In Relativity the Lagrangian of a free particle is \begin{align} \mathcal L=\sqrt{g_{ab}\frac{dx^a}{d\tau}\frac{dx^b}{d\tau}}\end{align} Since $\mathcal L$ is parameterization invariant we can always set $$\mathcal L=1.$$ In that case how can the Euler-Lagrange equation

\begin{align} \frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\partial L}{\partial \dot{x}^\mu}\right) - \frac{\partial L}{\partial x^\mu} &=0 \end{align} make sense? How can $\frac{\partial \mathcal L}{\partial \dot{x}^\mu}$ and $\frac{\partial \mathcal L}{\partial {x}^\mu}$ not be zero?

  • $\begingroup$ $$ f(x,y) = x^2+y^2=1 \quad \stackrel{???}{\Longrightarrow}\quad \left. \begin{cases} \dfrac{\partial f}{\partial x} & =2x=0\\ \dfrac{\partial f}{\partial y} & =2y=0 \end{cases} \right\} $$ $\endgroup$
    – Frobenius
    Aug 9, 2018 at 11:17

1 Answer 1


Recall that the parametrization interval, say $[a,b]$, is fixed in the principle of stationary action, and assumed common to all virtual paths.

If we choose unit parametrizations $L=1$ for all virtual paths, the parametrization interval $[a,b]$ would obvious depend on the virtual path (since not all paths have the same length). As a result, the principle of stationary action is no longer applicable.

TL;DR: We are not allowed to choose unit parametrizations $L=1$ before the variation.

Afterwards is another story, cf. my Phys.SE answer here.


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