- #1
crime9894
- 5
- 2
- Homework Statement
- Suppose for a certain 4-dimensional manifold, the Ricci tensor is given by ##R_{\mu}{\nu}=Cg_{\mu}{\nu}## .
Show C is constant (make necessary assumption)
- Relevant Equations
- Einstein's equation: ##R_{\mu\nu}-\frac{1}{2} Rg_{\mu\nu}=8\pi GT_{\mu\nu}##
This question wasn't particularly hard, so I assume metric compatibility and input Ricci tensor to the left side of Einstein's equation.
$$R_{\mu\nu}-\frac{1}{2} Rg_{\mu\nu}=Cg_{\mu\nu}-\frac{1}{2} (4C)g_{\mu\nu}=-Cg_{\mu\nu}$$
Then apply covariant derivative on both side:
$$\nabla^{\mu}(-Cg_{\mu\nu})=8\pi G\nabla^{\mu}T$$
From metric compatibility and conservation of energy-momentum tensor
$$\nabla_{\nu}C=0$$
Covariant derivative reduce to partial derivative when acted on scalar and thus conclude C is constant.
But now I thought of a different approach:
Maximally symmetric space has its Riemann tensor of the form:
$$R_{abcd}\propto g_{ac}g_{bd}-g_{ad}g_{bc}$$
Contracting both side once reduce it to Ricci tensor
$$R_{ab}\propto g_{ab}$$
So my question is: does it works the way around? Does ##R_{\mu\nu}## having the form ##Cg_{\mu\nu}## concludes the manifold is maximally symmetric?
Maximally symmetric space has constant Ricci scalar ##R## and contracting the Ricci tensor immediately yield desire conclusion.
But I'm not sure if such argument is valid at the first place
$$R_{\mu\nu}-\frac{1}{2} Rg_{\mu\nu}=Cg_{\mu\nu}-\frac{1}{2} (4C)g_{\mu\nu}=-Cg_{\mu\nu}$$
Then apply covariant derivative on both side:
$$\nabla^{\mu}(-Cg_{\mu\nu})=8\pi G\nabla^{\mu}T$$
From metric compatibility and conservation of energy-momentum tensor
$$\nabla_{\nu}C=0$$
Covariant derivative reduce to partial derivative when acted on scalar and thus conclude C is constant.
But now I thought of a different approach:
Maximally symmetric space has its Riemann tensor of the form:
$$R_{abcd}\propto g_{ac}g_{bd}-g_{ad}g_{bc}$$
Contracting both side once reduce it to Ricci tensor
$$R_{ab}\propto g_{ab}$$
So my question is: does it works the way around? Does ##R_{\mu\nu}## having the form ##Cg_{\mu\nu}## concludes the manifold is maximally symmetric?
Maximally symmetric space has constant Ricci scalar ##R## and contracting the Ricci tensor immediately yield desire conclusion.
But I'm not sure if such argument is valid at the first place
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