Difference between revisions of "EM Course Notes"

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

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

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

 \[
\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}\

  \]

\(E_{\parallel, 1} =E_{\parallel, 2} \)	\(D_{\perp, 1} - D_{\perp, 2} = \sigma_{f}\)
\(H_{\parallel, 1} -H_{\parallel, 2} =\mathbf{k_f} \times \mathbf{n}\)	\(B_{\perp, 1} =B_{\perp, 2} \)

\[
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, \[ R=\frac{|E_R H_R^*|}{E_IH_I^*} =\left(\frac{D}{A}\right)^2 = \left(\frac{1-\alpha}{1+\alpha}\right)^2 \\ T=\frac{|E_T H_T^*|}{E_IH_I^*} = \left(\frac{C}{A}\right)^2 \alpha = \frac{4\alpha}{(1+\alpha)^2} \]

\[
\frac{\partial u}{\partial t} + \nabla\cdot\mathbf{S} = - \mathbf{J}\cdot\mathbf{E}
\]

, where J is the total current density and the energy density u is \[ u = \frac{1}{2}\left(\varepsilon_0 \mathbf{E}^2 + \frac{1}{\mu_0}\mathbf{B}^2\right) \]

\[
  \omega_{c} = c \sqrt{\left(\frac{n \pi}{a}\right)^2 + \left(\frac{m \pi}{b}\right) ^2}, 
\]

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 \( n, m \ge 1 \).

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

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

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

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

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

\[ \boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}\] 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

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

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.