# Contraction of Einstein field equation

In the paper "Physical and Geometric Interpretations of the Riemann Tensor, Ricci Tensor, and Scalar Curvature" by Lee C. Loveridge we can find the following paragraph:

First we begin with Einstein’s equation written in terms of the Ricci tensor and scalar curvature $$R_{μν} − \frac{1}{2} Rg_{μν} = 8πGT_{μν}$$ and contract both sides to get $$R = −8πGT$$.

I do not known the meaning of "contract" in this context. If contract means the trace:

$$R_{νν} − \frac{1}{2} Rg_{νν} = R^v_νg_{νν} − \frac{1}{2} Rg_{νν} = Rg_{νν} − \frac{1}{2} Rg_{νν} = \frac{1}{2} Rg_{νν}$$

and:

$$8πGT_{νν} = 8πGT^ν_{ν}g_{νν} = 8πGTg_{νν}$$

that results in:

$$\frac{1}{2} R = 8πGT$$

with a -1/2 difference respect to the expected result.

• $R_{\nu\nu}$ is not a contraction and it isn’t a scalar. You have to contract an upper index with a lower index to get another tensor of smaller-by-2 rank. Commented Jan 9, 2021 at 20:41
• @G.Smith: thanks for your comment. About "it isn't a scalar", note steps $R_{\nu\nu}=R^\nu_\nu g_{\nu\nu}=Rg_{\nu\nu}$. Commented Jan 9, 2021 at 20:44
• Sorry, but that equation does not make sense. You cannot have two lower indices which are the same. And you can’t have four indices that are the same! Commented Jan 9, 2021 at 20:47
• It seems to me that you are trying to learn GR without first properly understanding SR in tensor notation. I suggest going back to Lorentz transformations, looking at how an arbitrary tensor $C_{\mu\nu}$ transforms, and convincing yourself that ${C^\mu}_\mu$ is Lorentz-invariant but $C_{\mu\mu}$ is not. You can perversely define $C_{\mu\mu}$ as $\sum_\mu C_{\mu\mu}$ but it has no physical significance because it is not a tensor of any rank; it does not transform like a tensor should. Commented Jan 9, 2021 at 21:02
• @G.Smith: no one learns to add until they try to multiply Commented Jan 11, 2021 at 0:12

In General Relativity, contraction of two indices is done using the metric tensor $$g_{\mu\nu}$$. In this case, you would hit both sides of the equality with the metric tensor noting that $$g_{\mu\nu}g^{\mu\nu} = D$$, where $$D$$ is the dimension of spacetime, in this case $$D=4$$.
$$g^{\mu\nu}R_{\mu\nu} -\frac12R g^{\mu\nu}g_{\mu\nu} = R -\frac{D}{2}R = 8\pi G g^{\mu\nu}T_{\mu\nu} = 8\pi G T$$