# Poyting theorem for a plane wave

I would like to apply and verify the Poynting theorem for a uniform plane wave but there is obviously something wrong in my demonstration.

The Poynting theorem expresses the conservation of energy: the change of energy within a volume V is equal to the power flow passing through its surrounding surface S. Written locally and without currents it gives:

$\nabla.\mathbf{R}=-\partial_{t}E_{m}$

Where the Poyting vector is : $\mathbf{R}=\mathbf{E}\times\mathbf{H}$ and the electromagnetic energy : $E_{m}=\frac{1}{2}\epsilon E^{2}+\frac{1}{2}\mu H^{2}$

In the case of a plane wave travelling along the x direction, the fields can be expressed as :

$\mathbf{E}=E_{y}\mathbf{e}_{y}=E_{0}cos(\omega t-kx)\mathbf{e}_{y}$

$\mathbf{H}=H_{z}\mathbf{e}_{z}=\frac{E_{0}}{\eta}sin(\omega t-kx)\mathbf{e}_{z}$

Which expresses that the wave is transverse electric magnetic, the magnetic field is $\pi/2$ out of phase with the electric field with the relation for the norms $E=\eta H$.

Consequently, the energy is : $E_{m}(x,t)=\frac{1}{2}\epsilon||\mathbf{E}||^{2}+\frac{1}{2}\mu||\mathbf{H}||^{2}=\frac{1}{2}\epsilon_{0}E_{0}^{2}(cos^{2}(\omega t-kx)+sin^{2}(\omega t-kx)) =\frac{1}{2}\epsilon_{0}E_{0}^{2}$ Which means that the wave is uniform both in space and time.

The poyting vector is only along the x direction since the wave travels along x:

$\mathbf{R}=E_{y}H_{z}\mathbf{e}_{x}$

$R_{x}(x,t)=\frac{1}{\eta}E_{0}^{2}cos(\omega t-kx)sin(\omega t-kx)=\frac{1}{2}\epsilon_{0}cE_{0}^{2}sin(2(\omega t-kx))$

Finally taking the divergence for the Poynting vector and the time derivative for the energy yields :

$\nabla.\mathbf{R}=\frac{dR_{x}}{dx}=-k\epsilon_{0}cE_{0}^{2}cos(2(\omega t-kx))$

and

$\partial_{t}E_{m}=0$

Which does not obey to the Poyting theorem

What is wrong in this demonstration ???

What's wrong is that you started with the waves $\pi/2$ out-of-phase. They should be in-phase, as this description shows, otherwise they will not obey the Maxwell Equations and you cannot use Poynting's Theorem (which itself is derived from the Maxwell Equations): Doing out the expression with $E \propto \cos(\omega t - k x)$, $B \propto \cos(\omega t - k x)$ yields an energy density that goes like $\epsilon_0 E_0^2 \cos^2(\omega t - k x)$ and a Poynting vector which goes likewise; in one case you take the derivative of $\cos^2$ with respect to $x$, in the other with respect to $t$, so you get a $\omega (\dots) = c~k (\dots)$ expression that should all work out if you get the coefficients right.