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Show that Einstein's equation $$G^{\mu\nu}=R^{\mu\nu}-\frac12\mathcal{R}g^{\mu\nu}=\frac{8\pi G}{c^4}T^{\mu\nu}\tag{1}$$ can be written in the form $$R^{\mu\nu}=\frac{8\pi G}{c^4}\left(T^{\mu\nu}-\frac12\mathcal{T}g^{\mu\nu}\right)\tag{2}$$ where $\mathcal{T}=g_{\alpha\beta}T^{\alpha\beta}=T_{\alpha}^\alpha$

Some quantities were not defined in this question. I have no doubt that most of you know what they are, but just in case, $R^{\mu\nu}$ is the Ricci tensor, and the scalar curvature $\mathcal{R}=g^{ab}R_{ab}$, $G^{\mu\nu}$ is the Einstein tensor and $T^{\mu\nu}$ is the energy-momentum tensor.

The way I would tackle this question is to multiply $(1)$ by $g_{\mu\nu}$ and contract, which is equivalent to taking the trace, from this approach $(2)$ can swiftly be obtained. However, that is not how the author presents the solution and is causing me confusion.

This is how the author tackled this question:

From Einsteins's equation $$G^{\mu\nu}=R^{\mu\nu}-\frac12\mathcal{R}g^{\mu\nu}=\frac{8\pi G}{c^4}T^{\mu\nu}\tag{A}$$ Lower the $\nu$ index: $$R^{\mu}_{{\,\,\,}\nu}-\frac12\mathcal{R}\delta_\nu^\mu=\frac{8\pi G}{c^4}T^{\mu}_{{\,\,\,}\nu}\tag{B}$$ Contract $\mu$ and $\nu$ indices $$\left(1-\frac12 4\right)\mathcal{R}=\frac{8\pi G}{c^4}T^\mu_{\,\mu}\tag{C}$$ $[\cdots]$

I don't need to write more of the solution since direct substitution yields the required $(2)$. My sole concern in this post is to figure out how the author obtained $(\mathrm{B})$. Put another way, how did the author "lower the $\nu$ index"?

The following manipulations are the only way I can think to get from $(\mathrm{A})$ to $(\mathrm{B})$, so starting with $(\mathrm{A})$, I'm trying to get factors of the form $R^\alpha_{\;\beta }$ (where the second index is now 'lowered'). To try to accomplish this I multiply first by the inverse metric and then by the type $(1,1)$ metric:

$$R^{\mu}_{{\,\,\,}\sigma}\,g^{\sigma\,\nu}-\frac12\mathcal{R}g^{\mu}_{{\,\,\,}\rho}\,g^{\rho\,\nu}=\frac{8\pi G}{c^4}T^{\mu}_{{\,\,\,}\gamma}\,g^{\gamma\,\nu}$$ $$R^{\mu}_{{\,\,\,}\sigma}\,g^{\sigma\,\epsilon}\,g_{\epsilon}^{\,\,\,\nu}-\frac12\mathcal{R}g^{\mu}_{{\,\,\,}\rho}\,g^{\rho\,\eta}\,g_{\eta}^{\,\,\,\nu}=\frac{8\pi G}{c^4}T^{\mu}_{{\,\,\,}\gamma}\,g^{\gamma\,\beta}\,g_{\beta}^{\,\,\,\nu}$$ $$\stackrel{\color{red}{?}}{\implies}R^{\mu}_{{\,\,\,}\sigma}\,\delta^{\sigma\,\nu}-\frac12\mathcal{R}g^{\mu}_{{\,\,\,}\rho}\,\delta^{\rho\nu}=\frac{8\pi G}{c^4}T^{\mu}_{{\,\,\,}\gamma}\,\delta^{\gamma\nu}$$ $$\implies R^{\mu}_{{\,\,\,}\nu}-\frac12\mathcal{R}g^{\mu}_{{\,\,\,}\nu}=\frac{8\pi G}{c^4}T^{\mu}_{{\,\,\,}\nu}$$

The last equation is the closest I can get to the author's eqn. $(\mathrm{B})$.

Continuing anyway, and contracting (setting $\mu=\nu$) yields: $$R^{\mu}_{{\,\,\,}\mu}-\frac12\mathcal{R}g^{\mu}_{{\,\,\,}\mu}=\frac{8\pi G}{c^4}T^{\mu}_{{\,\,\,}\mu}$$ and assuming $g^{\mu}_{{\,\,\,}\mu}=4$ by the Einstein summation convention since $\mu,\,\nu = 0,1,2,3$ then $$\mathcal{R}-\frac12\mathcal{R}\times 4=\frac{8\pi G}{c^4}\mathcal{T}\tag{3}$$

Where the last equation is the equivalent of eqn. $(\mathrm{C})$.

The problem is that I am doubtful that the manipulations used to reach equation $(3)$ are legitimate (flagged with a question mark above the implication sign).

In arriving at $(3)$ I have assumed the following:

  1. $g^{\sigma\,\epsilon}\,g_{\epsilon}^{\,\,\,\nu}=\delta^{\sigma\,\nu}$
  2. $g^{\rho\,\eta}\,g_{\eta}^{\,\,\,\nu}=\delta^{\rho\,\nu}$
  3. $g^{\gamma\,\beta}\,g_{\beta}^{\,\,\,\nu}=\delta^{\gamma\,\nu}$
  4. $g^{\mu}_{{\,\,\,}\mu}=4$

I'm questioning $1.$ to $3.$ since I am only acquainted with the fact that $g_{ab}\,g^{bc}=\delta^c_a$ - where the two metric tensors are of type $(0,2)$ or $(2,0)$, but for the implication (with the red question mark above) to hold this needs to be true for a type $(1,1)$ metric tensor. Is this true?

For $4.$ I am aware that $\delta^{\mu}_{{\,\,\,}\mu}=\delta^{0}_{{\,\,\,}0}+\delta^{1}_{{\,\,\,}1}+\delta^{2}_{{\,\,\,}2}+\delta^{3}_{{\,\,\,}3}=4$, but does the same logic hold for $g^{\mu}_{{\,\,\,}\mu}=4$?

Final remarks:

I'm sure you are itching to downvote this and/or request for it to be closed and migrated to Mathematics-StackExchange. So let me briefly justify why I purposely chose to ask this question here instead. The reason is that I am trying to avoid the geometrical formulation of tensors, as a physicist and a beginner to tensors I read this post and like the OP there I don't want to be bombarded with expressions like $(\mathbf{v} \otimes \mathbf{w})_{ij}$ which will only confuse me more than I already am.

Any help would be greatly appreciated.

2 Answers2

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You've almost got it, but there are a few small things you're missing. First, when a physicist says they're "lowering an index", what they mean is that if you start with the equation $A^{\mu\nu} = B^{\mu\nu}$, then we can rename $\nu$ to $\sigma$ and then contract that equation with $g_{\sigma \nu}$, to get $$A^{\mu\sigma} g_{\sigma \nu} = B^{\mu \sigma} g_{\sigma \nu}.$$ This equation is equivalent, by definition, to $$A^{\mu}_{\ \ \nu} = B^{\mu}_{\ \ \nu}.$$ Mechanically, this combination of steps has exactly the same effect as moving the $\nu$ from upstairs to downstairs, so that's how we think about it.

After performing this operation, we arrive at $$R^{\mu}_{{\,\,\,}\nu}-\frac12\mathcal{R} g^\mu_{\ \ \nu}=\frac{8\pi G}{c^4}T^{\mu}_{{\,\,\,}\nu},$$ where the metric with mixed indices is $$g^{\mu}_{\ \ \nu} \equiv g^{\mu \sigma} g_{\sigma \nu}.$$ The next "physicist" step is to note that the metric with mixed indices is constructed by contracting the metric with the inverse metric, so the elements of $g^{\mu}_{\ \ \nu}$ are simply those of the identity matrix, $$g^{\mu}_{\ \ \nu} = \begin{cases} 1 & \mu = \nu \\ 0 & \mathrm{otherwise}\end{cases}.$$ Since this is a very simple quantity, we define a separate symbol for it called the Kronecker delta, $$g^{\mu}_{\ \ \nu} \equiv \delta^\mu_\nu$$ where the up and down indices on the delta are aligned because it doesn't matter which one comes first. Upon substituting this in, you get the claimed equation (B).

knzhou
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8

Your assumption of $g^{\sigma\epsilon}g_\epsilon^{\;\nu} = \delta^{\sigma\nu}$ is not correct. We know that $g^{\mu\sigma}g_{\sigma\nu} = g_{\;\nu}^\mu$ by definition because that is what it means to lower the index of $g^{\mu\sigma}$. On the other hand, we also know that $g^{\mu\sigma}g_{\sigma\nu} = \delta_\nu^\mu$ because they are inverse matrices. Therefore, $g_{\;\nu}^\mu = \delta_\nu^\mu$ and thus $g^{\sigma\epsilon}g_\epsilon^{\;\nu} = g^{\sigma\epsilon}\delta_\epsilon^\nu = g^{\sigma\nu}$.

Vincent Thacker
  • 15,416