# Contracting Indices

Does anyone know how to get from (1) to (2) in the system

\begin{align} \mathrm{g}^{\mu\nu}_{,\rho}+ \mathrm{g}^{\sigma\nu}{{\Gamma}}^{\mu}_{\sigma\rho}+ \mathrm{g}^{\mu\sigma}{{\Gamma}}^{\nu}_{\rho\sigma} -\frac{1}{2}( {{\Gamma}}^{\sigma}_{\rho\sigma}+{{\Gamma}}^{\sigma}_{\sigma\rho} ) \mathrm{g}^{\mu\nu} &=0, \tag1 \\ \mathrm{g}^{[\mu\nu]}_{,\nu} +\frac{1}{2}( {{\Gamma}}^{\rho}_{\rho\nu}-{{\Gamma}}^{\rho}_{\nu\rho} ) \mathrm{g}^{(\mu\nu)} &=0, \tag2 \end{align}

by contracting equation (1) once with respect to ($\mu,\rho$), then with respect to ($\nu,\rho$)?

Where $\Gamma$ is not symmetric with respect to the lower Indices.

My attempt so far to solve this problem is: Well, when contracting with respect to μ and ρ I get: $$-\frac{1}{2} g^{\rho \nu } {\Gamma} _{a\rho}^{a}-\frac{1}{2} g^{\rho \nu} \Gamma _{\rho a}^{a}+g^{a\nu} \Gamma _{a\rho}^{\rho}+g^{\rho a} \Gamma _{\rho a}^{\nu}+g_{,\rho}^{\rho\nu}=0$$ and when contracting with respect to nu and ρ I get: $$-\frac{1}{2} g^{\mu \rho } \Gamma _{a\rho}^a-\frac{1}{2} g^{\mu \rho} \Gamma _{\rho a}^a+g_{}^{a\rho } \Gamma _{a\rho}^{\mu }+g^{\mu a} \Gamma _{\rho a}^{\rho }+g_{,\rho }^{\mu \rho }=0$$ when subtracting these two equations I get: $$\frac{1}{2} g_{}^{\mu \rho } \Gamma _{a\rho }^a-\frac{1}{2} g_{}^{\rho \nu } \Gamma _{a\rho }^a+\frac{1}{2} g_{}^{\mu \rho } \Gamma _{\rho a}^a-\frac{1}{2} g_{}^{\rho \nu } \Gamma _{\rho a}^a+g_{}^{a\nu } \Gamma _{a\rho }^{\rho }-g_{}^{a\rho } \Gamma _{a\rho }^{\mu }-g_{}^{\mu a} \Gamma _{\rho a}^{\rho }+g_{}^{\rho a} \Gamma _{\rho a}^{\nu }-g_{,\rho }^{\mu \rho }+g_{,\rho }^{\rho \nu }=0$$ I cant see how this is equal to equation (2)

• I think this is a valid question for practising physicists. @OP: Explain a little more about what you tried. In general, we discourage people from cogging HW solutions off this phy.SE. – Siva Jan 23 '14 at 16:11
• well, this is not a homework problem, but just an equation I found in Schrodinger's book "Space Time structure". – user38032 Jan 23 '14 at 17:17
• Hi user38032. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. – Qmechanic Jan 23 '14 at 19:00
• Look, I already said that this isn't a homework. – user38032 Jan 24 '14 at 15:20
• The homework-like question policy is not applied because the questions was assigned as homework, but because that question's main value appears to be pedagogical (that is they are the kind of question that would be given to student because they teach a particular skill or point). – dmckee Jan 25 '14 at 18:03

The attempt of obtaining

$$g^{[\mu\nu]}_{,\nu} +\frac{1}{2}( {{\Gamma}}^{\rho}_{\rho\nu}-{{\Gamma}}^{\rho}_{\nu\rho} ) g^{(\mu\nu)} =0,$$ was almost right! The only thing missing was a little care with relabeling indices. We will proceed in three main steps.

1.) So when contracting Eq. (1) with respect to $\mu$ and $\rho$, we get the identity:

$$-\frac{1}{2} g^{\rho \nu } {\Gamma} _{a\rho}^{a}-\frac{1}{2} g^{\rho \nu} \Gamma _{\rho a}^{a}+g^{a\nu} \Gamma _{a\rho}^{\rho}+g^{\rho a} \Gamma _{\rho a}^{\nu}+g_{,\rho}^{\rho\nu}=0.$$

Now by relabeling the dummy indices in the 3rd term as $a \leftrightarrow \rho$, we get that the 3rd term can be written as $g^{\rho \nu} \Gamma _{\rho a}^{a}$. Moreover, we can see that now the 2nd and 3rd term can be simplified: adding them together gives $+ \tfrac{1}{2}g^{\rho \nu} \Gamma _{\rho a}^{a}$. By a final index label change $\nu \to \mu$, we get that:

$$-\frac{1}{2} g^{\rho \mu } {\Gamma} _{a\rho}^{a}+\frac{1}{2} g^{\rho \mu} \Gamma _{\rho a}^{a}+g^{\rho a} \Gamma _{\rho a}^{\mu}+g_{,\rho}^{\rho\mu}=0. \; \; \; \; (A)$$

2.) When contracting Eq. (1) with respect to $\nu$ and $\rho$, we get the identity:

$$-\frac{1}{2} g^{\mu \rho } \Gamma _{a\rho}^a-\frac{1}{2} g^{\mu \rho} \Gamma _{\rho a}^a+g_{}^{a\rho } \Gamma _{a\rho}^{\mu }+g^{\mu a} \Gamma _{\rho a}^{\rho }+g_{,\rho }^{\mu \rho }=0.$$

Let us also rename the dummy indices in the 4th term as $a \leftrightarrow \rho$. We can see now that the 4th term is simply $g^{\mu \rho } \Gamma _{a\rho}^a$, and the 1st and 4th term hence together give $\frac{1}{2} g^{\mu \rho } \Gamma _{a\rho}^a$. Moreover, let us perform also the $a \leftrightarrow \rho$ "dummy index relabeling", yielding $g_{}^{\rho a} \Gamma _{\rho a}^{\mu }$ for the 3rd term. After these manipulations our identity reads as

$$\frac{1}{2} g^{\mu \rho } \Gamma _{a\rho}^a-\frac{1}{2} g^{\mu \rho} \Gamma _{\rho a}^a+g_{}^{\rho a} \Gamma _{\rho a}^{\mu }+g_{,\rho }^{\mu \rho }=0. \; \; \; \; (B)$$

3.) Now taking (B)-(A), we obtain:

$$g_{,\rho }^{\mu \rho } -g_{,\rho}^{\rho\mu} + \frac{1}{2}\left( g^{\mu \rho } + g^{\mu \rho }\right) \left( \Gamma _{a\rho}^a - \Gamma _{\rho a}^a \right)=0,$$ which after the $a \to \rho$ and $\rho \to \nu$ relabeling is exactly the same as the desired Eq. (2).

• I don't get why you relabeled $\nu \to \mu$ in you first step. – user38032 Feb 4 '14 at 3:05
• You don't understand why I did it or why I am allowed to do it? I will answer both. 1) Why did I do it? I calculated (B) beforehand, and I saw that after this $\nu \to \mu$ relabeling I get an Eq. (A) which subtracted from (B) gives me the equation you wanted to obtain. Why am I allowed to do it? Without the $\nu \to \mu$ relabeling I would get $-\tfrac{1}{2} g^{\rho\nu}\Gamma_{a\rho}^{a}+\tfrac{1}{2} g^{\rho\nu} \Gamma_{\rho a}^{a}+g^{\rho a} \Gamma_{\rho a}^{\nu}+g_{,\rho}^{\rho\nu}=0$, which I guess also you agree with. Now $\nu$ can be any index, so I simply choose $\nu=\mu$. – Zoltan Zimboras Feb 4 '14 at 3:29
• I don't understand why you're allowed to do it. Are you saying that equation (2) is only valid if ν=μ?. – user38032 Feb 4 '14 at 3:35
• No! Eq.(2) is also valid when $\nu\ne\mu$, you have to keep track of all the relabelings that I did. Let's go slower. Forget about all the previous equations, and answer to this: 1) Do you agree that if $-\tfrac{1}{2} g^{\rho\nu}\Gamma_{a\rho}^{a}+\tfrac{1}{2} g^{\rho\nu}\Gamma_{\rho a}^{a}+g^{\rho a} \Gamma_{\rho a}^{\nu}+g_{,\rho}^{\rho\nu}=0$ holds for any $\nu\in\{0,1,2,3\}$, then also $-\tfrac{1}{2} g^{\rho\mu}\Gamma_{a\rho}^{a}+\tfrac{1}{2} g^{\rho\mu} \Gamma_{\rho a}^{a}+g^{\rho a} \Gamma_{\rho a}^{\mu}+g_{,\rho}^{\rho\mu}=0$ holds for any $\mu\in\{0,1,2,3\}$? It's the same set of Eqs. – Zoltan Zimboras Feb 4 '14 at 3:46
• @user38032 So think over my previous question. After you answered, we'll make the next step. – Zoltan Zimboras Feb 4 '14 at 4:48