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In my classical electrodynamics class, we use the Lagrangian of the relativistic free massive particle as $$L = - mc\sqrt{\dot{r}\cdot\dot{r}}.$$ Where $\dot{r}^\mu = u^\mu = \frac{dr^\mu}{d \tau}$; $r^\mu = (ct, x, y, z)^\mu$; $\tau$ is the proper time; and we use the $(+,-,-,-)$ sign convention for the metric.

This recovers all of the correct equations of motion provided we only use the fact that $\dot{r}\cdot\dot{r} = c^2$ after we have applied Lagrange's equations of motion. This would make sense if the fact that $\dot{r}\cdot\dot{r} = c^2$ was a consequence of Lagrange's equations (as we always let the Lagrangian tell us the equations of motion, and never substitute them back into the Lagrangian). However, as far as I can see $\dot{r}\cdot\dot{r} = c^2$ is not a consequence of Lagrange's equations, it's just always true.

I believe $\dot{r}\cdot\dot{r} = c^2$ follows immediately from the definition of $\dot{r}$. I must be wrong?

Qmechanic
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Jack
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2 Answers2

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  1. Conceptionally, it may help to realize that the action for a relativistic massive point particle$^1$ $$\begin{align}S ~=~& \int_{\lambda_i}^{\lambda_f} \!d\lambda~ L, \qquad L~:=~- m_0c\sqrt{ \dot{x}^2}, \cr \dot{x}^2~:=~&g_{\mu\nu}~\dot{x}^{\mu}\dot{x}^{\nu}, \qquad \dot{x}^{\mu}~:=~\frac{dx^{\mu}}{d\lambda},\end{align} \tag{1} $$ has a world-line (WL) reparametrization gaugesymmetry $$\lambda\longrightarrow \lambda^{\prime}=f(\lambda),\tag{2}$$ where $\lambda$ is an arbitrary world-line parameter.

  2. A common gauge choice is to choose the static gauge, i.e. $\lambda=t$ equal to coordinate time $t$. In Minkowski spacetime this leads to a Lagrangian $$L~=~-\frac{m_0c^2}{\gamma({\bf v})}\tag{3}$$ with the Lorentz factor $\gamma({\bf v})$.

  3. Before applying the variational principle, it is inconsistent to impose that $\lambda=\tau$ is proper time $\tau$, since (among other things) then $\dot{x}^2$ becomes just $c^2$, and we have destroyed the variational principle, cf. OP's title question.

  4. However, after applying the variational principle to derive the Euler-Lagrange (= geodesic) equations, it is consistent to choose $\lambda=\tau$.

  5. See also e.g. my related Phys.SE answer here.

$^1$ In this answer we use the same metric sign convention $(+,-,-,-)$ as OP.

Qmechanic
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You are right that $u^\mu u_\mu=1$ (in natural units) follows from the definition of $u^\mu$, or more precisely from the fact that you chose the proper time $\tau$ to parametrize the trajectory $r^\mu(\tau)$. Indeed the action of a relativistic free particle is just proportional to its proper time,

$$ S = m \int_0^\tau \sqrt{u^\mu u_\mu}d\tau' = m \int_0^\tau d\tau' = m\tau$$

which tells you that the free particle moves in a trajectory of extremal (maximal in this case) proper time. But this is not particularly helpful if you want to express the trajectory in terms of the coordinates $(t,x,y,z)$ - in order to do that you need to express $\tau$ as a function of the coordinates $r^\mu$, and apply variations to the trajectory, which results in the Euler-Lagrange equations.

J. Delaney
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