Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Note: This question may be difficult or impossible to answer within the rules of these forums due to its philosophical nature. I will delete the question if I am violating the rules.

Onto the question! Recently I have been curious whether classical electromagnetism is fully solved (up to the divergences). Specifically, can we completely mathematically describe and then interpret the classical world of charges and currents and them the associated fields. Let us assume that our world is classical and that electromagnetism is a complete theory even though there are certain inconsistencies (self-interactions, infinite energy of point charges, etc). A common description of $\textbf{E}(\textbf{r}, t)$ and $\textbf{B}(\textbf{r}, t)$ among physics textbooks is that changing electric fields induce magnetic fields and vice-versa. This is assuming that there are no external fields or sources of field present. In symbols,

$$ \partial_t\textbf{E} \neq \textbf{0} \implies \exists \textbf{B} \quad (1) $$

$$ \partial_t\textbf{B} \neq \textbf{0} \implies \exists \textbf{E} \quad (2) $$

Many physicists come to this conclusion from Maxwell's equations. Specifically, they argue that Faraday's law,

$$ \nabla \times \textbf{E}(\textbf{r}, t) = -\partial_t\textbf{B}(\textbf{r},t), $$

implies (1), and that Ampere's law (with Maxwell's correction term and no currents),

$$ \nabla \times \textbf{B}(\textbf{r}, t) = \partial_t \textbf{E}(\textbf{r},t), $$

implies (2). Note that we are using natural units with $c = 1$. However, these equations do not have any obvious causal connection. While we may like to pretend that right implies left, this is purely notational convention. Who is to say from these equations alone that one field having a non-zero curl doesn't produce a changing dual field? One attempt at reconciling this problem seems to be in Jefimenko's equations. I will state the equations without derivation, but the fields can be solved for completely in terms of the source charges and currents (I'm lazy and the following equations are in mks units from Wikipedia):

$$ \textbf{E}(\textbf{r}, t) = \frac{1}{4 \pi \epsilon_0}\int \left[\frac{\rho(\textbf{r}', t_r)}{|\textbf{r} - \textbf{r}'|^3} + \frac{1}{c}\frac{\partial_t \rho(\textbf{r}', t_r)}{|\textbf{r} - \textbf{r}'|^2}\right] (\textbf{r} - \textbf{r}') - \frac{1}{c^2}\frac{\partial_t \textbf{J}(\textbf{r}', t_r)}{|\textbf{r} - \textbf{r}'|^2} d^3\textbf{r}', $$

$$ \textbf{B}(\textbf{r}, t) = \frac{\mu_0}{4 \pi}\int \left[\frac{\textbf{J}(\textbf{r}', t_r)}{|\textbf{r} - \textbf{r}'|^3} + \frac{1}{c}\frac{\partial_t \textbf{J}(\textbf{r}', t_r)}{|\textbf{r} - \textbf{r}'|^2}\right] \times (\textbf{r} - \textbf{r}' )d^3\textbf{r}' , $$

where $t_r = t - |\textbf{r} - \textbf{r}'|/c$ is the retarded time. These equations seem to imply that neither of the fields "causes" the other. Instead, Jefimenko's equations imply that only the source charges and currents generate the fields (without the presence of external charges, currents, or fields). My question is related to this approach. Is it valid? What are the arguments for and against? Is the matter settled in the classical context of electromagnetism, or are there certain subtleties I've skipped over?

As an extra question, is it instead better to consider $F_{\mu \nu}$, and treat it as one object arising solely from $J^\mu = (\rho, \textbf{J})$, instead of looking at the individual fields?

Thanks in advance for any and all answers!

share|cite|improve this question
I'm sorry, I'm not sure I understand. What is wrong with:$$ \frac{\partial \mathbf{B}}{\partial t} \neq 0 \implies \mathbf{\nabla} \times \mathbf{E} \neq 0 \implies \mathbf E \neq 0 $$ Where the first implication is Faraday's law and the second is because a zero field has no curl? I.e. the contrapositive of $$ \mathbf E = 0 \implies \mathbf \nabla \times \mathbf E = 0 $$ – alemi Aug 10 '14 at 6:18
My concern is that why is it not equally valid to say: $$ \nabla \times \textbf{E} \neq 0 \implies \partial_t \textbf{B} \neq 0 \implies \textbf{B} \neq 0,$$ which would imply that there is no obvious causal relationship in one direction. Rephrased, my question is thus: is it possible to decouple the electric and magnetic fields in such a way that it is obvious that one induces the other? OR, can we only treat them as one object arising from the four-current? – Ultima Aug 10 '14 at 6:30
up vote 3 down vote accepted

We could say: $$ \frac{\partial \mathbf{B}}{\partial t} \neq 0 \implies \mathbf{\nabla} \times \mathbf{E} \neq 0 \implies \mathbf E \neq 0 $$

Where the first implication follows from the transitivity of inequality and Faraday's law: $$ \mathbf \nabla \times \mathbf E = -\frac{\partial \mathbf B }{\partial t } $$ And the second implication is the contrapositive of $$ \mathbf E = 0 \implies \mathbf \nabla \times \mathbf E = 0 $$

You are right to point out that we could have just as well said: $$ \nabla \times \mathbf E \neq 0 \implies \frac{\partial \mathbf B}{\partial t} \neq 0 \implies \mathbf B \neq 0 $$

This doesn't present a problem. We tend to interpret any implication derived from physical law as causal, and while you will regularly find passages in books with things like: "The changing magnetic field causes an electric field", you will equally likely find other passages of the same book that say something like "The curl in the electric field causes a change in the magnetic field".

And I think calling these causal is justified, as we can set up an experiment where we modify a magnetic field and measure an electric field, or for the second one, you could perform an experiment where you rotate a capacitor and measure a magnetic field, for instance.

As for Jefimenko's equations, as far as I know they are completely equivalent to Maxwell's equations in their predictions. In fact, you can derive them from Maxwell's equations and vice versa. Similarly describing electromagnetism by use of the electromagnetic tensor $F_{\mu\nu}$ and four current is completely equivalent to the other two in its predictions. (In this way, we were both lucky and blind, to have developed a classical theory by 1865 that was completely relativistic, but not develop relativity until 1905.)

†: For instance, see section 6.5 of Jackson's Classical Electrodynamics, Third Edition

At that point, physics can no longer distinguish which is true, and you are free to believe whichever you like, or to switch beliefs between the three systems as you please, or as proves convenient. They are, in effect, three different, completely equivalent formulations of electromagnetism.

This is standard. For instance, Newtonian mechanics, Lagrangian mechanics, Hamiltonian mechanics, or Hamilton-Jacobi mechanics all give the same predictions, but all seem to tell a very different story about how the world works.

Each is useful for solving particular problems, and each gives you a different way to view the world, and each is, as far as science can ascertain, completely equivalent and valid ways to view the world.

Some people like to believe in one interpretation over another, but I think its healthy to keep in mind that they are all as valid as the other. But you are more than welcome to keep one as your favorite.

share|cite|improve this answer
Thanks, your comment makes sense and highlights my what I was missing. For some reason I thought that equalities could not show a causal relationship. Now I see that they are equivalent to showing an "if and only if" type of relationship. It now makes sense to me that saying which causes the other is only a matter of semantics, and that the correct way to describe such a relationship is that if one side of the equation holds true, so does the other. For example, a nonzero curl in E implies that there is a changing B, and vice-versa. Thanks for the help! – Ultima Aug 10 '14 at 7:42
These things aren't as equivalent as you make it seem. Maxwell allows nontrivial vacuum solutions that Jefimenko doesn't. Hamilton-Jaccobi also fails to have needed solutions see "Incompleteness of the Hamilton-Jacobi theory" by Nivaldo A. Lemos in Am. J. Phys. 82, 848 (2014) Basically, when going from Newton to Hamiltonian-Jacobi you divide by something that is sometimes zero, so you lose solutions. And while F=ma sometimes allows too many solutions, those aren't the ones you lose, you lose things like ciruclar orbits in favor of just all the elliptic ones. – Timaeus Mar 21 '15 at 20:28

As I understand them Jefimenko's equations integrate only over the past light cone of the field point being considered. Therefore it is OK to say that the right hand side "causes" the left hand side. I believe there is also a formulation involving the advanced time rather than the retarded time, for which this would not be true.

I admit that I can't say what the causal status of Maxwell's formulation is.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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