# Finding the EOM for a charged relativistic particle

For an exercise sheet of a course in general relativity I'm asked to derive the equations of motion for a charged particle in an EM-field given by a potential $A^\mu$. I am give the action: $$S = -m\int ds-e\int A_\mu dx^\mu$$ With parametrization in proper time I get: $$S = -\int d\tau\left(mg_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta+eA_\mu\dot{x}^\mu\right)$$ where $\dot{x}^\alpha =\frac{d x^\alpha}{d\tau}$. Euler-Lagrange gives: $$0=\frac{\delta L}{\delta x^\alpha}-\frac{d}{d\tau}\frac{\delta L}{\delta\dot{x}^\alpha}$$ The first term is zero, so: $$0 = \frac{d}{d\tau}\left(2mg_{\alpha\beta}\dot{x}^\beta+eA_\alpha\right)=2mg_{\alpha\beta}\ddot{x}^\beta+e\dot{A}_\alpha$$ where in the first equality we used the fact that $g$ is a symmetric $(0,2)$-tensor. Now I'm asked to write the equations of motion in terms of the field tensor $F_{\mu\nu}=\frac{\partial A_\nu}{\partial x^\mu}-\frac{\partial A_\mu}{\partial x^\nu}$. I have tried to use the identity $$\dot{A}_\alpha = \frac{\partial A_\alpha}{\partial x^\nu}\dot{x}^\nu$$ but I couldn't get anywhere. Can someone help me out see how to do it (it's probably some dumb algebra trick) or say me where I'm wrong (if I'm wrong somewhere in the steps above)?

• The first term is not zero, because $A_\mu$ depends on $x^\alpha$ – Trimok Sep 26 '13 at 15:21

You're wrong in $\frac{\partial L}{\partial x^{\mu}} = 0$. Let's look the equations for $\alpha = 1, 2, 3$: $$\frac{\partial L}{\partial \mathbf r} = q\frac{\partial A_{\mu}}{\partial \mathbf r}\dot {x}^{\mu} = \partial_{\mathbf r }(\mathbf A \cdot \mathbf v) + \partial_{\mathbf r}\varphi .$$
Also you made the mistake in the summand $mg_{\alpha \beta}\dot {x}^{\alpha}\dot {x}^{\beta}$. $ds$ is equal to $\sqrt{g_{\alpha \beta }dx^{\alpha}dx^{\beta}}$, not to $g_{\alpha \beta }dx^{\alpha}dx^{\beta}$. Also, your convolution $\sqrt{g_{\alpha \beta }\dot {x}^{\alpha}\dot {x}^{\beta}}$ is equal only to $1$. So you will get $$S = -\int (m + qA_{\mu}\dot {x}^{\mu})d\tau$$