EM Course Notes

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


210A Notes


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}


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}\



Disc 1 Disc 2

210B Notes


Boundary Conditions:
\(E_{\parallel, 1} =E_{\parallel, 2} \) \(D_{\perp, 1} - D_{\perp, 2} = \sigma_{f}\)
\(H_{\parallel, 1} -H_{\parallel, 2} =\mathbf{k_f} \times \mathbf{n}\) \(B_{\perp, 1} =B_{\perp, 2} \)
Find the transmission and reflection coefficients of light impinging normally on a boundary (repeat at an angle):
E_I(x, t) = A e^{-i\omega t + i\mathbf{k_1}.\mathbf{x} }\\
E_T(x, t) = C e^{-i\omega t + i\mathbf{k_2}.\mathbf{x} }\\
E_R(x, t) = D e^{-i\omega t - i\mathbf{k_1}.\mathbf{x} }\\

B_I(x, t) = \frac{1}{v_1} A e^{-i\omega t + i\mathbf{k_1}.\mathbf{x} }\\
B_T(x, t) =  \frac{1}{v_2} C e^{-i\omega t + i\mathbf{k_2}.\mathbf{x} }\\
B_R(x, t) = - \frac{1}{v_1} D  e^{-i\omega t - i\mathbf{k_1}.\mathbf{x} }\]

Where we have used[1] \[ i \omega B = ik E\]. Solving for the boundary conditions, \[ C = \frac{2A}{1+\alpha}, D = \frac{1-\alpha}{1+\alpha} \] Where we have used[2] $\alpha = \frac{v_1}{v_2}$. Therefore, \[ R=\frac{|E_R H_R^*|}{E_IH_I^*} =\left(\frac{D}{A}\right)^2 = \left(\frac{1-\alpha}{1+\alpha}\right)^2 \\ T=\frac{|E_T H_T^*|}{E_IH_I^*} = \left(\frac{C}{A}\right)^2 \alpha = \frac{4\alpha}{(1+\alpha)^2} \]

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


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

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


Q1: A pulsar emits bursts of radio waves at two frequencies $\omega_1$, $\omega_2$. The pulses arrive at different times $t_1$, $t_2$ due to interaction with the interstellar medium- a dilute Hydrogen plasma. Find the distance $s$ from pulsar to earth, given $\tau= t_2 - t_2$. See F'12Q12 for solutions.

Q2: Obtain the non-relativistic Larmor radiation equation from the relativistic one. (S'08Q10)

From Wikipedia: \[ P = \frac{2}{3}\frac{q^2}{c^3m^2}\left(\frac{d\vec{p}}{dt}\cdot\frac{d\vec{p}}{dt}\right). \]

Assume the generalisation;

\[ P = -\frac{2}{3}\frac{q^2}{m^2c^3}\frac{dP^{\mu}}{d\tau}\frac{dP_{\mu}}{d\tau}. \]

When we expand and rearrange the energy-momentum four vector product we get:

\[ \frac{dP^{\mu}}{d\tau}\frac{dP_{\mu}}{d\tau} = \frac{v^2}{c^2}\left(\frac{dp}{d\tau}\right)^2 - \left(\frac{d\vec{p}}{d\tau}\right)^2 \] where I have used the fact that \[ \frac{dE}{d\tau} = \frac{pc^2}{E}\frac{dp}{d\tau} = v\frac{dp}{d\tau} \] When you let \(\beta\) tend to zero, \(\gamma\) tends to one, so that \(d\tau\) tends to dt. Thus we recover the non relativistic case.

See S'08Q10.

General Identities



  1. Griffiths, 9.3.2 Reflection and Transmission at Normal Incidence, p.384.
  2. EM Lim #4041
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