Charge, velocity-dependent potentials and Lagrangian

Question

Given an electric charge $q$ of mass $m$ moving at a velocity ${\bf v}$ in a region containing both electric field ${\bf E}(t,x,y,z)$ and magnetic field ${\bf B}(t,x,y,z)$ (${\bf B}$ and ${\bf E}$ are derivable from a scalar potential $\phi(t, x, y, z) $and a vector potential ${\bf A}(t,x,y,z)$),

knowing that

1. ${\bf E}=- \nabla \phi - \frac{\partial {\bf A}} {\partial t}$
2. ${\bf B}= \nabla \times {\bf A} $
3. $U=q \phi - q {\bf A} \cdot{\bf v}$ ($U$ is the velocity-dependent potential)

the Lagrangian is $$L=(1/2) m v^2-U=(1/2) m v^2- q\phi + q{\bf A} \cdot{\bf v}.$$

Considering just the $x$-component of Lagrange's equation, how can I obtain $$m \ddot{x}=q\left (v_x \frac{\partial A_x}{\partial x} + v_y \frac {\partial A_y}{ \partial x} + v_z \frac{\partial A_z} {\partial x}\right )-q\left (\frac{\partial \phi }{\partial x} + \frac{d A_x}{dt}\right) ~?$$

Answer 1

Use the Euler-Lagrange equations (motivated in a previous post of mine). These are

$$ \partial_x L - \frac{\mathrm{d}}{\mathrm{d}t} \left( \partial_{v_x} L \right) = 0 $$

We therefore need $$ \partial_x L = -q \partial_x \Phi + q \left( \partial_x A_x v_x + \partial_x A_y v_y + \partial_x A_z v_z \right) $$ $$ \partial_{v_x} L = m v_x + q A_x $$ $$ \frac{\mathrm{d}}{\mathrm{d}t} \left( \partial_{v_x} L \right) = m \dot v_x + q \dot A_x $$

Putting this all together into the first equation and adding $\dot v_x m$ gives the equation for which you are looking.

Note especially that each component of $A$ is a function of all three coordinates: $A_i = A_i(x,y,z)$; the same goes for $\Phi = \Phi(x,y,z)$ (where usually $A_0$ is identified with $\Phi$ in relativistic electrodynamics).

Answer 2

Use the Euler-Lagrange equation:

$$\frac{\partial L}{\partial x} = \frac{d}{dt}\frac{\partial L}{\partial v_x}$$ $$-q \frac{\partial \phi}{\partial x} + q\frac{\partial}{\partial x}\left(\mathbf{A} \cdot \mathbf{v}\right) = m \dot{v_x} + q \frac{dA_x}{dt}$$