Galilean invariance of a subset of Maxwell equations

Question

I read in Feynman's proof of Maxwell equations the statement that the subset of Maxwell equations comming from the Bianchi identity:

$$ \nabla \cdot {\bf B} = 0, \quad \nabla \times {\bf E} + \frac{1}{c}\frac{\partial {\bf B} }{\partial t} = 0 $$ is actually invariant under Galilean transformations. I came up with this reasoning, is it correct, or am I missing something?

First $c$ has to be assumed a constant, otherwise it will not work, I think the author assumes this, though he does not say it.

A Galilean transformation, say on the $x$ axis, transforms $x' = x -ut$, $z'=z$, $y'=y$, $t' =t$. It does not mix electric and magnetic fields, so they only mix components amongst themselves. The transformation matrix for the electric field say, will be

$$ E'_i (x') = \frac{\partial x^k}{\partial x'^i} E_k(x) = E_{i}(x) $$ and something identical for the magnetic field. Also derivatives will be trivially transformed $\nabla =\nabla', \quad \frac{\partial {\bf } }{\partial t} = \frac{\partial {\bf } }{\partial t'}$ so invariance follows readily.

Then the author says the whole set of Maxwell equations is not invariant due to the displacement current term. I do not understand this statement. Presumably he reffers to the existence of electromagnetic waves propagating with speed $c$ and this will not be constant under Galilean transformations, other than that the equations would retain their same form due to the argument I gave above.

But if we started assuming $c$ is not a constant, then we could not even prove the invariance of the first two equations (the point is perhaps that in such case we cannot interpret $c$ as the speed of anything?).

Answer 1

Let's look to your own statements.

First, time derivative after transformations isn't equal to an "old" derivative: for $\mathbf r' = \mathbf r - \mathbf u t = \mathbf r - \mathbf u t' \Rightarrow \mathbf r = \mathbf r' + \mathbf u t'$ $$ \partial_{t'} = (\partial_{t'}\mathbf r )\partial_{\mathbf r} + (\partial_{t'}t) \partial_{\mathbf t} = (\mathbf u \cdot \nabla ) + \partial_{t}, \quad (\mathbf u \cdot \nabla ) = u^{i}\partial_{x_{i}} . $$ So, with $\nabla ' = \nabla$, "Bianchi" equations transforms to $$ (\nabla \cdot \mathbf B') = 0 , \quad [\nabla \times \mathbf E '] + \frac{1}{c}\partial_{t}\mathbf B' + \frac{1}{c}(\mathbf u \cdot \nabla)\mathbf B ' = 0. \qquad (.1) $$ Second, the form of $\mathbf {E}'(\mathbf r', t'), \mathbf B ' (\mathbf r ' , t')$ isn't equal to $\mathbf E (\mathbf r , t), \mathbf B (\mathbf r , t)$. Let's use the Lorentz force expression, $$ \mathbf F = q\mathbf E + \frac{q}{c}[\mathbf v \times \mathbf B ]. $$ It doesn't depend on acceleration, so the statement that $\mathbf F ' = \mathbf F$ under galilean transformation is true. It means that $$ \mathbf E + \frac{1}{c}[\mathbf v \times \mathbf B] = \mathbf E ' + \frac{1}{c}[\mathbf v ' \times \mathbf B']. $$ By using galilean transformation for speed, $\mathbf v' = \mathbf v - \mathbf u$, this equation can be rewritten as $$ \mathbf E + \frac{1}{c}[\mathbf v \times \mathbf B] = \mathbf E ' + \frac{1}{c}[\mathbf v \times \mathbf B '] - \frac{1}{c}[\mathbf u \times \mathbf B'], \qquad (.2) $$ so the statement that $\mathbf E = \mathbf E ' , \quad \mathbf B = \mathbf B '$ isn't correct. So you need to find expressions $\mathbf E ' $ and $\mathbf B'$ via $\mathbf E $, $\mathbf B$.

By rewriting $(.2)$, $$ \mathbf E + \frac{1}{c}[\mathbf v \times (\mathbf B - \mathbf B' )] = \mathbf E ' - \frac{1}{c}[\mathbf u \times \mathbf B '] , $$ in a reason of arbitrary $\mathbf u $ you can get the solution: $$ \mathbf B' = \mathbf B , \quad \mathbf E' = \mathbf E + \frac{1}{c}[\mathbf u \times \mathbf B ]. $$ By substitution these equations to $(.1)$ you will get $$ (\nabla \cdot \mathbf B)= 0 , \quad [\nabla \times \mathbf E] + \frac{1}{c}[\nabla \times [\mathbf u \times \mathbf B]] + \frac{1}{c}\partial_{t}\mathbf B + \frac{1}{c}(\mathbf u \cdot \nabla) \mathbf B = [\nabla \times \mathbf E] + \frac{1}{c}\partial_{t}\mathbf B = 0, $$ because for $\mathbf u = const$ $$ [\nabla \times [\mathbf u \times \mathbf B]] = \mathbf u (\nabla \cdot \mathbf B) - (\mathbf u \cdot \nabla )\mathbf B = - (\mathbf u \cdot \nabla )\mathbf B . $$ So the first pair of Maxwell's equations is clearly invariant under galilean transformations.

Let's look to the other pair of Maxwell's equations: $$ [\nabla \times \mathbf B] - \frac{1}{c}\partial_{t}\mathbf E = 0 , \quad (\nabla \cdot \mathbf E ) = 0 . \qquad (.3) $$ By using an expressions which were derived above, you can rewrite $(.3)$ as $$ [\nabla \times \mathbf B] - \frac{1}{c}\partial_{t}\mathbf E ' - \frac{1}{c}(\mathbf u \cdot \nabla)\mathbf E' = $$ $$ =[\nabla \times \mathbf B] - \frac{1}{c}\partial_{t}\mathbf E - \frac{1}{c}(\mathbf u \cdot \nabla)\mathbf E - \frac{1}{c^{2}}\partial_{t}[\mathbf u \times \mathbf B] - \frac{1}{c^{2}}(\mathbf u \cdot \nabla)[\mathbf u \times \mathbf B]= 0, $$ $$ (\nabla \cdot \mathbf E ) + \frac{1}{c}(\nabla \cdot [\mathbf u \times \mathbf B]) = (\nabla \cdot \mathbf E ) -\frac{1}{c}(\mathbf u \cdot [\nabla \times \mathbf B]) = 0 . $$ The requirement of galilean invariance of second equation leads to te state that $\frac{1}{c}(\mathbf u \cdot [\nabla \times \mathbf B])$, which isn't true in the general case. Analogically reasoning can be used for the first equation.

So the second pair of Maxwell's equations isn't invariant under Galilean transformations.

Answer 2

The answer already given is very neat, but I would like to address the following point.

The transformations

$\mathbf B' = \mathbf B , \quad \mathbf E' = \mathbf E + \frac{1}{c}[\mathbf u \times \mathbf B ]$

are obtained from

$\mathbf E + \frac{1}{c}[\mathbf v \times (\mathbf B - \mathbf B' )] = \mathbf E ' - \frac{1}{c}[\mathbf u \times \mathbf B ']$

through the following reasoning. $\mathbf v$ is the velocity of the charged particle, which is a test particle. As such, it cannot affect the fields, i.e. $\mathbf E$, $\mathbf E'$, $\mathbf B$ and $\mathbf B'$ do not depend on $\mathbf v$. Moreover, $\mathbf u$ is arbitrary and does not depend on $\mathbf v$. Hence, we must get rid of $\mathbf v$ in

$\mathbf E + \frac{1}{c}[\mathbf v \times (\mathbf B - \mathbf B' )] = \mathbf E ' - \frac{1}{c}[\mathbf u \times \mathbf B ']$

and this is obtained bt demanding that

$\mathbf B' = \mathbf B$

which is indeed one of our transformation, the other following immediately upon substitution.