2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ $$ f(x,y) = x^2+y^2=1 \quad \stackrel{???}{\Longrightarrow}\quad \left. \begin{cases} \dfrac{\partial f}{\partial x} & =2x=0\\ \dfrac{\partial f}{\partial y} & =2y=0 \end{cases} \right\} $$ $\endgroup$ – Frobenius Aug 9 '18 at 11:17
2
$\begingroup$

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 As a result, the principle of stationary action is no longer applicable.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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.