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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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.
