# Calculate the absorption of light using time averaged pointing vector (from RCWA simulation)

GOAL: I would like to compute the local absorption of light in material from an RCWA simulation. The RCWA tool can output the time-averaged (and input-normalized) Poynting vector.

The time-averaged Poynting vector is given as:

$$\textbf{S}=1/2*Re\big[\textbf{E(x,y,z,w)} \times \textbf{H*(x,y,z,w)}\big]$$

where * denotes the complex conjugate.

(and divided by Nc (refractive index of the cover medium) for normalization. Note that the E amplitude of the incident wave is 1.)

Now, I have seen in the literature, that the absorption per unit volume (@ a given wavelength w) and a given position (x,y,z) is:

$$A(x,y,z,w)/ V = -0.5*Re[\mathbf{\nabla}\cdot \textbf{P(x,y,z,w})]$$

where A is absorption, V is volume and P is the time-varying Poynting Vector.

Plugging in the defintion of the time varying Poynting vector:

$$A(x,y,z,w)/ V = -0.5*Re[\mathbf{\nabla}\cdot (\textbf{E(x,y,z,w)} \times \textbf{H(x,y,z,w}))]$$

What I would like to know: How can I compute the absorption per unit volume using the time-averaged Poynting vector, which I will call $$\textbf{S}$$?

Thanks a lot for your help!

EDIT:

I don't quite see how to get from the time-varying absorption to the time-averaged absorption?

$$$$\langle A/V \rangle = \langle -0.5*Re[\mathbf{\nabla}\cdot \textbf{P}]\rangle = -0.5*Re[\mathbf{\nabla}\cdot \langle\textbf{P}\rangle] = \mathbf{\nabla}\cdot \mathbf{S} \,.$$$$

Your equation with $$\nabla \mathbf{P}$$ makes sense if this is the divergence $$\mathbf{\nabla}\cdot \mathbf{P}$$. That is, the Poynting vector is the energy flux, i.e. the energy per unit are per unit time. To calculate the absorption, you want the net electromagnetic energy entering the volume. This is the surface integral of the Poynting vector, with a negative sign since you want the energy entering, and the area element points outward. Writing the surface integral and using the divergence theorem: $$$$A = -\int_S \mathbf{P}\cdot d\mathbf{S} = -\int_V \mathbf{\nabla}\cdot \mathbf{P} dV$$$$ Taking the volume to zero gives your $$A/V$$ equation.

For single frequency harmonic fields, the time average of $$\mathbf{P}$$ is $$\mathbf{S}$$, so the time averaged absorption per unit volume is $$$$\langle A/V \rangle = \mathbf{\nabla}\cdot \mathbf{S} \,.$$$$

If you have only $$\mathbf{S}(x,y,z,w)$$, then you can approximate the divergence as a central difference $$\begin{eqnarray} \mathbf{\nabla} \cdot \mathbf{S}(x,y,z,w) \simeq \frac{\mathbf{S}(x+\Delta x,y,z,w) -\mathbf{S}(x-\Delta x,y,z,w)}{2\Delta x} \nonumber\\ +\frac{\mathbf{S}(x,y+\Delta y,z,w) -\mathbf{S}(x,y-\Delta y,z,w)}{2\Delta x} \nonumber\\ +\frac{\mathbf{S}(x,y,z+\Delta z,w) -\mathbf{S}(x,y,z-\Delta z,w)}{2\Delta x} \end{eqnarray}$$ with whatever conveniently small $$\Delta x$$ your simulation gives you.

The $$\frac{1}{2}$$ and real part are the time average. That is with $$A(t) = {\rm Re} A e^{-i\omega t}$$, $$B(t)={\rm Re} B e^{-i\omega t}$$ where the real part of $$A$$ is half the sum of $$A$$ and its complex conjugate $$A^*$$. The product is $$$$A(t)B(t) = \frac{1}{4} \left [ AB^* +A^*B + ABe^{-i2\omega t}+A^*B^*e^{i2\omega t} \right ]$$$$ Time averaging gives just the first term $$$$\langle A(t)B(t)\rangle = \frac{1}{4} \left [ AB^* +A^*B \right ] = \frac{1}{2} {\rm Re} AB^*$$$$
Most people do not write the explicit time dependence when they write single frequency results. Here is the above time average result applied to the Poynting vector explicitly. That is, your time varying electric and magnetic fields for an angular frequency $$\omega$$ are $$\begin{eqnarray} \mathbf{E}(x,y,z,t) &=& {\rm Re} \left [ \mathbf{E}(x,y,z,\omega) e^{-i\omega t} \right ] =\frac{1}{2}\left [\mathbf{E}(x,y,z,\omega)e^{-i\omega t} +\mathbf{E^*}(x,y,z,\omega)e^{i\omega t} \right] \nonumber\\ \mathbf{H}(x,y,z,t) &=& {\rm Re} \left[ \mathbf{H}(x,y,z,\omega) e^{-i\omega t} \right] =\frac{1}{2}\left [\mathbf{H}(x,y,z,\omega)e^{-i\omega t} +\mathbf{H^*}(x,y,z,\omega)e^{i\omega t} \right] \nonumber\\ \end{eqnarray}$$ So the correct time varying Poynting vector is $$\begin{eqnarray} \mathbf{P}(x,y,z,t) &=& \mathbf{E}(x,y,z,t) \times \mathbf{H}(x,y,z,t) \nonumber\\ &=&\frac{1}{4} \left [ \mathbf{E}(x,y,z,\omega)\times \mathbf{H^*}(x,y,z,\omega) +\mathbf{E^*}(x,y,z,\omega)\times \mathbf{H}(x,y,z,\omega) \right] \nonumber\\ && +\frac{1}{4} \left [ \mathbf{E}(x,y,z,\omega)\times\mathbf{H}(x,y,z,\omega) e^{-i2\omega t} +\mathbf{E^*}(x,y,z,\omega)\times\mathbf{H^*}(x,y,z,\omega) e^{i2\omega t}\right] \nonumber\\ \end{eqnarray}$$ The time average of $$e^{\pm i 2\omega t}$$ is zero. If that is not obvious, you can just do the time average $$$$\left \langle e^{\pm 2i\omega t}\right\rangle = \frac{\omega}{2\pi} \int_0^{\frac{2\pi}{\omega}} dt e^{\pm i2\omega t} = \int_0^{2\pi} \frac{du}{2\pi} e^{\pm i2u} = 0$$$$ The first term of $$\mathbf{P}(x,y,z,t)$$ is independent of time so it equals its time average. The time averaged Poynting vector is $$\begin{eqnarray} \mathbf{S} &=&\frac{1}{4} \left [ \mathbf{E}(x,y,z,\omega)\times \mathbf{H^*}(x,y,z,\omega) +\mathbf{E^*}(x,y,z,\omega)\times \mathbf{H}(x,y,z,\omega) \right] \nonumber\\ &=& \frac{1}{2} {\rm Re} \left [ \mathbf{E}(x,y,z,\omega)\times \mathbf{H^*}(x,y,z,\omega)\right] \nonumber\\ &=& \frac{1}{2} {\rm Re} \left [\mathbf{E^*}(x,y,z,\omega)\times \mathbf{H}(x,y,z,\omega) \right ] \end{eqnarray}$$