Euler-Lagrange equations in relativity (Goldstein)

Question

In order to have a covariant formulation of special relativity, we stop using the time $t$ as a parameter and we choose some invariant parameter. In Goldstein (third edition), chapter $7.10$, it goes through this derivation making an argument about why proper time $\tau$ can't be the parameter we are looking for because of the constraint on 4-velocities. \begin{equation} u_\nu u^\nu=c^2 \tag{1} \end{equation} It then chooses another parameter $\theta$ and derives Euler Lagrange equations using this new parameter $\theta$. At the end he chooses \begin{equation}\theta=\tau\tag{*}\end{equation} Why does this allow us to ignore the constraint $(1)$?

We're basically using proper time the whole time but ignoring the constraint and using at the end.

I've read something similar on another book (which used $s=\int ds$ where $ds^2=dx^\nu dx_\nu$, thought) which said that to find the variation of action:

\begin{equation} S=\int Lds \end{equation}

We should also consider the variation of $ds$ with coordinates. Then it introduces a new parameter that is eventually replaced by $s$. How do you account for that?

Answer 1

1. Since the 4-velocity $$u^{\mu}~:=~\frac{dx^{\mu}}{d\tau} \tag{U}$$ by definition is the 4-position differentiated wrt. proper time $\tau$, the condition $$\begin{align} c^2\mathrm{d}\tau^2 ~=~&\pm \eta_{\mu\nu}\mathrm{d}x^{\mu} \mathrm{d}x^{\nu} \cr~\Updownarrow~&\cr u^{\mu} u_{\mu}~=~&\pm c^2 \end{align}\tag{C}$$ holds independently of any worldline (WL) parametrization $\theta$ [in Minkowski signature $(\pm,\mp,\mp,\mp)$, respectively].

2. Goldstein considers the action for a massive relativistic point particle $$\begin{align} I ~=~&\int_{\theta_i}^{\theta_f} \! d\theta ~\Lambda, \cr \Lambda~=~&-mc\sqrt{\pm x^{\prime}_{\mu}x^{\prime \mu}}, \cr x^{\prime \mu}~:=~\frac{dx^{\mu}}{d\theta},\end{align}\tag{I}$$ which is WL reparametrization invariant $\theta\to\tilde{\theta}=f(\theta)$.

3. It is important to realize that the 4 quantities $x^{\prime \mu}$ are not constrained by eq. (C). ($\leftarrow$ This is OP's main question.)

4. The corresponding Euler-Lagrange (EL) equation$^1$ $$ \frac{d}{d\theta} \left( \frac{mc x^{\prime}_{\mu}}{\sqrt{\pm x^{\prime}_{\mu}x^{\prime \mu}}}\right)~\approx~0\tag{EL} $$ is WL reparametrization covariant.

5. After the variation, it is legitimate to choose the parametrization $\theta=\tau$. The EL equation then simplifies [with the help of eq. (C)] to $$ \frac{d(m u_{\mu})}{d\tau} ~\approx~0,\tag{EL'}$$ i.e. the 4-acceleration is zero on-shell.

6. Goldstein makes the pragmatic observation that if we a priori choose $\theta=\tau$ in the Lagrangian $\Lambda$ [and somehow ignore the constraint (C)], then we can formally write down the correct EL-equation [with $\theta=\tau$].

7. Although Goldstein obtains the correct EL-equation by this dirty trick, it is conceptionally very misleading. In fact, if we literally choose the parametrization $\theta=\tau$ in the action prior to the variation then the action would become a constant off-shell: $$ \begin{align} I ~=~&\int_{\tau_i}^{\tau_f} \! d\tau~(-mc)\sqrt{\pm u_{\mu}u^{\mu}}\cr ~=~&-mc^2(\tau_f-\tau_i).\end{align}\tag{I'}$$ Phrased differently: all virtual paths would have the same value, i.e. the stationary action principle becomes ill-defined.

References:

1. H. Goldstein, Classical Mechanics, 2nd edition; Section 7.9.

2. H. Goldstein, Classical Mechanics, 3rd edition; Section 7.10.

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$^1$ The $\approx$ symbol means equality modulo EOM.

Answer 2

Proper time $\tau$ can be defined as the parametrization of $x^\mu(\tau)$ such that : $$u_\mu u^\mu = \frac{\text d x_\mu}{\text d\tau}\frac{\text d x^\mu}{\text d\tau} = c^2$$

If you choose an arbitrary parametrization $x^\mu(\theta)$, then you have : $$\frac{\text d x_\mu}{\text d\theta}\frac{\text d x^\mu}{\text d\theta} = \left(\frac{\text d\tau}{\text d\theta}\right)^2 \frac{\text d x_\mu}{\text d\tau}\frac{\text d x^\mu}{\text d\tau} = \left(\frac{\text d\tau}{\text d\theta}\right)^2 c^2$$

so contraint $(1)$ is indeed lifted.

Since the observables are $x^i$ in terms of $x^0$, the parametrization of the trajectory is irrelevant. The action and the equations of motion should be invariant under reparametrization. It is easier to work out the Euler-Lagrange equation with an arbitrary parametrization, and choose a sensible one (proper time) at the end.