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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?

\begin{equation} \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} \,. \end{equation}

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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: \begin{equation} A = -\int_S \mathbf{P}\cdot d\mathbf{S} = -\int_V \mathbf{\nabla}\cdot \mathbf{P} dV \end{equation} 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 \begin{equation} \langle A/V \rangle = \mathbf{\nabla}\cdot \mathbf{S} \,. \end{equation}

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.

Added:

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 \begin{equation} A(t)B(t) = \frac{1}{4} \left [ AB^* +A^*B + ABe^{-i2\omega t}+A^*B^*e^{i2\omega t} \right ] \end{equation} Time averaging gives just the first term \begin{equation} \langle A(t)B(t)\rangle = \frac{1}{4} \left [ AB^* +A^*B \right ] = \frac{1}{2} {\rm Re} AB^* \end{equation}

More additions:

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 \begin{equation} \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 \end{equation} 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}

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  • $\begingroup$ Thanks a lot for your answer. I corrected my mistake in the question about the divergence. You are saying that the time-averaged absorption per unit volume is the divergence of S. Would it be possible to derive this statement mathematically in your answer? $\endgroup$
    – henry
    Jun 8 '21 at 7:06
  • $\begingroup$ I thought I did. The net electromagnetic energy entering a volume is the surface integral I wrote. That is the difference of the electromagnetic energy entering minus that leaving. This is the definition of absorption. The divergence theorem then converts this to a volume integral of the divergence. When the volume becomes small enough, the divergence is sensibly constant over the volume and you can just multiply the divergence times the volume. The time average commutes with the divergence as you can see from my last equation, since the divergence is just differences. This gives the result. $\endgroup$
    – user200143
    Jun 9 '21 at 1:04
  • $\begingroup$ Sorry to bother you with this but it's not immediately obvious to me. I understand the derivation of the surface integral to get the absorption, but I don't understand the following few things: 1. My equation has a -0.5 and a Re[], I don't see how to get to this by simply taking the volume to zero ? 2. Then, it is not obvious to me how you derive the time-averaged absorption/area (and how come that there is no -0.5 and Re[]) ? I would greatly appreciate it if you could add a few more steps to your derivation. Thanks a lot ! $\endgroup$
    – henry
    Jun 9 '21 at 8:14
  • $\begingroup$ I have added a few sentences $\endgroup$
    – user200143
    Jun 9 '21 at 14:03
  • $\begingroup$ Please see my edited answer. it would be very helpful if you could help me show how to rigorously go from the time-varying power to the time-averaged power. This would be super helpful and I will gladly attribute you the 50 points if you can help me here. Thank you so much. $\endgroup$
    – henry
    Jun 12 '21 at 9:21

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