# Details on the magnetic field of a linearly polarized electric wave

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 $\nabla^2\vec{g}=\vec{0}$?

• 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? Oct 31, 2015 at 4:07
• 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! Oct 31, 2015 at 4:14
• 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. Oct 31, 2015 at 4:22

• @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. Oct 31, 2015 at 11:37