Physics Friday 137

Part 17: Momentum Transport and Radiation Pressure

Let us consider a linearly-polarized plane wave propagating in the z direction, with plane of polarization in the x direction; so we then have
electric and magnetic fields
$\mathbf{E}=E_0\cos\left[k(z-ct)\right]\mathbf{\hat{x}}$
$\mathbf{B}=B_0\cos\left[k(z-ct)\right]\mathbf{\hat{y}}$ ,
with $B_0=\frac{E_0}{c}$ .
Let us now compute the Maxwell stress tensor $\mathbb{T}$ . As you may recall from here, the components of this symmetric (Cartesian) tensor are given by
$T_{ij}=\epsilon_0\left(E_iE_j-\frac12E^2\delta_{ij}\right)+\frac1{\mu_0}\left(B_iB_j-\frac12B^2\delta_{ij}\right)$ .
First, we recognise that $E^2=E_0^2\cos^2\left[k(z-ct)\right]$ . Secondly, $E_2=E_3=0$ , and $E_1=E=E_0\cos\left[k(z-ct)\right]$ . Thus, for the electric parts, we see $E_iE_j=0$ except for when i=j=1, so the electrical component only gives diagonal terms; we have
$\epsilon_0\left(E_1E_1-\frac12E^2\right)=\epsilon_0\left(E_0^2\cos^2\left[k(z-ct)\right]-\frac{1}{2}E_0^2\cos^2\left[k(z-ct)\right]\right)=\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]$ ,
$\epsilon_0\left(E_2E_2-\frac12E^2\right)=-\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]$ ,
and
$\epsilon_0\left(E_3E_3-\frac12E^2\right)=-\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]$ ;
Next, for magnetic components, we have $B^2=B_0^2\cos^2\left[k(z-ct)\right]$ . Secondly, $B_1=B_3=0$ , and $B_1=B=B_0\cos\left[k(z-ct)\right]$ ,
so again, we only have diagonal terms,
$\frac1{\mu_0}\left(B_1B_1-\frac12B^2\right)=-\frac{1}{2\mu_0}B_0^2\cos^2\left[k(z-ct)\right]$
$\frac1{\mu_0}\left(B_2B_2-\frac12B^2\right)=\frac{1}{2\mu_0}B_0^2\cos^2\left[k(z-ct)\right]$
$\frac1{\mu_0}\left(B_3B_3-\frac12B^2\right)=-\frac{1}{2\mu_0}B_0^2\cos^2\left[k(z-ct)\right]$ ,
so our tensor’s off-diagonal terms are zero. As for the diagonal terms, we note that for T₁₁,
$\begin{eqnarray}T_{11}&=&\epsilon_0\left(E_1E_1-\frac12E^2\right)+\frac1{\mu_0}\left(B_1B_1-\frac12B^2\right)\\&=&\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]-\frac{1}{2\mu_0}B_0^2\cos^2\left[k(z-ct)\right]\\&=&\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]-\frac{1}{2\mu_0c^2}E_0^2\cos^2\left[k(z-ct)\right]\\&=&\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]-\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]\\&=&{0}\end{eqnarray}$ ,
where we used the facts that $B_0=\frac{E_0}{c}$ , and that $\epsilon_0\mu_0=\frac{1}{c^2}$ .
Similarly,
$\begin{eqnarray}T_{22}&=&\epsilon_0\left(E_2E_2-\frac12E^2\right)+\frac1{\mu_0}\left(B_2B_2-\frac12B^2\right)\\&=&-\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]+\frac{1}{2\mu_0}B_0^2\cos^2\left[k(z-ct)\right]\\&=&-\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]+\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]\\&=&{0}\end{eqnarray}$ .
Lastly,
$\begin{eqnarray}T_{33}&=&\epsilon_0\left(E_3E_3-\frac12E^2\right)+\frac1{\mu_0}\left(B_3B_3-\frac12B^2\right)\\&=&-\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]-\frac{1}{2\mu_0}B_0^2\cos^2\left[k(z-ct)\right]\\&=&-\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]-\frac{\epsilon_0}{2}E_0^2\cos^2\left[k(z-ct)\right]\\&=&-\epsilon_0E_0^2\cos^2\left[k(z-ct)\right]\end{eqnarray}$ .
Recalling that the flux of the ith component of momentum in the j direction is given by $-T_{ij}$ , we see that the only momentum flow is of the z component of momentum, p_z, flowing in the z direction; the wave carries momentum, aligned with its direction, in the direction of its motion. If we time-average over the cycle of the wave, we obtain $T_{33}=-\frac{\epsilon_0}{2}E_0^2=-\epsilon_0E_{rms}^2$ .
(This last statement, in terms of the root-mean-square average, is valid for elliptically-polarized light as well).

Now, suppose we have a flat surface, perpendicular to our wave (and thus it is normal to the z axis in our coordinates); and that this surface is a perfect absorber. As the wave arrives at this surface, we see that the average linear momentum per unit area flowing into the surface in unit time is the vector p with components
$p_i=\sum_{j}\left(-T_{ij}\right)n_j=-T_{i3}$ ;
as previously established, this is zero except for the z component:
$p_z=\epsilon_0E_{rms}^2$ .
Now, since the wave is being absorbed, this much momentum is lost from the field per unit time per unit of area of the surface. By conservation of the momentum, the surface must be gaining this momentum per unit time, and since change in linear momentum with time is a force (Newton’s second law); we see that the above gives the force on the surface per unit area; that is to say, a pressure equal to
$P=\epsilon_0E_{rms}^2$ ; this is radiation pressure.
Note also that we demonstrated here that the average energy density in the wave is
$\bar{U}=\epsilon_0{E_{rms}^2}$ ;
thus, we see that the radiation pressure on a perfect absorber due to a plane electromagnetic wave incident normally to the surface is equal to the energy density of that wave.

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Tags: Electricity & Magnetism, Friday Physics, Maxwell Stress Tensor, physics, Radiation Pressure

This entry was posted on September 24, 2010 at 12:10 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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Physics Friday 137

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