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## Homework Statement

We've got a line element [tex]ds^2 = f(x) du^2 + dx^2[/tex] From that we should find the geodesic equation

## Homework Equations

Line Element:

[tex]ds^2 = dq^j g_{jk} dq^k[/tex]

Geodesic Equation:

[tex]\ddot{q}^j = -\Gamma_{km}^j \dot{q}^k \dot{q}^m [/tex]

Christoffel Symbol:

[tex]\Gamma_{km}^j = \frac{g^{jl}}{2} \left( \frac{\partial g_{lk}}{\partial q^m} + \frac{\partial g_{lm}}{\partial q^k} - \frac{\partial g_{km}}{\partial q^l}\right)[/tex]

Lagrangian in this case (no potential, just free particle)[/B]: [tex]L = T = \frac{m}{2}\dot{q}^j g_{jk} \dot{q}^k[/tex]

**Coordinates: u and x**

3. The Attempt at a Solution

3. The Attempt at a Solution

I am stuck because I am not getting the same equations of motions for my coordinate u from the Lagrangian (with Euler-Lagrange) and the Geodesic equation, and I think, it is because I got the wrong inverse metric tensor.

My metric tensor looks like this: \begin{equation} \begin{pmatrix}f(x) & 0 \\ 0 & 1\end{pmatrix}\end{equation} .

The inverse: \begin{equation}\begin{pmatrix}1/f(x) & 0 \\ 0 & 1\end{pmatrix}\end{equation}

I got 8 possible Christoffel-Symbols, but only 2 will be non-zero:

\begin{equation}\Gamma_{ux}^{u} =\Gamma_{xu}^{u} = \frac{1}{f(x)} \frac{1}{2} \frac{\partial f(x)}{\partial x}\end{equation}

\begin{equation}\Gamma_{uu}^{x} = -\frac{1}{2} \frac{\partial f(x)}{\partial x}\end{equation}

If I plug those into the geodesic equation, I get 2 equations of motion. If I use the Euler-Lagrange-Equation to derive equations of motion from the lagrangian, I get the same equation of motion for x, but not for u.

Geodesic u:

\begin{equation}\ddot{u} + \frac{1}{2}\frac{1}{f(x)}\frac{\partial f(x)}{\partial x} \dot{u}\dot{x} = 0\end{equation}

Euler-Lagrange for u

**:**\begin{equation}\ddot{u}\dot{x}\frac{\partial f(x)}{\partial x} = 0 \end{equation}

I think it has to do with the inverse, cause for x I get the same equation for both ways:

\begin{equation}\ddot{x} - \dot{u}^2\frac{1}{2}\frac{\partial f(x)}{\partial x} = 0\end{equation}

So my question is, is the metric tensor and it's inverse correct? If yes, where am I wrong?

Thank you ^^!So my question is, is the metric tensor and it's inverse correct? If yes, where am I wrong?

Thank you ^^!