# Determining the partial derivative of a metric tensor

Im new to the Tensor Calculus and General Theory of Relativity, and I have one question. I want to determine the Christoffel symbols in FRW metric. This is the general equation of Christoffel symbols: $$\Gamma^{\mu}_{\hphantom{\mu}\alpha\beta}=\frac{1}{2}g^{\mu\nu}\bigg[\frac{\partial g_{\alpha\nu}}{\partial x^\beta}+\frac{\partial g_{\beta\nu}}{\partial x^\alpha}-\frac{\partial g_{\alpha\beta}}{\partial x^\nu}\bigg]$$ So the $$g_{\mu\nu}$$ in the expanding FRW universe equals to: $$g_{\mu\nu}=\begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & a^2(t) & 0 & 0 \\ 0 & 0 & a^2(t) & 0 \\ 0 & 0 & 0 & a^2(t) \end{bmatrix}$$ How can I calculate the partial derivatives of the metric with respect to coordinates, maybe time (In this situation It's $$g_{\alpha\nu,\beta}, g_{\beta\nu,\alpha}, g_{\alpha\beta,\nu}$$)? I did not find a normal explanation in the books. I am a beginner, for me it is quite difficult, I will be very glad to hear your feedback.

• Do you understand what the derivative of $a^2(t)$ with respect to $t$ is? If so, you know $g_{11,0}$, $g_{22,0}$, and $g_{33,0}$. – G. Smith Feb 17 at 20:18
• I am a beginner Have you used index notation before? – G. Smith Feb 17 at 20:22
• Yes, I am. But I'm a little confused with index notation. – AlexSok Feb 17 at 20:26
• No; $\alpha$, $\beta$, and $\mu$ can take any value from 0 to 3. There are 64 Christoffel vakues, from $\Gamma^0{}_{00}$ to $\Gamma^3{}_{33}$, including ones like $\Gamma^2{}_{13}$. To understand $\nu$, read this; you sum over $\nu$ from 0 to 3. – G. Smith Feb 17 at 20:41
• In general, Greek indices are usually spacetime indices that take the values 0, 1, 2, and 3. – G. Smith Feb 17 at 21:12

## 2 Answers

I'll show you how to compute one of the 64 Christoffel symbols, and you can compute the others.

Before computing the Christoffels, you need to know the inverse metric tensor, which has components $$g^{\mu\nu}$$. This is simply the inverse of the matrix for $$g_{\mu\nu}$$, so

$$g^{\mu\nu}=\begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & a^{-2}(t) & 0 & 0 \\ 0 & 0 & a^{-2}(t) & 0 \\ 0 & 0 & 0 & a^{-2}(t) \end{bmatrix}.$$

In the formula

$$\Gamma^{\mu}_{\hphantom{\mu}\alpha\beta}=\frac{1}{2}g^{\mu\nu}(g_{\alpha\nu,\beta}+g_{\beta\nu,\alpha}-g_{\alpha\beta,\nu})$$

for the Christoffel symbols, $$\mu$$, $$\alpha$$, and $$\beta$$ are "free indices", appearing once in each term. You can assign the values 0, 1, 2, or 3 to them to find the symbols 64 symbols $$\Gamma^0{}_{00}$$, $$\Gamma^0{}_{01}$$, $$\Gamma^0{}_{02}$$, ..., $$\Gamma^3{}_{33}$$.

The index $$\nu$$ is different. It appears twice in each term on the right, and is called a "dummy index", a "summation index", or a "contracted index". When you see such an index, there is an implied summation over the values 0, 1, 2, and 3, so the formula is actually

$$\Gamma^{\mu}_{\hphantom{\mu}\alpha\beta}=\sum_{\nu=0}^3\frac{1}{2}g^{\mu\nu}(g_{\alpha\nu,\beta}+g_{\beta\nu,\alpha}-g_{\alpha\beta,\nu}).$$

Let's compute the Christoffel symbol $$\Gamma^2{}_{02}$$. Setting $$\mu=2$$, $$\alpha=0$$, and $$\beta=2$$, we get

\begin{align} \Gamma^2{}_{02}&=\frac12g^{20}(g_{00,2}+g_{20,0}-g_{02,0})+\frac12g^{21}(g_{01,2}+g_{21,0}-g_{02,1})\\ &+\frac12g^{22}(g_{02,2}+g_{22,0}-g_{02,2})+\frac12g^{33}(g_{03,2}+g_{23,0}-g_{02,3}). \end{align}

Only one of the 12 terms is nonzero:

$$\Gamma^2{}_{02}=\frac12g^{22}g_{22,0}=\frac12[a^{-2}(t)]\frac{\partial}{\partial t}a^2(t)=\frac{\dot a(t)}{a(t)}.$$

I wrote out the 12 terms in detail because in the general case they all matter. But when you have a diagonal metric, you don't need to bother writing out the terms with $$g^{20}$$, $$g^{21}$$, and $$g^{23}$$ because they are zero.

Since the right side of the formula for the Christoffel symbols is symmetric in $$\alpha$$ and $$\beta$$, this means we have also just calculated $$\Gamma^2{}_{20}$$. So there are really only 40 calculations, not 64.

Many of the Christoffel symbols will turn out to be zero, because the metric tensor is relatively simple. Thirteen of the sixteen components of the metric tensor are constants, so their derivatives are zero; and the three components that are functions are only a function of $$t$$ and not of $$x$$, $$y$$, or $$z$$.

• Thank you. Now I understand. You are best. – AlexSok Feb 18 at 6:04
• I fixed an incorrect factor of 2 in the final result. Sorry about that! – G. Smith Feb 20 at 18:26

FRW metric is written as

$$-c^2 d\tau ^2 = -c^2 dt^2 +a(t)^2 d\textstyle{\sum}^2$$

where for simplicity take $$k=0$$ in Cartesian coordinates which refers to zero curvature then $$d\textstyle{\sum}^2 = dx^2 +dy^2 + dz^2$$. Where the metric tensor $$g_{\mu \nu}$$ is

$$g_{\mu\nu}=\begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & a^2(t) & 0 & 0 \\ 0 & 0 & a^2(t) & 0 \\ 0 & 0 & 0 & a^2(t) \end{bmatrix}$$

Keep in mind that scale factor $$a(t)$$ doesn't have any directional dependence since

All the off-diagonal components of the metric is zero, their derivatives as well. So you only need to consider the diagonal terms. For time derivatives you need to calculate $$g_{00,0}$$, $$g_{11,0}$$, $$g_{22,0}$$, $$g_{33,0}$$. For coordinate derivatives calculate the following:

$$g_{\mu \nu , j}$$

$$g_{00,1}$$, $$g_{00,2}$$, $$g_{00,3}$$, $$g_{11,1}$$, $$g_{11,2}$$, $$g_{11,3}$$, $$g_{22,1}$$, $$g_{22,2}$$, $$g_{22,3}$$, $$g_{33,1}$$, $$g_{33,2}$$, $$g_{33,3}$$ where Greek letters go from $$0,1,2,3$$ and Latin letter from $$1,2,3$$.

After comma it denotes w.r.t what you are taking the derivative for instance $$g_{22 , 0}$$ means you're taking the derivative of second row second column (count from zero not one) w.r.t time and $$g_{33 , 1}$$ means you're taking the derivative of third row third column (count from zero not one) w.r.t $$x-$$ component. Rest is just taking the partial derivative of the scale factor squared since $$g_{00}= -1$$.