Reconcile volumetric and surface energy

_psets/4.md

@@ -432,8 +432,8 @@ accurate model of that, which isn't practically possible.
 
 At the end of the day the point is to have the effects of the magnetic field cancel out, so it
 doesn't need to be perfect as it did in the previous definition of the ampere. In other words, we
-just want to measure a voltage and a current and not worry about the specifics of the magnetic
-field.
+just want to measure a voltage and a current and have that be enough. We don't want to worry about
+the specifics of the magnetic field, or the dynamic response of the mass balance, etc.
 
 
 ## (6.6)
@@ -445,66 +445,123 @@ Assume that sunlight has a power energy density of $$1 \si{kW/m^2}$$ (this is a
 typical average value in the continental USA is $$\approx 200 \si{W/m^2}$$). Estimate the electric
 field strength associated with this radiation.
 
-Update: this is wrong. I quickly jumped to using $$U$$ since I'd already used it in previous
-problems. But I need to take into account the Poynting vector.
+We need a relation between power, area, and electric field strength. We should be able to derive
+this from a volumetric integral over U, or a surface integral over P. Let's try both.
+
+### Using U
 
-The energy delivered per square meter in one second by sunlight is the field energy stored in a 1
-meter by 1 meter by 1 light second volume. So the average energy density of sunlight must be
+The energy density of light in a vacuum is
 
 $$
-\frac{10^3 \si{J}}{1 \si{m} \cdot 1 \si{m} \cdot \num{3e8} \si{m}} = \num{0.33e-5} \si{J/m^3}
+U = \frac{1}{2} (E \cdot D + B \cdot H)
 $$
 
-The energy density of light in a vacuum is
+For light propagating through free space, we know $$\vert E \vert / \vert H \vert =
+\sqrt{\mu_0/\epsilon_0}$$. So we can simplify to
 
 $$
 \begin{align*}
-U &= \frac{1}{2} (E \cdot D + B \cdot H) \\
-&= \frac{1}{2} \left( \epsilon_0 |E|^2 + \mu_0 |H|^2 \right) \\
+U &= \frac{1}{2} \left( \epsilon_0 |E|^2 + \mu_0 |H|^2 \right) \\
 &= \frac{1}{2} \left( \epsilon_0 |E|^2 + \epsilon_0 |E|^2 \right) \\
 &= \epsilon_0 |E|^2
 \end{align*}
 $$
 
-since $$\vert E \vert / \vert H \vert = \sqrt{\mu_0/\epsilon_0}$$.
+Since the strength of the electric field is oscillating sinusoidally, on average the squared field
+strength is one half the squared field strength squared:
 
-Thus the root mean square magnitude of the electric field must be
+$$
+\langle U \rangle = \epsilon_0 \langle |E|^2 \rangle = \frac{\epsilon_0}{2} E_\text{max}^2
+$$
+
+The energy delivered per square meter per second by light is the energy stored in a 1 meter by 1
+meter by 1 light second volume. So the power W (in watts) delivered to an area A (in square meters)
+is
 
 $$
-\frac{\num{0.33e-5} \si{J/m^3}}{\epsilon_0} = \num{3.8e5} \si{V^2/m^2}
+W = \frac{A c \epsilon_0}{2} E_\text{max}^2
 $$
 
-Since the electric field is a sinusoidal wave, the root mean square amplitude is half the maximum
-amplitude. So the peak field strength is $$\num{7.5e5} \si{V/m}$$.
+The speed of light is $$c = (\epsilon_0 \mu_0)^{-1/2}$$, so we can rewrite this as
 
+$$
+W = \frac{A}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_\text{max}^2
+$$
 
-### (b)
+### Using P
 
-{:.question}
-If 1 W of power is focused in a laser beam to a square millimeter, what is the field strength? What
-about if it is focused to the diffraction limit of $$\approx 1 \si{\mu m^2}$$?
+The [Poynting Vector](https://en.wikipedia.org/wiki/Poynting_vector) gives the direction and
+magnitude of energy flux due to the electric and magnetic fields. In MKS units it is measured in
+units of watts per square meter and is defined as $$P = E \times H$$.
+
+For light in free space, E and H are perpendicular to each other and to the direction of travel.
+Thus the Poynting vector points in the direction of travel. We also know that $$\vert E \vert /
+\vert H \vert = \sqrt{\mu_0/\epsilon_0}$$, so its magnitude is
+
+$$
+|P| = |E| |H| = \sqrt{\frac{\epsilon_0}{\mu_0}} |E|^2
+$$
+
+The E field oscillates sinusoidally, so the time average of its squared magnitude is one half its
+peak magnitude squared.
+
+$$
+\langle |P| \rangle = \sqrt{\frac{\epsilon_0}{\mu_0}} \langle |E|^2 \rangle
+= \frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_\text{max}^2
+$$
 
-Summarizing the previous section, we found that the root mean square field strength necessary to
-deliver $$x$$ watts of power to an area $$A$$ is
+So the average power W (in watts) delivered to an area A (in square meters) will be
 
 $$
-\frac{x \si{W}}{c \si{m/s}} \cdot \frac{1}{A \si{m^2}} \cdot \frac{1}{\epsilon_0 \si{F/m}}
+W = \frac{A}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_\text{max}^2
 $$
 
-So for one watt to be delivered to a square millimeter, the RMS field strength is
+as expected.
+
+### The field strength of sunlight
+
+Solving for the max field strength, we find that
 
 $$
-\frac{1 \si{W}}{\num{3e8} \si{m/s}} \cdot \frac{1}{10^{-6} \si{m^2}} \cdot \frac{1}{\num{8.85e-12} \si{F/m}}
-= \num{3.8e8} \si{V^2/m^2}
+E_\text{max}^2 = \frac{2 W}{A} \sqrt{\frac{\mu_0}{\epsilon_0}}
 $$
 
-which means the peak field strength is $$\num{7.5e8} \si{V/m}$$.
+So given that strong sunlight is one kilowatt per square meter, we can compute
 
-If it's focused down to a square micrometer, the RMS field strength is
+$$
+\begin{align*}
+E_\text{max}^2
+&= \frac{2 \cdot 10^3 \si{(W)}}{1 \si{(m^2)}} \sqrt{\frac{\mu_0}{\epsilon_0}} \si{(\ohm)} \\
+&= 753 \si{V^2/m^2} \\
+\end{align*}
+$$
+
+So the peak field strength is $$868 \si{V/m}$$.
+
+### (b)
+
+{:.question}
+If 1 W of power is focused in a laser beam to a square millimeter, what is the field strength? What
+about if it is focused to the diffraction limit of $$\approx 1 \si{\mu m^2}$$?
+
+For one watt in one square millimeter,
 
 $$
-\frac{1 \si{W}}{\num{3e8} \si{m/s}} \cdot \frac{1}{10^{-12} \si{m^2}} \cdot \frac{1}{\num{8.85e-12} \si{F/m}}
-= \num{3.8e14} \si{V^2/m^2}
+\begin{align*}
+E_\text{max}^2
+&= \frac{2 \cdot 1 \si{W}}{10^{-6} \si{m^2}} \sqrt{\frac{\mu_0}{\epsilon_0}} \\
+&= \num{7.5e8} \si{V^2/m^2} \\
+E_\text{max} &= 27 \si{kV/m}
+\end{align*}
 $$
 
-and the peak field strength is $$\num{7.5e14} \si{V/m}$$.
+For one watt in one square micrometer,
+
+$$
+\begin{align*}
+E_\text{max}^2
+&= \frac{2 \cdot 1 \si{W}}{10^{-12} \si{m^2}} \sqrt{\frac{\mu_0}{\epsilon_0}} \\
+&= \num{7.5e14} \si{V^2/m^2} \\
+E_\text{max} &= \num{2.7e7} \si{V/m}
+\end{align*}
+$$