The relativistic Lagrangian is given by
$$L = – m_0 c^2 \sqrt{1 – \frac{u^2}{c^2}} + \frac{q}{c} (\vec u \cdot \vec A) – q \Phi $$
I need to derive, $\displaystyle \frac{d\vec p}{dt} = q \left( \vec E + \frac 1 c (\vec u \times \vec B)\right)$. Everything (in my note) is fine until there is one expression which I don't understand. It writes,
$$ \frac 1 c \nabla (\vec u \cdot \vec A ) = \nabla \Phi $$
to make $\displaystyle \vec E = – \nabla \Phi – \frac 1 c \frac{\partial \vec A}{\partial t}$. Why would dot product of $\vec u$ and $\vec A$ be $c$ times the scalar potential?
Here is the derivation as requested:
The equation of motion is given by
$$\frac{d}{dt} \left( \frac{\partial L}{\partial u_i} \right ) – \frac{\partial L}{\partial x_i} = 0$$
Since the relativistic Lagrangian is transitionally invariant, there is no dependence on coordinates so the last term is zero. taking derivative w.r.t $u_i$
$$\frac{d}{dt} \left( \frac{m_0 u_i}{\sqrt{1 – \frac{u^2}{c^2}}} + \frac{q}{c} A_i \right ) = 0 $$
Adding $i$'s, we get
$$ \frac{d}{dt} \left( \gamma m_0 \vec u + \frac{q}{c} \vec A \right ) = 0$$
$$\implies \frac{d \vec p}{dt} + \frac{q}{c} \frac{d\vec A }{dt} = 0$$
\begin{align*}
\implies \frac{d \vec p}{dt} &= – \frac{q}{c} \left( \frac{\partial \vec A }{\partial t} + (\vec u \cdot \nabla)\vec A\right )\\
&= – \frac{q}{c} \left( \frac{\partial \vec A }{\partial t} + \nabla (\color{Red}{\vec u \cdot \vec A})- \vec u \times \left( \nabla \times \vec A \right )\right )\\
&= – \frac{q}{c} \left( \frac{\partial \vec A }{\partial t}+ \nabla \color{Red}{(c\Phi)} – \vec u \times \vec B \right )\\
&= q \left( -\nabla \Phi – \frac 1 c \frac{\partial \vec A }{\partial t} + \frac 1 c \vec (u \times \vec B) \right)\\
&= q\left ( \vec E + \frac 1 c \vec (u \times \vec B) \right )
\end{align*}
Best Answer
This is not correct: "Since the relativistic Lagrangian is transitionally invariant, there is no dependence on coordinates so the last term is zero." In fact, $$ \frac{\partial L}{\partial {\bf r}} = q\nabla\left[ \frac{1}{c} \left({\bf u} \cdot {\bf A}\right) - \phi \right] = q\left\{\frac{1}{c} \left[\left({\bf u} \cdot \nabla \right){\bf A} + {\bf u} \times \left(\nabla \times {\bf A}\right) \right]- \nabla\phi\right\}. $$ Setting this equal to $$ \frac{d}{dt}\left(\frac{\partial L}{\partial {\bf u}} \right)= \frac{d}{dt} \left[ \frac{\partial }{\partial {\bf u}}\left( -m c^2 \sqrt{1 - \frac{u^2}{c^2}} + \frac{q}{c} {\bf u} \cdot {\bf A}\right) \right]= \frac{d}{dt}\left(\gamma m {\bf u}\right) +\frac{q}{c} \frac{d{\bf A}}{dt}, $$ and noting that $$ \frac{d{\bf A}}{dt} = \frac{\partial {\bf A}}{\partial t} + \left({\bf u} \cdot \nabla\right) {\bf A}, $$ we get $$ \frac{d}{dt}\left(\gamma m {\bf u}\right) = q\left[- \nabla\phi - \frac{1}{c}\frac{\partial{\bf A}}{\partial t} + \frac{{\bf u}}{c} \times \left(\nabla \times {\bf A}\right) \right]. $$