# Question on the geometry concerning the Least Action principle for a free Lagrangian

Consider a continuum of particles lying on a 1d curve. As time evolves the curve traces out a worldsheet in spacetime. If there is no external force acting on this curve as it evolves in time I was wondering if the trajectories of the particles that lie on the curve are geodesics of the worldsheet traced out in spacetime. In other words is it correct to say that minimizing the action:

$$S[\chi, \gamma] = \iint ~\mathrm dx~\mathrm dt\left(\dot\chi(x,t)^2+\dot\gamma(x,t)^2\right)$$

Is the same as minimizing each particle's path length in space time:

$$S[\chi, \gamma] = \iint ~\mathrm dx~\mathrm dt~\sqrt{\left(\dot\chi(x,t)^2+\dot\gamma(x,t)^2\right)}~?$$

Where $\chi(x,t)$, for example, would be the x-coordinate of the particle originating at position $x$, and $\gamma(x,t)$ would be the $y$-coordinate of the particle originating at position $x$. This would seem to be trivially true to me but I was hoping for some extra insight.

Also, would it be possible to have these geodesics cross paths in space time if this were to be true?

• Essentially a duplicate of physics.stackexchange.com/q/149082/2451 and links therein. Dec 13, 2016 at 7:15
• Just to clarify, $\dot\chi$ is $\frac{d\chi}{d\tau}$ for some parameter $\tau$, right? (Sorry, I'm a novice.)
– Hrhm
Dec 13, 2016 at 17:17
• Here the dot notation would correspond to a partial derivative in t. Dec 13, 2016 at 18:47

Action has a specific unit (Energy$\cdot$seconds), so these are not equivalent on that basis alone.
If $\mathcal{L}_t=0$, $\mathcal{L}_x=0$, $\mathcal{L}\neq 0$, and the Euler-Lagrange equations for $\mathcal{L}^2$ are satisfied, then the Euler-Lagrange equations for $\mathcal{L}$ are satisfied. The Euler-Lagrange equations for $\mathcal{L}^2$ are as follows: $$0=\frac{\partial}{\partial x}\frac{\partial}{\partial\chi_x}\mathcal{L}^2+\frac{\partial}{\partial t}\frac{\partial}{\partial\chi_t}\mathcal{L}^2-\frac{\partial}{\partial\chi}\mathcal{L}^2$$ $$0=\frac{\partial}{\partial x}\frac{\partial}{\partial\gamma_x}\mathcal{L}^2+\frac{\partial}{\partial t}\frac{\partial}{\partial\gamma_t}\mathcal{L}^2-\frac{\partial}{\partial\gamma}\mathcal{L}^2$$ Expanding the two equations: $$0=2\frac{\partial}{\partial x}\left(\mathcal{L}\frac{\partial\mathcal{L}}{\partial\chi_x}\right)+2\frac{\partial}{\partial t}\left(\mathcal{L}\frac{\partial\mathcal{L}}{\partial\chi_t}\right)-2\mathcal{L}\frac{\partial\mathcal{L}}{\partial\chi}$$ $$0=2\frac{\partial}{\partial x}\left(\mathcal{L}\frac{\partial\mathcal{L}}{\partial\gamma_x}\right)+2\frac{\partial}{\partial t}\left(\mathcal{L}\frac{\partial\mathcal{L}}{\partial\gamma_t}\right)-2\mathcal{L}\frac{\partial\mathcal{L}}{\partial\gamma}$$ Dividing by $2$ and expanding even further: $$0=\mathcal{L}_x\frac{\partial\mathcal{L}}{\partial\chi_x}+\mathcal{L}\frac{\partial}{\partial x}\frac{\partial\mathcal{L}}{\partial\chi_x}+\mathcal{L}_t\frac{\partial\mathcal{L}}{\partial\chi_t}+\mathcal{L}\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\chi_t}-\mathcal{L}\frac{\partial\mathcal{L}}{\partial\chi}$$ $$0=\mathcal{L}_x\frac{\partial\mathcal{L}}{\partial\gamma_x}+\mathcal{L}\frac{\partial}{\partial x}\frac{\partial\mathcal{L}}{\partial\gamma_x}+\mathcal{L}_t\frac{\partial\mathcal{L}}{\partial\gamma_t}+\mathcal{L}\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\gamma_t}-\mathcal{L}\frac{\partial\mathcal{L}}{\partial\gamma}$$ Combining terms: $$0=\mathcal{L}\left(\frac{\partial}{\partial x}\frac{\partial \mathcal{L}}{\chi_x}+\frac{\partial}{\partial t}\frac{\partial \mathcal{L}}{\chi_t}-\frac{\partial\mathcal{L}}{\partial\chi}\right)+\mathcal{L}_x\frac{\partial\mathcal{L}}{\partial\chi_x}+\mathcal{L}_t\frac{\partial\mathcal{L}}{\partial\chi_t}$$ $$0=\mathcal{L}\left(\frac{\partial}{\partial x}\frac{\partial \mathcal{L}}{\gamma_x}+\frac{\partial}{\partial t}\frac{\partial \mathcal{L}}{\gamma_t}-\frac{\partial\mathcal{L}}{\partial\gamma}\right)+\mathcal{L}_x\frac{\partial\mathcal{L}}{\partial\gamma_x}+\mathcal{L}_t\frac{\partial\mathcal{L}}{\partial\gamma_t}$$ Using the fact that $\mathcal{L}_x=0$ and $\mathcal{L}_t=0$: $$0=\mathcal{L}\left(\frac{\partial}{\partial x}\frac{\partial \mathcal{L}}{\chi_x}+\frac{\partial}{\partial t}\frac{\partial \mathcal{L}}{\chi_t}-\frac{\partial\mathcal{L}}{\partial\chi}\right)$$ $$0=\mathcal{L}\left(\frac{\partial}{\partial x}\frac{\partial \mathcal{L}}{\gamma_x}+\frac{\partial}{\partial t}\frac{\partial \mathcal{L}}{\gamma_t}-\frac{\partial\mathcal{L}}{\partial\gamma}\right)$$ Using the fact that $\mathcal{L}\neq 0$: $$0=\frac{\partial}{\partial x}\frac{\partial \mathcal{L}}{\chi_x}+\frac{\partial}{\partial t}\frac{\partial \mathcal{L}}{\chi_t}-\frac{\partial\mathcal{L}}{\partial\chi}$$ $$0=\frac{\partial}{\partial x}\frac{\partial \mathcal{L}}{\gamma_x}+\frac{\partial}{\partial t}\frac{\partial \mathcal{L}}{\gamma_t}-\frac{\partial\mathcal{L}}{\partial\gamma}$$ $\blacksquare$