EM Course Notes

Here is a conglomerate of notes gathered for the graduate EM classes 210A and 210B.

221A Notes

Electrostatics

Image charge coordinates

 \[
 q' = - \frac{a}{y}q \text{     ,      }  y' = \frac{a^2}{y}
  \]

Current in conductor

 \[
I = nAve = \frac{\partial Q}{\partial t} 
  \]

Electric and magnetic fields from potentials

 \[
\vec{E} = -\vec{\nabla} \phi - \frac{\partial \vec{A}}{\partial t}   \]\[
\vec{B} = \vec{\nabla} \times \vec{A}
\]

Waves

Maxwell's equations in vacuum:

 \[
\nabla \cdot \mathbf{E} = \frac{\rho}{\mathcal{E_0}} , \nabla \cdot \mathbf{B} = 0 \\


\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}} {\partial t}\ , \nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t}\

  \]

210B Notes

Radiation

Poynting's theorem:

 \[
\frac{\partial u}{\partial t} + \nabla\cdot\mathbf{S} = - \mathbf{J}\cdot\mathbf{E},
where '''J''' is the ''total'' current density and the energy density ''u'' is
u = \frac{1}{2}\left(\varepsilon_0 \mathbf{E}^2 + \frac{1}{\mu_0}\mathbf{B}^2\right)

  \]

Cutoff frequency for a rectangular waveguide:

\[ \omega_{c} = c \sqrt{\left(\frac{n \pi}{a}\right)^2 + \left(\frac{m \pi}{b}\right) ^2}, \] where \(n,m \ge 0\) are the mode numbers and a and b the lengths of the sides of the rectangle. For TE modes \( n,m \ge 0\) and \( n \ne m \), while for TM modes \( n, m \ge 1 \).

Maxwell's Stress Tensor:

\[
\overset{\leftrightarrow  }{ \mathbf{T}}_{ij} \equiv \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0}  \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)\  
\]

Derivation of Maxwell's Stress Tensor:

Derivation from Wikipedia

1. Starting with the Lorentz force law \[\mathbf{F} = q(\mathbf{E} + \mathbf{v}\times\mathbf{B})\] the force per unit volume for an unknown charge distribution is \[ \mathbf{f} = \rho\mathbf{E} + \mathbf{J}\times\mathbf{B} \]
2. Next, ρ and J can be replaced by the fields E and B, using Gauss's law and Ampère's circuital law: \[ \mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}\, \]
3. The time derivative can be rewritten to something that can be interpreted physically, namely the Poynting vector. Using the product rule and Faraday's law of induction gives \[\frac{\partial}{\partial t} (\mathbf{E}\times\mathbf{B}) = \frac{\partial\mathbf{E}}{\partial t}\times \mathbf{B} + \mathbf{E} \times \frac{\partial\mathbf{B}}{\partial t} = \frac{\partial\mathbf{E}}{\partial t}\times \mathbf{B} - \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E})\,\] and we can now rewrite f as \[\mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right) - \epsilon_0 \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E})\,\], then collecting terms with E and B gives \[\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} - \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E}) \right] + \frac{1}{\mu_0} \left[ - \mathbf{B}\times\left(\boldsymbol{\nabla}\times \mathbf{B} \right) \right] - \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)\,\].
4. A term seems to be "missing" from the symmetry in E and B, which can be achieved by inserting (∇ • B)B because of Gauss' law for magnetism: \[\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} - \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E}) \right] + \frac{1}{\mu_0} \left[(\boldsymbol{\nabla}\cdot \mathbf{B} )\mathbf{B} - \mathbf{B}\times\left(\boldsymbol{\nabla}\times \mathbf{B} \right) \right] - \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)\,\]. Eliminating the curls (which are fairly complicated to calculate), using the vector calculus identity \[\tfrac{1}{2} \boldsymbol{\nabla} (\mathbf{A}\cdot\mathbf{A}) = \mathbf{A} \times (\boldsymbol{\nabla} \times \mathbf{A}) + (\mathbf{A} \cdot \boldsymbol{\nabla}) \mathbf{A} \], leads to: \[\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right] + \frac{1}{\mu_0} \left[(\boldsymbol{\nabla}\cdot \mathbf{B} )\mathbf{B} + (\mathbf{B}\cdot\boldsymbol{\nabla}) \mathbf{B} \right] - \frac{1}{2} \boldsymbol{\nabla}\left(\epsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) - \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)\,\].
5. This expression contains every aspect of electromagnetism and momentum and is relatively easy to compute. It can be written more compactly by introducing the Maxwell stress tensor, \[\overset{\leftrightarrow }{ \mathbf{T}}_{ij} \equiv \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0} \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)\,\],

Multipole

Electric Dipole Radiation:

\[\mathbf{B} = \frac{\omega^2}{4\pi\varepsilon_0 c^3} (\hat{\mathbf{r}} \times \mathbf{p}) \frac{e^{i\omega r/c}}{r}\] \[\mathbf{E} = c \mathbf{B} \times \hat{\mathbf{r}}\]

which produces a total time-average radiated power P given by

\[P = \frac{\omega^4}{12\pi\varepsilon_0 c^3} |\mathbf{p}|^2.\]

Torque on magnetic and electric dipoles:

\[ \boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}\] for an electric dipole moment p (in coulomb-meters), or

\[ \boldsymbol{\tau} = \mathbf{m} \times \mathbf{B}\] for a magnetic dipole moment m (in ampere-square meters).

The resulting torque will tend to align the dipole with the applied field, which in the case of an electric dipole, yields a potential energy of

\[ U = -\mathbf{p} \cdot \mathbf{E}\].

The energy of a magnetic dipole is similarly

\[ U = -\mathbf{m} \cdot \mathbf{B}\].

Exercises

A pulsar emits bursts of radio waves at two frequencies $\omega_1$ and $\omega_2$. The pulses arrive at different times $t_1$ and $t_2$ due to interaction with the interstellar medium, which is a dilute Hydrogen plasma. Find the distance from the pulsar to earth, given $\tau= t_2 - t_2$.

Method:

Need $v_g = \frac{\partial \omega}{\partial k}$, so we have to find $\omega(k)$.

Have $v_p = \frac{\omega}{k} = \frac{c}{n(\omega)} \approx \frac{c}{\sqrt{\epsilon_r}}$

$\rightarrow$ Need $\epsilon_r$.

Have $\epsilon_r =1+\chi_e = 1+\frac{n_e e\mathbf{x}}{\epsilon_0 \mathbf{E}}$, where $\mathbf{E}= \mathbf{E_0} e^{-i\omega t}$ and $n_e =$ electron number density.

$\rightarrow$ Need $\mathbf{x}$.

Have $\mathbf{F}= e\mathbf{E}= m_e \ddot{\mathbf{x}}$, so

$\ddot{\mathbf{x}} = \frac{e}{m_e} \mathbf{E_0}e^{-i\omega t}$.

Have $\ddot{\mathbf{x}} = -\omega \mathbf{x}$, so

\[\mathbf{x} = -\frac{e}{m_e \omega^2} \mathbf{E_0}e^{-i\omega t}UNIQ1dc8ff1824c900c1-MathJax-19-QINU\omega_p^2 = \frac{n_e e^2}{\epsilon_0 m_e}$. \[k = \frac{\omega \sqrt{\epsilon_r}}{c} = frac{ \sqrt{ \omega^2- \omega_p^2}}{c}\] \[\frac{1}{v_g} = frac{\partial k}{\partial \omega} = \frac{1}{c}\sqrt{1- \frac{\omega_p^2}{\omega^2}}}\] \[\tau= d \left[\frac{1}{v_{g,2}} - \frac{1}{v_{g,1}}\right] \approx \frac{d \omega_p^2}{2c} \left[\frac{1}{\omega_1^2} - \frac{1}{\omega_2^2}\right]\]

\rightarrow\[d = \frac{2c\tau}{\omega_p^2}\left[\frac{\omega_1^2\omega_2^2}{\omega_2^2-\omega_1^2}\right]\]

See F'12Q12.