Calculating Christoffel Symbols from a given line element

[Destroxia]

Homework Statement

Given some 2D line element, $ds^2 = -dt^2 +x^2 dx^2$, find the Christoffel Symbols, $\Gamma_{\beta \gamma}^{\alpha}$.

Homework Equations

$\Gamma_{\beta \gamma}^{\alpha} = \frac {1}{2} g^{\delta \alpha} (\frac{\partial g_{\alpha \beta}}{\partial x^\gamma} + \frac{\partial g_{\alpha \gamma}}{\partial x^{\beta}} - \frac{\partial g_{\beta \gamma}}{\partial x^\alpha})$

$ds^2 = g_{\alpha \beta} dx^\alpha dx^\beta$

The Attempt at a Solution

While I understand the theory behind what I'm doing, I'm lost when it comes to solving these Christoffel Symbols. The main issue for me is, although we know the equation $ds^2 = g_{\alpha \beta} dx^\alpha dx^\beta$, I don't know how to derive a metric from this for $ds^2 = -dt^2 +x^2 dx^2$, such that I could then simply use the Christoffel Symbol Equation to read off the coordinates.

What I assume I should do is somehow create a metric for this line element, and then proceed to relate the coefficients of this metric to the Christoffel Symbol Equation, so for example, we have some metric (not the correct one):

\begin{equation} g_{\alpha \beta} =
\left({\begin{array}{cc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}}\right)
\end{equation}

and then in order to solve these symbols, we need to follow the form of $\Gamma_{\beta \gamma}^{\alpha} = \frac {1}{2} g^{\delta \alpha} (\frac{\partial g_{\alpha \beta}}{\partial x^\gamma} + \frac{\partial g_{\alpha \gamma}}{\partial x^{\beta}} - \frac{\partial g_{\beta \gamma}}{\partial x^\alpha})$

Although, another hang up is index notation, I just have no idea how these $\alpha, \beta, and \gamma$ relate to each other. Any explanation, or resource for reading this index notation would be very helpful.

 

[Orodruin]

The expression $ds^2 = g_{\mu\nu} dx^\mu dx^\nu$ has repeated indices in the RHS so this indicates a sum. For a two-dimensional manifold with coordinates $t$ and $x$, writing out the sum would lead to
$$
ds^2 = g_{tt} dt^2 + g_{tx} dt\, dx + g_{xt} dx\, dt + g_{xx} dx^2.
$$
Now, in physics literature, you will typically have the order of $dx$ and $dt$ interchangeable (also, the metric is symmetric) and so this would be rewritten
$$
ds^2 = g_{tt} dt^2 + 2g_{tx} dt\, dx + g_{xx} dx^2.
$$
If you have an explicit expression for $ds^2$, you can just identify the components of the metric tensor from there.

However(!), note that using the expression you give for the Christoffel symbols in terms of the metric tensor is not the most efficient way of computing the Christoffel symbols of the Levi-Civita connection for a given metric. Instead, you can go back to how the expression is typically derived from the geodesic equations as derived from finding the stationary paths of
$$
S = \int g_{\mu\nu} \dot x^\mu \dot x^\nu d\tau.
$$
You will get the geodesic equations from which you can directly read off the Christoffel symbols. In effect, you will be doing the same calculations, but it will be much easier from a bookkeeping standpoint. (To get the integrand, replace any occurrence of $dx^\mu$ in the line element with $\dot x^\mu$.)

 

[Destroxia]

The expression $ds^2 = g_{\mu\nu} dx^\mu dx^\nu$ has repeated indices in the RHS so this indicates a sum. For a two-dimensional manifold with coordinates $t$ and $x$, writing out the sum would lead to
$$
ds^2 = g_{tt} dt^2 + g_{tx} dt\, dx + g_{xt} dx\, dt + g_{xx} dx^2.
$$
Now, in physics literature, you will typically have the order of $dx$ and $dt$ interchangeable (also, the metric is symmetric) and so this would be rewritten
$$
ds^2 = g_{tt} dt^2 + 2g_{tx} dt\, dx + g_{xx} dx^2.
$$
If you have an explicit expression for $ds^2$, you can just identify the components of the metric tensor from there.

However(!), note that using the expression you give for the Christoffel symbols in terms of the metric tensor is not the most efficient way of computing the Christoffel symbols of the Levi-Civita connection for a given metric. Instead, you can go back to how the expression is typically derived from the geodesic equations as derived from finding the stationary paths of
$$
S = \int g_{\mu\nu} \dot x^\mu \dot x^\nu d\tau.
$$
You will get the geodesic equations from which you can directly read off the Christoffel symbols. In effect, you will be doing the same calculations, but it will be much easier from a bookkeeping standpoint. (To get the integrand, replace any occurrence of $dx^\mu$ in the line element with $\dot x^\mu$.)

So going off this, just as a first step I could some kind of coefficient relation? Like $g_{tt} dt^2 + 2 g_{tx} dt dx + g_{xx} dx^2 = - dt^2 + x^2 dx^2$. And from that follows, $g_{tt} = -1, g_{tx} = 0,$ and $g_{xx} = x^2$ ? I'm still having trouble understanding how these metric components of t and x relate to the general components of the Christoffel Symbol's $\alpha, \beta,$ and $\gamma$.

 

[Orodruin]

So going off this, just as a first step I could some kind of coefficient relation? Like $g_{tt} dt^2 + 2 g_{tx} dt dx + g_{xx} dx^2 = - dt^2 + x^2 dx^2$. And from that follows, $g_{tt} = -1, g_{tx} = 0,$ and $g_{xx} = x^2$ ?

Yes.

I'm still having trouble understanding how these metric components of t and x relate to the general components of the Christoffel Symbol's $\alpha, \beta,$ and $\gamma$.

You can either use the expression for the Christoffel symbols in terms of the metric (remember that repeated indices imply a sum) or use the approach I suggested in #2. The second option is far easier from a bookkeeping standpoint so I suggest you try that.