Difference between revisions of "EM Course Notes"
Tom Neiser (Talk | contribs) |
Tom Neiser (Talk | contribs) |
||
Line 5: | Line 5: | ||
==221A Notes== | ==221A Notes== | ||
+ | |||
+ | |||
+ | |||
+ | '''Electrostatics''' | ||
+ | |||
+ | Image charge coordinates<div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
+ | \[ | ||
+ | q' = - \frac{a}{y}q \text{ , } y' = \frac{a^2}{y} | ||
+ | \] | ||
+ | </div> | ||
+ | |||
+ | Current in conductor<div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
+ | \[ | ||
+ | I = nAve = \frac{\partial Q}{\partial t} | ||
+ | \] | ||
+ | </div> | ||
+ | |||
+ | Electric and magnetic fields from potentials<div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
+ | \[ | ||
+ | \vec{E} = -\vec{\nabla} \phi - \frac{\partial \vec{A}}{\partial t} \]\[ | ||
+ | \vec{B} = \vec{\nabla} \times \vec{A} | ||
+ | \] | ||
+ | </div> | ||
+ | |||
+ | '''Waves''' | ||
+ | <!-- | ||
+ | This is a comment showing Maxwell's Eqn in Table Format: | ||
+ | |||
+ | :{| class="wikitable" | ||
+ | |- | ||
+ | | <math>\nabla \cdot \mathbf{E} = \frac {\rho} {\epsilon_0}</math> | ||
+ | | <math>\nabla \cdot \mathbf{B} = 0</math> | ||
+ | |- | ||
+ | | <math>\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}</math> | ||
+ | | <math>\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}} {\partial t}\ </math> | ||
+ | |} | ||
+ | --> | ||
+ | |||
+ | Maxwell's equations in vacuum:<div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
+ | \[ | ||
+ | \nabla \cdot \mathbf{E} = \frac{\rho}{\mathcal{E_0}} , \nabla \cdot \mathbf{B} = 0 \\ | ||
+ | |||
+ | |||
+ | \nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}} {\partial t}\ , \nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t}\ | ||
+ | |||
+ | \] | ||
+ | </div> | ||
+ | |||
<!-- | <!-- | ||
<div class="mw-collapsible mw-collapsed" style="width:750px"> | <div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
Line 25: | Line 73: | ||
</div> | </div> | ||
--> | --> | ||
− | + | Poynting's theorem:<div class="mw-collapsible mw-collapsed" style="width:750px"> | |
+ | \[ | ||
+ | \frac{\partial u}{\partial t} + \nabla\cdot\mathbf{S} = - \mathbf{J}\cdot\mathbf{E}, | ||
+ | where '''J''' is the ''total'' current density and the energy density ''u'' is | ||
+ | u = \frac{1}{2}\left(\varepsilon_0 \mathbf{E}^2 + \frac{1}{\mu_0}\mathbf{B}^2\right) | ||
+ | |||
+ | \] | ||
+ | </div> | ||
+ | |||
+ | Cutoff frequency for a rectangular waveguide: <div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
+ | :<math> | ||
+ | \omega_{c} = c \sqrt{\left(\frac{n \pi}{a}\right)^2 + \left(\frac{m \pi}{b}\right) ^2}, | ||
+ | </math> | ||
+ | where <math>n,m \ge 0</math> are the mode numbers and ''a'' and ''b'' the lengths of the sides of the rectangle. For TE modes <math> n,m \ge 0</math> and <math> n \ne m </math>, while for TM modes <math> n, m \ge 1 </math>. | ||
+ | </div> | ||
+ | |||
+ | Maxwell's Stress Tensor:<div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
+ | \[ | ||
+ | \overset{\leftrightarrow }{ \mathbf{T}}_{ij} \equiv \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0} \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)\ | ||
+ | \] | ||
+ | </div> | ||
+ | |||
+ | <div class="mw-collapsible mw-collapsed" style="width:700px"> | ||
+ | Derivation from [http://en.wikipedia.org/wiki/Maxwell_stress_tensor Wikipedia] | ||
+ | <ol> | ||
+ | <li>Starting with the Lorentz force law | ||
+ | :<math>\mathbf{F} = q(\mathbf{E} + \mathbf{v}\times\mathbf{B})</math> | ||
+ | the force per unit volume for an unknown charge distribution is | ||
+ | :<math> | ||
+ | \mathbf{f} = \rho\mathbf{E} + \mathbf{J}\times\mathbf{B} | ||
+ | </math></li> | ||
+ | <li>Next, ρ and '''J''' can be replaced by the fields '''E''' and '''B''', using Gauss's law and Ampère's circuital law: | ||
+ | :<math> | ||
+ | \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}\, | ||
+ | </math></li> | ||
+ | <li>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 | ||
+ | :<math>\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})\,</math> | ||
+ | and we can now rewrite '''f''' as | ||
+ | :<math>\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})\,</math>, | ||
+ | then collecting terms with '''E''' and '''B''' gives | ||
+ | :<math>\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)\,</math>.</li> | ||
+ | <li>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: | ||
+ | :<math>\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)\,</math>. | ||
+ | Eliminating the curls (which are fairly complicated to calculate), using the vector calculus identity | ||
+ | :<math>\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} </math>, | ||
+ | leads to: | ||
+ | :<math>\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)\,</math>.</li> | ||
+ | |||
+ | <li>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''', | ||
+ | :<math>\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)\,</math>, | ||
+ | </li> | ||
+ | </ol> | ||
+ | </div> | ||
+ | |||
+ | '''Radiation''' | ||
+ | Electric Dipole Radiation:<div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
+ | :<math>\mathbf{B} = \frac{\omega^2}{4\pi\varepsilon_0 c^3} (\hat{\mathbf{r}} \times \mathbf{p}) \frac{e^{i\omega r/c}}{r}</math> | ||
+ | :<math>\mathbf{E} = c \mathbf{B} \times \hat{\mathbf{r}}</math> | ||
+ | |||
+ | which produces a total time-average radiated power ''P'' given by | ||
+ | |||
+ | :<math>P = \frac{\omega^4}{12\pi\varepsilon_0 c^3} |\mathbf{p}|^2.</math> | ||
+ | </div> | ||
+ | |||
+ | '''Multipole''' | ||
+ | Torque on magnetic and electric dipoles:<div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
+ | |||
+ | :<math> \boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}</math> | ||
+ | for an [[electrical dipole moment|electric dipole moment]] '''p''' (in coulomb-meters), or | ||
+ | |||
+ | :<math> \boldsymbol{\tau} = \mathbf{m} \times \mathbf{B}</math> | ||
+ | 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 | ||
+ | |||
+ | :<math> U = -\mathbf{p} \cdot \mathbf{E}</math>. | ||
+ | |||
+ | The energy of a magnetic dipole is similarly | ||
+ | |||
+ | :<math> U = -\mathbf{m} \cdot \mathbf{B}</math>. | ||
+ | </div> | ||
+ | <br/> | ||
Revision as of 23:20, 14 April 2014
Here is a conglomerate of notes gathered for the graduate EM classes 210A and 210B.
Contents |
221A Notes
Electrostatics
Image charge coordinates\[ q' = - \frac{a}{y}q \text{ , } y' = \frac{a^2}{y} \]
\[ I = nAve = \frac{\partial Q}{\partial t} \]
\[ \vec{E} = -\vec{\nabla} \phi - \frac{\partial \vec{A}}{\partial t} \]\[ \vec{B} = \vec{\nabla} \times \vec{A} \]
Waves
Maxwell's equations in vacuum:\[ \nabla \cdot \mathbf{E} = \frac{\rho}{\mathcal{E_0}} , \nabla \cdot \mathbf{B} = 0 \\ \nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}} {\partial t}\ , \nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t}\ \]
Exercises
210B Notes
Poynting's theorem:\[ \frac{\partial u}{\partial t} + \nabla\cdot\mathbf{S} = - \mathbf{J}\cdot\mathbf{E}, where '''J''' is the ''total'' current density and the energy density ''u'' is u = \frac{1}{2}\left(\varepsilon_0 \mathbf{E}^2 + \frac{1}{\mu_0}\mathbf{B}^2\right) \]
\[ \omega_{c} = c \sqrt{\left(\frac{n \pi}{a}\right)^2 + \left(\frac{m \pi}{b}\right) ^2}, \] 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
- 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 \[ \mathbf{f} = \rho\mathbf{E} + \mathbf{J}\times\mathbf{B} \]
- Next, ρ and J can be replaced by the fields E and B, using Gauss's law and Ampère's circuital law: \[ \mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}\, \]
- 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 \[\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} - \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E}) \right] + \frac{1}{\mu_0} \left[ - \mathbf{B}\times\left(\boldsymbol{\nabla}\times \mathbf{B} \right) \right] - \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)\,\].
- 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: \[\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right] + \frac{1}{\mu_0} \left[(\boldsymbol{\nabla}\cdot \mathbf{B} )\mathbf{B} + (\mathbf{B}\cdot\boldsymbol{\nabla}) \mathbf{B} \right] - \frac{1}{2} \boldsymbol{\nabla}\left(\epsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) - \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)\,\].
- 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, \[\overset{\leftrightarrow }{ \mathbf{T}}_{ij} \equiv \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0} \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)\,\],
Radiation
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
\[P = \frac{\omega^4}{12\pi\varepsilon_0 c^3} |\mathbf{p}|^2.\]
Multipole
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
General Identities