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You could ask the same exact thing for the Lagrangian of a nonrelativistic particle. It's an exactly analogous situation. $$L = \frac{1}{2} m \dot x^2$$ Now, if we differentiate the Lagrangian with respect to time, we get \begin{align} \frac{d}{dt} L &= \frac{1}{2} m \frac{d}{dt} \dot x^2 \\ &= m \dot x \ddot x \\ &\neq 0 \end{align} It's not $... 14 OP's Lagrangian is indeed constant on-shell, but not necessarily off-shell. In contrast, the principle of stationary action compares various possibly off-shell paths. 1 Yes, the main point is that the arc length$^1$$$ds~=~cd\tau~=~\sqrt{-g_{\mu\nu}(x) dx^{\mu}dx^{\nu}}\tag{A}$$ in spacetime is the speed of light$c$times the proper time$d\tau$, so that $$c\dot{\tau}~=~\dot{s}~=~\sqrt{-g_{\mu\nu}(x) \dot{x}^{\mu}\dot{x}^{\nu}},\tag{B}$$ where dot mean differentiation wrt. the worldline (WL) parameter$\lambda$(which is ... 1 For a non varying Lagrangian surely the action will not vary with trajectory? The basic confusion is "constant in time" versus "constant over trajectories". The action is the time integral of the Lagrangian, and is a "functional" mapping an entire (hypothetical) trajectory to a number. Even if every trajectory had a constant-in-... 1 That seems to be a misunderstanding. The off-shell relation (3) with$\lambda\neq 0$is a sufficient (as opposed to a necessary) condition for the stationary action principles (1) & (2) to be equivalent, i.e. have the same stationary paths in phase space. These paths are solutions to Kamilton's and Hamilton's equations, respectively. For the various ... 1 Assuming that$\delta p$,$\delta q$are zero at$t=t_0, t_1$so that we can ignore integrated out bits, we have $$\delta S= \int_{t_0}^{t_1} \left\{\left(- \frac{d p}{dt} -\frac{\partial H}{\partial q}\right)\delta q(t) +\left(\frac{dq}{dt}- \frac{\partial H}{\partial p}\right)\delta p(t)\right\}dt$$ so stationarity needs$\$ \frac{d p}{dt} =-\frac{\...