Difference between revisions of "EM Course Notes"
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Tom Neiser (Talk | contribs) (→Exercises) |
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− | == | + | ==210A Notes== |
'''Electrostatics''' | '''Electrostatics''' | ||
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</div> | </div> | ||
===''<small>Exercises</small>''=== | ===''<small>Exercises</small>''=== | ||
+ | [http://physwiki.com/w/images/a/a3/210A_Comp_Q1_%26_Sol.pdf Disc 1] | ||
+ | [http://physwiki.com/w/images/e/e8/F11Q10.pdf Disc 2] | ||
==210B Notes== | ==210B Notes== | ||
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</div> | </div> | ||
--> | --> | ||
− | + | '''Radiation''' | |
− | + | ||
− | + | Boundary Conditions:<div class="mw-collapsible mw-collapsed" style="width:750px"> | |
− | + | :{| class="wikitable" | |
− | + | |- | |
+ | | <math>E_{\parallel, 1} =E_{\parallel, 2} </math> | ||
+ | | <math>D_{\perp, 1} - D_{\perp, 2} = \sigma_{f}</math> | ||
+ | |- | ||
+ | | <math>H_{\parallel, 1} -H_{\parallel, 2} =\mathbf{k_f} \times \mathbf{n}</math> | ||
+ | | <math>B_{\perp, 1} =B_{\perp, 2} </math> | ||
+ | |} | ||
− | + | </div> | |
+ | |||
+ | Find the transmission and reflection coefficients of light impinging normally on a boundary (repeat at an angle):<div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
+ | \[ | ||
+ | E_I(x, t) = A e^{-i\omega t + i\mathbf{k_1}.\mathbf{x} }\\ | ||
+ | E_T(x, t) = C e^{-i\omega t + i\mathbf{k_2}.\mathbf{x} }\\ | ||
+ | E_R(x, t) = D e^{-i\omega t - i\mathbf{k_1}.\mathbf{x} }\\ | ||
+ | |||
+ | B_I(x, t) = \frac{1}{v_1} A e^{-i\omega t + i\mathbf{k_1}.\mathbf{x} }\\ | ||
+ | B_T(x, t) = \frac{1}{v_2} C e^{-i\omega t + i\mathbf{k_2}.\mathbf{x} }\\ | ||
+ | B_R(x, t) = - \frac{1}{v_1} D e^{-i\omega t - i\mathbf{k_1}.\mathbf{x} }\] | ||
+ | Where we have used<ref>Griffiths, ''9.3.2 Reflection and Transmission at Normal Incidence'', p.384. </ref> \[ i \omega B = ik E\]. | ||
+ | Solving for the boundary conditions, | ||
+ | \[ | ||
+ | C = \frac{2A}{1+\alpha}, D = \frac{1-\alpha}{1+\alpha} | ||
+ | \] | ||
+ | Where we have used<ref>EM Lim #4041</ref> $\alpha = \frac{v_1}{v_2}$. Therefore, | ||
+ | \[ | ||
+ | R=\frac{|E_R H_R^*|}{E_IH_I^*} =\left(\frac{D}{A}\right)^2 = \left(\frac{1-\alpha}{1+\alpha}\right)^2 \\ | ||
+ | T=\frac{|E_T H_T^*|}{E_IH_I^*} = \left(\frac{C}{A}\right)^2 \alpha = \frac{4\alpha}{(1+\alpha)^2} | ||
+ | \] | ||
+ | |||
+ | </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> | </div> | ||
Cutoff frequency for a rectangular waveguide: <div class="mw-collapsible mw-collapsed" style="width:750px"> | Cutoff frequency for a rectangular waveguide: <div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
− | :<math> | + | :<math> |
\omega_{c} = c \sqrt{\left(\frac{n \pi}{a}\right)^2 + \left(\frac{m \pi}{b}\right) ^2}, | \omega_{c} = c \sqrt{\left(\frac{n \pi}{a}\right)^2 + \left(\frac{m \pi}{b}\right) ^2}, | ||
</math> | </math> | ||
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</div> | </div> | ||
− | <div class="mw-collapsible mw-collapsed" style="width: | + | Derivation of Maxwell's Stress Tensor:<div class="mw-collapsible mw-collapsed" style="width:750px"> |
Derivation from [http://en.wikipedia.org/wiki/Maxwell_stress_tensor Wikipedia] | Derivation from [http://en.wikipedia.org/wiki/Maxwell_stress_tensor Wikipedia] | ||
<ol> | <ol> | ||
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</div> | </div> | ||
− | ''' | + | '''Multipole''' |
Electric Dipole Radiation:<div class="mw-collapsible mw-collapsed" style="width:750px"> | 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>\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}}</math> | |
which produces a total time-average radiated power ''P'' given by | which produces a total time-average radiated power ''P'' given by | ||
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</div> | </div> | ||
− | |||
Torque on magnetic and electric dipoles:<div class="mw-collapsible mw-collapsed" style="width:750px"> | Torque on magnetic and electric dipoles:<div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
− | |||
:<math> \boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}</math> | :<math> \boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}</math> | ||
for an [[electrical dipole moment|electric dipole moment]] '''p''' (in coulomb-meters), or | for an [[electrical dipole moment|electric dipole moment]] '''p''' (in coulomb-meters), or | ||
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:<math> U = -\mathbf{m} \cdot \mathbf{B}</math>. | :<math> U = -\mathbf{m} \cdot \mathbf{B}</math>. | ||
</div> | </div> | ||
+ | |||
+ | |||
===''<small>Exercises</small>''=== | ===''<small>Exercises</small>''=== | ||
+ | '''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 [[Fall_2012#Question_12| F'12Q12]] for solutions. | ||
+ | |||
+ | '''Q2:''' Obtain the non-relativistic Larmor radiation equation from the relativistic one. (S'08Q10) <div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
+ | From [http://en.wikipedia.org/wiki/Larmor_formula Wikipedia]: | ||
+ | :<math> | ||
+ | P = \frac{2}{3}\frac{q^2}{c^3m^2}\left(\frac{d\vec{p}}{dt}\cdot\frac{d\vec{p}}{dt}\right). | ||
+ | </math> | ||
+ | |||
+ | Assume the generalisation; | ||
+ | |||
+ | :<math> | ||
+ | P = -\frac{2}{3}\frac{q^2}{m^2c^3}\frac{dP^{\mu}}{d\tau}\frac{dP_{\mu}}{d\tau}. | ||
+ | </math> | ||
+ | |||
+ | When we expand and rearrange the energy-momentum four vector product we get: | ||
+ | |||
+ | :<math> | ||
+ | \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 | ||
+ | </math> | ||
+ | where I have used the fact that | ||
+ | :<math> \frac{dE}{d\tau} = \frac{pc^2}{E}\frac{dp}{d\tau} = v\frac{dp}{d\tau} </math> | ||
+ | When you let <math>\beta</math> tend to zero, <math>\gamma</math> tends to one, so that <math>d\tau</math> tends to dt. Thus we recover the non relativistic case. | ||
+ | |||
+ | See [[Spring_2008#Question_10| S'08Q10]]. | ||
+ | </div> | ||
+ | <br/> | ||
==General Identities== | ==General Identities== |
Latest revision as of 03:53, 25 January 2015
Here is a conglomerate of notes gathered for the graduate EM classes 210A and 210B.
Contents |
[edit] 210A 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:\[ \nabla \cdot \mathbf{E} = \frac{\rho}{\mathcal{E_0}} , \nabla \cdot \mathbf{B} = 0 \\ \nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}} {\partial t}\ , \nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t}\ \]
[edit] Exercises
[edit] 210B Notes
Radiation
Boundary Conditions:\(E_{\parallel, 1} =E_{\parallel, 2} \) \(D_{\perp, 1} - D_{\perp, 2} = \sigma_{f}\) \(H_{\parallel, 1} -H_{\parallel, 2} =\mathbf{k_f} \times \mathbf{n}\) \(B_{\perp, 1} =B_{\perp, 2} \)
\[ E_I(x, t) = A e^{-i\omega t + i\mathbf{k_1}.\mathbf{x} }\\ E_T(x, t) = C e^{-i\omega t + i\mathbf{k_2}.\mathbf{x} }\\ E_R(x, t) = D e^{-i\omega t - i\mathbf{k_1}.\mathbf{x} }\\ B_I(x, t) = \frac{1}{v_1} A e^{-i\omega t + i\mathbf{k_1}.\mathbf{x} }\\ B_T(x, t) = \frac{1}{v_2} C e^{-i\omega t + i\mathbf{k_2}.\mathbf{x} }\\ B_R(x, t) = - \frac{1}{v_1} D e^{-i\omega t - i\mathbf{k_1}.\mathbf{x} }\]
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
- 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, \[\overset{\leftrightarrow }{ \mathbf{T}}_{ij} \equiv \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0} \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)\,\],
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
\[P = \frac{\omega^4}{12\pi\varepsilon_0 c^3} |\mathbf{p}|^2.\]
\[ \boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}\] 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
\[ U = -\mathbf{m} \cdot \mathbf{B}\].
[edit] Exercises
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.
[edit] General Identities