Proving ##C## is constant in 4-dim ##R_{\mu\nu}=Cg_{\mu\nu}##

AI Thread Summary
The discussion revolves around proving that the constant C in the equation R_{\mu\nu} = Cg_{\mu\nu} is indeed a constant. By applying metric compatibility and the Ricci tensor in Einstein's equation, it is shown that the covariant derivative leads to the conclusion that C is constant. A different approach considers whether the form of R_{\mu\nu} implies that the manifold is maximally symmetric, noting that maximally symmetric spaces have constant Ricci scalars. The validity of this argument is questioned, particularly regarding the implications of the constant C and the notation used. Ultimately, the conversation clarifies that the constant C is unrelated to the speed of light.
crime9894
Messages
5
Reaction score
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
 
Last edited:
Physics news on Phys.org
What is C? Also, the notation doesn't seem to be working in LaTeX.
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top