# Generalizing the harmonic wave: plane waves

We can generalize the harmonic wave solution to Maxwell’s Equations a bit here.  We required that the electric field only depend on position coordinate $z$, which in practice restricted the electromagnetic wave to move in the $+z$ or $-z$ directions.  But we live in a 3d world (actually maybe more than that, according to our string theorist friends) and we know in practice that electromagnetic waves can travel in any direction.  The harmonic waves we saw so far are called plane waves, because the surfaces of constant phase are planes; specifically, constant- $z$ planes.  (Phase, in this context, refers to the argument of the cosine function).  A more general plane wave solution to Maxwell’s Equations can propagate in any arbitrary direction, and can also have more than one vector component.  This solution would have to satisfy the vector Helmholtz equation: $\displaystyle \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) \vec{E} - \epsilon \mu \frac{\partial^2 \vec{E}}{\partial t^2} = 0$

The plane wave solutions using cosine waves have the form $\displaystyle \vec{E} = A \hat{a}_p \cos{(k_x x + k_y y + k_z z - \omega t + \phi)}$

The $\phi$ in the equation above is the starting phase of the wave, which merely shifts the wave oscillations by a constant amount. $k_x$, $k_y$, and $k_z$ are the three vector components of the wavevector $\vec{k}$.  By substituting this solution back into the vector Helmholtz equation, you can prove its validity, subject to the condition that $\displaystyle k_x^2 + k_y^2 + k_z^2 = \frac{\omega ^2 }{v_p^2}$

or $\displaystyle \sqrt{k_x^2 + k_y^2 + k_z^2} = \frac{\omega}{v_p} = k$

where $k$ is our old friend, the wavenumber.  Thus, the magnitude of the wavevector is equal to the wavenumber.  The direction that the wavevector points is the direction that the wave is propagating.  If we return to the old simple case of a wave propagating in the $+z$ direction, we see that the only nonzero component of the wavevector would be $k_z$, which is consistent with the above.

The unit vector $\hat{a}_p$ is the direction that the electric field is pointing, not to be confused with the direction that the wave is propagating.  In fact, if we apply the fact that $\displaystyle \nabla \cdot \vec{E} = 0$

in charge-free, homogeneous regions, we can prove that $\vec{k} \cdot \hat{a}_p = 0$

in other words, the direction the electric field is pointing (the polarization direction) must be perpendicular to the direction the wave is propagating.  For the simpler electromagnetic wave we saw in an earlier section, the direction of polarization was $\hat{a}_p = \hat{a}_x$.

Up until now, we’ve only discussed the electric field of the plane wave, but the plane wave will also have an associated magnetic field (it must, as the time-varying electric and magnetic fields support one another).  To find this, we can call upon our old friend Faraday’s law: $\displaystyle \nabla \times \vec{E} = - \frac{\partial \vec{B} }{\partial t}$

By plugging in, you can show that the following magnetic field will satisfy this equation: Note that the direction that the magnetic field points is obtained by taking the cross product of the wavevector $\vec{k}$ and the polarization vector $\hat {a}_p$, and therefore is perpendicular to both.  Also note that the magnetic field is in phase with the electric field; i.e. the maxima of the magnetic field occur at the some points in space and time as the maxima of the electric field.  There are lots of nice illustrations available of the electric and magnetic fields of plane waves on the Wikipedia page for electromagnetic waves, linked to below.