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

Question

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 http://physics.stackexchange.com/q/149082/2451 and links therein. — Qmechanic, Dec 13 '16 at 07:15
Just to clarify, $\dot\chi$ is $\frac{d\chi}{d\tau}$ for some parameter $\tau$, right? (Sorry, I'm a novice.) — Hrhm, Dec 13 '16 at 17:17
Here the dot notation would correspond to a partial derivative in t. — I.E.P., Dec 13 '16 at 18:47

score 0 · Answer 1 · answered Dec 13 '16 at 04:44

Action has a specific unit (Energy$\cdot$seconds), so these are not equivalent on that basis alone.

The Lagrangian is not the same as the trajectory. Rather, physical trajectories may be obtained from the Lagrangian through the application of the Euler-Lagrange equation, which is a result of minimizing the action functional.

Geodesics may not cross in spacetime. If they crossed, it would imply that two different particles could exist at the same place and time. To convince yourself that this is trouble, try to run the motion backwards: what happens when you reach a branch point where two geodesics have crossed? How do you know which particle is associated with the correct path? Classically acceptable trajectories should be time-reversible, and that cannot happen if geodesics cross in such a fashion.

score 0 · Answer 2 · answered Dec 18 '16 at 01:49

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 $\mathcal{L}\neq 0$: $$0=\frac{\partial}{\partial x}\frac{\partial \mathcal{L}}{\chi_x}+\frac{\partial}{\partial t}\frac{\partial \mathcal{L}}{\chi_t}-\frac{\partial\mathcal{L}}{\partial\chi}$$ $$0=\frac{\partial}{\partial x}\frac{\partial \mathcal{L}}{\gamma_x}+\frac{\partial}{\partial t}\frac{\partial \mathcal{L}}{\gamma_t}-\frac{\partial\mathcal{L}}{\partial\gamma}$$ $\blacksquare$

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

2 Answers2