Since the refractive index of vacuum is real, this is highly unlikely. However, it seems that it does following the absorption per unit volume defined in terms of the Poynting vector:

\begin{equation} \text{Absorption} = -0.5 \nabla\cdot\vec{S}. \end{equation}

Using $E = E_0\cos(k_yy-\omega t)\hat{z}$ and $H = (E_0/\eta)\cos(k_yy-\omega t)\hat{x}$ it follows that the Poynting vector is:

\begin{equation} \vec{S} = \vec{E} \times \vec{H} = (E_0^2/\eta)\cos^2(k_yy-\omega t)\hat{y}, \end{equation}

leading to the absorption expression

\begin{equation} \text{Absorption} = -(k_yE_0^2/\eta)\sin(\omega t-k_yy)\cos(\omega t -k_yy), \end{equation}

which obviously is non-zero for vacuum. The average of the absorption, however, is zero. Is this expression correct, and if it is, what is its physical interpretation?

Thank you kindly in advance, Ian


The divergence of the Poynting vector is a measure of the rate at which energy is arriving or departing from a given region. As a sinusoidal plane wave propagates along the $y$ axis, wave crests wash over any given point, arriving and departing. This doesn't mean that energy is being absorbed. Absorption would be occurring if the time average of the divergence were nonzero.

  • $\begingroup$ In that case, would it not make more sense to define absorption by $-0.5\nabla\cdot\langle\vec{S}\rangle$? $\endgroup$ – Ian Berkman Nov 6 '19 at 0:52

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