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Is there an analytical mechanics with SR? Of course you can write down the Lagrangian and Hamiltonian of a free particle. What about non-free? Are there any problems? To be specific: what would the Lagrangian and Hamiltonian look like for a spherical pendulum considering SR?

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  • $\begingroup$ A pendulum requires gravity, so now you are talking about general relativity. Nothing stops you from solving the equations of motion of a pendulum in a weak (or strong) gravity metric caused by a small or large gravitating body. $\endgroup$ – CuriousOne Jun 27 '15 at 9:33
  • $\begingroup$ Thanks. Lets stick to flat spacetime. So in that case wouldnt an infinite charged plane and the pendulum with an opposite charge lead to a spherical pendulum without gravity. $\endgroup$ – lalala Jun 27 '15 at 12:53
  • $\begingroup$ Gravity means that spacetime is not flat, no matter how much you wish that it were. $\endgroup$ – CuriousOne Jun 27 '15 at 15:58
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A Lagrangian can easily be written down for a relativistic particle in a curved spacetime (i.e., under the influence of gravity.) Specifically, the "action" is the proper time between two events along a particle's world-line, and the particle's trajectory will extremize the proper time between these events: $$ S = \tau = \int \sqrt{ - g_{\mu \nu} dx^\mu dx^\nu } $$ Here, $s$ is a parameter along the particle's worldline, and $x^\mu$ are a set of spacetime coordinates.

In particular, if we want to look at a test particle moving in a weak gravitational field, then the metric is such that $$ S = \int \sqrt{ \left( 1 + \frac{2 \Phi}{c^2} \right) dt^2 - \left( 1 - \frac{2 \Phi}{c^2} \right) d\vec{r}^2 } = \int \sqrt{ \left( 1 + \frac{2 \Phi}{c^2} \right) - \left( 1 - \frac{2 \Phi}{c^2} \right) \vec{v}^2 } dt $$ where $\Phi (\vec{r})$ is the Newtonian gravitational potential and $\vec{v} = d\vec{r}/dt$ is the coordinate velocity of the particle. Extremizing this integral over all paths $\vec{r}(t)$ will yield the equations of motion for the particle.1 You can also define a Hamiltonian from this "Lagrangian" (i.e., the integrand above) by taking a Legendre transform in the usual way.

Free bonus Lagrangian: If you want to add a charge to your relativistic particle, you can do that too; the Lagrangian becomes $$ S = \int \sqrt{ - g_{\mu \nu} dx^\mu dx^\nu } + q \int A_\mu dx^\mu $$ where $A_\mu$ is the relativistic four-vector potential.


1 This assumes that $t$ is a valid parameter for the particle's trajectory, which is true in the case of weak gravitational fields but may not be so in stronger gravitational fields (e.g., black holes.)

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    $\begingroup$ Thanks. How do u add the rope constraint this Lagrangian. $\endgroup$ – lalala Jun 27 '15 at 13:41
  • $\begingroup$ Same way you would in the non-relativistic case: either set $|\vec{r}| = L$ in the Lagrangian and use $\theta$ and $\phi$ as generalized coordinates, or use a Lagrange multiplier to enforce the constraint. $\endgroup$ – Michael Seifert Jun 27 '15 at 14:29
  • $\begingroup$ A little addendum to the answer: if you pass to a system of particles instead and want to maintain Lorentz covariance then you have to introduce fields interactions, otherwise the invariance cannot hold (which is the basis of why you need field theories instead of classical mechanics). $\endgroup$ – gented Jun 27 '15 at 17:16

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