Derive the equation of motion from Lagrangian of a particle moving in an electromagnetic field

Question

I really don't even know where to start with this question any help would go very very far. Thank you.

A particle with charge $q$ moving in an electromagnetic field is described by the Lagrangian $$L=\frac{m\mathbf{v}^2}2+\frac qc\mathbf{v}\cdot\mathbf{A}(\mathbf{r},t)-q\phi(\mathbf{r},t).$$ The electric and magnetic fields $(\mathbf{E}(\mathbf{r},t)$ and $\mathbf{B}(\mathbf{r},t)$ correspondingly$)$ are related to 'potentials' $\mathbf{A}(\mathbf{r},t)$ and $\phi(\mathbf{r},t)$ introduced above as follows: $$\mathbf{E}=-\frac1c\frac{\partial\mathbf{A}}{\partial t}-\nabla\phi,\qquad\mathbf{B}=\nabla\times\mathbf{A}.$$ Using the identity $$\mathbf{v}\times(\nabla\times\mathbf{A})=\nabla(\mathbf{v}\cdot\mathbf{A})-(\mathbf{v}\cdot\nabla)\mathbf{A}$$

(a) find the equation of motion of the charged particle from the Lagrangian.

(b) express the force acting on the particle in terms of electric and magnetic fields only (i.e. the equation of motion should have the form of the Newton's second law and contain fields $\mathbf{E}$ and $\mathbf{B}$ but not the 'potentials' $\mathbf{A}$ and $\phi$).

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $$ \totald{}{t}\bracks{m{\bf v} + {q \over c}\,{\bf A}\pars{{\bf r},t}} = \nabla\bracks{% {q \over c}\,{\bf v}\cdot{\bf A}\pars{{\bf r},t} - q\phi\pars{{\bf r},t}} = \nabla\bracks{% {q \over c}\,{\bf v}\cdot{\bf A}\pars{{\bf r},t} - q\phi\pars{{\bf r},t}} $$ Now, use vectorial identities.

Answer 2

This is an effort to get this question off from the unanswered queue.

I would like to share my two cents: asking a seemingly homework-like question with the name of Professor Monk, the book written by whom I had learned a lot from, I would say this looks very inappropriate.

Anyway I just would like extend Felix Marin's incomplete answer to a full one. There is one most important feature of Euler-Lagrange equation he forgot to mention.

Another relevant question I remembered answering not very long time ago: Euler-Lagrange equations of the Lagrangian related to Maxwell's equations

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Find the equation of motion of the charged particle from the Lagrangian.

The equation of motion obtained from the Lagrangian basically is the Euler-Lagrange equation: $\newcommand{\b}{\mathbf}$ $$ \frac{\partial L}{\partial {\mathbf r}} - \frac{\mathrm d }{\mathrm dt}\left( \frac{\partial L}{\partial {\mathbf r'}}\right) =0, \tag{1} $$ whereas the $\dfrac{\mathrm d }{\mathrm dt}$ represents taking the total derivative (time derivative, plus the convective derivative w.r.t. the spacial variable, or so to speak). Yet prime only stand for the time derivative usually.

Plugging $L$ into (1) we have: (the usual postulate is that $\mathbf v$ is not explicitly a function of $\mathbf r$) $$ \frac{\partial L}{\partial {\mathbf r}} = \nabla \left( \frac{q}{c} \b{v}\cdot \b{A} - q\phi\right) \\ = q\left(\frac{1}{c} \nabla(\b{v}\cdot \b{A}) - \nabla \phi \right) \\ = q\left(\frac{1}{c} \b{v}\times( \nabla\times\b{A}) + \frac{1}{c} (\b{v}\cdot \nabla) \b{A} - \nabla \phi \right). $$ And $$ \frac{\partial L}{\partial {\mathbf r'}} = \frac{\partial L}{\partial {\mathbf v}} = m\b{v} + \frac{q}{c}\b{A}. $$ Now taking the total derivative: $$ \frac{\mathrm d }{\mathrm dt}\left( \frac{\partial L}{\partial {\mathbf r'}}\right) = \frac{\partial }{\partial t}\left( \frac{\partial L}{\partial {\mathbf r'}}\right) + \left(\frac{\mathrm d \b{r} }{\mathrm dt}\cdot\nabla\right)\left( \frac{\partial L}{\partial {\mathbf r'}}\right). $$ Notice the total derivative of $\b{r} = (x(t),y(t),z(t))$ is $\b{r}'$, hence: $$ \frac{\mathrm d }{\mathrm dt}\left( \frac{\partial L}{\partial {\mathbf r'}}\right) = \frac{\partial }{\partial t}\left(m\b{v} + \frac{q}{c}\b{A}\right) +(\b{v}\cdot \nabla ) \left(m\b{v} + \frac{q}{c}\b{A}\right) \\ = m\frac{\partial^2 \b{r}}{\partial t^2} + \frac{q}{c}\frac{\partial \b{A}}{\partial t} + \frac{q}{c}(\b{v}\cdot \nabla )\b{A}, $$ where we used that the convective derivative of $\b{v} = \b{v}(t)$ is zero.

Now put everything back to (1), we have: $$ m\frac{\partial^2 \b{r}}{\partial t^2} = - \frac{q}{c}\frac{\partial \b{A}}{\partial t} +\frac{q}{c} \b{v}\times( \nabla\times\b{A} )- q\nabla \phi. \tag{2} $$ Equation (2) is basically the motion equation of the Lagrangian.

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Express the force acting on the particle in terms of electric and magnetic fields only (i.e. the equation of motion should have the form of the Newton's second law and contain fields $\mathbf{E}$ and $\mathbf{B}$ but not the 'potentials' $\mathbf{A}$ and $\phi$).

If the potentials are replaced using the $\b{E}$ and $\b{B}$ respectively, (2) will become: $$ m\frac{\partial^2 \b{r}}{\partial t^2} = q(\b{E} + \b{v}\times \b{B}). \tag{3} $$ This is nothing but the famous Lorentz force formula, a.k.a. the Newton Second Law for a particle moving through an electromagnetic field.