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I try to get the momenta $$p_{\sigma} = \frac{\partial L}{\partial \dot{x}^{\sigma}}$$ from the free one particle Lagrangian $$L = -mc\sqrt{-\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}.$$ I got to the equation

$$p_{\sigma} = \frac{mc\dot{x}_{\sigma}}{\sqrt{-\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}}$$

but I don't know how to solve for the velocities $\dot{x}^{\sigma}$ so that I can put this into the definition of the Hamiltonian. I know the solution to this from a Wikipedia article, but the steps how to solve it are not given. Can somebody please show them to me?

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TL;DR: OP's $\dot{x} \leftrightarrow p$ relation is not directly invertible, i.e. the Legendre transformation is singular. This reflects an underlying world-line (WL) reparametrization symmetry $$ \tau\quad\longrightarrow\quad\tau^{\prime}~=~f(\tau) \tag{1}$$ of the action.

In more details:

  1. The action for a massive relativistic point particle is$^1$ $$\begin{align} S_0~=~&\int \! d\tau~ L_0,\cr L_0~:=~& -m\sqrt{-\dot{x}^2}, \cr \dot{x}^2~:=~&g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~<~0,\cr \dot{x}^{\mu} ~:=~&\frac{dx^{\mu}}{d\tau}, \end{align}\tag{2}$$ where $\tau$ is the world-line (WL) parameter $\tau$ (which does not have to be the proper time).

  2. The components of the Lagrangian 4-momentum $$ p_{\mu}~=~\frac{\partial L_0}{\partial \dot{x}^{\mu}}~\stackrel{(2)}{=}~\frac{m\dot{x}_{\mu}}{\sqrt{-\dot{x}^2}} \tag{3} $$ are not all independent since its length-square satisfies a primary constraint $$ p^2~\stackrel{(3)}{=}~-m^2 ,\tag{4}$$ aka. the mass-shell condition.

  3. The bare Hamiltonian vanishes $$ H_0~=~p_{\mu}\dot{x}^{\mu}-L_0~\stackrel{(2)+(3)}{=}~0,\tag{5}$$ cf. e.g. this related Phys.SE post.

  4. The total Hamiltonian becomes of the form "Lagrange multiplier times constraint" $$H~=~ \frac{e}{2}(p^2+m^2),\tag{6} $$ cf. e.g. this related Phys.SE post.

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$^1$ In this answer we work in units where the speed-of-light $c=1$ is one, and we use the Minkowski sign convention $(−,+,+,+)$.

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