I want to prove that $$ \vec{F}=d\vec{p}/dt=q\vec{E}+(q/c) \cdot v\times \vec{B} $$ in CGS system, using

$$ L=-mc^{2}/\gamma-q\phi+(q/c)\cdot \vec{v}\cdot \vec{A} \hspace{10mm} \tag 1 $$


$$ \nabla (\vec{v}\cdot\vec{A})=(\vec{v}\cdot\nabla)\vec{A}+\vec{v}\times\vec{B} \hspace{10mm} \tag 2 $$

Starting with (1) I can find that

$$ d\vec{p}/dt=dL/dx_{i}=-q\nabla\phi+(q/c)\cdot \nabla(\vec{v}\cdot\vec{A}) $$

And then introducing at this equation (2),

$$ d\vec{p}/dt=-q\nabla\phi+(q/c)\cdot [(\vec{v}\cdot\nabla)\vec{A}+\vec{v}\times\vec{B}]=-q\nabla\phi+(q/c)\cdot [d\vec{A}/dt-\partial\vec{A}/\partial t]+(q/c)\cdot\vec{v}\times\vec{B} $$

Finally the result is $$ d\vec{p}/dt=q\vec{E}+ (q/c)\cdot d\vec{A}/dt+(q/c)\cdot\vec{v}\times\vec{B} $$

What is wrong?

  • 4
    $\begingroup$ $p$ is canonical momentum, not mechanical momentum. Find an expression for it and you will see that it includes a term with $A$. $\endgroup$ – Javier Oct 26 '16 at 10:55
  • $\begingroup$ then $d\vec{p}'/dt-(q/c)\cdot d\vec{A}/dt$ is the force? And $p'$ the canonical momentumm? $\endgroup$ – Sergi Oct 26 '16 at 11:52

@Javier has answered your question in some sense. But let me do it more detailed.

Let us begin with the above Lagrangian: $$L = -mc^2\sqrt{1-\beta^2} - q\phi + \frac{q}{c}\vec{v}\cdot\vec{A}$$ We can write then equations of motion: $$\frac{d}{dt}\frac {\partial{L}}{\partial{\dot{x_i}}}\vec{e_i} = \nabla_i{L}\vec{e_i}$$. Note that in the RHS it is partial derivatives. Substituting Lagrangian function we have: $$\frac{d}{dt}\bigl({p_i} + \frac{q}{c}A_i\bigr) =-q\nabla_i\phi+(q/c)\cdot [\frac{dA_i}{dt}-\frac{\partial A_i}{\partial t}]+(q/c)\cdot{[\vec{v}\times\vec{B}}]_i$$ So you see that the term with $\frac{dA}{dt}$ is canceled out.


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