Euler-Lagrange partial with respect to $r$ of a dot product involing velocity and a vector potential [closed]

Question

I will outline what I believe to a correct way to go from a Lagrangian of a charged particle in a EM field to Lorentz force via the Euler-Lagrange equations. At the very beginning when I use the EL equation, I will bold what my concern is and return to it after I finish the derivation.

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The Lagrangian (in SI units) reads

\begin{equation} L = \frac{1}{2} m\dot{r}^2 - q \phi + q \dot{r}\cdot A \end{equation} where both the scalar potential $\phi$ and $A$ depend of space and time $\phi(t,r)$, $A(t,r)$.

Euler Lagrange gives us

\begin{align} \frac{\partial L}{\partial r} - \frac{d}{dt}\frac{\partial L}{\partial \dot{r}} &= 0 \\ -q \frac{\partial }{\partial r} \phi + \mathbf{q \frac{\partial}{\partial r} (\dot{r}\cdot A)} - \frac{d}{dt}(m\dot{r}+\mathbf{qA}) &=0 \\ q\big[-\nabla \phi + \nabla(\dot{r}\cdot A) - \frac{d A}{dt}\big] &= m\ddot{r} \end{align}

then using the following expression for a total time derivative of some function of space and time $f(t,x_1,x_2,...)$,

\begin{align} \frac{d}{dt} f &= \big[\frac{\partial}{\partial t} + \sum \frac{d x_i}{dt} \frac{\partial}{\partial x_i}\big]f \\ &= \big[\frac{\partial}{\partial t} + (\dot{r}\cdot \nabla)\big]f \end{align}

on the $\frac{d A}{dt}$ term we have

\begin{align} q\big[-\nabla \phi + \nabla(\dot{r}\cdot A) - \frac{d A}{dt}\big] &= m\ddot{r} \\ q\big[-\nabla \phi + \nabla(\dot{r}\cdot A) - \frac{\partial A}{\partial t} - (\dot{r}\cdot \nabla) A \big]&= m\ddot{r} \end{align}

then using the ``bac-cab" rule of vector calculus with $\dot{r}=v$ and the magnetic field $B=\nabla\times A$,

\begin{align} a\times b\times c &= b(a\cdot c) - c(a\cdot b) \\ &=b(a\cdot c) - (a\cdot b)c \\ v \times \nabla \times A &= \nabla (v\cdot A) - (v\cdot \nabla)A \\ v \times B &= \nabla (\dot{r} \cdot A) - (\dot{r} \cdot \nabla)A \end{align}

on the $\nabla(\dot{r}\cdot A)$ and $- (\dot{r}\cdot \nabla) A$ terms, we have

\begin{align} q\big[-\nabla \phi + \nabla(\dot{r}\cdot A) - \frac{\partial A}{\partial t} - (\dot{r}\cdot \nabla) A\big] &= m\ddot{r} \\ q\big[-\nabla \phi - \frac{\partial A}{\partial t} + v\times B\big] &= m\ddot{r} \\ \end{align}

which, using $E=-\nabla \phi - \frac{\partial A}{\partial t}$, we have

\begin{align} m\ddot{r} &= q[E+v \times B] \\ F &= q[E+v \times B] \end{align}

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Now, why is the $q \frac{\partial}{\partial r} (\dot{r}\cdot A)$ from the $\frac{\partial L}{\partial r}$ term in the EL not simply $\dot{r} \cdot \nabla A$? We take the derivative of the Lagrangian with respect to velocities in the second term, why do we take the partial with respect to r of the velocity in this instance?

If we follow this precedent, why isn't the $\frac{d}{dt}(qA)$ from the from the $\frac{d}{dt}\frac{\partial L}{\partial \dot{r}}$ term in the EL equation $\frac{d}{dt}(\dot{r}\cdot A)$?

Answer 1

$\frac{\partial}{\partial\mathbf{r}}$ is the same as $\nabla$ operator (https://en.wikipedia.org/wiki/Del). Using the properties of $\nabla$: $$ \nabla(\mathbf{\dot{r}}\mathbf{A})=\mathbf{\dot{r}}\times(\nabla\times\mathbf{A})+\mathbf{A}\times(\nabla\times\mathbf{\dot{r}})+(\mathbf{\dot{r}}\nabla)\mathbf{A}+(\mathbf{A}\nabla)\mathbf{\dot{r}} $$ $\mathbf{\dot{r}}$ doesn't depend on $\mathbf{r}$, that's why $$ \nabla(\mathbf{\dot{r}}\mathbf{A})=\mathbf{\dot{r}}\times(\nabla\times\mathbf{A})+(\mathbf{\dot{r}}\nabla)\mathbf{A} $$

$\nabla(\mathbf{\dot{r}}\mathbf{A})$ is not equal to $\mathbf{\dot{r}}(\nabla\mathbf{A})$ because $\mathbf{A}$ and $\mathbf{\dot{r}}$ are non-collinear vectors (the same as $\mathbf{a}(\mathbf{b}\mathbf{c})\neq\mathbf{b}(\mathbf{a}\mathbf{c})$ in general).

Answer 2

[W]hy is the $q \frac{\partial}{\partial r} (\dot{r}\cdot A)$ from the term in the EL not simply $[q] \dot{r} \cdot \nabla A$?

In a liberal sense, this term is equal to $\dot{r} \cdot \nabla A$, so long as you're clear about what you mean by the $\cdot$ symbol. The catch is that $\nabla A$ is a rank-2 tensor, and there are two different ways to contract a rank-2 tensor with a vector.

This becomes much clearer when you write things out using indices & Einstein summation. The term you're concerned about is $$ \nabla (\dot{r} \cdot A) \to \partial_i (\dot{r}_j A_j) = \dot{r}_j (\partial_i A_j), $$ where we can take the second step because (as you correctly note) $\dot{r}_j$ can be treated as independent of $r_j$. Meanwhile, the similar term in the EM equations, arising from the total derivative of $\vec{A}$ with respect to $t$, is $$ -(\dot{r} \cdot \nabla) A \to -\dot{r}_j (\partial_j A_i) $$ Essentially, in the first expression, the vector $\dot{\vec{r}}$ is contracted with one of the indices of $\vec{A}$; while in the second, it's contracted with one of the indices of $\vec{\nabla}$. This means that the expressions are not equivalent; and, in fact, they can be combined to yield $$ \dot{r}_j \left( \partial_i A_j - \partial_j A_i \right) = \dot{r}_j \epsilon_{ijk} B_k = (\vec{v} \times \vec{B})_i. $$