For relativistic particle dynamics, there are two different approaches to choosing a Lagrangian that give the same equations of motion:(adsbygoogle = window.adsbygoogle || []).push({});

The quadratic form is:

[itex]\mathcal{L} = \frac{m}{2} g_{\mu \nu} U^\mu U^\nu[/itex]

where [itex]U^\mu = \frac{d x^\mu}{d \tau}[/itex]

This is for the action that involves integration over proper time:

[itex]\mathcal{A} = \int d\tau \mathcal{L}[/itex]

But there is also an integral that is directly in terms of the invariant interval:

[itex]L = \sqrt{g_{\mu \nu} U^\mu U^\nu}[/itex]

where [itex]U^\mu = \frac{dx^\mu}{ds}[/itex], and [itex]s[/itex] is an arbitrary path parameter, which we can choose to be coordinate time, leading to a Lagrangian:

[itex]L = \sqrt{g_{00} + 2 g_{0i} v^i + g_{ij} v^i v^j}[/itex]

or [itex]L = \sqrt{1 - v^2}[/itex] (using inertial coordinates).

These two Lagrangians give the same equations of motion, which makes me think that they must be related in some way. The one with the square-root is a lot messier to work with, but its interpretation is a lot more direct: you're getting the equations of motion by extremizing the proper time, while the meaning of the quadratic form is obscure to me.

The quadratic form is a lot nicer when you want to add interactions:

[itex]\mathcal{L} = \frac{1}{2} m g_{\mu \nu} U^\mu U^\nu + q g_{\mu \nu} U^\mu A^\nu + \lambda \Phi[/itex]

where [itex]A^\nu[/itex] is the electromagnetic vector potential, and [itex]\Phi[/itex] is a scalar field (not to be confused with [itex]A^0[/itex]). It's not clear to me where to put the vector and scalar potentials into the square-root form of the Lagrangian.

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# I Relationship between quadratic and square-root lagrangians

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