Here's an exercise from Jackson:

A plane wave is incident normally on a perfectly absorbing flat screen. From the law of conservation of linear momentum show that the pressure exerted from the screen is equal to the field energy per unit volume in the wave.

Suppose the screen is parallel to the $zy$ plane at $x=0$. For simplicity let the fields be:

$$\textbf{E}=E_0e^{i(kx-\omega t+\delta)} \hat{\textbf{y}}$$ $$\textbf{B}=\frac{E_0}{c}e^{i(kx-\omega t+\delta)} \hat{\textbf{z}}$$

The force exerted on a volume $dV$ is therefore:

$$d\textbf{f}=\textbf{n}\cdot TdS+\frac{1}{c^2}\frac{\partial \textbf{S}}{\partial t}dSdx$$ Where $T$ is the Maxwell stress tensor and $\textbf{n}$ is the normal vector to the surface.

Since the screen is perfectly flat we let $dx \rightarrow 0$ and we are left with

$$\frac{d\textbf{f}}{dS}=\textbf{n}\cdot T$$

After some calculation the result is:

$$\frac{d\textbf{f}}{dS}=\epsilon_0E_0^2\cos^2(kx-\omega t+ \delta)\hat{\textbf{x}}$$

Taking the module and mediating in time the result is there, but this left me a bit baffled. I don't see where I used the fact that the screen is perfectly absorbing! (And, honestly, I wouldn't know how to mathematically formulate such a propriety). Moreover, why is the mean in time usually taken? Do you have a more rigorous approach to this?

• Have you tried computing the momentum of the incident wave?
– rob
Apr 17, 2015 at 17:51
• Mmmh. If we interpret the integral $$\frac{1}{c^2}\int_{V}\textbf{S}d^3x$$ as the wave momentum in the volume V then since the screen is flat the momentum is zero. Is this what you meant? Apr 17, 2015 at 18:05
• You did not have a reflected component of light in your equation - that's where you used the fact that the screen was perfectly absorbing... Apr 17, 2015 at 18:55
• Some questions: 1) Can I infer directly from Maxwell equations that there must be a reflected and refracted wave? 2) What would the treatment be, in the spirit of my solution, if there were a reflected component? 3) What does it mean, mathematically, that the surface is completely absorbent? Apr 17, 2015 at 19:21

You assume that the screen is perfectly absorbing when you use this field. By uses this field you are assuming that the electromagnetic field pass through the material without reflection. Furthermore, you are assuming that you don't have field in the another side of the surface. $$\textbf{E}=E_0e^{i(kx-\omega t+\delta)} \hat{\textbf{y}}, \ \ x<0$$ $$\textbf{B}=\frac{E_0}{c}e^{i(kx-\omega t+\delta)} \hat{\textbf{z}}, \ \ x<0$$ $$\textbf{E}=0 \ \ and \ \ \textbf{B}=0, \ \ when \ \ x>0$$ For example, if your screen is perfectly reflective, then you need to use the superposition of the incoming field and the outgoing field (reflected), getting twice of your result. If your screen is perfectly transparent, then you need to use the field in the two side of the surface, getting zero.