Space With Schwarzschild Metric

[e-pie]

This is a problem from Tensor Calculus:Barry Spain on # 69

Prove that a space with Schwarzschild's metric is an Einstein space, but not a space of constant curvature.

The metric as given in the book is $$d\sigma^2=-\bigg(1-\frac{2m}{c^2r}\bigg)^{-1}dr^2-r^2d\theta^2-r^2\sin^2 \theta d\psi^2-c^2\bigg(1-\frac{2m}{c^2r}\bigg)dt^2$$

 

[PeterDonis]

Thread has been moved to Advanced Physics Homework. @e-pie please show some attempt at a solution to the problem. (Hint: since the problem asks about curvature, how would you check the curvature?)

 

[e-pie]

Thanks I was about to give some attempt,I worked out that the line element of a space with constant curvature is $$ds^2=dt^2-\frac{[S(t)]^2}{1+\frac{1}{4}Kr^2}^2[dr^2+r^2(d\theta^2+sin^2(\theta)d\phi^2]$$ I also have the radius and some other things,but how do I show it as an Einstein space? I am thinking that "Any manifold with constant sectional curvature is an Einstein manifold"(Wikipedia) So do I need to show that $$\nabla R=0$$ Also I know that a Riemannian manifold with constant sectional curvature is Einstein. but the converse is true for dimensions $$\leq 3$$

Am I on the right track?I think these concepts should link to the answer,but I can't seem to fimd how...

 

[PeterDonis]

I worked out that the line element of a space with constant curvature is

This requires a particular choice of coordinates, which might not be the same as the coordinates in which you wrote down the Schwarzschild metric. So just looking at this line element and comparing it with the Schwarzschild metric line element won't necessarily tell you whether the Schwarzschild metric does or doesn't have constant curvature. But there are other ways of checking that.

I am thinking that "Any manifold with constant sectional curvature is an Einstein manifold"(Wikipedia)

But the converse is not true: not every Einstein manifold has constant sectional curvature. So this doesn't help either.

The general definition of an Einstein manifold is a manifold in which the Ricci tensor is some constant $k$ times the metric tensor. Can you see how that implies that the Schwarzschild metric (or, in fact, any vacuum solution) is an Einstein manifold?

 

[e-pie]

An Einstein manifold in dimension 3 has a constant sectional curvature.

To show that any given metric is curved we need to calculate curvature invariant. The Ricci curvature
$$R= R^i{}_i = R^{ki}{}_{ki}$$

The general definition of an Einstein manifold is a manifold in which the Ricci tensor is some constant $k$ times the metric tensor.

Can you show some math on this part?

 

[PeterDonis]

An Einstein manifold in dimension 3 has a constant sectional curvature.

But we're not talking about 3 dimensions, we're talking about 4.

To show that any given metric is curved we need to calculate curvature invariant. The Ricci curvature

That doesn't help for the Schwarzschild metric, because it vanishes (as it does for any vacuum spacetime). But there is another curvature invariant that doesn't.

Can you show some math on this part?

You mean $R_{ab} = k g_{ab}$? That's the math version of what I said.

 

[e-pie]

Are you talking about the Kretschmann Invariant
$$R_{abcd} \, R^{abcd} = C_{abcd} \, C^{abcd} +\frac{4}{d-2} R_{ab}\, R^{ab} - \frac{2}{(d-1)(d-2)}R^2$$

You mean $R_{ab} = k g_{ab}$? That's the math version of what I said.

No I meant how that relates to the Schwarzschilde Metric

 

[PeterDonis]

Are you talking about the Kertschmann Invariant?

Yes. What is it for the Schwarzschild metric?

I meant how that relates to the Schwarzschilde Metric

I've already said what the Ricci tensor is for any vacuum metric. Can you see how that relates to a vacuum metric being an Einstein manifold?

 

[e-pie]

Yes. What is it for the Schwarzschild metric?

$$R_{abcd} R^{abcd} = \frac{12 r_s^2}{r^6}$$ is the Kretschmann scalar for Schwarzschild metric and the curvature of Schwarzschild Solution is $$
\frac{12r_s^2}{r^6}$$

 

[PeterDonis]

the Kretschmann scalar for Schwarzschild metric and the curvature of Schwarzschild Solution

Yes, you have these right. So is this curvature constant?

 

[e-pie]

I calculate it at $$R=0$$.
The scalar curvature is the Ricci tensor contracted and the Ricci tensor is null, the cuvature should be zero.

And vacuum metrics are Einstein metrics,Schwarzschild metric is the only spherically symmetric vacuum solution of Einstein's field

 

[PeterDonis]

The scalar curvature is the Ricci tensor contracted

The Ricci scalar is one scalar curvature, but not the only one. "Constant curvature" means all curvature scalars are constant. Is that true for the Schwarzschild metric?

vacuum metrics are Einstein metrics

Yes, that's right.

 

[e-pie]

The Schwarzschild metrics should be the only complete, spherically symmetric 3-manifolds with zero scalar curvature(except for the flat metric on $$\mathbb{R^3}$$) Am I right?

 

[PeterDonis]

The Schwarzschild metrics should be the only complete, spherically symmetric 3-manifolds

As I said before, we are not talking about 3-manifolds, we are talking about 4-manifolds. The Schwarzschild metric is a 4-manifold.

 

[e-pie]

No,then I don't get it,I am guessing it should be constant? Can you show me?

 

[PeterDonis]

I don't get it,I am guessing it should be constant? Can you show me?

You have already shown yourself; you have given answers to both questions that you asked in the OP:

Prove that a space with Schwarzschild's metric is an Einstein space

You said:

vacuum metrics are Einstein metrics

Which answers that question. And then:

but not a space of constant curvature

You said:

the curvature of Schwarzschild Solution is
$$
\frac{12r_s^2}{r^6}
$$

Which answers that question. What's left to show?

 

[e-pie]

I am fooling myself! This is what happens when I try to cram GR!

The curvature is not constant,I guess?

 

[PeterDonis]

The curvature is not constant,I guess?

Go look at your statement of the curvature, that I quoted in my last post. Is it constant?

 

[e-pie]

Go look at your statement of the curvature, that I quoted in my last post. Is it constant?

No,$$
K \propto 1/r^6.
$$

 

[PeterDonis]

No

Well, there you go.

 

[e-pie]

Thanks...Marking this as solved.