How does the Lagrangian work with a magnetic field?

Question

I was told the following Lagrangian is for a charged particle with spin moving in a constant magnetic field: $$ L = \frac{\left ( \vec p \right ) ^2}{2m} + \vec \mu \cdot \vec B$$

Let's just say $B$ is just constant in the $z$ direction, and not dependent on position. Then wouldn't the Lagrange equation give us $\dot p_x=0$ because the Lagrangian does not depend on $x$? This is obviously wrong because the particle will accelerate due to the Lorentz force. What mistake am I making? Or is this Lagrangian just wrong (the ones I've seen before use a vector potential)?

Answer 1

As was said in the commentary by @knzhou, what you have written is the Lagrangian for a particle with magnetic moment and no charge (e.g. like neutron). Then your logic is correct: as long as magnetic field is uniform, the particle will experience no force. The Lagrangian of a $charged$ particle in magnetic field on the other hand will read: $$L(\vec{v}, \vec{x}) = \frac{(m\vec{v}-e\vec{A}(\vec{x}))^2}{2m} + \vec{\mu}\cdot \vec{B}$$ where $\vec{A}$ is the vector potential such that $\vec{\nabla} \times \vec{A} = \vec{B}$. For example, for a uniform magnetic field $\vec{B} = (0,0,B)$, the vector potential can be chosen to be $\vec{A} = (0,-Bx,0)$. As you see, now the Lagrangian is explicitly coordinate-dependent, so $\dot{p_x} \neq 0$.