# Lorentz force with Lagrangian

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

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

and

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

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?

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

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.