I have trouble understanding the following. If $$ \mathcal{L}(x,\dot{x})=\sqrt{\eta_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}}$$ why is then $$\frac{\partial\mathcal{L}}{\partial x^{\sigma}}=0~?$$
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Possible duplicates: https://physics.stackexchange.com/q/885/2451 and links therein. – Qmechanic Jul 28 '17 at 21:24
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Well, the Lagrangian does not explicitly depend on the position, therefore their corresponding derivatives vanish. And no, the velocity does not depend on the position either (outside the solution of the equation of motion). – gented Jul 28 '17 at 22:44
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This question is kind of simple and does not directly refer to Relativity. Because $\mathcal{L}(x,\dot{x})=\sqrt{\eta_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}}$ does not explicitly depend on $x$ but only on $\frac{dx^{\nu}}{d\tau}$, thus $$\mathcal{L}(x,\dot{x})=\mathcal{L}(\frac{dx^{\nu}}{d\tau}).$$ $$\frac{\partial\mathcal{L}}{\partial x^{\sigma}}=\frac{\partial\mathcal{L}(\frac{dx^{\nu}}{d\tau})}{\partial x^{\sigma}}=0$$ You can try to derive equations of motion from here.
