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?).

  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/hep-ph/0106235 $\endgroup$
    – Qmechanic
    Oct 5, 2013 at 10:36
  • $\begingroup$ I was not aware that it should be done that way, apologies. $\endgroup$ Oct 5, 2013 at 11:15

2 Answers 2


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.

  • $\begingroup$ Nice answer, very clear. $\endgroup$ Oct 6, 2013 at 2:27
  • $\begingroup$ @PhysiXxx Really, really nice answer. If I may suggest something, I would be happy to see the Galilean transformation of the sources $j^{\prime}=j$ and $\rho^{\prime}=\rho + v\cdot j$, since the Ampère-Maxwell and Gauß laws are equations with sources. Adding the sources will of course not change at all neither your reasoning nor your conclusion. An other remark possibly interesting for Rogelio Molina: the Galilean invariance of the Faraday law is used to demonstrate it in the Jackson's book, in the chapter about induction (section 5.15 titled Farday law). $\endgroup$
    – FraSchelle
    Nov 19, 2013 at 20:47

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.