# Analytical mechanics with SR

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?

• 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. – CuriousOne Jun 27 '15 at 9:33
• 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. – lalala Jun 27 '15 at 12:53
• Gravity means that spacetime is not flat, no matter how much you wish that it were. – CuriousOne Jun 27 '15 at 15:58

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.)
• 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. – Michael Seifert Jun 27 '15 at 14:29