{\alpha}_{\mu \alpha \nu}[/math]

 

Repeated indices means you automatically sum over these indices. The lowercase [math]\mu \alpha[/math] actually describe some rotation plain for a very small area displacement [math](dx^{\nu}, dx^{\mu})[/math]

 

You can also contract using the metric, for instance

 

[math]R_{\lambda \sigma \mu \nu} = g_{\lambda \rho} R^{\rho}_{\sigma \mu \nu}[/math]

 

Can you guess which one is contracted? If you said [math]\rho[/math], you'd be right. What is [math]g_{\mu \nu}[/math] contracted with [math]R^{\mu \nu}[/math]? It's just [math]R[/math] is the answer. You would get the curvature scalar by contracted the Ricci Tensor [math]R^{\mu \nu}[/math] and has this following form

 

[math]\nabla_{\mu} R^{\mu \nu} = \frac{1}{2} g^{\mu \nu} \partial_{\mu} R[/math]

 

where we call [math]\nabla_{\mu}[/math] the covariant derivative. I think the covariant derivative originally came from work on fibre bundles. The property of a covariant derivative just has this form:

 

[math]\nabla_{\mu}AB = A\nabla B + (\nabla A) B[/math]

 

The covariant derivative of [math]g_{\mu \nu}[/math] is actually zero.

 

[math]\nabla_{\mu} R^{\mu \nu} = \frac{1}{2} \nabla_{\mu}g^{\mu \nu} R[/math]

 

[math]\nabla [R^{\mu \nu} - \frac{1}{2}g^{\mu \nu} R ]= 0[/math]

 

This can be rewritten as a short-hand

 

[math]R^{\mu \nu} - \frac{1}{2}g^{\mu \nu} R = G^{\mu \nu}[/math]

 

so

 

[math]\nabla_{\mu}G^{\mu \nu} = 0[/math]

 

This is the local continuity equation for gravitational energy. As I said before, [math]g^{\mu \nu}[/math] derivative is zero, so what we have is

 

[math]R - 2R = 0[/math]

 

and [math]R=0[/math] when there is no energy-momentum present.

 

So we learned the ''Einstein Tensor'' [math]\nabla_{\mu} G^{\mu \nu}=0[/math]

 

The right hand side of [math]\nabla_{\mu} R^{\mu \nu} = \frac{1}{2} \nabla_{\mu}g^{\mu \nu} R[/math] describes the matter in a universe. Even when matter is zero, it does not mean that curvature has to be zero. Gravitational waves for instance and other forms of energy can cause curvature in a vacuum which is an interesting facet of the theory to keep in mind. Now all this stuff is related to Einsteins equations because they can either derive the equations or be derived from his field equations. They are what you call a rank 2 tensor, and If I have time later, I will come back and explain in some detail what a rank 2 tensor is by introducing a new thing for you to try and understand, called contravariant and covariant tensors.

Edited June 19, 2012 by Aethelwulf

[Aethelwulf]

Right, so I said I would try and explain what a rank 2 tensor is. It's difficult trying to explain this stuff to anyone - even when you are trying to teach something to someone in a very short post - but I think I have a way that can help you understand it.

 

Some simple equations to consider might be

 

[math]A_{(y)}^{m} = \frac{\partial y^m}{\partial x^y} A_{(x)}^{r}[/math]

 

The upper and lower indices here [math]A_{(x)}^{r}[/math] are called the ''components of the vector.'' Consider a second one as well

 

[math]B_{(y)}^{n} = \frac{\partial y^n}{\partial x^s} B_{(x)}^{s}[/math]

 

How would we write our first tensor mixed in with our second tensor? You would have to mix them in an appropriate way:

 

[math]A_{(y)}^{m}B_{(y)}^{n} = \frac{\partial y^m}{\partial x^y}\frac{\partial y^n}{\partial x^s}A_{(x)}^{r} B_{(x)}^{s}[/math]

 

And that is us, wasn't too hard eh? This is what you call a mixed tensor of ''second rank''. We can identify it by changing it slightly

 

[math]T_{(y)}^{mn} = \frac{\partial y^m}{\partial x^y}\frac{\partial y^n}{\partial x^s} T_{(x)}^{rs}[/math]

 

Think of these upper indices as counting the rank of your tensor, so the upper indices [math]T_{(y)}^{mn}[/math] would count as a second rank tensor. The tensor above is in fact a contravariant tensor because the important indices just spoke about are on the uppercase. If they are lowercase, they are covariant tensors:

 

[math]T^{(y)}_{mn} = \frac{\partial y^r}{\partial x^m}\frac{\partial y^s}{\partial x^n} T^{(x)}_{rs}[/math]

 

And viola! That's you done.

 

If you need to know anything else, I will be happy to try and help.

[pmb]

How do you work out the space-time curvature according to general relativity? Iv seen the special relativity equations, but never the general

It depends on the information which you have in front of you. If you have the metric tensor you can calculate the spacetime curvature straight from that. If you're given the stress-energy-momentum tensor T then you plug that into Einstein's equation and then solve for the metric tensor, Once you have the metric tensor you can calculate the curvature tensor.

Edited June 25, 2012 by pmb