# Lagrangian for a particle in a uniform magnetic field

1. Jul 1, 2012

### AbigailM

Any advice on deriving the lagrangian for a particle in a magnetic field?

$L=\frac{1}{2}mv^{2} + \frac{q}{c}\mathbf{v}\mathbf{.} \mathbf{A}$

I've been searching through Griffith's, Jackson, and google to no avail.
Can we start from the lorentz force and work backwards?

Thanks for the help.

2. Jul 5, 2012

### gabbagabbahey

The closest I've seen to a strict derivation of the E-M Lagrangian is to use a little vector calculus to rewrite the Lorentz Force Law (in Gaussian units) as

$$\mathbf{F} = -\nabla\left[qV - \frac{q}{c}\mathbf{v}\cdot\mathbf{A}\right] -q\frac{d\mathbf{A}}{dt}$$

And from there, postulate a generalized potential of $U(q,\dot{q},t) = qV - \frac{q}{c}\mathbf{v}\cdot\mathbf{A}$ and then (and this is very important!) check that the Euler-Lagrange equations result in the correct force law, regardless of your gauge choice.

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