I was trying to derive Lagrangian for the electromagnetic field from Lorentz's force formula $$\mathbf{F}= q(\mathbf{E} + \mathbf{v}\times \mathbf{B})$$ I have to find the potential by using the line integral of the force. I have written the above equation in terms of scalar and vector potential which is given by. $$F=q(-\nabla\phi-\frac{1}{c}\frac{d\mathbf{A}}{dt}+\frac{1}{c}\nabla(\mathbf{v}\cdot \mathbf{A}))$$ now I have to find potential by using line integral $$V=-\int \mathbf{F}\cdot d\mathbf{r}$$ now I have to show that the integral of the second term is zero so that I can get exact potential which is given by$$V=q(\phi-\frac{1}{c}\mathbf{v}\cdot \mathbf{A})$$ So my question is how to show that the integration of second term is zero?
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$\begingroup$ Related : Deriving Lagrangian density for electromagnetic field. $\endgroup$– FrobeniusCommented Oct 28, 2020 at 19:43
2 Answers
The potential $$U~=~q(\phi - {\bf v}\cdot {\bf A})\tag{1}$$ for the Lorentz force $${\bf F}~=~ q({\bf E} + {\bf v}\times {\bf B}) \tag{2}$$ is a velocity-dependent generalized potential. Such force $\leftrightarrow$ potential interrelation is more subtle: $$ {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}.\tag{3} $$
References:
- Herbert Goldstein, Classical Mechanics, Chapter 1.
Let $$ F_i(x,\dot x,t)=q(E_i(x,t)+\epsilon_{ijk}\dot x_j B_k(x,t)) $$ be the Lorentz force. During the derivation one will have to use $E_i=-\partial_i\phi-\partial_tA_i$ and $B_i=\epsilon_{ijk}\partial_jA_k$, eg. the decomposition into potentials.
If the Lorentz force comes from a velocity-dependent potential, then there is a function $U(x,\dot x,t)$ such that $$ F_i=\frac{\mathrm d}{\mathrm dt}\frac{\partial U}{\partial \dot x_i}-\frac{\partial U}{\partial x_i}, $$ or alternatively $$ F_i(x(t),\dot x(t),t)=-\frac{\delta S_U}{\delta x_i(t)}, $$ where $ S_U=\int U(x(t),\dot x(t),t)\mathrm dt$ and $\delta/\delta x_i(t)$ is the functional derivative with respect to the trajectory $x_i(t)$.
It is relatively easy to see that if an expression $E_i[x](t)=E_i(x(t),\dot x(t),...,x^{(k)}(t),t)$ comes from a functional derivative, then it satisfies $$\frac{\delta E_i[x](t)}{\delta x_j(s)}-\frac{\delta E_j[x](s)}{\delta x_i(t)}=0. $$
This is also the integrability condition for the functional differential equation $$ E_i[x](t)=\frac{\delta S[x]}{\delta x_i(t)}, $$ and if it is satisfied we can find the functional $S$ (which is nonunique) as $$ S[x]=\int_0^1\mathrm d\lambda \int_{t_0}^{t_1}\mathrm dtE_i[x\lambda](t)x_i(t). $$
Thus, for the Lorentz force, $S_U$ sill be given by $S_U=-\int\mathrm dt\int_0^1\mathrm d\lambda\ F_i[\lambda x](t)x_i(t)$, which, when properly expanded and algebraically manipulated will produce the familiar $$ S_U=\int\mathrm dt\ q\big(\phi(x(t),t)-\dot x_i(t) A_i(x(t),t)\big). $$
