Suppose we are in vacuum and we have an electric field $\vec{E}$ which we assume is simple harmonic wave that propagates through $z$ and is linearly polarized in the $x$-$y$ plane along $x$ i.e. $\vec{E}(t,x,y,z)=E_0\cos(\omega t-kz)\hat{x}$. This function obviously satisfies $\vec{\nabla}\cdot\vec{E}=0$. Now note that $$\vec{\nabla}\times\vec{E}(t,x,y,z)=\frac{\partial}{\partial z}\left( E_0\cos(\omega t-kz) \right)\hat{y}=kE_0\sin(\omega t-kz)\hat{y}=-\frac{\partial \vec{B}}{\partial t}(t,x,y,z)$$ If we integrate this, we get $\vec{B}(t,x,y,z)=B_0\cos(\omega t-kz)\hat{y}+\vec{g}(x,y,z)$ for some $\vec{g}:\mathbb{R}^3\rightarrow\mathbb{R}^3$ and where $B_0=\frac{k}{\omega}E_0$. Now note that $B_0\cos(\omega t-kz)\hat{y}$ has no gradient and therefore $\vec{\nabla}\cdot\vec{B}=0$ implies $\vec{\nabla}\cdot\vec{g}=0$. On the other hand $\vec{\nabla}\times (B_0\cos(\omega t-kz)\hat{y})=\mu_0\epsilon_0\frac{\partial\vec{E}}{\partial t}(t,x,y,z)$, which implies $\vec{\nabla}\times\vec{g}=\vec{0}$. Finally, clearly $B_0\cos(\omega t-kz)\hat{y}$ satisfies the wave equation, which means that $\vec{g}$ must do it too. Since $\vec{g}$ has no time dependence, $\nabla^2\vec{g}=\vec{0}$.

On every source I have seen, the vector field $\vec{g}$ has been taken to be null. From this assumption things such as the perpendicularity between the fields and the direction of propagation are explained. Non the less, non of the three restrictions imply that $\vec{g}$ be null. In particular, $\vec{g}$ could be a constant vector field with any direction and still satisfy Maxwell's equations.

Is there any way to show that in general E&M waves must be perpendicular and therefore show that $\vec{g}=\vec{0}$? In the case there isn't, what can we say about $\vec{g}$ knowing $\vec{\nabla}\cdot\vec{g}=0$, $\vec{\nabla}\times\vec{g}=\vec{0}$ and $\nabla^2\vec{g}=\vec{0}$?

  • $\begingroup$ I am not sure I am understanding the problem. Maxwell's equation are satisfied by any electromagnetic field, not just waves. A wave superimposed on a constant field is a perfectly valid solution, of course and therefor the magnetic component and the electric component do not have to be perpendicular for a general field. Does any particular textbook claim that they have to be? $\endgroup$
    – CuriousOne
    Commented Oct 31, 2015 at 4:07
  • $\begingroup$ Well, my problem is that when e&m waves are studied you never take into account this $\vec{g}$. I am interested in how this field looks since I cant think of a reason for it to be null. In particular, I want to point out that a varying SHW electric field can produce a magnetic field with a constant component! $\endgroup$ Commented Oct 31, 2015 at 4:14
  • $\begingroup$ For any two valid solutions to Maxwell's equations any linear combination of those solutions is also a solution. In general we can therefor describe any em-field with its Fourier transform. This is a direct consequence of the perfect linearity of the theory and it doesn't impact the analysis of an em-wave with a single frequency. The constant field component in your example is not produced by the wave but it's just an integration constant. $\endgroup$
    – CuriousOne
    Commented Oct 31, 2015 at 4:22

1 Answer 1


There is no reason for the field to be null. You have correctly identified the properties of another magnetic field that could be the solution of Maxwell's equations that is consistent with electric field you started with - namely a time-independent magnetic field.

The situation is exactly that by which you are reading this answer. EM waves propagate into your eyes, but between you and the screen the space is filled with the Earth's (roughly) time-independent magnetic field.

As CuriousOne completely describes, the reason this works is that you can superpose any solutions to Maxwell's equations and a time-independent magnetic field can be a solution with zero electric field and a Laplacian of zero.

The reason books ignore this, is because you can ignore it. The time-dependent EM wave fields can be considered in isolation from any time-independent E- and B-fields in the "background". Saying that a time-dependent E-field can "produce" a time-independent B-field is not a correct way to think about electromagnetism - the fields co-exist. No SHM E-field is associated with a time-independent B-field - as you confirmed by showing that its curl was zero.

  • $\begingroup$ Is there a way to show that the time dependent electric field, time dependent magnetic field and the propagation are perpendicular without assuming the form of the electric field? Thanks! $\endgroup$ Commented Oct 31, 2015 at 11:33
  • $\begingroup$ @IvánMauricioBurbano What, you mean just say ${\bf E} = {\bf E_0} f({\bf k}\cdot {\bf r} - \omega t)$, where ${\bf E_{0}}$ is an arbitrary vector? Yes - Gauss's law shows that ${\bf k}\cdot {\bf E_0} = 0$, then take the curl and integrate to show that the time-dependent B-field is perpendicular to ${\bf E_0}$ (and ${\bf k}$). This is standard bookwork. $\endgroup$
    – ProfRob
    Commented Oct 31, 2015 at 11:37
  • $\begingroup$ Yeah, but you (and I in my question) assumed a particular form of the electric field. I want to show this without this particular form for the electric field. In that way I can convince myself that circularly polarized electric fields are also perpendicular to the magnetic fields. $\endgroup$ Commented Oct 31, 2015 at 11:40
  • $\begingroup$ @IvánMauricioBurbano Sorry? Where did I assume a particular form? I assume that the form is an electromagnetic wave, thats all. The relationships between E- and B-fields for an EM wave are not generally true for all solutions of Maxwells equations! A circ. pol. wave is just the sum of two EM waves plane polarised at right angles, with a 90 degree lag. Therefore if the relationships are true for any plane polarised EM wave then they are also true for circ. pol. waves, because, as stated, any linear combinations of solutions to Maxwell's eqs. are also solutions $\endgroup$
    – ProfRob
    Commented Oct 31, 2015 at 11:42

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.