Action is independent of parametrization

Question

Consider the action (in the context of calculus of variation) given by $$S=\int_a^b \Bigr|\frac{d\gamma}{dt} \Bigr|^2dt$$ where $\gamma:[a,b]\to M$ is a curve on a manifold $M$ parametrized by $t$. I want to show that if $\gamma'=\gamma\circ \tau$, where $\tau:[a,b]\to [a',b']$ is a new parametrization of the curve $\gamma$, the action is independent of this parametrization.

The chain rule works well if this is a problem on arc-length: this is because both the chain rule and the Jacobian will work nicely to cancel the $d\tau/dt$ term that appear when we change parameterization. However I could not get this for this action: I seem to have one extra copy of $d\tau/dt$ because the integrand is a square.

On the other hand, if I work this from the perspective of metric tensor, it seems to work: I simply have to think of something like $$g_{\bar a\bar b }\frac{dx^{\bar a}}{d\tau}\frac{dx^{\bar b}}{d\tau}$$ and the two copies of $d\tau/dt$ will be absorbed by the metric tensor (or is it?): what goes wrong here?

I do have suspicion that the action may be in fact not equal. If so, are there relationships between them?

Answer 1

The action depends on the parametrization. For instance consider the parametrization $c:[0,1]\to \bf R$, $c(t)=t$, and $d(t)=t^2$, then the action of $c$ is $1$, the action of $d$ is $4/3$

Answer 2

$\tau:[a',b']\to [a,b]$ so that $\gamma \circ \tau$ makes sense. I'll denote by $()'$ the derivative of $()$, but $a'$ and $b'$ are still just real numbers. Let $\tilde\gamma = \gamma \circ \tau$. The goal is to say something about $\int_{a'}^{b'}\|\tilde\gamma'\|^2$.

\begin{align} \int_{a'}^{b'}\|\tilde\gamma'\|^2 &= \int_{a'}^{b'}\|\gamma'\circ \tau \cdot \tau'\|^2 = \int_{\tau^{-1}(a)}^{\tau^{-1}(b)}\|\gamma'\circ \tau\|^2 (\tau')^2 \\ &= \int_a^b\|\gamma'\circ \tau \circ \tau^{-1}\|^2 (\tau' \circ \tau^{-1})^2 [\tau^{-1}]^{'} = \int_a^b\|\gamma'\|^2 \left(\frac 1 {[\tau^{-1}]'}\right)^2 [\tau^{-1}]^{'} \\&= \int_a^b \frac {\|\gamma'\|^2} {[\tau^{-1}]'} \end{align}

This would imply that $S$ is not independent of $\tau$ ... but I can't find the mistake in it, so I'm posting it.