# Contracting Indices

1. Jan 24, 2014

### direct99

Hello,

I was reading a book about general relativity and I came across these two equations

\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}_{\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}

and it says that by contracting equation (1) once with respect to ($\mu,\rho$), then with respect to ($\nu,\rho$), then subtracting the resulting equations we can get equation (2), however I can't see how that's possible.
Also, This connection is not symmetric with respect to the lower Indices

Last edited: Jan 24, 2014
2. Jan 24, 2014

### bcrowell

Staff Emeritus
It breaks down into two parts, one part for the terms with only the derivatives of the metric, and one for the terms that involve the metric contracted with the Christoffel symbols. I did the first part and it worked out. Want to try that part and post what you get? Then we could move on to the second part. Remember that when an index is summed over, you're free to rename it to whatever you like.

3. Jan 24, 2014

### direct99

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).

Last edited: Jan 25, 2014
4. Jan 25, 2014

### direct99

How to get equation (2) from

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

5. Jan 25, 2014

### bcrowell

Staff Emeritus
For now let's concentrate on the terms that have derivatives of the metric in them. The way you've done it, you end up with something ungrammatical. In your expression

$$-g_{,\rho }^{\mu \rho }+g_{,\rho }^{\rho \nu }=0$$

you have a term with an upper μ index and a term with an upper ν index. You can't add terms that are written in this way. This tells you that you should rename your ρ indices before doing the subtraction.

6. Jan 25, 2014

### direct99

so are you saying that

$$-g_{,\rho }^{\mu \rho }+g_{,\rho }^{\rho \nu }=0$$
should be something like
$$-g_{,\sigma }^{\mu \sigma }+g_{,\rho }^{\rho \nu }=0$$
If so does this mean that its equivalent to
$$g_{,\nu }^{[\mu \nu ]}=0$$

Last edited: Jan 26, 2014