I have a hard time to fully understand (classical) electromagnetic field theory with respect to Helmholtz's decomposition. Let me start from Helmholtz's theorem:

Any vector field of class $C^{\infty}$ in $R^3$ can be docomposed into sum of >two other fields: one curl-free and one divergence free.


but (due to some vector operator identities) we can rewrite $F_1$ and $F_2$ to

$\bf{F_1}=-\nabla F_3$



$F_3$,$\bf{F_4}$ are scalar and vector fields respectively

Now going to electrodynamic we know that in stationary case




It fits very well so we can write that electromagnetic field is equal


or can we? Why in none of my books nor in the net there is written that EM field is just $\bf{E+B}$? For example wikipedia states that EM is combination of $\bf{E}$ and $\bf{B}$. Yes, of course it is combination (from Maxwell equations) but that is not precise statement. Obviously nowhere I could find any equation for EM field (treated as one single vector field).

So, can someone please elaborate what this EM field is with respect to $\bf{E}$ and $\bf{B}$ in the context of Helmholtz decomposition?


6 Answers 6


Let me try this more clearly than the other answers, which aren't wrong. You ask:

So, can someone please elaborate what this EM field is with respect to $\vec E$ and $\vec B$ in the context of Helmholtz decomposition?

There is no "EM field in the context of Helmholtz decomposition". Helmholtz just says that every vector field $\vec V$ is decomposable as curl and gradient of two other fields, i.e.

$$\vec V = \vec \nabla \phi + \vec \nabla \times \vec A $$

You can do this for the electric or the magnetic field, of course, but this isn't particularly enlightening as to the nature of "the EM field". A field should behave nicely under transformations, and special relativity with its action on the electric and magnetic fields shows us that we should not add them together, but seek a quantity that transforms nicely under Lorentz transformations instead:

"The electromagnetic field" is equivalently the gauge four-potential $A$ (consisting of the scalar electrostatic potential in the temporal and the magnetic vector potential in the spatial entries) or its derivative, the field strength tensor $F = \mathrm{d}A$. Electric and magnetic fields become part of the tensor as \begin{align} F^{0i} & = E^i \\ F^{ij} & = \sum_k\epsilon^{ijk}B^k \end{align} This is "the EM field", but it has nothing to do with Helmholtz decomposition, since electromagnetism is properly looked at in the four-dimensional setting of special relativity, for which only the general Hodge decomposition may be applied, of which Helmholtz is a special case, but even this has nothing to do with it.

This EM field acts on the four-velocity, reproducing the Lorentz force by

$$ \frac{\mathrm{d}p}{\mathrm{d}t} = q F(u)$$

where $u$ is the four-velocity, and $(F(u))_\mu = F_{\mu\nu}u^\nu$.


If the field is not stationary, curl of $\vec{E}$ does not vanish. So generally you cannot identify electromagnetic field with the curl-free part of the decomposition.

However, you can indeed introduce a complex vector combination of electric and magnetic field, in a certain system of units it is $\vec{E}+i\vec{H}$. This is the so-called Riemann-Silberstein vector (http://en.wikipedia.org/wiki/Riemann%E2%80%93Silberstein_vector ). It is sometimes very useful (for example, I used it in my recent articles (http://arxiv.org/abs/1502.02351 and http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (published in J. Math. Phys.)). However, it is a vector only under the transformations of the rotation group, not of the entire Lorentz group.

  • $\begingroup$ OK, but in stationary case E is curl-free so I can write that EM=E+B, or not? On the other hand can we define other then $E$ field that is always curl-free to fit to Helmholtz theory? Or maybe assumptions about EM field in this decomposition are not valid? $\endgroup$ Commented Mar 15, 2015 at 23:45
  • $\begingroup$ @WeSenseASoulInSearchOfAnswers: As I wrote in a comment to another answer here: ask yourself what vector field you are actually trying to decompose. Yes, in the stationary case, you can introduce vector field $\vec{E}+\vec{H}$, but I have no idea why it would be useful in physics. $\endgroup$
    – akhmeteli
    Commented Mar 15, 2015 at 23:51
  • $\begingroup$ @WeSenseASoulInSearchOfAnswers I wonder if you read my answer? E and B act on different objects. E on electric charges q and B on magnetic dipoles or magnetic fields generated by currents. So, in whichever way you write them, they are different things. $\endgroup$
    – Sofia
    Commented Mar 15, 2015 at 23:55
  • $\begingroup$ @akhmeteli I don't know if it would or wouldn't be useful, but if something can be done it should be done (or at least tried), so if Helmholtz states that one can decompose any smooth field to curl-free and div-free then we should do that and study properties of those components. This is how I see it, and in fact I'm surprised it is so hard to understand what I mean (some answers are completely of topic here). $\endgroup$ Commented Mar 16, 2015 at 0:13
  • 2
    $\begingroup$ @Sofia: I am not sure they act on different objects: as parts of the electromagnetic tensor, they act on the four-velocity to produce the Lorentz force (en.wikipedia.org/wiki/… ) $\endgroup$
    – akhmeteli
    Commented Mar 16, 2015 at 0:21

If fits very well so we can write that electromagnetic field is equal


or can we?

No! For the love of god, no!

Do not just add those fields together... it's not a useful quantity.

In the SI system $E$ and $B$ have different units. Another good indicator that you don't want to just add them together....

Additionally, the electric field is not always longitudinal (i.e., not always equal to the gradient of a scalar $-\nabla \phi$). In general it can have a transverse component: $$ \vec E = -\nabla \phi - \frac{\partial \vec A}{\partial t} $$

  • 5
    $\begingroup$ But in CGS they have the same unit, where certainly $E^2$ is added to $B^2$ all the time. Even in SI you can just preface one or the other with enough $\mu$'s and $\epsilon$'s to make the units work. $\endgroup$
    – user10851
    Commented Mar 15, 2015 at 23:13
  • $\begingroup$ So let me reformulate the question - if EM is not a sum of E and B, then what fields do sum up to EM? According to Helmholtz there are such fields (if conditions are meet) and moreover those two fields (curl-free and divergence-free) should uniquely define EM. $\endgroup$ Commented Mar 15, 2015 at 23:20
  • 2
    $\begingroup$ @WeSense: What do you mean "sum up to EM"? EM is, in non-relativistic notation, described by the two vector fields $\vec E$ and $\vec B$. Electromagnetism is a theory, not a vector any thing can sum up to. In relativistic notation, you unify this description into the EM field strength tensor. And that's it. $\endgroup$
    – ACuriousMind
    Commented Mar 15, 2015 at 23:36
  • $\begingroup$ @WeSenseASoulInSearchOfAnswers: See my answer. You should also ask yourself what vector field you actually decompose in the Helmholtz decomposition. It can be any vector field, for example $\vec{E}$. $\endgroup$
    – akhmeteli
    Commented Mar 15, 2015 at 23:38
  • $\begingroup$ @ChrisWhite, yes you can go ahead and add those fields if you want... but you ought not. Why not just add in $\vec A$ itself to the sum? I'm sure you can finagle the units by introducing some dimensionful constants so that the terms with A,B, and E all match. Why not do this? Because it is pointless. $\endgroup$
    – hft
    Commented Mar 16, 2015 at 3:07

I will find the Maxwell word in the Oersted Medal Lecture 2002: Reforming the Mathematical Language of Physics by Hestenes, on pages 25/26

That formidable text presents a better math formalism for physics, imo.

Starting with $ F(x,t) = E(x,t) + i B(x,t) $ ...
The 4 equations of Maxwell (64..67) that describe two viewpoints ( E and B ) of a single entity 'EM' field can be expressed with only one equation (63):

$$(\frac{1}{c}\partial_t+\nabla) F= \rho - \frac{1}{c}J$$

The word Helmholtz , etc, etc, is not present in that formalism. Complex numbers, vectors, matrices, tensors, etc, are particular viewpoints that Geometric Algebra integrates.


Actually the electromagnetic field can be seen as a tensor. The combination Wikipedia talks about is this, $E$ and $B$ are organised in an antisymmetric matrix $F_{\mu\nu}$ with $\mu,\nu = 0,\ldots, 4$ so the number of independent components is $6$.


You are free to define the vector $\bf{F_{EM}}$, but I don't believe this vector would have any value. It wouldn't obey any simple laws, and it would not be found to have any practical use in the lab.


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