Introduction to Electrodynamics/Chapter 7/5/2
In chapter 2, we figured out the energy stored in an electric field for the electrostatic case -- E squared.
In chapter 5, we figured out the energy stored in a magnetic field for the magnetostatic case -- B squared.
Let's derive the energy stored in electrodynamic fields. Maybe we'll see terms that look like the static case, and maybe we'll pick up some new, interesting terms along the way.
To calculate work, we need to figure out how much force is being applied on the particles in the system, and in what direction they are moving. dW = F . dl
We know the force --- the Lorentz Force Law gives us everything we need. F is the small charge dq times E + v x B, while dl is dimply v dt.
dW = F . dl = dq (E + v x B) . v dt
Since the magnetic field never does any work (it's always perpendicular to the velocity), we can forget about it.
dW = E . v dq dt
dq is rho dtau.
dW = E . v rho dtau dt
v rho is the volume current density, J.
dW = E . J dtau dt
So, dW/dt of a volume is integral over the volume of E . J dtau.
E . J is our interesting quantity. It represents the change in work over time at a point (rather, an infinitesimally small volume) in space.
Let's substitute in Ampere's Law with Maxwell's Correction to get an expression in terms of the fields alone.
<math> \begin{align} \vec{E} \cdot \vec{J} &= \vec{E} \cdot \left( {1 \over \mu_0} \nabla \times \vec{B} - \epsilon_0 {\partial \vec{E} \over \partial t} \right) \\
&= {1 \over \mu_0} \vec{E} \cdot (\nabla \times \vec{B}) - \epsilon_0 \vec{E} \cdot {\partial \vec{E} \over \partial t} \\
\end{align} </math>
This E dot dE by dt business can be simplified. Consider what the time derivative of E squared is. It's the time derivative of E dot E, which is E dot the time derivative of E plus the time derivative of E, dot E. dot products are commutative, so it is simple 2 E dot dE by dt, so 1 E dot dE by dt is the same as 1/2 d E squared by dt.
<math> \vec{E} \cdot \vec{J} = {1 \over \mu_0} \vec{E} \cdot (\nabla \times \vec{B}) - {\epsilon_0 \over 2} {\partial E^2 \over \partial t} </math>
Let's apply product rule 6 with E and B:
- <math>
\nabla \cdot (\vec{E} \times \vec{B}) = \vec{B} \cdot (\nabla \times \vec{E}) - \vec{E} \cdot (\nabla \times \vec{B}) </math>
Faraday's Law:
- <math>
\nabla \cdot (\vec{E} \times \vec{B}) = \vec{B} \cdot {\partial \vec{B} \over \partial t} - \vec{E} \cdot (\nabla \times \vec{B}) </math>
Now we have B dot dB by dt. Applying the same logic, we can rewrite that as 1/2 dB squared by dt.
- <math>
\nabla \cdot (\vec{E} \times \vec{B}) = {1 \over 2} {\partial B^2 \over \partial t} - \vec{E} \cdot (\nabla \times \vec{B}) </math>
Plugging it all in, and collecting the d square by dt togethter:
- <math>
\vec{E} \cdot \vec{J} = - {1 \over 2} {\partial \over \partial t} \left( \epsilon_0 E^2 + {1 \over \mu_0} B^2 \right) - {1 \over \mu_0} \nabla \cdot (\vec{E} \times \vec{B}) </math>
Let's stick this back into our dW/dt equation integral:
- <math>
{dW \over dt} = \int_V \left[- {1 \over 2} {\partial \over \partial t} \left( \epsilon_0 E^2 + {1 \over \mu_0} B^2 \right) - {1 \over \mu_0} \nabla \cdot (\vec{E} \times \vec{B})\right] d\tau </math>
We can apply Gauss's Theorem to the right hand side.
- <math>
{dW \over dt} = -{d \over dt} \int_V {1 \over 2} \left( \epsilon_0 E^2 + {1 \over \mu_0} B^2 \right)d\tau - {1 \over \mu_0} \oint_S (\vec{E} \times \vec{B}) \cdot d\vec{a} </math>
V is any volume, S is the surface of that volume, and dW/dt is the change in energy over time in that volume.
That was Poynting's Theorem. It's a particularly elegant answer to the question of how much energy is in a charge configuration, or rather, how is the energy changing over time, which is actually more interesting.
The left hand side is familiar. It's the old energy stored in the fields from our static cases. If the fields are decreasing, then work is being done on the system. If increasing, then work is being done BY the system.
The right hand side is new. It's weird. It's saying that if ExB flows out of the volume, then work is being done by the system, but if the flux is negative, ExB is flowing in, then work is being done on the system.
The ExB vector is named the Poynting Vector S. It represents the energy per unit time per unit area transported by the E and B field.
We can now rewrite the change in energy in terms of the static case plus the Poynting Vector flux.
- <math>
{dW \over dt} = -{d W_{EB}\over dt} - \oint_S \vec{S} \cdot d\vec{a} </math>
If we rewrite the work done over time as the change in mechanical energy density U of the charges,
- <math>
{dW \over dt} = {d \over dt} \int_V U_M\ d\tau = -{d W_{EB}\over dt} - \oint_S \vec{S} \cdot d\vec{a} </math>
We'll call the energy density of the E and B fields U_EB.
- <math>
U_{EB} = {1 \over 2} \left( \epsilon_0 E^2 + {1 \over \mu_0} B^2 \right) </math>
So now we can move things around, rewrite the S term according to the divergence theorem, and get that the divergence of S is the partial time derivative of the change of mechanical energy and energy density of the fields.
THis looks a lot like the equation for the divergence of J, and so we can think of S as the flow of energy.
(TODO: Equations)
Example 15
We calculated the Joule heating of a wire earlier in this chapter in 7.1. Let's do the same but using E and B fields and Poynting vectors.
Let's calculate the S vector on the surface of the wire.
The E field is simple V/L.
The B field is u0 I / 2 pi R.
E cross B = V I / 2 pi R L
The direction of E is down the wire, along the direction of J, and the B is tangential to the surface, so we get a direction of S pointing into the wire.
The change in energy over time is the integral of S . da along the surface. S is uniform, the area of the surface is 2 pi R L, so the answer is VI.
This seems like a rather roundabout way to calculate the energy, and it is, but it shows that the S vector is very useful, and somewhat simple to calculate.
