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Posts Tagged ‘magnetic field’

An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 2013

Here’s my second update of my notes compilation for this course, including all of the following:

March 27, 2013 Fermi gas

March 26, 2013 Fermi gas thermodynamics

March 26, 2013 Fermi gas thermodynamics

March 23, 2013 Relativisitic generalization of statistical mechanics

March 21, 2013 Kittel Zipper problem

March 18, 2013 Pathria chapter 4 diatomic molecule problem

March 17, 2013 Gibbs sum for a two level system

March 16, 2013 open system variance of N

March 16, 2013 probability forms of entropy

March 14, 2013 Grand Canonical/Fermion-Bosons

March 13, 2013 Quantum anharmonic oscillator

March 12, 2013 Grand canonical ensemble

March 11, 2013 Heat capacity of perturbed harmonic oscillator

March 10, 2013 Langevin small approximation

March 10, 2013 Addition of two one half spins

March 10, 2013 Midterm II reflection

March 07, 2013 Thermodynamic identities

March 06, 2013 Temperature

March 05, 2013 Interacting spin

plus everything detailed in the description of my first update and before.

Midterm II reflection

Posted by peeterjoot on March 10, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Here’s some reflection about this Thursday’s midterm, redoing the problems without the mad scramble. I don’t think my results are too different from what I did in the midterm, even doing them casually now, but I’ll have to see after grading if these solutions are good.

Question: Magnetic field spin level splitting (2013 midterm II p1)

A particle with spin $S$ has $2 S + 1$ states $-S, -S + 1, \cdots S-1, S$. When exposed to a magnetic field, state splitting results in energy $E_m = \hbar m B$. Calculate the partition function, and use this to find the temperature specific magnetization. A “sum the geometric series” hint was given.

Our partition function is

\begin{aligned}Z &= \sum_{m = -S}^S e^{-\hbar \beta m B} \\ &= e^{-\hbar \beta S B}\sum_{m = -S}^S e^{-\hbar \beta (m + S) B} \\ &= e^{\hbar \beta S B}\sum_{n = 0}^{2 S} e^{-\hbar \beta n B}.\end{aligned} \hspace{\stretch{1}}(1.0.1)

Writing

\begin{aligned}a = e^{-\hbar \beta B},\end{aligned} \hspace{\stretch{1}}(1.0.2)

that is

\begin{aligned}Z &= a^{-S}\sum_{n = 0}^{2 S} a^n \\ &= a^{-S} \frac{ a^{2 S + 1} - 1 }{a - 1} \\ &= \frac{ a^{S + 1} - a^{-S} }{a - 1} \\ &= \frac{ a^{S + 1/2} - a^{-S - 1/2} }{a^{1/2} - a^{-1/2}}.\end{aligned} \hspace{\stretch{1}}(1.0.3)

Substitution of $a$ gives us

\begin{aligned}\boxed{Z = \frac{ \sinh( \hbar \beta B (S + 1/2) ) }{ \sinh( \hbar \beta B /2 ) }.}\end{aligned} \hspace{\stretch{1}}(1.0.4)

To calculate the magnetization $M$, I used

\begin{aligned}M = -\left\langle{{H}}\right\rangle/B.\end{aligned} \hspace{\stretch{1}}(1.0.5)

As [1] defines magnetization for a spin system. It was pointed out to me after the test that magnetization was defined differently in class as

\begin{aligned}\mu = \frac{\partial {B}}{\partial {F}}.\end{aligned} \hspace{\stretch{1}}(1.0.6)

These are, up to a sign, identical, at least in this case, since we have $\beta$ and $B$ travelling together in the partition function. In terms of the average energy

\begin{aligned}M &= -\frac{\left\langle{{H}}\right\rangle}{B} \\ &= \frac{1}{{B}} \frac{\partial {}}{\partial {\beta}} \ln Z(\beta B) \\ &= \frac{1}{{Z B}} \frac{\partial {}}{\partial {\beta}}Z(\beta B) \\ &= \frac{1}{{Z}} \frac{\partial {}}{\partial {(\beta B)}} Z(\beta B)\end{aligned} \hspace{\stretch{1}}(1.0.7)

Compare this to the in-class definition of magnetization

\begin{aligned}\mu &= \frac{\partial {F}}{\partial {B}} \\ &= \frac{\partial {}}{\partial {B}} \left( - k_{\mathrm{B}} T \ln Z(\beta B) \right) \\ &= -\frac{\partial {}}{\partial {B}} \frac{\ln Z (\beta B)}{\beta} \\ &= -\frac{1}{{\beta Z}} \frac{\partial {}}{\partial {B}} Z(\beta B) \\ &= -\frac{1}{{Z}} \frac{\partial {}}{\partial {(\beta B)}} Z(\beta B).\end{aligned} \hspace{\stretch{1}}(1.0.8)

For this derivative we have

\begin{aligned}\frac{\partial {}}{\partial {(\beta B)}} \ln Z \\ &= \frac{\partial {}}{\partial {(\beta B)}} \ln \frac{ \sinh( \hbar \beta B (S + 1/2) ) }{ \sinh( \hbar \beta B /2 ) } \\ &= \frac{\partial {}}{\partial {(\beta B)}} \left( \ln \sinh( \hbar \beta B (S + 1/2) ) - \ln \sinh( \hbar \beta B /2 ) \right) \\ &= \frac{\hbar }{2}\left( (2 S + 1) \coth( \hbar \beta B (S + 1/2) ) - \coth( \hbar \beta B /2 ) \right).\end{aligned} \hspace{\stretch{1}}(1.0.9)

This gives us

\begin{aligned}\mu &= -\frac{1}{{Z}} \frac{\hbar }{2}\left( (2 S + 1) \coth( \hbar \beta B (S + 1/2) ) - \coth( \hbar \beta B /2 ) \right) \\ &= -\frac{ \sinh( \hbar \beta B /2 ) }{ \sinh( \hbar \beta B (S + 1/2) ) }\frac{\hbar }{2}\left( (2 S + 1) \coth( \hbar \beta B (S + 1/2) ) - \coth( \hbar \beta B /2 ) \right)\end{aligned} \hspace{\stretch{1}}(1.0.10)

After some simplification (done offline in \nbref{midtermTwoQ1FinalSimplificationMu.nb}) we get

\begin{aligned}\boxed{\mu = \hbar \frac{(s+1) \sinh(\hbar \beta B s)-s \sinh(\hbar \beta B (s+1))}{\cosh(\hbar \beta B(2 s+1)) - 1}.}\end{aligned} \hspace{\stretch{1}}(1.0.11)

I got something like this on the midterm, but recall doing it somehow much differently.

Question: Pertubation of classical harmonic oscillator (2013 midterm II p2)

Consider a single particle perturbation of a classical simple harmonic oscillator Hamiltonian

\begin{aligned}H = \frac{1}{{2}} m \omega^2 \left( x^2 + y^2 \right) + \frac{1}{{2 m}} \left( p_x^2 + p_y^2 \right) + a x^4 + b y^6\end{aligned} \hspace{\stretch{1}}(1.0.12)

Calculate the canonical partition function, mean energy and specific heat of this system.

There were some instructions about the form to put the integrals in.

The canonical partition function is

\begin{aligned}Z &= \int dx dy dp_x dp_y e^{-\beta H} \\ &= \int dx e^{-\beta \left( \frac{1}{{2}} m \omega^2 x^2 + a x^4 \right)}\int dy e^{-\beta \left( \frac{1}{{2}} m \omega^2 y^2 + b y^6 \right)}\int dp_x dp_y e^{-\beta p_x^2/2 m} e^{-\beta p_y^2/2 m}.\end{aligned} \hspace{\stretch{1}}(1.0.13)

With

\begin{aligned}u = \sqrt{\frac{\beta}{2m}} p_x\end{aligned} \hspace{\stretch{1}}(1.0.14a)

\begin{aligned}v = \sqrt{\frac{\beta}{2m}} p_y,\end{aligned} \hspace{\stretch{1}}(1.0.14b)

the momentum integrals are

\begin{aligned}\int dp_x dp_y e^{-\beta p_x^2/2 m} e^{-\beta p_y^2/2 m} \\ &= \frac{2m}{\beta}\int du du e^{- u^2 - v^2} \\ &= \frac{m}{\beta}2 \pi\int 2 r dr e^{- r^2} \\ &= \frac{2 \pi m}{\beta}.\end{aligned} \hspace{\stretch{1}}(1.0.15)

Writing

\begin{aligned}f(x) = \frac{1}{{2}} m \omega^2 x^2 + a x^4\end{aligned} \hspace{\stretch{1}}(1.0.16a)

\begin{aligned}g(x) = \frac{1}{{2}} m \omega^2 y^2 + b y^4,\end{aligned} \hspace{\stretch{1}}(1.0.16b)

we have

\begin{aligned}\boxed{Z = \frac{2 \pi m}{\beta}\int dx e^{- \beta f(x)}\int dy e^{- \beta g(y)}.}\end{aligned} \hspace{\stretch{1}}(1.0.17)

The mean energy is

\begin{aligned}\left\langle{{H}}\right\rangle &= \frac{\int H e^{-\beta H}}{\int e^{-\beta H}} \\ &= -\frac{\partial {}}{\partial {\beta}} \ln \int e^{-\beta H} \\ &= \frac{\partial {}}{\partial {\beta}} \left( \ln \beta -\ln \int dx e^{- \beta f(x)} -\ln \int dy e^{- \beta g(y)} \right) \\ &= \frac{1}{{\beta}} + \frac{\int dx f(x) e^{- \beta f(x)}}{\int dx e^{- \beta f(x)}}+ \frac{\int dy g(y) e^{- \beta g(y)}}{\int dy e^{- \beta g(y)}}.\end{aligned} \hspace{\stretch{1}}(1.0.18)

The specific heat follows by differentiating once more

\begin{aligned}C_{\mathrm{V}} &= \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {T}} \\ &= \frac{\partial {\beta}}{\partial {T}}\frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {\beta}} \\ &= -\frac{1}{{k_{\mathrm{B}} T^2}}\frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {\beta}} \\ &= -k_{\mathrm{B}} \beta^2\frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {\beta}} \\ &= - k_{\mathrm{B}} \beta^2\left( -\frac{1}{{\beta^2}} + \frac{\partial {}}{\partial {\beta}} \left( \frac{ \int dx f(x) e^{- \beta f(x)} } { \int dx e^{- \beta f(x)} } + \frac{ \int dy g(y) e^{- \beta g(y)} } { \int dy e^{- \beta g(y)} } \right) \right).\end{aligned} \hspace{\stretch{1}}(1.0.19)

Differentiating the integral terms we have, for example,

\begin{aligned}\frac{\partial {}}{\partial {\beta}} \frac{\int dx f(x) e^{- \beta f(x)}}{\int dx e^{- \beta f(x)}}=-\frac{\int dx f^2(x) e^{- \beta f(x)}}{\int dx e^{- \beta f(x)}}-\left( \frac{ \int dx f(x) e^{- \beta f(x)} } { \int dx e^{- \beta f(x)} } \right)^2,\end{aligned} \hspace{\stretch{1}}(1.0.20)

so that the specific heat is

\begin{aligned}\boxed{C_{\mathrm{V}} =k_{\mathrm{B}} \left(1 + \frac{\int dx f^2(x) e^{- \beta f(x)}}{\int dx e^{- \beta f(x)}}+\left( \frac{ \int dx f(x) e^{- \beta f(x)} } { \int dx e^{- \beta f(x)} } \right)^2+ \frac{\int dy g^2(y) e^{- \beta g(y)}}{\int dy e^{- \beta g(y)}}+\left( \frac{ \int dy g(y) e^{- \beta g(y)} } { \int dy e^{- \beta g(y)} } \right)^2\right).}\end{aligned} \hspace{\stretch{1}}(1.0.21)

That’s as far as I took this problem. There was a discussion after the midterm with Eric about Taylor expansion of these integrals. That’s not something that I tried.

References

[1] C. Kittel and H. Kroemer. Thermal physics. WH Freeman, 1980.

PHY450HS1. Relativistic electrodynamics. Problem Set 5.

Posted by peeterjoot on April 7, 2011

Problem 1. Sinusoidal current density on an infinite flat conducting sheet.

Statement

An infinitely thin flat conducting surface lying in the $x-z$ plane carries a surface current density:

\begin{aligned}\boldsymbol{\kappa} = \mathbf{e}_3 \theta(t) \kappa_0 \sin\omega t\end{aligned} \hspace{\stretch{1}}(1.1)

Here $\mathbf{e}_3$ is a unit vector in the $z$ direction, $\kappa_0$ is the peak value of the current density, and $\theta(t)$ is the theta function: $\theta(t 0) = 1$.

\begin{enumerate}
\item Write down the equations determining the electromagnetic potentials. Specify which gauge you choose to work in.
\item Find the electromagnetic potentials outside the plane.
\item Find the electric and magnetic fields outside the plane.
\item Give a physical interpretation of the results of the previous section. Do they agree with your qualitative expectations?
\item Find the direction and magnitude of the energy flux outside the plane.
\item Consider a point at some distance from the plane. Sketch the intensity of the electromagnetic field near this point as a function of time. Explain physically.
\item Consider now a point near the plane. Are the electric and magnetic fields you found continuous across the conducting plane? Explain.
\end{enumerate}

1-2. Determining the electromagnetic potentials.

Augmenting the surface current density with a delta function we can form the current density for the system

\begin{aligned}\mathbf{J} = \delta(y) \boldsymbol{\kappa} = \mathbf{e}_3 \theta(t) \delta(y) \kappa_0 \sin\omega t.\end{aligned} \hspace{\stretch{1}}(1.2)

With only a current distribution specified use of the Coulomb gauge allows for setting the scalar potential on the surface equal to zero, so that we have

\begin{aligned}\square \mathbf{A} &= \frac{4 \pi \mathbf{J}}{c} \\ \mathbf{E} &= - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \\ \mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{B}\end{aligned} \hspace{\stretch{1}}(1.3)

Utilizing our Green’s function

\begin{aligned}G(\mathbf{x}, t) = \frac{\delta(t - {\left\lvert{\mathbf{x}}\right\rvert}/c)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert}} = \delta^3(\mathbf{x})\delta(t),\end{aligned} \hspace{\stretch{1}}(1.6)

we can invert our vector potential equation, solving for $\mathbf{A}$

\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= \int d^3 \mathbf{x}' dt' \square_{\mathbf{x}', t'} G(\mathbf{x} - \mathbf{x}', t - t') \mathbf{A}(\mathbf{x}', t') \\ &= \int d^3 \mathbf{x}' dt' G(\mathbf{x} - \mathbf{x}', t - t') \frac{4 \pi \mathbf{J}(\mathbf{x}', t')}{c} \\ &= \int d^3 \mathbf{x}' dt' \frac{\delta(t - t' - {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}/c)}{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}\frac{4 \pi \mathbf{J}(\mathbf{x}', t')}{c} \\ &= \int d^3 \mathbf{x}' \frac{\mathbf{J}(\mathbf{x}', t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c}{c {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \\ &= \frac{1}{{c}} \int dx' dy' dz'\mathbf{e}_3 \theta(t - {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}/c) \delta(y) \kappa_0 \sin(\omega(t - {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}/c))\frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}} \\ &= \frac{\mathbf{e}_3 \kappa_0}{c} \int dx' dz'\theta(t - {\left\lvert{\mathbf{x} -(x', 0, z')}\right\rvert}/c) \sin(\omega(t - {\left\lvert{\mathbf{x} -(x', 0, z')}\right\rvert}/c))\frac{1}{{{\left\lvert{\mathbf{x} - (x', 0, z')}\right\rvert}}} \\ &= \frac{\mathbf{e}_3 \kappa_0}{c} \int dx' dz'\theta\left(t - \frac{1}{{c}} \sqrt{(x-x')^2 + y^2 + (z-z')^2}\right) \frac{\sin\left(\omega\left(t - \frac{1}{{c}} \sqrt{(x-x')^2 + y^2 + (z-z')^2}\right)\right)}{\sqrt{(x-x')^2 + y^2 + (z-z')^2}}\end{aligned}

Now a switch to polar coordinates makes sense. Let’s use

\begin{aligned}x' - x &= r \cos\alpha \\ z' - z &= r \sin\alpha \end{aligned} \hspace{\stretch{1}}(1.7)

This gives us

\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= \frac{\mathbf{e}_3 \kappa_0}{c} \int_{r=0}^\infty \int_{\alpha=0}^{2\pi} r dr d\alpha\theta\left(t - \frac{1}{{c}} \sqrt{r^2 + y^2}\right) \frac{\sin\left(\omega\left(t - \frac{1}{{c}} \sqrt{r^2 + y^2 }\right)\right)}{\sqrt{r^2 + y^2}} \\ &= \frac{2 \pi \mathbf{e}_3 \kappa_0}{c} \int_{r=0}^\infty r dr \theta\left(t - \frac{1}{{c}} \sqrt{r^2 + y^2}\right) \frac{\sin\left(\omega\left(t - \frac{1}{{c}} \sqrt{r^2 + y^2 }\right)\right)}{\sqrt{r^2 + y^2}} \\ \end{aligned}

Since the theta function imposes a

\begin{aligned}t - \frac{1}{{c}} \sqrt{r^2 + y^2 } > 0\end{aligned} \hspace{\stretch{1}}(1.9)

constraint, equivalent to

\begin{aligned}c^2 t^2 > r^2 + y^2,\end{aligned} \hspace{\stretch{1}}(1.10)

we can reduce the upper range of the integral and drop the theta function explicitly

\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \frac{2 \pi \mathbf{e}_3 \kappa_0}{c} \int_{r=0}^{\sqrt{c^2 t^2 - y^2}} r dr \frac{\sin\left(\omega\left(t - \frac{1}{{c}} \sqrt{r^2 + y^2 }\right)\right)}{\sqrt{r^2 + y^2}} \end{aligned} \hspace{\stretch{1}}(1.11)

Here I got lazy and used Mathematica to help evaluate this integral, for an end result of

\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \frac{2 \pi \kappa_0 \omega}{c^2} \mathbf{e}_3 (1 - \cos(\omega(t - {\left\lvert{y}\right\rvert}/c))).\end{aligned} \hspace{\stretch{1}}(1.12)

3. Find the electric and magnetic fields outside the plane.

Our electric field can be calculated by inspection

\begin{aligned}\mathbf{E} = -\frac{1}{{c}}\frac{\partial {\mathbf{A}}}{\partial {t}}= -\frac{2 \pi \kappa_0 \omega^2}{c^3} \mathbf{e}_3 \sin(\omega(t - {\left\lvert{y}\right\rvert}/c)).\end{aligned} \hspace{\stretch{1}}(1.13)

For the magnetic field we have

\begin{aligned}\mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{A} \\ &= -\frac{2 \pi \kappa_0 \omega}{c^2} \mathbf{e}_3 \times \boldsymbol{\nabla} (1 -\cos(\omega(t - {\left\lvert{y}\right\rvert}/c)))) \\ &= \frac{2 \pi \kappa_0 \omega}{c^2} (-\sin(\omega(t - {\left\lvert{y}\right\rvert}/c))\mathbf{e}_3 \times \boldsymbol{\nabla} \omega(t - {\left\lvert{y}\right\rvert}/c) \\ &= \frac{2 \pi \kappa_0 \omega^2}{c^3} \sin(\omega(t - {\left\lvert{y}\right\rvert}/c))\mathbf{e}_3 \times \boldsymbol{\nabla} {\left\lvert{y}\right\rvert} \\ &= \frac{2 \pi \kappa_0 \omega^2}{c^3} \sin(\omega(t - {\left\lvert{y}\right\rvert}/c)) \mathbf{e}_3 \times \mathbf{e}_2,\end{aligned}

which gives us

\begin{aligned}\mathbf{B} = -\frac{2 \pi \kappa_0 \omega^2}{c^3}\mathbf{e}_1 \sin(\omega(t - {\left\lvert{y}\right\rvert}/c) \end{aligned} \hspace{\stretch{1}}(1.14)

4. Give a physical interpretation of the results of the previous section.

It was expected that the lack of boundary on the conducting sheet would make the potential away from the plane only depend on the $y$ components of the spatial distance, and this is precisely what we find performing the grunt work of the integration.

Given that we had a sinusoidal forcing function for our wave equation, it seems logical that we also find our non-homogeneous solution to the wave equation has sinusoidal dependence. We find that the sinusoidal current results in sinusoidal potentials and fields very much like one has in the electric circuits problem that we solve with phasors in engineering applications.

We find that the electric and magnetic fields are oriented parallel to the plane containing the surface current density, with the electric field in the direction of the current, and the magnetic field perpendicular to that, but having energy propagate outwards from the plane.

It’s kind of curious that despite introducing a step function in time for the current, that the end result appears to have no constraints on time $t > 0$, and that we have fields at all points in space and time from this current distribution, even the $t < 0$ case (ie: we have a non-causal system). I'd have expected a transient response to switching on the current. Perhaps this is because instantaneously inducing a current on all points in an infinite sheet is not physically realizable?

5. Find the direction and magnitude of the energy flux outside the plane.

Our energy flux, the Poynting vector, is

\begin{aligned}\mathbf{S} = \frac{c}{4\pi}\left( \frac{2 \pi \kappa_0 \omega^2}{c^3} \right)^2 \sin^2(\omega(t - {\left\lvert{y}\right\rvert}/c) \mathbf{e}_3 \times \mathbf{e}_1.\end{aligned} \hspace{\stretch{1}}(1.15)

This is

\begin{aligned}\mathbf{S} = \frac{ \pi \kappa_0^2 \omega^4 }{c^5} \sin^2(\omega(t - {\left\lvert{y}\right\rvert}/c) \mathbf{e}_2= \frac{ \pi \kappa_0^2 \omega^4 }{2 c^5} (1 - \cos( 2 \omega(t - {\left\lvert{y}\right\rvert}/c) ) ) \mathbf{e}_2.\end{aligned} \hspace{\stretch{1}}(1.16)

This energy flux is directed outwards along the $y$ axis, with magnitude oscillating around an average value of

\begin{aligned}{\left\lvert{\left\langle{{S}}\right\rangle}\right\rvert} = \frac{ \pi \kappa_0^2 \omega^4 }{2 c^5}.\end{aligned} \hspace{\stretch{1}}(1.17)

6. Sketch the intensity of the electromagnetic field far from the plane.

I’m assuming here that this question does not refer to the flux intensity $\left\langle{\mathbf{S}}\right\rangle$, since that is constant, and boring to sketch.

The time varying portion of either the electric or magnetic field is proportional to

\begin{aligned}\sin( \omega t - \omega {\left\lvert{y}\right\rvert}/c )\end{aligned} \hspace{\stretch{1}}(1.18)

We have a sinusoid as a function of time, of period $T = 2 \pi/\omega$ where the phase is adjusted at each position by the factor $\omega {\left\lvert{y}\right\rvert}/c$. Every increase of $\Delta y = 2 \pi c/\omega$ shifts the waveform back.

A sketch is attached.

7. Continuity across the plane?

It is sufficient to consider either the electric or magnetic field for the continuity question since the continuity is dictated by the sinusoidal term for both fields.

The point in time only changes the phase, so let’s consider the electric field at $t=0$, and an infinitesimal distance $y = \pm \epsilon c/\omega$. At either point we have

\begin{aligned}\mathbf{E}(0, \pm \epsilon c/\omega, 0, 0) = \frac{ 2 \pi \kappa_0 \omega^2 }{c^3} \mathbf{e}_3 \epsilon\end{aligned} \hspace{\stretch{1}}(1.19)

In the limit as $\epsilon \rightarrow 0$ the field strength matches on either side of the plane (and happens to equal zero for this $t= 0$ case).

We have a discontinuity in the spatial derivative of either field near the plate, but not for the fields themselves. A plot illustrates this nicely

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{phy450ps5P1Q7}
\caption{$\sin(t - {\left\lvert{y}\right\rvert})$}
\end{figure}

Problem 2. Fields generated by an arbitrarily moving charge.

Statement

Show that for a particle moving on a worldline parametrized by $(ct, \mathbf{x}_c(t))$, the retarded time $t_r$ with respect to an arbitrary space time point $(ct, \mathbf{x})$, defined in class as:

\begin{aligned}{\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert} = c(t - t_r)\end{aligned} \hspace{\stretch{1}}(2.20)

obeys

\begin{aligned}\boldsymbol{\nabla} t_r = -\frac{\mathbf{x} - \mathbf{x}_c(t_r)}{c {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert} = c(t - t_r) - \mathbf{v}_c(t_r) \cot (\mathbf{x} - \mathbf{x}_c(t_r))}\end{aligned} \hspace{\stretch{1}}(2.21)

and

\begin{aligned}\frac{\partial {t_r}}{\partial {t}} = \frac{c {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert}}{c {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert} = c(t - t_r) - \mathbf{v}_c(t_r) \cot (\mathbf{x} - \mathbf{x}_c(t_r))}\end{aligned} \hspace{\stretch{1}}(2.22)

\begin{enumerate}
\item Then, use these to derive the expressions for $\mathbf{E}$ and $\mathbf{B}$ given in the book (and in the class notes).
\item Finally, re-derive the already familiar expressions for the EM fields of a particle moving with uniform velocity.
\end{enumerate}

0. Solution. Gradient and time derivatives of the retarded time function.

Let’s use notation something like our text [1], where the solution to this problem is outlined in section 63, and write

\begin{aligned}\mathbf{R}(t_r) &= \mathbf{x} - \mathbf{x}_c(t_r) \\ R &= {\left\lvert{\mathbf{R}}\right\rvert}\end{aligned} \hspace{\stretch{1}}(2.23)

where

\begin{aligned}\frac{\partial {\mathbf{R}}}{\partial {t_r}} = - \mathbf{v}_c.\end{aligned} \hspace{\stretch{1}}(2.25)

From $R^2 = \mathbf{R} \cdot \mathbf{R}$ we also have

\begin{aligned}2 R \frac{\partial {R}}{\partial {t_r}} = 2 \mathbf{R} \cdot \frac{\partial {\mathbf{R}}}{\partial {t_r}},\end{aligned} \hspace{\stretch{1}}(2.26)

so if we write

\begin{aligned}\hat{\mathbf{R}} = \frac{\mathbf{R}}{R},\end{aligned} \hspace{\stretch{1}}(2.27)

we have

\begin{aligned}R'(t_r) = -\hat{\mathbf{R}} \cdot \mathbf{v}_c.\end{aligned} \hspace{\stretch{1}}(2.28)

Proceeding in the manner of the text, we have

\begin{aligned}\frac{\partial {R}}{\partial {t}} = \frac{\partial {R}}{\partial {t_r}} \frac{\partial {t_r}}{\partial {t}} = -\hat{\mathbf{R}} \cdot \mathbf{v}_c \frac{\partial {t_r}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(2.29)

From 2.20 we also have

\begin{aligned}R = {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert} = c(t - t_r),\end{aligned} \hspace{\stretch{1}}(2.30)

so

\begin{aligned}\frac{\partial {R}}{\partial {t}} = c\left(1 - \frac{\partial {t_r}}{\partial {t}}\right).\end{aligned} \hspace{\stretch{1}}(2.31)

This and 2.29 gives us

\begin{aligned}\boxed{\frac{\partial {t_r}}{\partial {t}} = \frac{1}{{ 1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} }}}\end{aligned} \hspace{\stretch{1}}(2.32)

For the gradient we operate on the implicit equation 2.30 again. This gives us

\begin{aligned}\boldsymbol{\nabla} R = \boldsymbol{\nabla} (c t - c t_r) = - c \boldsymbol{\nabla} t_r.\end{aligned} \hspace{\stretch{1}}(2.33)

However, we can also use the spatial definition of $R = {\left\lvert{\mathbf{x} - \mathbf{x}_c(t')}\right\rvert}$. Note that this distance $R = R(t_r)$ is a function of space and time, since $t_r = t_r(\mathbf{x}, t)$ is implicitly a function of the spatial and time positions at which the retarded time is to be measured.

\begin{aligned}\boldsymbol{\nabla} R &=\boldsymbol{\nabla} \sqrt{(\mathbf{x} - \mathbf{x}_c(t_r))^2} \\ &=\frac{1}{{2R}} \boldsymbol{\nabla} (\mathbf{x} - \mathbf{x}_c(t_r))^2 \\ &=\frac{1}{{R}} (x^\beta - x_c^\beta) \mathbf{e}_\alpha \partial_\alpha (x^\beta - x_c^\beta(t_r)) \\ &=\frac{1}{{R}} (\mathbf{R})_\beta \mathbf{e}_\alpha ({\delta_\alpha}^\beta - \partial_\alpha x_c^\beta(t_r)) \\ \end{aligned}

We have only this bit $\partial_\alpha x_c^\beta(t_r)$ to expand, but that’s just going to require a chain rule expansion. This is easier to see in a more generic form

\begin{aligned}\frac{\partial {f(g)}}{\partial {x^\alpha}} = \frac{\partial {f}}{\partial {g}} \frac{\partial {g}}{\partial {x^\alpha}},\end{aligned} \hspace{\stretch{1}}(2.34)

so we have

\begin{aligned}\frac{\partial {x_c^\beta(t_r)}}{\partial {x^\alpha}} = \frac{\partial {x_c^\beta(t_r)}}{\partial {t_r}} \frac{\partial {t_r}}{\partial {x^\alpha}},\end{aligned} \hspace{\stretch{1}}(2.35)

which gets us close to where we want to be

\begin{aligned}\boldsymbol{\nabla} R&=\frac{1}{{R}} \left(\mathbf{R} - (\mathbf{R})_\beta \frac{\partial {x_c^\beta(t_r)}}{\partial {t_r}} \mathbf{e}_\alpha \frac{\partial {t_r}}{\partial {x^\alpha}} \right) \\ &=\frac{1}{{R}} \left(\mathbf{R} - \mathbf{R} \cdot \frac{\partial {\mathbf{x}_c^\beta(t_r)}}{\partial {t_r}} \boldsymbol{\nabla} t_r \right)\end{aligned}

Putting the pieces together we have only minor algebra left since we can now equate the two expansions of $\boldsymbol{\nabla} R$

\begin{aligned}- c \boldsymbol{\nabla} t_r = \hat{\mathbf{R}} - \hat{\mathbf{R}} \cdot \mathbf{v}_c(t_r) \boldsymbol{\nabla} t_r.\end{aligned} \hspace{\stretch{1}}(2.36)

This is given in the text, but these in between steps are left for us and for our homework assignments! From this point we can rearrange to find the desired result

\begin{aligned}\boxed{\boldsymbol{\nabla} t_r = -\frac{1}{{c}} \frac{\hat{\mathbf{R}} }{ 1 - \hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} } = - \frac{\hat{\mathbf{R}}}{c} \frac{\partial {t_r}}{\partial {t}}}\end{aligned} \hspace{\stretch{1}}(2.37)

1. Solution. Computing the EM fields from the Lienard-Wiechert potentials.

Now we are ready to derive the values of $\mathbf{E}$ and $\mathbf{B}$ that arise from the Lienard-Wiechert potentials. We have for the electric field.

We’ll evaluate

\begin{aligned}\mathbf{E} &= -\frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} - \boldsymbol{\nabla} \phi \\ \mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{B}\end{aligned}

For the electric field we’ll use the chain rule on the vector potential

\begin{aligned}\frac{\partial {\mathbf{A}}}{\partial {t}} = \frac{\partial {t_r}}{\partial {t}} \frac{\partial {\mathbf{A}}}{\partial {t_r}}.\end{aligned} \hspace{\stretch{1}}(2.38)

Similarly for the gradient of the scalar potential we have

\begin{aligned}\boldsymbol{\nabla} \phi &=\mathbf{e}_\alpha \frac{\partial {\phi}}{\partial {x^\alpha}} \\ &=\mathbf{e}_\alpha \frac{\partial {\phi}}{\partial {t_r}} \frac{\partial {t_r}}{\partial {x^\alpha}} \\ &=\frac{\partial {\phi}}{\partial {t_r}} \boldsymbol{\nabla} t_r \\ &=- \frac{\hat{\mathbf{R}}}{c} \frac{\partial {t_r}}{\partial {t}}.\end{aligned}

Our electric field is thus

\begin{aligned}\mathbf{E} = - \frac{\partial {t_r}}{\partial {t}} \left( \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t_r}} - \frac{\hat{\mathbf{R}}}{c} \frac{\partial {\phi}}{\partial {t_r}} \right)\end{aligned} \hspace{\stretch{1}}(2.39)

For the magnetic field we have

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{A}&=\mathbf{e}_\alpha \times \frac{\partial {\mathbf{A}}}{\partial {x^\alpha}} \\ &=\mathbf{e}_\alpha \times \frac{\partial {\mathbf{A}}}{\partial {t_r}} \frac{\partial {t_r}}{\partial {x^\alpha}}.\end{aligned}

The magnetic field will therefore be found by evaluating

\begin{aligned}\mathbf{B} = (\boldsymbol{\nabla} t_r) \times \frac{\partial {\mathbf{A}}}{\partial {t_r}} = - \frac{\partial {t_r}}{\partial {t}} \frac{\hat{\mathbf{R}}}{c} \times \frac{\partial {\mathbf{A}}}{\partial {t_r}} \end{aligned} \hspace{\stretch{1}}(2.40)

Let’s compare this to $\hat{\mathbf{R}} \times \mathbf{E}$

\begin{aligned}\hat{\mathbf{R}} \times \mathbf{E} &= \hat{\mathbf{R}} \times \left( - \frac{\partial {t_r}}{\partial {t}} \left( \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t_r}} - \frac{\hat{\mathbf{R}}}{c} \frac{\partial {\phi}}{\partial {t_r}} \right) \right) \\ &= \hat{\mathbf{R}} \times \left( - \frac{\partial {t_r}}{\partial {t}} \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t_r}} \right)\end{aligned}

This equals 2.40, verifying that we have

\begin{aligned}\mathbf{B} = \hat{\mathbf{R}} \times \mathbf{E},\end{aligned} \hspace{\stretch{1}}(2.41)

something that we can determine even without fully evaluating $\mathbf{E}$.

We are now left to evaluate the retarded time derivatives found in 2.39. Our potentials are

\begin{aligned}\phi(\mathbf{x}, t) &= \frac{e}{R(t_r)} \frac{\partial {t_r}}{\partial {t}} \\ \mathbf{A}(\mathbf{x}, t) &= \frac{e \mathbf{v}_c(t_r)}{c R(t_r)} \frac{\partial {t_r}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(2.42)

It’s clear that the quantity ${\partial {t_r}}/{\partial {t}}$ is going to show up all over the place, so let’s label it $\gamma_{t_r}$. This is justified by comparing to a particle’s boosted rest frame worldline

\begin{aligned}\begin{bmatrix}c t' \\ x'\end{bmatrix}=\gamma\begin{bmatrix}1 & -\beta \\ -\beta & 1\end{bmatrix}\begin{bmatrix}c t \\ 0\end{bmatrix}= \begin{bmatrix}\gamma c t \\ -\gamma \beta c t\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.44)

where we have ${\partial {t'}}/{\partial {t}} = \gamma$, so for the remainder of this part of this problem we’ll write

\begin{aligned}\gamma_{t_r} \equiv \frac{\partial {t_r}}{\partial {t}} = \frac{1}{{ 1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} }}.\end{aligned} \hspace{\stretch{1}}(2.45)

Using primes to denote partial derivatives with respect to the retarded time $t_r$ we have

\begin{aligned}\phi' &= e \left( -\frac{R'}{R^2} \gamma_{t_r} + \frac{\gamma_{t_r}'}{R} \right) \\ \mathbf{A}' &= e \frac{\mathbf{v}_c}{c} \left( -\frac{R'}{R^2} \gamma_{t_r} + \frac{\gamma_{t_r}'}{R} \right)+ e \frac{\mathbf{a}_c}{c} \frac{\gamma_{t_r}}{R},\end{aligned} \hspace{\stretch{1}}(2.46)

so the electric field is

\begin{aligned}\mathbf{E} &= - \gamma_{t_r} \left( \frac{1}{{c}} \mathbf{A}' - \frac{\hat{\mathbf{R}}}{c} \phi' \right) \\ &= - \frac{e \gamma_{t_r}}{c} \left( \frac{\mathbf{v}_c}{c} \left( -\frac{R'}{R^2} \gamma_{t_r} + \frac{\gamma_{t_r}'}{R} \right)+ \frac{\mathbf{a}_c}{c} \frac{\gamma_{t_r}}{R}- \hat{\mathbf{R}} \left( -\frac{R'}{R^2} \gamma_{t_r} + \frac{\gamma_{t_r}'}{R} \right) \right) \\ &= - \frac{e \gamma_{t_r}}{c} \left( \frac{\mathbf{v}_c}{c} \left( \frac{c}{R^2} \gamma_{t_r} + \frac{\gamma_{t_r}'}{R} \right)+ \frac{\mathbf{a}_c}{c} \frac{\gamma_{t_r}}{R}- \hat{\mathbf{R}} \left( \frac{c}{R^2} \gamma_{t_r} + \frac{\gamma_{t_r}'}{R} \right) \right) \\ &= - \frac{e \gamma_{t_r}}{c R} \left( \gamma_{t_r} \left( \frac{\mathbf{a}_c}{c} +\frac{\mathbf{v}_c}{R} - \frac{\hat{\mathbf{R}} c}{R}\right)+ \gamma_{t_r}' \left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right).\end{aligned}

Here’s where things get slightly messy.

\begin{aligned}\gamma_{t_r}' &= \frac{\partial {}}{\partial {t_r}} \frac{1}{{1 - \frac{\mathbf{v}_c}{c} \cdot \hat{\mathbf{R}} }} \\ &= -\gamma_{t_r}^2 \frac{\partial {}}{\partial {t_r}} \left( 1 - \frac{\mathbf{v}_c}{c} \cdot \hat{\mathbf{R}} \right) \\ &= \gamma_{t_r}^2 \left( \frac{\mathbf{a}_c}{c} \cdot \hat{\mathbf{R}} + \frac{\mathbf{v}_c}{c} \cdot \hat{\mathbf{R}}' \right),\end{aligned}

and messier

\begin{aligned}\hat{\mathbf{R}}' &= \frac{\partial {}}{\partial {t_r}} \frac{ \mathbf{R} }{ R } \\ &= \frac{ \mathbf{R}' }{ R } - \frac{\mathbf{R} R'}{R^2} \\ &= -\frac{ \mathbf{v}_c }{ R } - \frac{\hat{\mathbf{R}} (-c)}{R} \\ &= \frac{1}{{R}} \left( -\mathbf{v}_c + c \hat{\mathbf{R}} \right),\end{aligned}

then a bit unmessier

\begin{aligned}\gamma_{t_r}'&= \gamma_{t_r}^2 \left( \frac{\mathbf{a}_c}{c} \cdot \hat{\mathbf{R}} + \frac{\mathbf{v}_c}{c} \cdot \hat{\mathbf{R}}' \right) \\ &= \gamma_{t_r}^2 \left( \frac{\mathbf{a}_c}{c} \cdot \hat{\mathbf{R}} + \frac{\mathbf{v}_c}{c R} \cdot (-\mathbf{v}_c + c \hat{\mathbf{R}}) \right) \\ &= \gamma_{t_r}^2 \left( \hat{\mathbf{R}} \cdot \left( \frac{\mathbf{a}_c}{c} + \frac{\mathbf{v}_c}{R} \right) - \frac{\mathbf{v}_c^2}{c R} \right).\end{aligned}

Now we are set to plug this back into our electric field expression and start grouping terms

\begin{aligned}\mathbf{E}&= - \frac{e \gamma_{t_r}^2}{c R} \left( \frac{\mathbf{a}_c}{c} +\frac{\mathbf{v}_c}{R} - \frac{\hat{\mathbf{R}} c}{R}+ \gamma_{t_r} \left( \hat{\mathbf{R}} \cdot \left( \frac{\mathbf{a}_c}{c} + \frac{\mathbf{v}_c}{R} \right) - \frac{\mathbf{v}_c^2}{c R} \right)\left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right) \\ &= - \frac{e \gamma_{t_r}^3}{c R} \left( \left(\frac{\mathbf{a}_c}{c} +\frac{\mathbf{v}_c}{R} - \frac{\hat{\mathbf{R}} c}{R}\right) \left(1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} \right)+ \left( \hat{\mathbf{R}} \cdot \left( \frac{\mathbf{a}_c}{c} + \frac{\mathbf{v}_c}{R} \right) - \frac{\mathbf{v}_c^2}{c R} \right)\left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right) \\ &= - \frac{e \gamma_{t_r}^3}{c^2 R} \left( \mathbf{a}_c\left(1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} \right)+\hat{\mathbf{R}} \cdot \mathbf{a}_c \left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right) \\ &\quad - \frac{e \gamma_{t_r}^3}{c R} \left( \left(\frac{\mathbf{v}_c}{R} - \frac{\hat{\mathbf{R}} c}{R}\right) \left(1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} \right)+ \left( \hat{\mathbf{R}} \cdot \left( \frac{\mathbf{v}_c}{R} \right) - \frac{\mathbf{v}_c^2}{c R} \right)\left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right) \\ \end{aligned}

Using

\begin{aligned}\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{b} (\mathbf{a} \cdot \mathbf{c}) - \mathbf{c} (\mathbf{a} \cdot \mathbf{b})\end{aligned} \hspace{\stretch{1}}(2.48)

We can verify that

\begin{aligned}- \left( \mathbf{a}_c\left(1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} \right)+\hat{\mathbf{R}} \cdot \mathbf{a}_c \left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right) &=-\mathbf{a}_c + \mathbf{a} \hat{\mathbf{R}} \cdot \frac{\mathbf{v}}{c} - \hat{\mathbf{R}} \cdot \mathbf{a}_c \frac{\mathbf{v}_c}{c} + \hat{\mathbf{R}} \cdot \mathbf{a}_c \hat{\mathbf{R}} \\ &= \hat{\mathbf{R}} \times \left( \left(\hat{\mathbf{R}} - \frac{\mathbf{v}_c}{c} \right) \times \mathbf{a}_c \right),\end{aligned}

which gets us closer to the desired end result

\begin{aligned}\mathbf{E}= \frac{e \gamma_{t_r}^3}{c^2 R} \hat{\mathbf{R}} \times \left( \left(\hat{\mathbf{R}} - \frac{\mathbf{v}_c}{c} \right) \times \mathbf{a}_c \right)- \frac{e \gamma_{t_r}^3}{c R^2} \left( \left(\mathbf{v}_c- \hat{\mathbf{R}} c\right) \left(1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} \right)+ \left( \hat{\mathbf{R}} \cdot \mathbf{v}_c - \frac{\mathbf{v}_c^2}{c} \right)\left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right).\end{aligned} \hspace{\stretch{1}}(2.49)

It is also easy to show that the remaining bit reduces nicely, since all the dot product terms conveniently cancel

\begin{aligned}- \left( \left(\mathbf{v}_c- \hat{\mathbf{R}} c\right) \left(1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} \right)+ \left( \hat{\mathbf{R}} \cdot \mathbf{v}_c - \frac{\mathbf{v}_c^2}{c} \right)\left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right) = c \left( 1 - \frac{\mathbf{v}_c^2}{c^2} \right) \left( \hat{\mathbf{R}} - \frac{\mathbf{v}}{c} \right)\end{aligned} \hspace{\stretch{1}}(2.50)

This completes the exercise, leaving us with

\begin{aligned}\boxed{\begin{aligned}\mathbf{E}&= \frac{e \gamma_{t_r}^3}{c^2 R} \hat{\mathbf{R}} \times \left( \left(\hat{\mathbf{R}} - \frac{\mathbf{v}_c}{c} \right) \times \mathbf{a}_c \right)+\frac{e \gamma_{t_r}^3}{R^2} \left( 1 - \frac{\mathbf{v}_c^2}{c^2} \right) \left( \hat{\mathbf{R}} - \frac{\mathbf{v}_c}{c} \right) \\ \mathbf{B} &= \hat{\mathbf{R}} \times \mathbf{E}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(2.51)

Looking back to 2.45 where $\gamma_{t_r}$ was defined, we see that this compares to (63.8-9) in the text.

2. Solution. EM fields from a uniformly moving source.

For a uniform source moving in space at constant velocity

\begin{aligned}\mathbf{x}_c(t) = \mathbf{v} t,\end{aligned} \hspace{\stretch{1}}(2.52)

our retarded time measured from the spacetime point $(ct, \mathbf{x})$ is defined implicitly by

\begin{aligned}R = {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r) }\right\rvert} = c (t - t_r).\end{aligned} \hspace{\stretch{1}}(2.53)

Squaring this we have

\begin{aligned}\mathbf{x}^2 + \mathbf{v}^2 t_r^2 - 2 t_r \mathbf{x} \cdot \mathbf{v} = c^2 t^2 + c^2 t_r^2 - 2 c t t_r,\end{aligned} \hspace{\stretch{1}}(2.54)

or

\begin{aligned}( c^2 -\mathbf{v}^2) t_r^2 + 2 t_r ( - c t + \mathbf{x} \cdot \mathbf{v} ) = \mathbf{x}^2 - c^2 t^2.\end{aligned} \hspace{\stretch{1}}(2.55)

Rearranging to complete the square we have

\begin{aligned}\left( \sqrt{c^2 - \mathbf{v}^2} t_r - \frac{ t c^2 - \mathbf{x} \cdot \mathbf{v} }{\sqrt{c^2 - \mathbf{v}^2}} \right)^2 &= \mathbf{x}^2 - c^2 t^2 +\frac{ (t c^2 - \mathbf{x} \cdot \mathbf{v})^2 }{c^2 - \mathbf{v}^2} \\ &= \frac{ (\mathbf{x}^2 - c^2 t^2)( c^2 - \mathbf{v}^2) + (t c^2 - \mathbf{x} \cdot \mathbf{v})^2}{ c^2 - \mathbf{v}^2} \\ &= \frac{ \mathbf{x}^2 c^2 - \mathbf{x}^2 \mathbf{v}^2 - {c^4 t^2} + c^2 t^2 \mathbf{v}^2 + {t^2 c^4} + (\mathbf{x} \cdot \mathbf{v})^2 - 2 t c^2 (\mathbf{x} \cdot \mathbf{v}) }{ c^2 - \mathbf{v}^2 } \\ &= \frac{ c^2 ( \mathbf{x}^2 + t^2 \mathbf{v}^2 -2 t (\mathbf{x} \cdot \mathbf{v})) - \mathbf{x}^2 \mathbf{v}^2 + (\mathbf{x} \cdot \mathbf{v})^2 }{ c^2 - \mathbf{v}^2 } \\ &= \frac{ c^2 ( \mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \mathbf{v})^2 }{ c^2 - \mathbf{v}^2 } \\ \end{aligned}

Taking roots (and keeping the negative so that we have $t_r = t - {\left\lvert{\mathbf{x}}\right\rvert}/c$ for the $\mathbf{v} = 0$ case, we have

\begin{aligned}\sqrt{1 - \frac{\mathbf{v}^2}{c^2}} c t_r =\frac{1}{{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}}} \left(c t - \mathbf{x} \cdot \frac{\mathbf{v}}{c} - \sqrt{ \left( \mathbf{x} - \mathbf{v} t \right)^2 - \left(\mathbf{x} \times \frac{\mathbf{v}}{c}\right)^2 }\right),\end{aligned} \hspace{\stretch{1}}(2.56)

or with $\boldsymbol{\beta} = \mathbf{v}/c$, this is

\begin{aligned}c t_r = \frac{1}{{1 - \boldsymbol{\beta}^2}} \left( c t - \mathbf{x} \cdot \boldsymbol{\beta} - \sqrt{ \left( \mathbf{x} - \mathbf{v} t \right)^2 - \left(\mathbf{x} \times \boldsymbol{\beta}\right)^2 } \right).\end{aligned} \hspace{\stretch{1}}(2.57)

What’s our retarded distance $R = c t - c t_r$? We get

\begin{aligned}R = \frac{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 }}{ 1 - \boldsymbol{\beta}^2 }.\end{aligned} \hspace{\stretch{1}}(2.58)

For the vector distance we get (with $\boldsymbol{\beta} \cdot (\mathbf{x} \wedge \boldsymbol{\beta}) = (\boldsymbol{\beta} \cdot \mathbf{x}) \boldsymbol{\beta} - \mathbf{x} \boldsymbol{\beta}^2$)

\begin{aligned}\mathbf{R} = \frac{\mathbf{x} -\mathbf{v} t + \boldsymbol{\beta} \cdot (\mathbf{x} \wedge \boldsymbol{\beta}) + \boldsymbol{\beta} \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 }}{ 1 - \boldsymbol{\beta}^2 }.\end{aligned} \hspace{\stretch{1}}(2.59)

For the unit vector $\hat{\mathbf{R}} = \mathbf{R}/R$ we have

\begin{aligned}\hat{\mathbf{R}} = \frac{\mathbf{x} - \mathbf{v} t + \boldsymbol{\beta} \cdot (\mathbf{x} \wedge \boldsymbol{\beta}) + \boldsymbol{\beta} \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 }}{ \boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 } }.\end{aligned} \hspace{\stretch{1}}(2.60)

The acceleration term in the electric field is zero, so we are left with just

\begin{aligned}\mathbf{E}= \frac{e \gamma_{t_r}^3}{R^2} \left( 1 - \frac{\mathbf{v}_c^2}{c^2} \right) \left( \hat{\mathbf{R}} - \frac{\mathbf{v}_c}{c} \right).\end{aligned} \hspace{\stretch{1}}(2.61)

Leading to $\gamma_{t_r}$, we have

\begin{aligned}\hat{\mathbf{R}} \cdot \boldsymbol{\beta} = \frac{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t + R^{*} \boldsymbol{\beta})}{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + R^{*}},\end{aligned} \hspace{\stretch{1}}(2.62)

where, following section 38 of the text we write

\begin{aligned}R^{*} = \sqrt{(\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 }\end{aligned} \hspace{\stretch{1}}(2.63)

This gives us

\begin{aligned}\gamma_{t_r} = \frac{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + R^{*}}{R^{*}(1 - \boldsymbol{\beta}^2)}.\end{aligned} \hspace{\stretch{1}}(2.64)

Observe that this equals one when $\boldsymbol{\beta} = 0$ as expected.

We can also compute

\begin{aligned}\hat{\mathbf{R}} - \boldsymbol{\beta} &= \frac{\mathbf{x} + \boldsymbol{\beta} \cdot (\mathbf{x} \wedge \boldsymbol{\beta}) - \mathbf{v} t + \boldsymbol{\beta} \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 }}{ \boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 } } - \boldsymbol{\beta} \\ &=\frac{(\mathbf{x} - \mathbf{v} t)(1 -\boldsymbol{\beta}^2)}{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 } }.\end{aligned}

Our long and messy expression for the field is therefore

\begin{aligned}\mathbf{E} &=e \gamma_{t_r}^3 \frac{1}{{R^2}} (1 - \boldsymbol{\beta}^2)(\hat{\mathbf{R}} - \boldsymbol{\beta}) \\ &=e \left( \frac{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + R^{*}}{R^{*}(1 - \boldsymbol{\beta}^2)}\right)^3\frac{(1 - \boldsymbol{\beta}^2)^2 }{ (\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + R^{*})^2 } (1 -\boldsymbol{\beta}^2)\frac{(\mathbf{x} - \mathbf{v} t)(1 -\boldsymbol{\beta}^2)}{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + R^{*}} \\ \end{aligned}

This gives us our final result

\begin{aligned}\mathbf{E} =e \frac{1}{{(R^{*})^3}}(1 -\boldsymbol{\beta}^2)(\mathbf{x} - \mathbf{v} t)\end{aligned} \hspace{\stretch{1}}(2.65)

As a small test we observe that we get the expected result

\begin{aligned}\mathbf{E} = e \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}^3}\end{aligned} \hspace{\stretch{1}}(2.66)

for the $\boldsymbol{\beta} = 0$ case.

When $\mathbf{v} = V \mathbf{e}_1$ this also recovers equation (38.6) from the text as desired, and if we switch to primed coordinates

\begin{aligned}x' &= \gamma( x - v t) \\ y' &= y \\ z' &= z \\ (1 - \beta^2) {r'}^2 &= (x - v t)^2 + (y^2 + z^2)(1 - \beta^2),\end{aligned} \hspace{\stretch{1}}(2.67)

we recover the field equation derived twice before in previous problem sets

\begin{aligned}\mathbf{E} = \frac{e}{(r')^3} ( x', \gamma y', \gamma z')\end{aligned} \hspace{\stretch{1}}(2.71)

2. Solution. EM fields from a uniformly moving source along x axis.

Initially I had errors in the vector treatment above, so tried with the simpler case using uniform velocity $v$ along the $x$ axis instead. Comparison of the two showed where my errors were in the vector algebra, and that’s now also fixed up.

Performing all the algebra to solve for $t_r$ in

\begin{aligned}{\left\lvert{\mathbf{x} - v t_r \mathbf{e}_1}\right\rvert} = c(t - t_r),\end{aligned} \hspace{\stretch{1}}(2.72)

I get

\begin{aligned}c t_r = \frac{c t - x \beta - \sqrt{ (x- v t)^2 + (y^2 + z^2)(1-\beta^2) } }{ 1 - \beta^2 } = - \gamma (\beta x' + r' )\end{aligned} \hspace{\stretch{1}}(2.73)

This matches the vector expression from 2.57 with the special case of $\mathbf{v} = v \mathbf{e}_1$ so we at least started off on the right foot.

For the retarded distance $R = ct - c t_r$ we get

\begin{aligned}R = \frac{ \beta( x - v t) + \sqrt{ (x- v t)^2 + (y^2 + z^2)(1-\beta^2) } }{ 1 - \beta^2 } = \gamma( \beta x' + r' )\end{aligned} \hspace{\stretch{1}}(2.74)

This also matches 2.58, so things still seem okay with the vector approach. What’s our vector retarded distance

\begin{aligned}\mathbf{R} &= \mathbf{x} - \beta c t_r \mathbf{e}_1 \\ &= (x - \beta c t_r, y, z) \\ &= \left( \frac{ x - v t + \beta \sqrt{ (x- v t)^2 + (y^2 + z^2)(1-\beta^2) } }{ 1 - \beta^2 }, y, z \right) \\ &= \left( \gamma (x' + \beta r'), y', z' \right)\end{aligned}

So

\begin{aligned}\hat{\mathbf{R}} &= \frac{1}{{ \gamma (\beta x' + r') }} \left( \gamma(x' + \beta r'), y', z' \right) \\ &= \frac{1}{{ \beta x' + r' }} \left( x' + \beta r', \frac{y'}{\gamma}, \frac{z'}{\gamma} \right)\end{aligned}

\begin{aligned}\hat{\mathbf{R}} -\boldsymbol{\beta}&= \frac{1}{{ \gamma (\beta x' + r') }} \left( \gamma(x' + \beta r'), y', z' \right) - (\beta, 0, 0) \\ &= \frac{1}{{ \beta x' + r' }} \left( x'(1- \beta^2), \frac{y'}{\gamma}, \frac{z'}{\gamma} \right) \\ &= \frac{1}{{\gamma (\beta x' + r')}} ( x - v t, y, z)\end{aligned}

For ${\partial {t_r}}/{\partial {t}}$, using $\hat{\mathbf{R}}$ calculated above, or from 2.73 calculating directly I get

\begin{aligned}\frac{\partial {t_r}}{\partial {t}} = \frac{r' + \beta x'}{r'(1 - \beta^2)} = \frac{\gamma( r' + \beta x') }{R^{*}},\end{aligned} \hspace{\stretch{1}}(2.75)

where, as in section 38 of the text, we write

\begin{aligned}R^{*} = \sqrt{ (x - v t)^2 + (y^2 + z^2)(1-\beta^2) }.\end{aligned} \hspace{\stretch{1}}(2.76)

Putting all the pieces together I get

\begin{aligned}\mathbf{E} &= e (1 -\beta^2) \frac{(x - v t, y, z)}{{\gamma( \beta x' + r'})} \left( \frac{{\gamma( r' + \beta x')}}{R^{*}} \right)^3 \frac{1}{{{ \gamma^2 (\beta x' + r')^2 } }} \\ \end{aligned}

so we have

\begin{aligned}\mathbf{E} = e \frac{1 -\beta^2}{(R^{*})^3} (x - v t, y, z) \end{aligned} \hspace{\stretch{1}}(2.77)

This matches equation (38.6) in the text.

Problem 3.

FIXME: TODO.

Only the first question was graded (I lost 1.5 marks). I got my units wrong when I integrated 1.11, and my $\omega/c$ factor in the result is circled. Should have also done a dimensional analysis check. There was also a remark that the integral is zero if $t < y/c$, which introduces a $\theta$ function. I think that I incorrectly dropped my $\theta$ function, and should have retained.

FIXME: Revisit both of these and make sure to understand where exactly I went wrong.

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

PHY450H1S (relativistic electrodynamics) Problem Set 3.

Posted by peeterjoot on March 2, 2011

Disclaimer.

This problem set is as yet ungraded (although only the second question will be graded).

Problem 1. Fun with $\epsilon_{\alpha\beta\gamma}$, $\epsilon^{ijkl}$, $F_{ij}$, and the duality of Maxwell’s equations in vacuum.

1. Statement. rank 3 spatial antisymmetric tensor identities.

Prove that

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \nu \gamma}=\delta_{\alpha\mu} \delta_{\beta\nu}-\delta_{\alpha\nu} \delta_{\beta\mu}\end{aligned} \hspace{\stretch{1}}(2.1)

and use it to find the familiar relation for

\begin{aligned}(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D})\end{aligned} \hspace{\stretch{1}}(2.2)

Also show that

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \beta \gamma}=2 \delta_{\alpha\mu}.\end{aligned} \hspace{\stretch{1}}(2.3)

(Einstein summation implied all throughout this problem).

1. Solution

We can explicitly expand the (implied) sum over indexes $\gamma$. This is

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \nu \gamma}=\epsilon_{\alpha \beta 1} \epsilon_{\mu \nu 1}+\epsilon_{\alpha \beta 2} \epsilon_{\mu \nu 2}+\epsilon_{\alpha \beta 3} \epsilon_{\mu \nu 3}\end{aligned} \hspace{\stretch{1}}(2.4)

For any $\alpha \ne \beta$ only one term is non-zero. For example with $\alpha,\beta = 2,3$, we have just a contribution from the $\gamma = 1$ part of the sum

\begin{aligned}\epsilon_{2 3 1} \epsilon_{\mu \nu 1}.\end{aligned} \hspace{\stretch{1}}(2.5)

The value of this for $(\mu,\nu) = (\alpha,\beta)$ is

\begin{aligned}(\epsilon_{2 3 1})^2\end{aligned} \hspace{\stretch{1}}(2.6)

whereas for $(\mu,\nu) = (\beta,\alpha)$ we have

\begin{aligned}-(\epsilon_{2 3 1})^2\end{aligned} \hspace{\stretch{1}}(2.7)

Our sum has value one when $(\alpha, \beta)$ matches $(\mu, \nu)$, and value minus one for when $(\mu, \nu)$ are permuted. We can summarize this, by saying that when $\alpha \ne \beta$ we have

\begin{aligned}\boxed{\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \nu \gamma}=\delta_{\alpha\mu} \delta_{\beta\nu}-\delta_{\alpha\nu} \delta_{\beta\mu}.}\end{aligned} \hspace{\stretch{1}}(2.8)

However, observe that when $\alpha = \beta$ the RHS is

\begin{aligned}\delta_{\alpha\mu} \delta_{\alpha\nu}-\delta_{\alpha\nu} \delta_{\alpha\mu} = 0,\end{aligned} \hspace{\stretch{1}}(2.9)

as desired, so this form works in general without any $\alpha \ne \beta$ qualifier, completing this part of the problem.

\begin{aligned}(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D})&=(\epsilon_{\alpha \beta \gamma} \mathbf{e}^\alpha A^\beta B^\gamma ) \cdot(\epsilon_{\mu \nu \sigma} \mathbf{e}^\mu C^\nu D^\sigma ) \\ &=\epsilon_{\alpha \beta \gamma} A^\beta B^\gamma\epsilon_{\alpha \nu \sigma} C^\nu D^\sigma \\ &=(\delta_{\beta \nu} \delta_{\gamma\sigma}-\delta_{\beta \sigma} \delta_{\gamma\nu} )A^\beta B^\gammaC^\nu D^\sigma \\ &=A^\nu B^\sigmaC^\nu D^\sigma-A^\sigma B^\nuC^\nu D^\sigma.\end{aligned}

This gives us

\begin{aligned}\boxed{(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D})=(\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot \mathbf{D})-(\mathbf{A} \cdot \mathbf{D})(\mathbf{B} \cdot \mathbf{C}).}\end{aligned} \hspace{\stretch{1}}(2.10)

We have one more identity to deal with.

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \beta \gamma}\end{aligned} \hspace{\stretch{1}}(2.11)

We can expand out this (implied) sum slow and dumb as well

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \beta \gamma}&=\epsilon_{\alpha 1 2} \epsilon_{\mu 1 2}+\epsilon_{\alpha 2 1} \epsilon_{\mu 2 1} \\ &+\epsilon_{\alpha 1 3} \epsilon_{\mu 1 3}+\epsilon_{\alpha 3 1} \epsilon_{\mu 3 1} \\ &+\epsilon_{\alpha 2 3} \epsilon_{\mu 2 3}+\epsilon_{\alpha 3 2} \epsilon_{\mu 3 2} \\ &=2 \epsilon_{\alpha 1 2} \epsilon_{\mu 1 2}+ 2 \epsilon_{\alpha 1 3} \epsilon_{\mu 1 3}+ 2 \epsilon_{\alpha 2 3} \epsilon_{\mu 2 3}\end{aligned}

Now, observe that for any $\alpha \in (1,2,3)$ only one term of this sum is picked up. For example, with no loss of generality, pick $\alpha = 1$. We are left with only

\begin{aligned}2 \epsilon_{1 2 3} \epsilon_{\mu 2 3}\end{aligned} \hspace{\stretch{1}}(2.12)

This has the value

\begin{aligned}2 (\epsilon_{1 2 3})^2 = 2\end{aligned} \hspace{\stretch{1}}(2.13)

when $\mu = \alpha$ and is zero otherwise. We can therefore summarize the evaluation of this sum as

\begin{aligned}\boxed{\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \beta \gamma}= 2\delta_{\alpha\mu},}\end{aligned} \hspace{\stretch{1}}(2.14)

completing this problem.

2. Statement. Determinant of three by three matrix.

Prove that for any $3 \times 3$ matrix ${\left\lVert{A_{\alpha\beta}}\right\rVert}$: $\epsilon_{\mu\nu\lambda} A_{\alpha \mu} A_{\beta\nu} A_{\gamma\lambda} = \epsilon_{\alpha \beta \gamma} \text{Det} A$ and that $\epsilon_{\alpha\beta\gamma} \epsilon_{\mu\nu\lambda} A_{\alpha \mu} A_{\beta\nu} A_{\gamma\lambda} = 6 \text{Det} A$.

2. Solution

In class Simon showed us how the first identity can be arrived at using the triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \text{Det}(\mathbf{a} \mathbf{b} \mathbf{c})$. It occurred to me later that I’d seen the identity to be proven in the context of Geometric Algebra, but hadn’t recognized it in this tensor form. Basically, a wedge product can be expanded in sums of determinants, and when the dimension of the space is the same as the vector, we have a pseudoscalar times the determinant of the components.

For example, in $\mathbb{R}^{2}$, let’s take the wedge product of a pair of vectors. As preparation for the relativistic $\mathbb{R}^{4}$ case We won’t require an orthonormal basis, but express the vector in terms of a reciprocal frame and the associated components

\begin{aligned}a = a^i e_i = a_j e^j\end{aligned} \hspace{\stretch{1}}(2.15)

where

\begin{aligned}e^i \cdot e_j = {\delta^i}_j.\end{aligned} \hspace{\stretch{1}}(2.16)

When we get to the relativistic case, we can pick (but don’t have to) the standard basis

\begin{aligned}e_0 &= (1, 0, 0, 0) \\ e_1 &= (0, 1, 0, 0) \\ e_2 &= (0, 0, 1, 0) \\ e_3 &= (0, 0, 0, 1),\end{aligned} \hspace{\stretch{1}}(2.17)

for which our reciprocal frame is implicitly defined by the metric

\begin{aligned}e^0 &= (1, 0, 0, 0) \\ e^1 &= (0, -1, 0, 0) \\ e^2 &= (0, 0, -1, 0) \\ e^3 &= (0, 0, 0, -1).\end{aligned} \hspace{\stretch{1}}(2.21)

Anyways. Back to the problem. Let’s examine the $\mathbb{R}^{2}$ case. Our wedge product in coordinates is

\begin{aligned}a \wedge b=a^i b^j (e_i \wedge e_j)\end{aligned} \hspace{\stretch{1}}(2.25)

Since there are only two basis vectors we have

\begin{aligned}a \wedge b=(a^1 b^2 - a^2 b^1) e_1 \wedge e_2 = \text{Det} {\left\lVert{a^i b^j}\right\rVert} (e_1 \wedge e_2).\end{aligned} \hspace{\stretch{1}}(2.26)

Our wedge product is a product of the determinant of the vector coordinates, times the $\mathbb{R}^{2}$ pseudoscalar $e_1 \wedge e_2$.

This doesn’t look quite like the $\mathbb{R}^{3}$ relation that we want to prove, which had an antisymmetric tensor factor for the determinant. Observe that we get the determinant by picking off the $e_1 \wedge e_2$ component of the bivector result (the only component in this case), and we can do that by dotting with $e^2 \cdot e^1$. To get an antisymmetric tensor times the determinant, we have only to dot with a different pseudoscalar (one that differs by a possible sign due to permutation of the indexes). That is

\begin{aligned}(e^t \wedge e^s) \cdot (a \wedge b)&=a^i b^j (e^t \wedge e^s) \cdot (e_i \wedge e_j) \\ &=a^i b^j\left( {\delta^{s}}_i {\delta^{t}}_j-{\delta^{t}}_i {\delta^{s}}_j \right) \\ &=a^i b^j{\delta^{[t}}_j {\delta^{s]}}_i \\ &=a^i b^j{\delta^{t}}_{[j} {\delta^{s}}_{i]} \\ &=a^{[i} b^{j]}{\delta^{t}}_{j} {\delta^{s}}_{i} \\ &=a^{[s} b^{t]}\end{aligned}

Now, if we write $a^i = A^{1 i}$ and $b^j = A^{2 j}$ we have

\begin{aligned}(e^t \wedge e^s) \cdot (a \wedge b)=A^{1 s} A^{2 t} -A^{1 t} A^{2 s}\end{aligned} \hspace{\stretch{1}}(2.27)

We can write this in two different ways. One of which is

\begin{aligned}A^{1 s} A^{2 t} -A^{1 t} A^{2 s} =\epsilon^{s t} \text{Det} {\left\lVert{A^{ij}}\right\rVert}\end{aligned} \hspace{\stretch{1}}(2.28)

and the other of which is by introducing free indexes for $1$ and $2$, and summing antisymmetrically over these. That is

\begin{aligned}A^{1 s} A^{2 t} -A^{1 t} A^{2 s}=A^{a s} A^{b t} \epsilon_{a b}\end{aligned} \hspace{\stretch{1}}(2.29)

So, we have

\begin{aligned}\boxed{A^{a s} A^{b t} \epsilon_{a b} =A^{1 i} A^{2 j} {\delta^{[t}}_j {\delta^{s]}}_i =\epsilon^{s t} \text{Det} {\left\lVert{A^{ij}}\right\rVert},}\end{aligned} \hspace{\stretch{1}}(2.30)

This result hold regardless of the metric for the space, and does not require that we were using an orthonormal basis. When the metric is Euclidean and we have an orthonormal basis, then all the indexes can be dropped.

The $\mathbb{R}^{3}$ and $\mathbb{R}^{4}$ cases follow in exactly the same way, we just need more vectors in the wedge products.

For the $\mathbb{R}^{3}$ case we have

\begin{aligned}(e^u \wedge e^t \wedge e^s) \cdot ( a \wedge b \wedge c)&=a^i b^j c^k(e^u \wedge e^t \wedge e^s) \cdot (e_i \wedge e_j \wedge e_k) \\ &=a^i b^j c^k{\delta^{[u}}_k{\delta^{t}}_j{\delta^{s]}}_i \\ &=a^{[s} b^t c^{u]}\end{aligned}

Again, with $a^i = A^{1 i}$ and $b^j = A^{2 j}$, and $c^k = A^{3 k}$ we have

\begin{aligned}(e^u \wedge e^t \wedge e^s) \cdot ( a \wedge b \wedge c)=A^{1 i} A^{2 j} A^{3 k}{\delta^{[u}}_k{\delta^{t}}_j{\delta^{s]}}_i\end{aligned} \hspace{\stretch{1}}(2.31)

and we can choose to write this in either form, resulting in the identity

\begin{aligned}\boxed{\epsilon^{s t u} \text{Det} {\left\lVert{A^{ij}}\right\rVert}=A^{1 i} A^{2 j} A^{3 k}{\delta^{[u}}_k{\delta^{t}}_j{\delta^{s]}}_i=\epsilon_{a b c} A^{a s} A^{b t} A^{c u}.}\end{aligned} \hspace{\stretch{1}}(2.32)

The $\mathbb{R}^{4}$ case follows exactly the same way, and we have

\begin{aligned}(e^v \wedge e^u \wedge e^t \wedge e^s) \cdot ( a \wedge b \wedge c \wedge d)&=a^i b^j c^k d^l(e^v \wedge e^u \wedge e^t \wedge e^s) \cdot (e_i \wedge e_j \wedge e_k \wedge e_l) \\ &=a^i b^j c^k d^l{\delta^{[v}}_l{\delta^{u}}_k{\delta^{t}}_j{\delta^{s]}}_i \\ &=a^{[s} b^t c^{u} d^{v]}.\end{aligned}

This time with $a^i = A^{0 i}$ and $b^j = A^{1 j}$, and $c^k = A^{2 k}$, and $d^l = A^{3 l}$ we have

\begin{aligned}\boxed{\epsilon^{s t u v} \text{Det} {\left\lVert{A^{ij}}\right\rVert}=A^{0 i} A^{1 j} A^{2 k} A^{3 l}{\delta^{[v}}_l{\delta^{u}}_k{\delta^{t}}_j{\delta^{s]}}_i=\epsilon_{a b c d} A^{a s} A^{b t} A^{c u} A^{d v}.}\end{aligned} \hspace{\stretch{1}}(2.33)

This one is almost the identity to be established later in problem 1.4. We have only to raise and lower some indexes to get that one. Note that in the Minkowski standard basis above, because $s, t, u, v$ must be a permutation of $0,1,2,3$ for a non-zero result, we must have

\begin{aligned}\epsilon^{s t u v} = (-1)^3 (+1) \epsilon_{s t u v}.\end{aligned} \hspace{\stretch{1}}(2.34)

So raising and lowering the identity above gives us

\begin{aligned}-\epsilon_{s t u v} \text{Det} {\left\lVert{A_{ij}}\right\rVert}=\epsilon^{a b c d} A_{a s} A_{b t} A_{c u} A_{d u}.\end{aligned} \hspace{\stretch{1}}(2.35)

No sign changes were required for the indexes $a, b, c, d$, since they are paired.

Until we did the raising and lowering operations here, there was no specific metric required, so our first result 2.33 is the more general one.

There’s one more part to this problem, doing the antisymmetric sums over the indexes $s, t, \cdots$. For the $\mathbb{R}^{2}$ case we have

\begin{aligned}\epsilon_{s t} \epsilon_{a b} A^{a s} A^{b t}&=\epsilon_{s t} \epsilon^{s t} \text{Det} {\left\lVert{A^{ij}}\right\rVert} \\ &=\left( \epsilon_{1 2} \epsilon^{1 2} +\epsilon_{2 1} \epsilon^{2 1} \right)\text{Det} {\left\lVert{A^{ij}}\right\rVert} \\ &=\left( 1^2 + (-1)^2\right)\text{Det} {\left\lVert{A^{ij}}\right\rVert}\end{aligned}

We conclude that

\begin{aligned}\boxed{\epsilon_{s t} \epsilon_{a b} A^{a s} A^{b t} = 2! \text{Det} {\left\lVert{A^{ij}}\right\rVert}.}\end{aligned} \hspace{\stretch{1}}(2.36)

For the $\mathbb{R}^{3}$ case we have the same operation

\begin{aligned}\epsilon_{s t u} \epsilon_{a b c} A^{a s} A^{b t} A^{c u}&=\epsilon_{s t u} \epsilon^{s t u} \text{Det} {\left\lVert{A^{ij}}\right\rVert} \\ &=\left( \epsilon_{1 2 3} \epsilon^{1 2 3} +\epsilon_{1 3 2} \epsilon^{1 3 2} + \cdots\right)\text{Det} {\left\lVert{A^{ij}}\right\rVert} \\ &=(\pm 1)^2 (3!)\text{Det} {\left\lVert{A^{ij}}\right\rVert}.\end{aligned}

So we conclude

\begin{aligned}\boxed{\epsilon_{s t u} \epsilon_{a b c} A^{a s} A^{b t} A^{c u}= 3! \text{Det} {\left\lVert{A^{ij}}\right\rVert}.}\end{aligned} \hspace{\stretch{1}}(2.37)

It’s clear what the pattern is, and if we evaluate the sum of the antisymmetric tensor squares in $\mathbb{R}^{4}$ we have

\begin{aligned}\epsilon_{s t u v} \epsilon_{s t u v}&=\epsilon_{0 1 2 3} \epsilon_{0 1 2 3}+\epsilon_{0 1 3 2} \epsilon_{0 1 3 2}+\epsilon_{0 2 1 3} \epsilon_{0 2 1 3}+ \cdots \\ &= (\pm 1)^2 (4!),\end{aligned}

So, for our SR case we have

\begin{aligned}\boxed{\epsilon_{s t u v} \epsilon_{a b c d} A^{a s} A^{b t} A^{c u} A^{d v}= 4! \text{Det} {\left\lVert{A^{ij}}\right\rVert}.}\end{aligned} \hspace{\stretch{1}}(2.38)

This was part of question 1.4, albeit in lower index form. Here since all indexes are matched, we have the same result without major change

\begin{aligned}\boxed{\epsilon^{s t u v} \epsilon^{a b c d} A_{a s} A_{b t} A_{c u} A_{d v}= 4! \text{Det} {\left\lVert{A_{ij}}\right\rVert}.}\end{aligned} \hspace{\stretch{1}}(2.39)

The main difference is that we are now taking the determinant of a lower index tensor.

3. Statement. Rotational invariance of 3D antisymmetric tensor

Use the previous results to show that $\epsilon_{\mu\nu\lambda}$ is invariant under rotations.

3. Solution

We apply transformations to coordinates (and thus indexes) of the form

\begin{aligned}x_\mu \rightarrow O_{\mu\nu} x_\nu\end{aligned} \hspace{\stretch{1}}(2.40)

With our tensor transforming as its indexes, we have

\begin{aligned}\epsilon_{\mu\nu\lambda} \rightarrow \epsilon_{\alpha\beta\sigma} O_{\mu\alpha} O_{\nu\beta} O_{\lambda\sigma}.\end{aligned} \hspace{\stretch{1}}(2.41)

We’ve got 2.32, which after dropping indexes, because we are in a Euclidean space, we have

\begin{aligned}\epsilon_{\mu \nu \lambda} \text{Det} {\left\lVert{A_{ij}}\right\rVert} = \epsilon_{\alpha \beta \sigma} A_{\alpha \mu} A_{\beta \nu} A_{\sigma \lambda}.\end{aligned} \hspace{\stretch{1}}(2.42)

Let $A_{i j} = O_{j i}$, which gives us

\begin{aligned}\epsilon_{\mu\nu\lambda} \rightarrow \epsilon_{\mu\nu\lambda} \text{Det} A^\text{T}\end{aligned} \hspace{\stretch{1}}(2.43)

but since $\text{Det} O = \text{Det} O^\text{T}$, we have shown that $\epsilon_{\mu\nu\lambda}$ is invariant under rotation.

4. Statement. Rotational invariance of 4D antisymmetric tensor

Use the previous results to show that $\epsilon_{i j k l}$ is invariant under Lorentz transformations.

4. Solution

This follows the same way. We assume a transformation of coordinates of the following form

\begin{aligned}(x')^i &= {O^i}_j x^j \\ (x')_i &= {O_i}^j x_j,\end{aligned} \hspace{\stretch{1}}(2.44)

where the determinant of ${O^i}_j = 1$ (sanity check of sign: ${O^i}_j = {\delta^i}_j$).

Our antisymmetric tensor transforms as its coordinates individually

\begin{aligned}\epsilon_{i j k l} &\rightarrow \epsilon_{a b c d} {O_i}^a{O_j}^b{O_k}^c{O_l}^d \\ &= \epsilon^{a b c d} O_{i a}O_{j b}O_{k c}O_{l d} \\ \end{aligned}

Let $P_{ij} = O_{ji}$, and raise and lower all the indexes in 2.46 for

\begin{aligned}-\epsilon_{s t u v} \text{Det} {\left\lVert{P_{ij}}\right\rVert}=\epsilon^{a b c d} P_{a s} P_{b t} P_{c u} P_{d v}.\end{aligned} \hspace{\stretch{1}}(2.46)

We have

\begin{aligned}\epsilon_{i j k l} &= \epsilon^{a b c d} P_{a i}P_{a j}P_{a k}P_{a l} \\ &=-\epsilon_{i j k l} \text{Det} {\left\lVert{P_{ij}}\right\rVert} \\ &=-\epsilon_{i j k l} \text{Det} {\left\lVert{O_{ij}}\right\rVert} \\ &=-\epsilon_{i j k l} \text{Det} {\left\lVert{g_{im} {O^m}_j }\right\rVert} \\ &=-\epsilon_{i j k l} (-1)(1) \\ &=\epsilon_{i j k l}\end{aligned}

Since $\epsilon_{i j k l} = -\epsilon^{i j k l}$ both are therefore invariant under Lorentz transformation.

5. Statement. Sum of contracting symmetric and antisymmetric rank 2 tensors

Show that $A^{ij} B_{ij} = 0$ if $A$ is symmetric and $B$ is antisymmetric.

5. Solution

We swap indexes in $B$, switch dummy indexes, then swap indexes in $A$

\begin{aligned}A^{i j} B_{i j} &= -A^{i j} B_{j i} \\ &= -A^{j i} B_{i j} \\ &= -A^{i j} B_{i j} \\ \end{aligned}

Our result is the negative of itself, so must be zero.

6. Statement. Characteristic equation for the electromagnetic strength tensor

Show that $P(\lambda) = \text{Det} {\left\lVert{F_{i j} - \lambda g_{i j}}\right\rVert}$ is invariant under Lorentz transformations. Consider the polynomial of $P(\lambda)$, also called the characteristic polynomial of the matrix ${\left\lVert{F_{i j}}\right\rVert}$. Find the coefficients of the expansion of $P(\lambda)$ in powers of $\lambda$ in terms of the components of ${\left\lVert{F_{i j}}\right\rVert}$. Use the result to argue that $\mathbf{E} \cdot \mathbf{B}$ and $\mathbf{E}^2 - \mathbf{B}^2$ are Lorentz invariant.

6. Solution

The invariance of the determinant

Let’s consider how any lower index rank 2 tensor transforms. Given a transformation of coordinates

\begin{aligned}(x^i)' &= {O^i}_j x^j \\ (x_i)' &= {O_i}^j x^j ,\end{aligned} \hspace{\stretch{1}}(2.47)

where $\text{Det} {\left\lVert{ {O^i}_j }\right\rVert} = 1$, and ${O_i}^j = {O^m}_n g_{i m} g^{j n}$. Let’s reflect briefly on why this determinant is unit valued. We have

\begin{aligned}(x^i)' (x_i)'= {O_i}^a x^a {O^i}_b x^b = x^b x_b,\end{aligned} \hspace{\stretch{1}}(2.49)

which implies that the transformation product is

\begin{aligned}{O_i}^a {O^i}_b = {\delta^a}_b,\end{aligned} \hspace{\stretch{1}}(2.50)

the identity matrix. The identity matrix has unit determinant, so we must have

\begin{aligned}1 = (\text{Det} \hat{G})^2 (\text{Det} {\left\lVert{ {O^i}_j }\right\rVert})^2.\end{aligned} \hspace{\stretch{1}}(2.51)

Since $\text{Det} \hat{G} = -1$ we have

\begin{aligned}\text{Det} {\left\lVert{ {O^i}_j }\right\rVert} = \pm 1,\end{aligned} \hspace{\stretch{1}}(2.52)

which is all that we can say about the determinant of this class of transformations by considering just invariance. If we restrict the transformations of coordinates to those of the same determinant sign as the identity matrix, we rule out reflections in time or space. This seems to be the essence of the $SO(1,3)$ labeling.

Why dwell on this? Well, I wanted to be clear on the conventions I’d chosen, since parts of the course notes used $\hat{O} = {\left\lVert{O^{i j}}\right\rVert}$, and $X' = \hat{O} X$, and gave that matrix unit determinant. That $O^{i j}$ looks like it is equivalent to my ${O^i}_j$, except that the one in the course notes is loose when it comes to lower and upper indexes since it gives $(x')^i = O^{i j} x^j$.

I’ll write

\begin{aligned}\hat{O} = {\left\lVert{{O^i}_j}\right\rVert},\end{aligned} \hspace{\stretch{1}}(2.53)

and require this (not ${\left\lVert{O^{i j}}\right\rVert}$) to be the matrix with unit determinant. Having cleared the index upper and lower confusion I had trying to reconcile the class notes with the rules for index manipulation, let’s now consider the Lorentz transformation of a lower index rank 2 tensor (not necessarily antisymmetric or symmetric)

We have, transforming in the same fashion as a lower index coordinate four vector (but twice, once for each index)

\begin{aligned}A_{i j} \rightarrow A_{k m} {O_i}^k{O_j}^m.\end{aligned} \hspace{\stretch{1}}(2.54)

The determinant of the transformation tensor ${O_i}^j$ is

\begin{aligned}\text{Det} {\left\lVert{ {O_i}^j }\right\rVert} = \text{Det} {\left\lVert{ g^{i m} {O^m}_n g^{n j} }\right\rVert} = (\text{Det} \hat{G}) (1) (\text{Det} \hat{G} ) = (-1)^2 (1) = 1.\end{aligned} \hspace{\stretch{1}}(2.55)

We see that the determinant of a lower index rank 2 tensor is invariant under Lorentz transformation. This would include our characteristic polynomial $P(\lambda)$.

Expanding the determinant.

Utilizing 2.39 we can now calculate the characteristic polynomial. This is

\begin{aligned}\text{Det} {\left\lVert{F_{ij} - \lambda g_{ij} }\right\rVert}&= \frac{1}{{4!}}\epsilon^{s t u v} \epsilon^{a b c d} (F_{ a s } - \lambda g_{a s}) (F_{ b t } - \lambda g_{b t}) (F_{ c u } - \lambda g_{c u}) (F_{ d v } - \lambda g_{d v}) \\ &=\frac{1}{{24}}\epsilon^{s t u v} \epsilon_{a b c d} ({F^a}_s - \lambda {g^a}_s) ({F^b}_t - \lambda {g^b}_t) ({F^c}_u - \lambda {g^c}_u) ({F^d}_v - \lambda {g^d}_v) \\ \end{aligned}

However, ${g^a}_b = g_{b c} g^{a c}$, or ${\left\lVert{{g^a}_b}\right\rVert} = \hat{G}^2 = I$. This means we have

\begin{aligned}{g^a}_b = {\delta^a}_b,\end{aligned} \hspace{\stretch{1}}(2.56)

and our determinant is reduced to

\begin{aligned}\begin{aligned}P(\lambda) &=\frac{1}{{24}}\epsilon^{s t u v} \epsilon_{a b c d} \Bigl({F^a}_s {F^b}_t - \lambda( {\delta^a}_s {F^b}_t + {\delta^b}_t {F^a}_s ) + \lambda^2 {\delta^a}_s {\delta^b}_t \Bigr) \\ &\times \qquad \qquad \Bigl({F^c}_u {F^d}_v - \lambda( {\delta^c}_u {F^d}_v + {\delta^d}_v {F^c}_u ) + \lambda^2 {\delta^c}_u {\delta^d}_v \Bigr) \end{aligned}\end{aligned} \hspace{\stretch{1}}(2.57)

If we expand this out we have our powers of $\lambda$ coefficients are

\begin{aligned}\lambda^0 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} {F^a}_s {F^b}_t {F^c}_u {F^d}_v \\ \lambda^1 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} \Bigl(- ({\delta^c}_u {F^d}_v + {\delta^d}_v {F^c}_u ) {F^a}_s {F^b}_t - ({\delta^a}_s {F^b}_t + {\delta^b}_t {F^a}_s ) {F^c}_u {F^d}_v \Bigr) \\ \lambda^2 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} \Bigl({\delta^c}_u {\delta^d}_v {F^a}_s {F^b}_t +( {\delta^a}_s {F^b}_t + {\delta^b}_t {F^a}_s ) ( {\delta^c}_u {F^d}_v + {\delta^d}_v {F^c}_u ) + {\delta^a}_s {\delta^b}_t {F^c}_u {F^d}_v \Bigr) \\ \lambda^3 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} \Bigl(- ( {\delta^a}_s {F^b}_t + {\delta^b}_t {F^a}_s ) {\delta^c}_u {\delta^d}_v - {\delta^a}_s {\delta^b}_t ( {\delta^c}_u {F^d}_v + {\delta^d}_v {F^c}_u ) \Bigr) \\ \lambda^4 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} \Bigl({\delta^a}_s {\delta^b}_t {\delta^c}_u {\delta^d}_v \Bigr) \\ \end{aligned}

By 2.39 the $\lambda^0$ coefficient is just $\text{Det} {\left\lVert{F_{i j}}\right\rVert}$.

The $\lambda^3$ terms can be seen to be zero. For example, the first one is

\begin{aligned}-\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} {\delta^a}_s {F^b}_t {\delta^c}_u {\delta^d}_v &=-\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{s b u v} {F^b}_t \\ &=-\frac{1}{{12}} \delta^{t}_b {F^b}_t \\ &=-\frac{1}{{12}} {F^b}_b \\ &=-\frac{1}{{12}} F^{bu} g_{ub} \\ &= 0,\end{aligned}

where the final equality to zero comes from summing a symmetric and antisymmetric product.

Similarly the $\lambda$ coefficients can be shown to be zero. Again the first as a sample is

\begin{aligned}-\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} {\delta^c}_u {F^d}_v {F^a}_s {F^b}_t &=-\frac{1}{{24}} \epsilon^{u s t v} \epsilon_{u a b d} {F^d}_v {F^a}_s {F^b}_t \\ &=-\frac{1}{{24}} \delta^{[s}_a\delta^{t}_b\delta^{v]}_d{F^d}_v {F^a}_s {F^b}_t \\ &=-\frac{1}{{24}} {F^a}_{[s}{F^b}_{t}{F^d}_{v]} \\ \end{aligned}

Disregarding the $-1/24$ factor, let’s just expand this antisymmetric sum

\begin{aligned}{F^a}_{[a}{F^b}_{b}{F^d}_{d]}&={F^a}_{a}{F^b}_{b}{F^d}_{d}+{F^a}_{d}{F^b}_{a}{F^d}_{b}+{F^a}_{b}{F^b}_{d}{F^d}_{a}-{F^a}_{a}{F^b}_{d}{F^d}_{b}-{F^a}_{d}{F^b}_{b}{F^d}_{a}-{F^a}_{b}{F^b}_{a}{F^d}_{d} \\ &={F^a}_{d}{F^b}_{a}{F^d}_{b}+{F^a}_{b}{F^b}_{d}{F^d}_{a} \\ \end{aligned}

Of the two terms above that were retained, they are the only ones without a zero ${F^i}_i$ factor. Consider the first part of this remaining part of the sum. Employing the metric tensor, to raise indexes so that the antisymmetry of $F^{ij}$ can be utilized, and then finally relabeling all the dummy indexes we have

\begin{aligned}{F^a}_{d}{F^b}_{a}{F^d}_{b}&=F^{a u}F^{b v}F^{d w}g_{d u}g_{a v}g_{b w} \\ &=(-1)^3F^{u a}F^{v b}F^{w d}g_{d u}g_{a v}g_{b w} \\ &=-(F^{u a}g_{a v})(F^{v b}g_{b w} )(F^{w d}g_{d u})\\ &=-{F^u}_v{F^v}_w{F^w}_u\\ &=-{F^a}_b{F^b}_d{F^d}_a\\ \end{aligned}

This is just the negative of the second term in the sum, leaving us with zero.

Finally, we have for the $\lambda^2$ coefficient ($\times 24$)

\begin{aligned}&\epsilon^{s t u v} \epsilon_{a b c d} \Bigl({\delta^c}_u {\delta^d}_v {F^a}_s {F^b}_t +{\delta^a}_s {F^b}_t {\delta^c}_u {F^d}_v +{\delta^b}_t {F^a}_s {\delta^d}_v {F^c}_u \\ &\qquad +{\delta^b}_t {F^a}_s {\delta^c}_u {F^d}_v +{\delta^a}_s {F^b}_t {\delta^d}_v {F^c}_u + {\delta^a}_s {\delta^b}_t {F^c}_u {F^d}_v \Bigr) \\ &=\epsilon^{s t u v} \epsilon_{a b u v} {F^a}_s {F^b}_t +\epsilon^{s t u v} \epsilon_{s b u d} {F^b}_t {F^d}_v +\epsilon^{s t u v} \epsilon_{a t c v} {F^a}_s {F^c}_u \\ &\qquad +\epsilon^{s t u v} \epsilon_{a t u d} {F^a}_s {F^d}_v +\epsilon^{s t u v} \epsilon_{s b c v} {F^b}_t {F^c}_u + \epsilon^{s t u v} \epsilon_{s t c d} {F^c}_u {F^d}_v \\ &=\epsilon^{s t u v} \epsilon_{a b u v} {F^a}_s {F^b}_t +\epsilon^{t v s u } \epsilon_{b d s u} {F^b}_t {F^d}_v +\epsilon^{s u t v} \epsilon_{a c t v} {F^a}_s {F^c}_u \\ &\qquad +\epsilon^{s v t u} \epsilon_{a d t u} {F^a}_s {F^d}_v +\epsilon^{t u s v} \epsilon_{b c s v} {F^b}_t {F^c}_u + \epsilon^{u v s t} \epsilon_{c d s t} {F^c}_u {F^d}_v \\ &=6\epsilon^{s t u v} \epsilon_{a b u v} {F^a}_s {F^b}_t \\ &=6 (2){\delta^{[s}}_a{\delta^{t]}}_b{F^a}_s {F^b}_t \\ &=12{F^a}_{[a} {F^b}_{b]} \\ &=12( {F^a}_{a} {F^b}_{b} - {F^a}_{b} {F^b}_{a} ) \\ &=-12 {F^a}_{b} {F^b}_{a} \\ &=-12 F^{a b} F_{b a} \\ &=12 F^{a b} F_{a b}\end{aligned}

Therefore, our characteristic polynomial is

\begin{aligned}\boxed{P(\lambda) = \text{Det} {\left\lVert{F_{i j}}\right\rVert} + \frac{\lambda^2}{2} F^{a b} F_{a b} + \lambda^4.}\end{aligned} \hspace{\stretch{1}}(2.58)

Observe that in matrix form our strength tensors are

\begin{aligned}{\left\lVert{ F^{ij} }\right\rVert} &= \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix} \\ {\left\lVert{ F_{ij} }\right\rVert} &= \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.59)

From these we can compute $F^{a b} F_{a b}$ easily by inspection

\begin{aligned}F^{a b} F_{a b} = 2 (\mathbf{B}^2 - \mathbf{E}^2).\end{aligned} \hspace{\stretch{1}}(2.61)

Computing the determinant is not so easy. The dumb and simple way of expanding by cofactors takes two pages, and yields eventually

\begin{aligned}\text{Det} {\left\lVert{ F^{i j} }\right\rVert} = (\mathbf{E} \cdot \mathbf{B})^2.\end{aligned} \hspace{\stretch{1}}(2.62)

That supplies us with a relation for the characteristic polynomial in $\mathbf{E}$ and $\mathbf{B}$

\begin{aligned}\boxed{P(\lambda) = (\mathbf{E} \cdot \mathbf{B})^2 + \lambda^2 (\mathbf{B}^2 - \mathbf{E}^2) + \lambda^4.}\end{aligned} \hspace{\stretch{1}}(2.63)

Observe that we found this for the special case where $\mathbf{E}$ and $\mathbf{B}$ were perpendicular in homework 2. Observe that when we have that perpendicularity, we can solve for the eigenvalues by inspection

\begin{aligned}\lambda \in \{ 0, 0, \pm \sqrt{ \mathbf{E}^2 - \mathbf{B}^2 } \},\end{aligned} \hspace{\stretch{1}}(2.64)

and were able to diagonalize the matrix ${F^{i}}_j$ to solve the Lorentz force equation in parametric form. When ${\left\lvert{\mathbf{E}}\right\rvert} > {\left\lvert{\mathbf{B}}\right\rvert}$ we had real eigenvalues and an orthogonal diagonalization when $\mathbf{B} = 0$. For the ${\left\lvert{\mathbf{B}}\right\rvert} > {\left\lvert{\mathbf{E}}\right\rvert}$, we had a two purely imaginary eigenvalues, and when $\mathbf{E} = 0$ this was a Hermitian diagonalization. For the general case, when one of $\mathbf{E}$, or $\mathbf{B}$ was zero, things didn’t have the same nice closed form solution.

In general our eigenvalues are

\begin{aligned}\lambda = \pm \frac{1}{{\sqrt{2}}} \sqrt{ \mathbf{E}^2 - \mathbf{B}^2 \pm \sqrt{ (\mathbf{E}^2 - \mathbf{B}^2)^2 - 4 (\mathbf{E} \cdot \mathbf{B})^2 }}.\end{aligned} \hspace{\stretch{1}}(2.65)

For the purposes of this problem we really only wish to show that $\mathbf{E} \cdot \mathbf{B}$ and $\mathbf{E}^2 - \mathbf{B}^2$ are Lorentz invariants. When $\lambda = 0$ we have $P(\lambda) = (\mathbf{E} \cdot \mathbf{B})^2$, a Lorentz invariant. This must mean that $\mathbf{E} \cdot \mathbf{B}$ is itself a Lorentz invariant. Since that is invariant, and we require $P(\lambda)$ to be invariant for any other possible values of $\lambda$, the difference $\mathbf{E}^2 - \mathbf{B}^2$ must also be Lorentz invariant.

7. Statement. Show that the pseudoscalar invariant has only boundary effects.

Use integration by parts to show that $\int d^4 x \epsilon^{i j k l} F_{ i j } F_{ k l }$ only depends on the values of $A^i(x)$ at the “boundary” of spacetime (e.g. the “surface” depicted on page 105 of the notes) and hence does not affect the equations of motion for the electromagnetic field.

7. Solution

This proceeds in a fairly straightforward fashion

\begin{aligned}\int d^4 x \epsilon^{i j k l} F_{ i j } F_{ k l }&=\int d^4 x \epsilon^{i j k l} (\partial_i A_j - \partial_j A_i) F_{ k l } \\ &=\int d^4 x \epsilon^{i j k l} (\partial_i A_j) F_{ k l } -\epsilon^{j i k l} (\partial_i A_j) F_{ k l } \\ &=2 \int d^4 x \epsilon^{i j k l} (\partial_i A_j) F_{ k l } \\ &=2 \int d^4 x \epsilon^{i j k l} \left( \frac{\partial {}}{\partial {x^i}}(A_j F_{ k l }-A_j \frac{\partial { F_{ k l } }}{\partial {x^i}}\right)\\ \end{aligned}

Now, observe that by the Bianchi identity, this second term is zero

\begin{aligned}\epsilon^{i j k l} \frac{\partial { F_{ k l } }}{\partial {x^i}}=-\epsilon^{j i k l} \partial_i F_{ k l } = 0\end{aligned} \hspace{\stretch{1}}(2.66)

Now we have a set of perfect differentials, and can integrate

\begin{aligned}\int d^4 x \epsilon^{i j k l} F_{ i j } F_{ k l }&= 2 \int d^4 x \epsilon^{i j k l} \frac{\partial {}}{\partial {x^i}}(A_j F_{ k l })\\ &= 2 \int dx^j dx^k dx^l\epsilon^{i j k l} {\left.{{(A_j F_{ k l })}}\right\vert}_{{\Delta x^i}}\\ \end{aligned}

We are left with a only contributions to the integral from the boundary terms on the spacetime hypervolume, three-volume normals bounding the four-volume integration in the original integral.

8. Statement. Electromagnetic duality transformations.

Show that the Maxwell equations in vacuum are invariant under the transformation: $F_{i j} \rightarrow \tilde{F}_{i j}$, where $\tilde{F}_{i j} = \frac{1}{{2}} \epsilon_{i j k l} F^{k l}$ is the dual electromagnetic stress tensor. Replacing $F$ with $\tilde{F}$ is known as “electric-magnetic duality”. Explain this name by considering the transformation in terms of $\mathbf{E}$ and $\mathbf{B}$. Are the Maxwell equations with sources invariant under electric-magnetic duality transformations?

8. Solution

Let’s first consider the explanation of the name. First recall what the expansions are of $F_{i j}$ and $F^{i j}$ in terms of $\mathbf{E}$ and $\mathbf{E}$. These are

\begin{aligned}F_{0 \alpha} &= \partial_0 A_\alpha - \partial_\alpha A_0 \\ &= -\frac{1}{{c}} \frac{\partial {A^\alpha}}{\partial {t}} - \frac{\partial {\phi}}{\partial {x^\alpha}} \\ &= E_\alpha\end{aligned}

with $F^{0 \alpha} = -E^\alpha$, and $E^\alpha = E_\alpha$.

The magnetic field components are

\begin{aligned}F_{\beta \alpha} &= \partial_\beta A_\alpha - \partial_\alpha A_\beta \\ &= -\partial_\beta A^\alpha + \partial_\alpha A^\beta \\ &= \epsilon_{\alpha \beta \sigma} B^\sigma\end{aligned}

with $F^{\beta \alpha} = \epsilon^{\alpha \beta \sigma} B_\sigma$ and $B_\sigma = B^\sigma$.

Now let’s expand the dual tensors. These are

\begin{aligned}\tilde{F}_{0 \alpha} &=\frac{1}{{2}} \epsilon_{0 \alpha i j} F^{i j} \\ &=\frac{1}{{2}} \epsilon_{0 \alpha \beta \sigma} F^{\beta \sigma} \\ &=\frac{1}{{2}} \epsilon_{0 \alpha \beta \sigma} \epsilon^{\sigma \beta \mu} B_\mu \\ &=-\frac{1}{{2}} \epsilon_{0 \alpha \beta \sigma} \epsilon^{\mu \beta \sigma} B_\mu \\ &=-\frac{1}{{2}} (2!) {\delta_\alpha}^\mu B_\mu \\ &=- B_\alpha \\ \end{aligned}

and

\begin{aligned}\tilde{F}_{\beta \alpha} &=\frac{1}{{2}} \epsilon_{\beta \alpha i j} F^{i j} \\ &=\frac{1}{{2}} \left(\epsilon_{\beta \alpha 0 \sigma} F^{0 \sigma} +\epsilon_{\beta \alpha \sigma 0} F^{\sigma 0} \right) \\ &=\epsilon_{0 \beta \alpha \sigma} (-E^\sigma) \\ &=\epsilon_{\alpha \beta \sigma} E^\sigma\end{aligned}

Summarizing we have

\begin{aligned}F_{0 \alpha} &= E^\alpha \\ F^{0 \alpha} &= -E^\alpha \\ F^{\beta \alpha} &= F_{\beta \alpha} = \epsilon_{\alpha \beta \sigma} B^\sigma \\ \tilde{F}_{0 \alpha} &= - B_\alpha \\ \tilde{F}^{0 \alpha} &= B_\alpha \\ \tilde{F}_{\beta \alpha} &= \tilde{F}^{\beta \alpha} = \epsilon_{\alpha \beta \sigma} E^\sigma\end{aligned} \hspace{\stretch{1}}(2.67)

Is there a sign error in the $\tilde{F}_{0 \alpha} = - B_\alpha$ result? Other than that we have the same sort of structure for the tensor with $E$ and $B$ switched around.

Let’s write these in matrix form, to compare

\begin{aligned}\begin{array}{l l l l}{\left\lVert{ \tilde{F}_{i j} }\right\rVert} &= \begin{bmatrix}0 & -B_x & -B_y & -B_z \\ B_x & 0 & -E_z & E_y \\ B_y & E_z & 0 & E_x \\ B_z & -E_y & -E_x & 0 \\ \end{bmatrix} ^{i j} }\right\rVert} &= \begin{bmatrix}0 & B_x & B_y & B_z \\ -B_x & 0 & -E_z & E_y \\ -B_y & E_z & 0 & -E_x \\ -B_z & -E_y & E_x & 0 \\ \end{bmatrix} \\ {\left\lVert{ F^{ij} }\right\rVert} &= \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix} }\right\rVert} &= \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0\end{bmatrix}.\end{array}\end{aligned} \hspace{\stretch{1}}(2.73)

From these we can see by inspection that we have

\begin{aligned}\tilde{F}^{i j} F_{ij} = \tilde{F}_{i j} F^{ij} = 4 (\mathbf{E} \cdot \mathbf{B})\end{aligned} \hspace{\stretch{1}}(2.74)

This is consistent with the stated result in [1] (except for a factor of $c$ due to units differences), so it appears the signs above are all kosher.

Now, let’s see if the if the dual tensor satisfies the vacuum equations.

\begin{aligned}\partial_j \tilde{F}^{i j}&=\partial_j \frac{1}{{2}} \epsilon^{i j k l} F_{k l} \\ &=\frac{1}{{2}} \epsilon^{i j k l} \partial_j (\partial_k A_l - \partial_l A_k) \\ &=\frac{1}{{2}} \epsilon^{i j k l} \partial_j \partial_k A_l - \frac{1}{{2}} \epsilon^{i j l k} \partial_k A_l \\ &=\frac{1}{{2}} (\epsilon^{i j k l} - \epsilon^{i j k l} \partial_k A_l \\ &= 0 \qquad\square\end{aligned}

So the first checks out, provided we have no sources. If we have sources, then we see here that Maxwell’s equations do not hold since this would imply that the four current density must be zero.

How about the Bianchi identity? That gives us

\begin{aligned}\epsilon^{i j k l} \partial_j \tilde{F}_{k l} &=\epsilon^{i j k l} \partial_j \frac{1}{{2}} \epsilon_{k l a b} F^{a b} \\ &=\frac{1}{{2}} \epsilon^{k l i j} \epsilon_{k l a b} \partial_j F^{a b} \\ &=\frac{1}{{2}} (2!) {\delta^i}_{[a} {\delta^j}_{b]} \partial_j F^{a b} \\ &=\partial_j (F^{i j} - F^{j i} ) \\ &=2 \partial_j F^{i j} .\end{aligned}

The factor of two is slightly curious. Is there a mistake above? If there is a mistake, it doesn’t change the fact that Maxwell’s equation

\begin{aligned}\partial_k F^{k i} = \frac{4 \pi}{c} j^i\end{aligned} \hspace{\stretch{1}}(2.75)

Gives us zero for the Bianchi identity under source free conditions of $j^i = 0$.

Problem 2. Transformation properties of $\mathbf{E}$ and $\mathbf{B}$, again.

1. Statement

Use the form of $F^{i j}$ from page 82 in the class notes, the transformation law for ${\left\lVert{ F^{i j} }\right\rVert}$ given further down that same page, and the explicit form of the $SO(1,3)$ matrix $\hat{O}$ (say, corresponding to motion in the positive $x_1$ direction with speed $v$) to derive the transformation law of the fields $\mathbf{E}$ and $\mathbf{B}$. Use the transformation law to find the electromagnetic field of a charged particle moving with constant speed $v$ in the positive $x_1$ direction and check that the result agrees with the one that you obtained in Homework 2.

1. Solution

Given a transformation of coordinates

\begin{aligned}{x'}^i \rightarrow {O^i}_j x^j\end{aligned} \hspace{\stretch{1}}(3.76)

our rank 2 tensor $F^{i j}$ transforms as

\begin{aligned}F^{i j} \rightarrow {O^i}_aF^{a b}{O^j}_b.\end{aligned} \hspace{\stretch{1}}(3.77)

Introducing matrices

\begin{aligned}\hat{O} &= {\left\lVert{{O^i}_j}\right\rVert} \\ \hat{F} &= {\left\lVert{F^{ij}}\right\rVert} = \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix} \end{aligned} \hspace{\stretch{1}}(3.78)

and noting that $\hat{O}^\text{T} = {\left\lVert{{O^j}_i}\right\rVert}$, we can express the electromagnetic strength tensor transformation as

\begin{aligned}\hat{F} \rightarrow \hat{O} \hat{F} \hat{O}^\text{T}.\end{aligned} \hspace{\stretch{1}}(3.80)

The class notes use ${x'}^i \rightarrow O^{ij} x^j$, which violates our conventions on mixed upper and lower indexes, but the end result 3.80 is the same.

\begin{aligned}{\left\lVert{{O^i}_j}\right\rVert} =\begin{bmatrix}\cosh\alpha & -\sinh\alpha & 0 & 0 \\ -\sinh\alpha & \cosh\alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.81)

Writing

\begin{aligned}C &= \cosh\alpha = \gamma \\ S &= -\sinh\alpha = -\gamma \beta,\end{aligned} \hspace{\stretch{1}}(3.82)

we can compute the transformed field strength tensor

\begin{aligned}\hat{F}' &=\begin{bmatrix}C & S & 0 & 0 \\ S & C & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix} \begin{bmatrix}C & S & 0 & 0 \\ S & C & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix} \\ &=\begin{bmatrix}C & S & 0 & 0 \\ S & C & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}- S E_x & -C E_x & -E_y & -E_z \\ C E_x & S E_x & -B_z & B_y \\ C E_y + S B_z & S E_y + C B_z & 0 & -B_x \\ C E_z - S B_y & S E_z - C B_y & B_x & 0 \end{bmatrix} \\ &=\begin{bmatrix}0 & -E_x & -C E_y - S B_z & - C E_z + S B_y \\ E_x & 0 & -S E_y - C B_z & - S E_z + C B_y \\ C E_y + S B_z & S E_y + C B_z & 0 & -B_x \\ C E_z - S B_y & S E_z - C B_y & B_x & 0\end{bmatrix} \\ &=\begin{bmatrix}0 & -E_x & -\gamma(E_y - \beta B_z) & - \gamma(E_z + \beta B_y) \\ E_x & 0 & - \gamma (-\beta E_y + B_z) & \gamma( \beta E_z + B_y) \\ \gamma (E_y - \beta B_z) & \gamma(-\beta E_y + B_z) & 0 & -B_x \\ \gamma (E_z + \beta B_y) & -\gamma(\beta E_z + B_y) & B_x & 0\end{bmatrix}.\end{aligned}

As a check we have the antisymmetry that is expected. There is also a regularity to the end result that is aesthetically pleasing, hinting that things are hopefully error free. In coordinates for $\mathbf{E}$ and $\mathbf{B}$ this is

\begin{aligned}E_x &\rightarrow E_x \\ E_y &\rightarrow \gamma ( E_y - \beta B_z ) \\ E_z &\rightarrow \gamma ( E_z + \beta B_y ) \\ B_z &\rightarrow B_x \\ B_y &\rightarrow \gamma ( B_y + \beta E_z ) \\ B_z &\rightarrow \gamma ( B_z - \beta E_y ) \end{aligned} \hspace{\stretch{1}}(3.84)

Writing $\boldsymbol{\beta} = \mathbf{e}_1 \beta$, we have

\begin{aligned}\boldsymbol{\beta} \times \mathbf{B} = \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ \beta & 0 & 0 \\ B_x & B_y & B_z\end{vmatrix} = \mathbf{e}_2 (-\beta B_z) + \mathbf{e}_3( \beta B_y ),\end{aligned} \hspace{\stretch{1}}(3.90)

which puts us enroute to a tidier vector form

\begin{aligned}E_x &\rightarrow E_x \\ E_y &\rightarrow \gamma ( E_y + (\boldsymbol{\beta} \times \mathbf{B})_y ) \\ E_z &\rightarrow \gamma ( E_z + (\boldsymbol{\beta} \times \mathbf{B})_z ) \\ B_z &\rightarrow B_x \\ B_y &\rightarrow \gamma ( B_y - (\boldsymbol{\beta} \times \mathbf{E})_y ) \\ B_z &\rightarrow \gamma ( B_z - (\boldsymbol{\beta} \times \mathbf{E})_z ).\end{aligned} \hspace{\stretch{1}}(3.91)

For a vector $\mathbf{A}$, write $\mathbf{A}_\parallel = (\mathbf{A} \cdot \hat{\mathbf{v}})\hat{\mathbf{v}}$, $\mathbf{A}_\perp = \mathbf{A} - \mathbf{A}_\parallel$, allowing a compact description of the field transformation

\begin{aligned}\mathbf{E} &\rightarrow \mathbf{E}_\parallel + \gamma \mathbf{E}_\perp + \gamma (\boldsymbol{\beta} \times \mathbf{B})_\perp \\ \mathbf{B} &\rightarrow \mathbf{B}_\parallel + \gamma \mathbf{B}_\perp - \gamma (\boldsymbol{\beta} \times \mathbf{E})_\perp.\end{aligned} \hspace{\stretch{1}}(3.97)

Now, we want to consider the field of a moving particle. In the particle’s (unprimed) rest frame the field due to its potential $\phi = q/r$ is

\begin{aligned}\mathbf{E} &= \frac{q}{r^2} \hat{\mathbf{r}} \\ \mathbf{B} &= 0.\end{aligned} \hspace{\stretch{1}}(3.99)

Coordinates for a “stationary” observer, who sees this particle moving along the x-axis at speed $v$ are related by a boost in the $-v$ direction

\begin{aligned}\begin{bmatrix}ct' \\ x' \\ y' \\ z'\end{bmatrix}\begin{bmatrix}\gamma & \gamma (v/c) & 0 & 0 \\ \gamma (v/c) & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}ct \\ x \\ y \\ z\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.101)

Therefore the fields in the observer frame will be

\begin{aligned}\mathbf{E}' &= \mathbf{E}_\parallel + \gamma \mathbf{E}_\perp - \gamma \frac{v}{c}(\mathbf{e}_1 \times \mathbf{B})_\perp = \mathbf{E}_\parallel + \gamma \mathbf{E}_\perp \\ \mathbf{B}' &= \mathbf{B}_\parallel + \gamma \mathbf{B}_\perp + \gamma \frac{v}{c}(\mathbf{e}_1 \times \mathbf{E})_\perp = \gamma \frac{v}{c}(\mathbf{e}_1 \times \mathbf{E})_\perp \end{aligned} \hspace{\stretch{1}}(3.102)

More explicitly with $\mathbf{E} = \frac{q}{r^3}(x, y, z)$ this is

\begin{aligned}\mathbf{E}' &= \frac{q}{r^3}(x, \gamma y, \gamma z) \\ \mathbf{B}' &= \gamma \frac{q v}{c r^3} ( 0, -z, y )\end{aligned} \hspace{\stretch{1}}(3.104)

Comparing to Problem 3 in Problem set 2, I see that this matches the result obtained by separately transforming the gradient, the time partial, and the scalar potential. Actually, if I am being honest, I see that I made a sign error in all the coordinates of $\mathbf{E}'$ when I initially did (this ungraded problem) in problem set 2. That sign error should have been obvious by considering the $v=0$ case which would have mysteriously resulted in inversion of all the coordinates of the observed electric field.

2. Statement

A particle is moving with velocity $\mathbf{v}$ in perpendicular $\mathbf{E}$ and $\mathbf{B}$ fields, all given in some particular “stationary” frame of reference.

\begin{enumerate}
\item Show that there exists a frame where the problem of finding the particle trajectory can be reduced to having either only an electric or only a magnetic field.
\item Explain what determines which case takes place.
\item Find the velocity $\mathbf{v}_0$ of that frame relative to the “stationary” frame.
\end{enumerate}

2. Solution

\paragraph{Part 1 and 2:} Existence of the transformation.

In the single particle Lorentz trajectory problem we wish to solve

\begin{aligned}m c \frac{du^i}{ds} = \frac{e}{c} F^{i j} u_j,\end{aligned} \hspace{\stretch{1}}(3.106)

which in matrix form we can write as

\begin{aligned}\frac{d U}{ds} = \frac{e}{m c^2} \hat{F} \hat{G} U.\end{aligned} \hspace{\stretch{1}}(3.107)

where we write our column vector proper velocity as $U = {\left\lVert{u^i}\right\rVert}$. Under transformation of coordinates ${u'}^i = {O^i}_j x^j$, with $\hat{O} = {\left\lVert{{O^i}_j}\right\rVert}$, this becomes

\begin{aligned}\hat{O} \frac{d U}{ds} = \frac{e}{m c^2} \hat{O} \hat{F} \hat{O}^\text{T} \hat{G} \hat{O} U.\end{aligned} \hspace{\stretch{1}}(3.108)

Suppose we can find eigenvectors for the matrix $\hat{O} \hat{F} \hat{O}^\text{T} \hat{G}$. That is for some eigenvalue $\lambda$, we can find an eigenvector $\Sigma$

\begin{aligned}\hat{O} \hat{F} \hat{O}^\text{T} \hat{G} \Sigma = \lambda \Sigma.\end{aligned} \hspace{\stretch{1}}(3.109)

Rearranging we have

\begin{aligned}(\hat{O} \hat{F} \hat{O}^\text{T} \hat{G} - \lambda I) \Sigma = 0\end{aligned} \hspace{\stretch{1}}(3.110)

and conclude that $\Sigma$ lies in the null space of the matrix $\hat{O} \hat{F} \hat{O}^\text{T} \hat{G} - \lambda I$ and that this difference of matrices must have a zero determinant

\begin{aligned}\text{Det} (\hat{O} \hat{F} \hat{O}^\text{T} \hat{G} - \lambda I) = -\text{Det} (\hat{O} \hat{F} \hat{O}^\text{T} - \lambda \hat{G}) = 0.\end{aligned} \hspace{\stretch{1}}(3.111)

Since $\hat{G} = \hat{O} \hat{G} \hat{O}^\text{T}$ for any Lorentz transformation $\hat{O}$ in $SO(1,3)$, and $\text{Det} ABC = \text{Det} A \text{Det} B \text{Det} C$ we have

\begin{aligned}\text{Det} (\hat{O} \hat{F} \hat{O}^\text{T} - \lambda G)= \text{Det} (\hat{F} - \lambda \hat{G}).\end{aligned} \hspace{\stretch{1}}(3.112)

In problem 1.6, we called this our characteristic equation $P(\lambda) = \text{Det} (\hat{F} - \lambda \hat{G})$. Observe that the characteristic equation is Lorentz invariant for any $\lambda$, which requires that the eigenvalues $\lambda$ are also Lorentz invariants.

In problem 1.6 of this problem set we computed that this characteristic equation expands to

\begin{aligned}P(\lambda) = \text{Det} (\hat{F} - \lambda \hat{G}) = (\mathbf{E} \cdot \mathbf{B})^2 + \lambda^2 (\mathbf{B}^2 - \mathbf{E}^2) + \lambda^4.\end{aligned} \hspace{\stretch{1}}(3.113)

The eigenvalues for the system, also each necessarily Lorentz invariants, are

\begin{aligned}\lambda = \pm \frac{1}{{\sqrt{2}}} \sqrt{ \mathbf{E}^2 - \mathbf{B}^2 \pm \sqrt{ (\mathbf{E}^2 - \mathbf{B}^2)^2 - 4 (\mathbf{E} \cdot \mathbf{B})^2 }}.\end{aligned} \hspace{\stretch{1}}(3.114)

Observe that in the specific case where $\mathbf{E} \cdot \mathbf{B} = 0$, as in this problem, we must have $\mathbf{E}' \cdot \mathbf{B}'$ in all frames, and the two non-zero eigenvalues of our characteristic polynomial are simply

\begin{aligned}\lambda = \pm \sqrt{\mathbf{E}^2 - \mathbf{B}^2}.\end{aligned} \hspace{\stretch{1}}(3.115)

These and $\mathbf{E} \cdot \mathbf{B} = 0$ are the invariants for this system. If we have $\mathbf{E}^2 > \mathbf{B}^2$ in one frame, we must also have ${\mathbf{E}'}^2 > {\mathbf{B}'}^2$ in another frame, still maintaining perpendicular fields. In particular if $\mathbf{B}' = 0$ we maintain real eigenvalues. Similarly if $\mathbf{B}^2 > \mathbf{E}^2$ in some frame, we must always have imaginary eigenvalues, and this is also true in the $\mathbf{E}' = 0$ case.

While the problem can be posed as a pure diagonalization problem (and even solved numerically this way for the general constant fields case), we can also work symbolically, thinking of the trajectories problem as simply seeking a transformation of frames that reduce the scope of the problem to one that is more tractable. That does not have to be the linear transformation that diagonalizes the system. Instead we are free to transform to a frame where one of the two fields $\mathbf{E}'$ or $\mathbf{B}'$ is zero, provided the invariants discussed are maintained.

\paragraph{Part 3:} Finding the boost velocity that wipes out one of the fields.

Let’s now consider a Lorentz boost $\hat{O}$, and seek to solve for the boost velocity that wipes out one of the fields, given the invariants that must be maintained for the system

To make things concrete, suppose that our perpendicular fields are given by $\mathbf{E} = E \mathbf{e}_2$ and $\mathbf{B} = B \mathbf{e}_3$.

Let also assume that we can find the velocity $\mathbf{v}_0$ for which one or more of the transformed fields is zero. Suppose that velocity is

\begin{aligned}\mathbf{v}_0 = v_0 (\alpha_1, \alpha_2, \alpha_3) = v_0 \hat{\mathbf{v}}_0,\end{aligned} \hspace{\stretch{1}}(3.116)

where $\alpha_i$ are the direction cosines of $\mathbf{v}_0$ so that $\sum_i \alpha_i^2 = 1$. We will want to compute the components of $\mathbf{E}$ and $\mathbf{B}$ parallel and perpendicular to this velocity.

Those are

\begin{aligned}\mathbf{E}_\parallel &= E \mathbf{e}_2 \cdot (\alpha_1, \alpha_2, \alpha_3) (\alpha_1, \alpha_2, \alpha_3) \\ &= E \alpha_2 (\alpha_1, \alpha_2, \alpha_3) \\ \end{aligned}

\begin{aligned}\mathbf{E}_\perp &= E \mathbf{e}_2 - \mathbf{E}_\parallel \\ &= E (-\alpha_1 \alpha_2, 1 - \alpha_2^2, -\alpha_2 \alpha_3) \\ &= E (-\alpha_1 \alpha_2, \alpha_1^2 + \alpha_3^2, -\alpha_2 \alpha_3) \\ \end{aligned}

For the magnetic field we have

\begin{aligned}\mathbf{B}_\parallel &= B \alpha_3 (\alpha_1, \alpha_2, \alpha_3),\end{aligned}

and

\begin{aligned}\mathbf{B}_\perp &= B \mathbf{e}_3 - \mathbf{B}_\parallel \\ &= B (-\alpha_1 \alpha_3, -\alpha_2 \alpha_3, \alpha_1^2 + \alpha_2^2) \\ \end{aligned}

Now, observe that $(\boldsymbol{\beta} \times \mathbf{B})_\parallel \propto ((\mathbf{v}_0 \times \mathbf{B}) \cdot \mathbf{v}_0) \mathbf{v}_0$, but this is just zero. So we have $(\boldsymbol{\beta} \times \mathbf{B})_\parallel = \boldsymbol{\beta} \times \mathbf{B}$. So our cross products terms are just

\begin{aligned}\hat{\mathbf{v}}_0 \times \mathbf{B} &= \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ \alpha_1 & \alpha_2 & \alpha_3 \\ 0 & 0 & B \end{vmatrix} = B (\alpha_2, -\alpha_1, 0) \\ \hat{\mathbf{v}}_0 \times \mathbf{E} &= \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ \alpha_1 & \alpha_2 & \alpha_3 \\ 0 & E & 0 \end{vmatrix} = E (-\alpha_3, 0, \alpha_1)\end{aligned}

We can now express how the fields transform, given this arbitrary boost velocity. From 3.97, this is

\begin{aligned}\mathbf{E} &\rightarrow E \alpha_2 (\alpha_1, \alpha_2, \alpha_3) + \gamma E (-\alpha_1 \alpha_2, \alpha_1^2 + \alpha_3^2, -\alpha_2 \alpha_3) + \gamma \frac{v_0^2}{c^2} B (\alpha_2, -\alpha_1, 0) \\ \mathbf{B} &\rightarrowB \alpha_3 (\alpha_1, \alpha_2, \alpha_3)+ \gamma B (-\alpha_1 \alpha_3, -\alpha_2 \alpha_3, \alpha_1^2 + \alpha_2^2) - \gamma \frac{v_0^2}{c^2} E (-\alpha_3, 0, \alpha_1)\end{aligned} \hspace{\stretch{1}}(3.117)

Zero Electric field case.

Let’s tackle the two cases separately. First when ${\left\lvert{\mathbf{B}}\right\rvert} > {\left\lvert{\mathbf{E}}\right\rvert}$, we can transform to a frame where $\mathbf{E}'=0$. In coordinates from 3.117 this supplies us three sets of equations. These are

\begin{aligned}0 &= E \alpha_2 \alpha_1 (1 - \gamma) + \gamma \frac{v_0^2}{c^2} B \alpha_2 \\ 0 &= E \alpha_2^2 + \gamma E (\alpha_1^2 + \alpha_3^2) - \gamma \frac{v_0^2}{c^2} B \alpha_1 \\ 0 &= E \alpha_2 \alpha_3 (1 - \gamma).\end{aligned} \hspace{\stretch{1}}(3.119)

With an assumed solution the $\mathbf{e}_3$ coordinate equation implies that one of $\alpha_2$ or $\alpha_3$ is zero. Perhaps there are solutions with $\alpha_3 = 0$ too, but inspection shows that $\alpha_2 = 0$ nicely kills off the first equation. Since $\alpha_1^2 + \alpha_2^2 + \alpha_3^2 = 1$, that also implies that we are left with

\begin{aligned}0 = E - \frac{v_0^2}{c^2} B \alpha_1 \end{aligned} \hspace{\stretch{1}}(3.122)

Or

\begin{aligned}\alpha_1 &= \frac{E}{B} \frac{c^2}{v_0^2} \\ \alpha_2 &= 0 \\ \alpha_3 &= \sqrt{1 - \frac{E^2}{B^2} \frac{c^4}{v_0^4} }\end{aligned} \hspace{\stretch{1}}(3.123)

Our velocity was $\mathbf{v}_0 = v_0 (\alpha_1, \alpha_2, \alpha_3)$ solving the problem for the ${\left\lvert{\mathbf{B}}\right\rvert}^2 > {\left\lvert{\mathbf{E}}\right\rvert}^2$ case up to an adjustable constant $v_0$. That constant comes with constraints however, since we must also have our cosine $\alpha_1 \le 1$. Expressed another way, the magnitude of the boost velocity is constrained by the relation

\begin{aligned}\frac{\mathbf{v}_0^2}{c^2} \ge {\left\lvert{\frac{E}{B}}\right\rvert}.\end{aligned} \hspace{\stretch{1}}(3.126)

It appears we may also pick the equality case, so one velocity (not unique) that should transform away the electric field is

\begin{aligned}\boxed{\mathbf{v}_0 = c \sqrt{{\left\lvert{\frac{E}{B}}\right\rvert}} \mathbf{e}_1 = \pm c \sqrt{{\left\lvert{\frac{E}{B}}\right\rvert}} \frac{\mathbf{E} \times \mathbf{B}}{{\left\lvert{\mathbf{E}}\right\rvert} {\left\lvert{\mathbf{B}}\right\rvert}}.}\end{aligned} \hspace{\stretch{1}}(3.127)

This particular boost direction is perpendicular to both fields. Observe that this highlights the invariance condition ${\left\lvert{\frac{E}{B}}\right\rvert} < 1$ since we see this is required for a physically realizable velocity. Boosting in this direction will reduce our problem to one that has only the magnetic field component.

Zero Magnetic field case.

Now, let’s consider the case where we transform the magnetic field away, the case when our characteristic polynomial has strictly real eigenvalues $\lambda = \pm \sqrt{\mathbf{E}^2 - \mathbf{B}^2}$. In this case, if we write out our equations for the transformed magnetic field and require these to separately equal zero, we have

\begin{aligned}0 &= B \alpha_3 \alpha_1 ( 1 - \gamma ) + \gamma \frac{v_0^2}{c^2} E \alpha_3 \\ 0 &= B \alpha_2 \alpha_3 ( 1 - \gamma ) \\ 0 &= B (\alpha_3^2 + \gamma (\alpha_1^2 + \alpha_2^2)) - \gamma \frac{v_0^2}{c^2} E \alpha_1.\end{aligned} \hspace{\stretch{1}}(3.128)

Similar to before we see that $\alpha_3 = 0$ kills off the first and second equations, leaving just

\begin{aligned}0 = B - \frac{v_0^2}{c^2} E \alpha_1.\end{aligned} \hspace{\stretch{1}}(3.131)

We now have a solution for the family of direction vectors that kill the magnetic field off

\begin{aligned}\alpha_1 &= \frac{B}{E} \frac{c^2}{v_0^2} \\ \alpha_2 &= \sqrt{ 1 - \frac{B^2}{E^2} \frac{c^4}{v_0^4} } \\ \alpha_3 &= 0.\end{aligned} \hspace{\stretch{1}}(3.132)

In addition to the initial constraint that ${\left\lvert{\frac{B}{E}}\right\rvert} < 1$, we have as before, constraints on the allowable values of $v_0$

\begin{aligned}\frac{\mathbf{v}_0^2}{c^2} \ge {\left\lvert{\frac{B}{E}}\right\rvert}.\end{aligned} \hspace{\stretch{1}}(3.135)

Like before we can pick the equality $\alpha_1^2 = 1$, yielding a boost direction of

\begin{aligned}\boxed{\mathbf{v}_0 = c \sqrt{{\left\lvert{\frac{B}{E}}\right\rvert}} \mathbf{e}_1 = \pm c \sqrt{{\left\lvert{\frac{B}{E}}\right\rvert}} \frac{\mathbf{E} \times \mathbf{B}}{{\left\lvert{\mathbf{E}}\right\rvert} {\left\lvert{\mathbf{B}}\right\rvert}}.}\end{aligned} \hspace{\stretch{1}}(3.136)

Again, we see that the invariance condition ${\left\lvert{\mathbf{B}}\right\rvert} < {\left\lvert{\mathbf{E}}\right\rvert}$ is required for a physically realizable velocity if that velocity is entirely perpendicular to the fields.

Problem 3. Continuity equation for delta function current distributions.

Statement

Show explicitly that the electromagnetic 4-current $j^i$ for a particle moving with constant velocity (considered in class, p. 100-101 of notes) is conserved $\partial_i j^i = 0$. Give a physical interpretation of this conservation law, for example by integrating $\partial_i j^i$ over some spacetime region and giving an integral form to the conservation law ($\partial_i j^i = 0$ is known as the “continuity equation”).

Solution

First lets review. Our four current was defined as

\begin{aligned}j^i(x) = \sum_A c e_A \int_{x(\tau)} dx_A^i(\tau) \delta^4(x - x_A(\tau)).\end{aligned} \hspace{\stretch{1}}(4.137)

If each of the trajectories $x_A(\tau)$ represents constant motion we have

\begin{aligned}x_A(\tau) = x_A(0) + \gamma_A \tau ( c, \mathbf{v}_A ).\end{aligned} \hspace{\stretch{1}}(4.138)

The spacetime split of this four vector is

\begin{aligned}x_A^0(\tau) &= x_A^0(0) + \gamma_A \tau c \\ \mathbf{x}_A(\tau) &= \mathbf{x}_A(0) + \gamma_A \tau \mathbf{v},\end{aligned} \hspace{\stretch{1}}(4.139)

with differentials

\begin{aligned}dx_A^0(\tau) &= \gamma_A d\tau c \\ d\mathbf{x}_A(\tau) &= \gamma_A d\tau \mathbf{v}_A.\end{aligned} \hspace{\stretch{1}}(4.141)

Writing out the delta functions explicitly we have

\begin{aligned}\begin{aligned}j^i(x) = \sum_A &c e_A \int_{x(\tau)} dx_A^i(\tau) \delta(x^0 - x_A^0(0) - \gamma_A c \tau) \delta(x^1 - x_A^1(0) - \gamma_A v_A^1 \tau) \\ &\delta(x^2 - x_A^2(0) - \gamma_A v_A^2 \tau) \delta(x^3 - x_A^3(0) - \gamma_A v_A^3 \tau)\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.143)

So our time and space components of the current can be written

\begin{aligned}j^0(x) &= \sum_A c^2 e_A \gamma_A \int_{x(\tau)} d\tau\delta(x^0 - x_A^0(0) - \gamma_A c \tau)\delta^3(\mathbf{x} - \mathbf{x}_A(0) - \gamma_A \mathbf{v}_A \tau) \\ \mathbf{j}(x) &= \sum_A c e_A \mathbf{v}_A \gamma_A \int_{x(\tau)} d\tau\delta(x^0 - x_A^0(0) - \gamma_A c \tau)\delta^3(\mathbf{x} - \mathbf{x}_A(0) - \gamma_A \mathbf{v}_A \tau).\end{aligned} \hspace{\stretch{1}}(4.144)

Each of these integrals can be evaluated with respect to the time coordinate delta function leaving the distribution

\begin{aligned}j^0(x) &= \sum_A c e_A \delta^3(\mathbf{x} - \mathbf{x}_A(0) - \frac{\mathbf{v}_A}{c} (x^0 - x_A^0(0))) \\ \mathbf{j}(x) &= \sum_A e_A \mathbf{v}_A \delta^3(\mathbf{x} - \mathbf{x}_A(0) - \frac{\mathbf{v}_A}{c} (x^0 - x_A^0(0)))\end{aligned} \hspace{\stretch{1}}(4.146)

With this more general expression (multi-particle case) it should be possible to show that the four divergence is zero, however, the problem only asks for one particle. For the one particle case, we can make things really easy by taking the initial point in space and time as the origin, and aligning our velocity with one of the coordinates (say $x$).

Doing so we have the result derived in class

\begin{aligned}j = e \begin{bmatrix}c \\ v \\ 0 \\ 0 \end{bmatrix}\delta(x - v x^0/c)\delta(y)\delta(z).\end{aligned} \hspace{\stretch{1}}(4.148)

Our divergence then has only two portions

\begin{aligned}\frac{\partial {j^0}}{\partial {x^0}} &= e c (-v/c) \delta'(x - v x^0/c) \delta(y) \delta(z) \\ \frac{\partial {j^1}}{\partial {x}} &= e v \delta'(x - v x^0/c) \delta(y) \delta(z).\end{aligned} \hspace{\stretch{1}}(4.149)

and these cancel out when summed. Note that this requires us to be loose with our delta functions, treating them like regular functions that are differentiable.

For the more general multiparticle case, we can treat the sum one particle at a time, and in each case, rotate coordinates so that the four divergence only picks up one term.

As for physical interpretation via integral, we have using the four dimensional divergence theorem

\begin{aligned}\int d^4 x \partial_i j^i = \int j^i dS_i\end{aligned} \hspace{\stretch{1}}(4.151)

where $dS_i$ is the three-volume element perpendicular to a $x^i = \text{constant}$ plane. These volume elements are detailed generally in the text [2], however, they do note that one special case specifically $dS_0 = dx dy dz$, the element of the three-dimensional (spatial) volume “normal” to hyperplanes $ct = \text{constant}$.

Without actually computing the determinants, we have something that is roughly of the form

\begin{aligned}0 = \int j^i dS_i=\int c \rho dx dy dz+\int \mathbf{j} \cdot (\mathbf{n}_x c dt dy dz + \mathbf{n}_y c dt dx dz + \mathbf{n}_z c dt dx dy).\end{aligned} \hspace{\stretch{1}}(4.152)

This is cheating a bit to just write $\mathbf{n}_x, \mathbf{n}_y, \mathbf{n}_z$. Are there specific orientations required by the metric. To be precise we’d have to calculate the determinants detailed in the text, and then do the duality transformations.

Per unit time, we can write instead

\begin{aligned}\frac{\partial {}}{\partial {t}} \int \rho dV= -\int \mathbf{j} \cdot (\mathbf{n}_x dy dz + \mathbf{n}_y dx dz + \mathbf{n}_z dx dy)\end{aligned} \hspace{\stretch{1}}(4.153)

Rather loosely this appears to roughly describe that the rate of change of charge in a volume must be matched with the “flow” of current through the surface within that amount of time.

References

[1] Wikipedia. Electromagnetic tensor — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 27-February-2011]. http://en.wikipedia.org/w/index.php?title=Electromagnetic_tensor&oldid=414989505.

[2] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

PHY450H1S. Relativistic Electrodynamics Lecture 11 (Taught by Prof. Erich Poppitz). Unpacking Lorentz force equation. Lorentz transformations of the strength tensor, Lorentz field invariants, Bianchi identity, and first half of Maxwell’s.

Posted by peeterjoot on February 24, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Covering chapter 3 material from the text [1].

Covering lecture notes pp. 74-83: Lorentz transformation of the strength tensor (82) [Tuesday, Feb. 8] [extra reading for the mathematically minded: gauge field, strength tensor, and gauge transformations in differential form language, not to be covered in class (83)]

Covering lecture notes pp. 84-102: Lorentz invariants of the electromagnetic field (84-86); Bianchi identity and the first half of Maxwell’s equations (87-90)

Chewing on the four vector form of the Lorentz force equation.

After much effort, we arrived at

\begin{aligned}\frac{d{{(m c u_l) }}}{ds} = \frac{e}{c} \left( \partial_l A_i - \partial_i A_l \right) u^i\end{aligned} \hspace{\stretch{1}}(2.1)

or

\begin{aligned}\frac{d{{ p_l }}}{ds} = \frac{e}{c} F_{l i} u^i\end{aligned} \hspace{\stretch{1}}(2.2)

Elements of the strength tensor

\paragraph{Claim}: there are only 6 independent elements of this matrix (tensor)

\begin{aligned}\begin{bmatrix}0 & . & . & . \\ & 0 & . & . \\ & & 0 & . \\ & & & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.3)

This is a no-brainer, for we just have to mechanically plug in the elements of the field strength tensor

Recall

\begin{aligned}A^i &= (\phi, \mathbf{A}) \\ A_i &= (\phi, -\mathbf{A})\end{aligned} \hspace{\stretch{1}}(2.4)

\begin{aligned}F_{0\alpha} &= \partial_0 A_\alpha - \partial_\alpha A_0 \\ &= -\partial_0 (\mathbf{A})_\alpha - \partial_\alpha \phi \\ \end{aligned}

\begin{aligned}F_{0\alpha} = E_\alpha\end{aligned} \hspace{\stretch{1}}(2.6)

For the purely spatial index combinations we have

\begin{aligned}F_{\alpha\beta} &= \partial_\alpha A_\beta - \partial_\beta A_\alpha \\ &= -\partial_\alpha (\mathbf{A})_\beta + \partial_\beta (\mathbf{A})_\alpha \\ \end{aligned}

Written out explicitly, these are

\begin{aligned}F_{1 2} &= \partial_2 (\mathbf{A})_1 -\partial_1 (\mathbf{A})_2 \\ F_{2 3} &= \partial_3 (\mathbf{A})_2 -\partial_2 (\mathbf{A})_3 \\ F_{3 1} &= \partial_1 (\mathbf{A})_3 -\partial_3 (\mathbf{A})_1 .\end{aligned} \hspace{\stretch{1}}(2.7)

We can compare this to the elements of $\mathbf{B}$

\begin{aligned}\mathbf{B} = \begin{vmatrix}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \partial_1 & \partial_2 & \partial_3 \\ A_x & A_y & A_z\end{vmatrix}\end{aligned} \hspace{\stretch{1}}(2.10)

We see that

\begin{aligned}(\mathbf{B})_z &= \partial_1 A_y - \partial_2 A_x \\ (\mathbf{B})_x &= \partial_2 A_z - \partial_3 A_y \\ (\mathbf{B})_y &= \partial_3 A_x - \partial_1 A_z\end{aligned} \hspace{\stretch{1}}(2.11)

So we have

\begin{aligned}F_{1 2} &= - (\mathbf{B})_3 \\ F_{2 3} &= - (\mathbf{B})_1 \\ F_{3 1} &= - (\mathbf{B})_2.\end{aligned} \hspace{\stretch{1}}(2.14)

These can be summarized as simply

\begin{aligned}F_{\alpha\beta} = - \epsilon_{\alpha\beta\gamma} B_\gamma.\end{aligned} \hspace{\stretch{1}}(2.17)

This provides all the info needed to fill in the matrix above

\begin{aligned}{\left\lVert{ F_{i j} }\right\rVert} = \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0.\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.18)

Index raising of rank 2 tensor

To raise indexes we compute

\begin{aligned}F^{i j} = g^{i l} g^{j k} F_{l k}.\end{aligned} \hspace{\stretch{1}}(2.19)

Justifying the raising operation.

To justify this consider raising one index at a time by applying the metric tensor to our definition of $F_{l k}$. That is

\begin{aligned}g^{a l} F_{l k} &=g^{a l} (\partial_l A_k - \partial_k A_l) \\ &=\partial^a A_k - \partial_k A^a.\end{aligned}

Now apply the metric tensor once more

\begin{aligned}g^{b k} g^{a l} F_{l k} &=g^{b k} (\partial^a A_k - \partial_k A^a) \\ &=\partial^a A^b - \partial^b A^a.\end{aligned}

This is, by definition $F^{a b}$. Since a rank 2 tensor has been defined as an object that transforms like the product of two pairs of coordinates, it makes sense that this particular tensor raises in the same fashion as would a product of two vector coordinates (in this case, it happens to be an antisymmetric product of two vectors, and one of which is an operator, but we have the same idea).

Consider the components of the raised $F_{i j}$ tensor.

\begin{aligned}F^{0\alpha} &= -F_{0\alpha} \\ F^{\alpha\beta} &= F_{\alpha\beta}.\end{aligned} \hspace{\stretch{1}}(2.20)

\begin{aligned}{\left\lVert{ F^{i j} }\right\rVert} = \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.22)

Back to chewing on the Lorentz force equation.

\begin{aligned}m c \frac{d{{ u_i }}}{ds} = \frac{e}{c} F_{i j} u^j\end{aligned} \hspace{\stretch{1}}(2.23)

\begin{aligned}u^i &= \gamma \left( 1, \frac{\mathbf{v}}{c} \right) \\ u_i &= \gamma \left( 1, -\frac{\mathbf{v}}{c} \right)\end{aligned} \hspace{\stretch{1}}(2.24)

For the spatial components of the Lorentz force equation we have

\begin{aligned}m c \frac{d{{ u_\alpha }}}{ds} &= \frac{e}{c} F_{\alpha j} u^j \\ &= \frac{e}{c} F_{\alpha 0} u^0+ \frac{e}{c} F_{\alpha \beta} u^\beta \\ &= \frac{e}{c} (-E_{\alpha}) \gamma+ \frac{e}{c} (- \epsilon_{\alpha\beta\gamma} B_\gamma ) \frac{v^\beta}{c} \gamma \end{aligned}

But

\begin{aligned}m c \frac{d{{ u_\alpha }}}{ds} &= -m \frac{d{{(\gamma \mathbf{v}_\alpha)}}}{ds} \\ &= -m \frac{d(\gamma \mathbf{v}_\alpha)}{c \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} dt} \\ &= -\gamma \frac{d(m \gamma \mathbf{v}_\alpha)}{c dt}.\end{aligned}

Canceling the common $-\gamma/c$ terms, and switching to vector notation, we are left with

\begin{aligned}\frac{d( m \gamma \mathbf{v}_\alpha)}{dt} = e \left( E_\alpha + \frac{1}{{c}} (\mathbf{v} \times \mathbf{B})_\alpha \right).\end{aligned} \hspace{\stretch{1}}(2.26)

Now for the energy term. We have

\begin{aligned}m c \frac{d{{u_0}}}{ds} &= \frac{e}{c} F_{0\alpha} u^\alpha \\ &= \frac{e}{c} E_{\alpha} \gamma \frac{v^\alpha}{c} \\ \frac{d{{ m c \gamma }}}{ds} &=\end{aligned}

Putting the final two lines into vector form we have

\begin{aligned}\frac{d{{ (m c^2 \gamma)}}}{dt} = e \mathbf{E} \cdot \mathbf{v},\end{aligned} \hspace{\stretch{1}}(2.27)

or

\begin{aligned}\frac{d{{ \mathcal{E} }}}{dt} = e \mathbf{E} \cdot \mathbf{v}\end{aligned} \hspace{\stretch{1}}(2.28)

Transformation of rank two tensors in matrix and index form.

Transformation of the metric tensor, and some identities.

With

\begin{aligned}\hat{G} = {\left\lVert{ g_{i j} }\right\rVert} = {\left\lVert{ g^{i j} }\right\rVert}\end{aligned} \hspace{\stretch{1}}(3.29)

\paragraph{We claim:}
The rank two tensor $\hat{G}$ transforms in the following sort of sandwich operation, and this leaves it invariant

\begin{aligned}\hat{G} \rightarrow \hat{O} \hat{G} \hat{O}^\text{T} = \hat{G}.\end{aligned} \hspace{\stretch{1}}(3.30)

To demonstrate this let’s consider a transformed vector in coordinate form as follows

\begin{aligned}{x'}^i &= O^{i j} x_j = {O^i}_j x^j \\ {x'}_i &= O_{i j} x^j = {O_i}^j x_j.\end{aligned} \hspace{\stretch{1}}(3.31)

We can thus write the equation in matrix form with

\begin{aligned}X &= {\left\lVert{x^i}\right\rVert} \\ X' &= {\left\lVert{{x'}^i}\right\rVert} \\ \hat{O} &= {\left\lVert{{O^i}_j}\right\rVert} \\ X' &= \hat{O} X\end{aligned} \hspace{\stretch{1}}(3.33)

Our invariant for the vector square, which is required to remain unchanged is

\begin{aligned}{x'}^i {x'}_i &= (O^{i j} x_j)(O_{i k} x^k) \\ &= x^k (O^{i j} O_{i k}) x_j.\end{aligned}

This shows that we have a delta function relationship for the Lorentz transform matrix, when we sum over the first index

\begin{aligned}O^{a i} O_{a j} = {\delta^i}_j.\end{aligned} \hspace{\stretch{1}}(3.37)

It appears we can put 3.37 into matrix form as

\begin{aligned}\hat{G} \hat{O}^\text{T} \hat{G} \hat{O} = I\end{aligned} \hspace{\stretch{1}}(3.38)

Now, if one considers that the transpose of a rotation is an inverse rotation, and the transpose of a boost leaves it unchanged, the transpose of a general Lorentz transformation, a composition of an arbitrary sequence of boosts and rotations, must also be a Lorentz transformation, and must then also leave the norm unchanged. For the transpose of our Lorentz transformation $\hat{O}$ lets write

\begin{aligned}\hat{P} = \hat{O}^\text{T}\end{aligned} \hspace{\stretch{1}}(3.39)

For the action of this on our position vector let’s write

\begin{aligned}{x''}^i &= P^{i j} x_j = O^{j i} x_j \\ {x''}_i &= P_{i j} x^j = O_{j i} x^j\end{aligned} \hspace{\stretch{1}}(3.40)

so that our norm is

\begin{aligned}{x''}^a {x''}_a &= (O_{k a} x^k)(O^{j a} x_j) \\ &= x^k (O_{k a} O^{j a} ) x_j \\ &= x^j x_j \\ \end{aligned}

We must then also have an identity when summing over the second index

\begin{aligned}{\delta_{k}}^j = O_{k a} O^{j a} \end{aligned} \hspace{\stretch{1}}(3.42)

Armed with these facts on the products of $O_{i j}$ and $O^{i j}$ we can now consider the transformation of the metric tensor.

The rule (definition) supplied to us for the transformation of an arbitrary rank two tensor, is that this transforms as its indexes transform individually. Very much as if it was the product of two coordinate vectors and we transform those coordinates separately. Doing so for the metric tensor we have

\begin{aligned}g^{i j} &\rightarrow {O^i}_k g^{k m} {O^j}_m \\ &= ({O^i}_k g^{k m}) {O^j}_m \\ &= O^{i m} {O^j}_m \\ &= O^{i m} (O_{a m} g^{a j}) \\ &= (O^{i m} O_{a m}) g^{a j}\end{aligned}

However, by 3.42, we have $O_{a m} O^{i m} = {\delta_a}^i$, and we prove that

\begin{aligned}g^{i j} \rightarrow g^{i j}.\end{aligned} \hspace{\stretch{1}}(3.43)

Finally, we wish to put the above transformation in matrix form, look more carefully at the very first line

\begin{aligned}g^{i j}&\rightarrow {O^i}_k g^{k m} {O^j}_m \\ \end{aligned}

which is

\begin{aligned}\hat{G} \rightarrow \hat{O} \hat{G} \hat{O}^\text{T} = \hat{G}\end{aligned} \hspace{\stretch{1}}(3.44)

We see that this particular form of transformation, a sandwich between $\hat{O}$ and $\hat{O}^\text{T}$, leaves the metric tensor invariant.

Lorentz transformation of the electrodynamic tensor

Having identified a composition of Lorentz transformation matrices, when acting on the metric tensor, leaves it invariant, it is a reasonable question to ask how this form of transformation acts on our electrodynamic tensor $F^{i j}$?

\paragraph{Claim:} A transformation of the following form is required to maintain the norm of the Lorentz force equation

\begin{aligned}\hat{F} \rightarrow \hat{O} \hat{F} \hat{O}^\text{T} ,\end{aligned} \hspace{\stretch{1}}(3.45)

where $\hat{F} = {\left\lVert{F^{i j}}\right\rVert}$. Observe that our Lorentz force equation can be written exclusively in upper index quantities as

\begin{aligned}m c \frac{d{{u^i}}}{ds} = \frac{e}{c} F^{i j} g_{j l} u^l\end{aligned} \hspace{\stretch{1}}(3.46)

Because we have a vector on one side of the equation, and it transforms by multiplication with by a Lorentz matrix in SO(1,3)

\begin{aligned}\frac{du^i}{ds} \rightarrow \hat{O} \frac{du^i}{ds} \end{aligned} \hspace{\stretch{1}}(3.47)

The LHS of the Lorentz force equation provides us with one invariant

\begin{aligned}(m c)^2 \frac{d{{u^i}}}{ds} \frac{d{{u_i}}}{ds}\end{aligned} \hspace{\stretch{1}}(3.48)

so the RHS must also provide one

\begin{aligned}\frac{e^2}{c^2} F^{i j} g_{j l} u^lF_{i k} g^{k m} u_m=\frac{e^2}{c^2} F^{i j} u_jF_{i k} u^k.\end{aligned} \hspace{\stretch{1}}(3.49)

Let’s look at the RHS in matrix form. Writing

\begin{aligned}U = {\left\lVert{u^i}\right\rVert},\end{aligned} \hspace{\stretch{1}}(3.50)

we can rewrite the Lorentz force equation as

\begin{aligned}m c \dot{U} = \frac{e}{c} \hat{F} \hat{G} U.\end{aligned} \hspace{\stretch{1}}(3.51)

In this matrix formalism our invariant 3.49 is

\begin{aligned}\frac{e^2}{c^2} (\hat{F} \hat{G} U)^\text{T} G \hat{F} \hat{G} U=\frac{e^2}{c^2} U^\text{T} \hat{G} \hat{F}^\text{T} G \hat{F} \hat{G} U.\end{aligned} \hspace{\stretch{1}}(3.52)

If we compare this to the transformed Lorentz force equation we have

\begin{aligned}m c \hat{O} \dot{U} = \frac{e}{c} \hat{F'} \hat{G} \hat{O} U.\end{aligned} \hspace{\stretch{1}}(3.53)

Our invariant for the transformed equation is

\begin{aligned}\frac{e^2}{c^2} (\hat{F'} \hat{G} \hat{O} U)^\text{T} G \hat{F'} \hat{G} \hat{O} U&=\frac{e^2}{c^2} U^\text{T} \hat{O}^\text{T} \hat{G} \hat{F'}^\text{T} G \hat{F'} \hat{G} \hat{O} U \\ \end{aligned}

Thus the transformed electrodynamic tensor $\hat{F}'$ must satisfy the identity

\begin{aligned}\hat{O}^\text{T} \hat{G} \hat{F'}^\text{T} G \hat{F'} \hat{G} \hat{O} = \hat{G} \hat{F}^\text{T} G \hat{F} \hat{G} \end{aligned} \hspace{\stretch{1}}(3.54)

With the substitution $\hat{F}' = \hat{O} \hat{F} \hat{O}^\text{T}$ the LHS is

\begin{aligned}\hat{O}^\text{T} \hat{G} \hat{F'}^\text{T} \hat{G} \hat{F'} \hat{G} \hat{O} &= \hat{O}^\text{T} \hat{G} ( \hat{O} \hat{F} \hat{O}^\text{T})^\T \hat{G} (\hat{O} \hat{F} \hat{O}^\text{T}) \hat{G} \hat{O} \\ &= (\hat{O}^\text{T} \hat{G} \hat{O}) \hat{F}^\text{T} (\hat{O}^\text{T} \hat{G} \hat{O}) \hat{F} (\hat{O}^\text{T} \hat{G} \hat{O}) \\ \end{aligned}

We’ve argued that $\hat{P} = \hat{O}^\text{T}$ is also a Lorentz transformation, thus

\begin{aligned}\hat{O}^\text{T} \hat{G} \hat{O}&=\hat{P} \hat{G} \hat{O}^\text{T} \\ &=\hat{G}\end{aligned}

This is enough to make both sides of 3.54 match, verifying that this transformation does provide the invariant properties desired.

Direct computation of the Lorentz transformation of the electrodynamic tensor.

We can construct the transformed field tensor more directly, by simply transforming the coordinates of the four gradient and the four potential directly. That is

\begin{aligned}F^{i j} = \partial^i A^j - \partial^j A^i&\rightarrow {O^i}_a {O^j}_b \left( \partial^a A^b - \partial^b A^a \right) \\ &={O^i}_a F^{a b} {O^j}_b \end{aligned}

By inspection we can see that this can be represented in matrix form as

\begin{aligned}\hat{F} \rightarrow \hat{O} \hat{F} \hat{O}^\text{T}\end{aligned} \hspace{\stretch{1}}(3.55)

Four vector invariants

For three vectors $\mathbf{A}$ and $\mathbf{B}$ invariants are

\begin{aligned}\mathbf{A} \cdot \mathbf{B} = A^\alpha B_\alpha\end{aligned} \hspace{\stretch{1}}(4.56)

For four vectors $A^i$ and $B^i$ invariants are

\begin{aligned}A^i B_i = A^i g_{i j} B^j \end{aligned} \hspace{\stretch{1}}(4.57)

For $F_{i j}$ what are the invariants? One invariant is

\begin{aligned}g^{i j} F_{i j} = 0,\end{aligned} \hspace{\stretch{1}}(4.58)

but this isn’t interesting since it is uniformly zero (product of symmetric and antisymmetric).

The two invariants are

\begin{aligned}F_{i j}F^{i j}\end{aligned} \hspace{\stretch{1}}(4.59)

and

\begin{aligned}\epsilon^{i j k l} F_{i j}F_{k l}\end{aligned} \hspace{\stretch{1}}(4.60)

where

\begin{aligned}\epsilon^{i j k l} =\left\{\begin{array}{l l}0 & \quad \mbox{if any two indexes coincide} \\ 1 & \quad \mbox{for even permutations ofi j k l=0123$} \\ -1 & \quad \mbox{for odd permutations of$i j k l=0123} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(4.61)

We can show (homework) that

\begin{aligned}F_{i j}F^{i j} \propto \mathbf{E}^2 - \mathbf{B}^2\end{aligned} \hspace{\stretch{1}}(4.62)

\begin{aligned}\epsilon^{i j k l} F_{i j}F_{k l} \propto \mathbf{E} \cdot \mathbf{B}\end{aligned} \hspace{\stretch{1}}(4.63)

This first invariant serves as the action density for the Maxwell field equations.

There’s some useful properties of these invariants. One is that if the fields are perpendicular in one frame, then will be in any other.

From the first, note that if ${\left\lvert{\mathbf{E}}\right\rvert} > {\left\lvert{\mathbf{B}}\right\rvert}$, the invariant is positive, and must be positive in all frames, or if ${\left\lvert{\mathbf{E}}\right\rvert} {\left\lvert{\mathbf{B}}\right\rvert}$ in one frame, we can transform to a frame with only $\mathbf{E}'$ component, solve that, and then transform back. Similarly if ${\left\lvert{\mathbf{E}}\right\rvert} < {\left\lvert{\mathbf{B}}\right\rvert}$ in one frame, we can transform to a frame with only $\mathbf{B}'$ component, solve that, and then transform back.

The first half of Maxwell’s equations.

\paragraph{Claim: } The source free portions of Maxwell’s equations are a consequence of the definition of the field tensor alone.

Given

\begin{aligned}F_{i j} = \partial_i A_j - \partial_j A_i,\end{aligned} \hspace{\stretch{1}}(5.64)

where

\begin{aligned}\partial_i = \frac{\partial {}}{\partial {x^i}}\end{aligned} \hspace{\stretch{1}}(5.65)

This alone implies half of Maxwell’s equations. To show this we consider

\begin{aligned}e^{m k i j} \partial_k F_{i j} = 0.\end{aligned} \hspace{\stretch{1}}(5.66)

This is the Bianchi identity. To demonstrate this identity, we’ll have to swap indexes, employ derivative commutation, and then swap indexes once more

\begin{aligned}e^{m k i j} \partial_k F_{i j} &= e^{m k i j} \partial_k (\partial_i A_j - \partial_j A_i) \\ &= 2 e^{m k i j} \partial_k \partial_i A_j \\ &= 2 e^{m k i j} \frac{1}{{2}} \left( \partial_k \partial_i A_j + \partial_i \partial_k A_j \right) \\ &= e^{m k i j} \partial_k \partial_i A_j e^{m i k j} \partial_k \partial_i A_j \\ &= (e^{m k i j} - e^{m k i j}) \partial_k \partial_i A_j \\ &= 0 \qquad \square\end{aligned}

This is the 4D analogue of

\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} f) = 0\end{aligned} \hspace{\stretch{1}}(5.67)

i.e.

\begin{aligned}e^{\alpha\beta\gamma} \partial_\beta \partial_\gamma f = 0\end{aligned} \hspace{\stretch{1}}(5.68)

Let’s do this explicitly, starting with

\begin{aligned}{\left\lVert{ F_{i j} }\right\rVert} = \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0.\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(5.69)

For the $m= 0$ case we have

\begin{aligned}\epsilon^{0 k i j} \partial_k F_{i j}&=\epsilon^{\alpha \beta \gamma} \partial_\alpha F_{\beta \gamma} \\ &= \epsilon^{\alpha \beta \gamma} \partial_\alpha (-\epsilon_{\beta \gamma \delta} B_\delta) \\ &= -\epsilon^{\alpha \beta \gamma} \epsilon_{\delta \beta \gamma }\partial_\alpha B_\delta \\ &= - 2 {\delta^\alpha}_\delta \partial_\alpha B_\delta \\ &= - 2 \partial_\alpha B_\alpha \end{aligned}

We must then have

\begin{aligned}\partial_\alpha B_\alpha = 0.\end{aligned} \hspace{\stretch{1}}(5.70)

This is just Gauss’s law for magnetism

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{B} = 0.\end{aligned} \hspace{\stretch{1}}(5.71)

Let’s do the spatial portion, for which we have three equations, one for each $\alpha$ of

\begin{aligned}e^{\alpha j k l} \partial_j F_{k l}&=e^{\alpha 0 \beta \gamma} \partial_0 F_{\beta \gamma}+e^{\alpha 0 \gamma \beta} \partial_0 F_{\gamma \beta}+e^{\alpha \beta 0 \gamma} \partial_\beta F_{0 \gamma}+e^{\alpha \beta \gamma 0} \partial_\beta F_{\gamma 0}+e^{\alpha \gamma 0 \beta} \partial_\gamma F_{0 \beta}+e^{\alpha \gamma \beta 0} \partial_\gamma F_{\beta 0} \\ &=2 \left( e^{\alpha 0 \beta \gamma} \partial_0 F_{\beta \gamma}+e^{\alpha \beta 0 \gamma} \partial_\beta F_{0 \gamma}+e^{\alpha \gamma 0 \beta} \partial_\gamma F_{0 \beta}\right) \\ &=2 e^{0 \alpha \beta \gamma} \left(-\partial_0 F_{\beta \gamma}+\partial_\beta F_{0 \gamma}- \partial_\gamma F_{0 \beta}\right)\end{aligned}

This implies

\begin{aligned}0 =-\partial_0 F_{\beta \gamma}+\partial_\beta F_{0 \gamma}- \partial_\gamma F_{0 \beta}\end{aligned} \hspace{\stretch{1}}(5.72)

Referring back to the previous expansions of 2.6 and 2.17, we have

\begin{aligned}0 =\partial_0 \epsilon_{\beta\gamma\mu} B_\mu+\partial_\beta E_\gamma- \partial_\gamma E_{\beta},\end{aligned} \hspace{\stretch{1}}(5.73)

or

\begin{aligned}\frac{1}{{c}} \frac{\partial {B_\alpha}}{\partial {t}} + (\boldsymbol{\nabla} \times \mathbf{E})_\alpha = 0.\end{aligned} \hspace{\stretch{1}}(5.74)

These are just the components of the Maxwell-Faraday equation

\begin{aligned}0 = \frac{1}{{c}} \frac{\partial {\mathbf{B}}}{\partial {t}} + \boldsymbol{\nabla} \times \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(5.75)

Transposition of mixed index tensor.

Is the transpose of a mixed index object just a substitution of the free indexes? This wasn’t obvious to me that it would be the case, especially since I’d made an error in some index gymnastics that had me temporarily convinced differently. However, working some examples clears the fog. For example let’s take the transpose of 3.37.

\begin{aligned}{\left\lVert{ {\delta^i}_j }\right\rVert}^\text{T} &= {\left\lVert{ O^{a i} O_{a j} }\right\rVert}^\text{T} \\ &= \left( {\left\lVert{ O^{j i} }\right\rVert} {\left\lVert{ O_{i j} }\right\rVert} \right)^\text{T} \\ &={\left\lVert{ O_{i j} }\right\rVert}^\text{T}{\left\lVert{ O^{j i} }\right\rVert}^\text{T} \\ &={\left\lVert{ O_{j i} }\right\rVert}{\left\lVert{ O^{i j} }\right\rVert} \\ &={\left\lVert{ O_{a i} O^{a j} }\right\rVert} \\ \end{aligned}

If the transpose of a mixed index tensor just swapped the indexes we would have

\begin{aligned}{\left\lVert{ {\delta^i}_j }\right\rVert}^\text{T} = {\left\lVert{ O_{a i} O^{a j} }\right\rVert} \end{aligned} \hspace{\stretch{1}}(6.76)

From this it does appear that all we have to do is switch the indexes and we will write

\begin{aligned}{\delta^j}_i = O_{a i} O^{a j} \end{aligned} \hspace{\stretch{1}}(6.77)

We can consider a more general operation

\begin{aligned}{\left\lVert{{A^i}_j}\right\rVert}^\text{T}&={\left\lVert{ A^{i m} g_{m j} }\right\rVert}^\text{T} \\ &={\left\lVert{ g_{i j} }\right\rVert}^\text{T}{\left\lVert{ A^{i j} }\right\rVert}^\text{T} \\ &={\left\lVert{ g_{i j} }\right\rVert}{\left\lVert{ A^{j i} }\right\rVert} \\ &={\left\lVert{ g_{i m} A^{j m} }\right\rVert} \\ &={\left\lVert{ {A^{j}}_i }\right\rVert}\end{aligned}

So we see that we do just have to swap indexes.

Transposition of lower index tensor.

We’ve saw above that we had

\begin{aligned}{\left\lVert{ {A^{i}}_j }\right\rVert}^\text{T} &= {\left\lVert{ {A_{j}}^i }\right\rVert} \\ {\left\lVert{ {A_{i}}^j }\right\rVert}^\text{T} &= {\left\lVert{ {A^{j}}_i }\right\rVert} \end{aligned} \hspace{\stretch{1}}(6.78)

which followed by careful treatment of the transposition in terms of $A^{i j}$ for which we defined a transpose operation. We assumed as well that

\begin{aligned}{\left\lVert{ A_{i j} }\right\rVert}^\text{T} = {\left\lVert{ A_{j i} }\right\rVert}.\end{aligned} \hspace{\stretch{1}}(6.80)

However, this does not have to be assumed, provided that $g^{i j} = g_{i j}$, and $(AB)^\text{T} = B^\text{T} A^\text{T}$. We see this by expanding this transposition in products of $A^{i j}$ and $\hat{G}$

\begin{aligned}{\left\lVert{ A_{i j} }\right\rVert}^\text{T}&= \left( {\left\lVert{g_{i j}}\right\rVert} {\left\lVert{ A^{i j} }\right\rVert} {\left\lVert{g_{i j}}\right\rVert} \right)^\text{T} \\ &= \left( {\left\lVert{g^{i j}}\right\rVert} {\left\lVert{ A^{i j} }\right\rVert} {\left\lVert{g^{i j}}\right\rVert} \right)^\text{T} \\ &= {\left\lVert{g^{i j}}\right\rVert}^\text{T} {\left\lVert{ A^{i j}}\right\rVert}^\text{T} {\left\lVert{g^{i j}}\right\rVert}^\text{T} \\ &= {\left\lVert{g^{i j}}\right\rVert} {\left\lVert{ A^{j i}}\right\rVert} {\left\lVert{g^{i j}}\right\rVert} \\ &= {\left\lVert{g_{i j}}\right\rVert} {\left\lVert{ A^{i j}}\right\rVert} {\left\lVert{g_{i j}}\right\rVert} \\ &= {\left\lVert{ A_{j i}}\right\rVert} \end{aligned}

It would be worthwhile to go through all of this index manipulation stuff and lay it out in a structured axiomatic form. What is the minimal set of assumptions, and how does all of this generalize to non-diagonal metric tensors (even in Euclidean spaces).

Translating the index expression of identity from Lorentz products to matrix form

A verification that the matrix expression 3.38, matches the index expression 3.37 as claimed is worthwhile. It would be easy to guess something similar like $\hat{O}^\text{T} \hat{G} \hat{O} \hat{G}$ is instead the matrix representation. That was in fact my first erroneous attempt to form the matrix equivalent, but is the transpose of 3.38. Either way you get an identity, but the indexes didn’t match.

Since we have $g^{i j} = g_{i j}$ which do we pick to do this verification? This appears to be dictated by requirements to match lower and upper indexes on the summed over index. This is probably clearest by example, so let’s expand the products on the LHS explicitly

\begin{aligned}{\left\lVert{ g^{i j} }\right\rVert} {\left\lVert{ {O^{i}}_j }\right\rVert} ^\text{T}{\left\lVert{ g_{i j} }\right\rVert}{\left\lVert{ {O^{i}}_j }\right\rVert} &=\left( {\left\lVert{ {O^{i}}_j }\right\rVert} {\left\lVert{ g^{i j} }\right\rVert} \right) ^\text{T}{\left\lVert{ g_{i j} }\right\rVert}{\left\lVert{ {O^{i}}_j }\right\rVert} \\ &=\left( {\left\lVert{ {O^{i}}_k g^{k j} }\right\rVert} \right) ^\text{T}{\left\lVert{ g_{i m} {O^{m}}_j }\right\rVert} \\ &={\left\lVert{ O^{i j} }\right\rVert} ^\text{T}{\left\lVert{ O_{i j} }\right\rVert} \\ &={\left\lVert{ O^{j i} }\right\rVert} {\left\lVert{ O_{i j} }\right\rVert} \\ &={\left\lVert{ O^{k i} O_{k j} }\right\rVert} \\ \end{aligned}

This matches the ${\left\lVert{{\delta^i}_j}\right\rVert}$ that we have on the RHS, and all is well.

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

PHY450H1S. Relativistic Electrodynamics Lecture 9 (Taught by Prof. Erich Poppitz). Dynamics in a vector field.

Posted by peeterjoot on February 3, 2011

Covering chapter 2 material from the text [1].

Covering lecture notes pp. 56.1-72: comments on mass, energy, momentum, and massless particles (56.1-58); particles in external fields: Lorentz scalar field (59-62); reminder of a vector field under spatial rotations (63) and a Lorentz vector field (64-65) [Tuesday, Feb. 1]; the action for a relativistic particle in an external 4-vector field (65-66); the equation of motion of a relativistic particle in an external electromagnetic (4-vector) field (67,68,73) [Wednesday, Feb. 2]; mathematical interlude: (69-72): on 3×3 antisymmetric matrices, 3-vectors, and totally antisymmetric 3-index tensor – please read by yourselves, preferably by Wed., Feb. 2 class! (this is important, well also soon need the 4-dimensional generalization)

More on the action.

Action for a relativistic particle in an external 4-scalar field

\begin{aligned}S = -m c \int ds - g \int ds \phi(x)\end{aligned} \hspace{\stretch{1}}(2.1)

Unfortunately we have no 4-vector scalar fields (at least for particles that are long lived and stable).

PICTURE: 3-vector field, some arrows in various directions.

PICTURE: A vector $\mathbf{A}$ in an $x,y$ frame, and a rotated (counterclockwise by angle $\alpha$) $x', y'$ frame with the components in each shown pictorially.

We have

\begin{aligned}A_x'(x', y') &= \cos\alpha A_x(x,y) + \sin\alpha A_y(x,y) \\ A_y'(x', y') &= -\sin\alpha A_x(x,y) + \cos\alpha A_y(x,y) \end{aligned} \hspace{\stretch{1}}(2.2)

\begin{aligned}\begin{bmatrix}A_x'(x', y') \\ A_y'(x', y')\end{bmatrix}=\begin{bmatrix}\cos\alpha A_x(x,y) & \sin\alpha A_y(x,y) \\ -\sin\alpha A_x(x,y) & \cos\alpha A_y(x,y) \end{bmatrix}\begin{bmatrix}A_x(x, y) \\ A_y(x, y)\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.4)

More generally we have

\begin{aligned}\begin{bmatrix}A_x'(x', y', z') \\ A_y'(x', y', z') \\ A_z'(x', y', z')\end{bmatrix}=\hat{O}\begin{bmatrix}A_x(x, y, z) \\ A_y(x, y, z) \\ A_z(x, y, z)\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.5)

Here $\hat{O}$ is an $SO(3)$ matrix rotating $x \rightarrow x'$

\begin{aligned}\mathbf{A}(\mathbf{x}) \cdot \mathbf{y} = \mathbf{A}'(\mathbf{x}') \cdot \mathbf{y}'\end{aligned} \hspace{\stretch{1}}(2.6)

\begin{aligned}\mathbf{A} \cdot \mathbf{B} = \text{invariant}\end{aligned} \hspace{\stretch{1}}(2.7)

A four vector field is $A^i(x)$, with $x = x^i, i = 0,1,2,3$ and we’d write

\begin{aligned}\begin{bmatrix}(x^0)' \\ (x^1)' \\ (x^2)' \\ (x^3)'\end{bmatrix}=\hat{O}\begin{bmatrix}x^0 \\ x^1 \\ x^2 \\ x^3\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.8)

Now $\hat{O}$ is an $SO(1,3)$ matrix. Our four vector field is then

\begin{aligned}\begin{bmatrix}(A^0)' \\ (A^1)' \\ (A^2)' \\ (A^3)'\end{bmatrix}=\hat{O}\begin{bmatrix}A^0 \\ A^1 \\ A^2 \\ A^3\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.9)

We have

\begin{aligned}A^i g_{ij} x^i = \text{invariant} = {A'}^i g_{ij} {x'}^i \end{aligned} \hspace{\stretch{1}}(2.10)

From electrodynamics we know that we have a scalar field, the electrostatic potential, and a vector field

What’s a plausible action?

\begin{aligned}\int ds x^i g_{ij} A^j\end{aligned} \hspace{\stretch{1}}(2.11)

This isn’t translation invariant.

\begin{aligned}\int ds x^i g_{ij} A^j\end{aligned} \hspace{\stretch{1}}(2.12)

Next simplest is

\begin{aligned}\int ds u^i g_{ij} A^j\end{aligned} \hspace{\stretch{1}}(2.13)

Could also do

\begin{aligned}\int ds A^i g_{ij} A^j\end{aligned} \hspace{\stretch{1}}(2.14)

but it turns out that this isn’t gauge invariant (to be defined and discussed in detail).

Note that the convention for this course is to write

\begin{aligned}u^i = \left( \gamma, \gamma \frac{\mathbf{v}}{c} \right) = \frac{dx^i}{ds}\end{aligned} \hspace{\stretch{1}}(2.15)

Where $u^i$ is dimensionless ($u^i u_i = 1$). Some authors use

\begin{aligned}u^i = \left( \gamma c, \gamma \mathbf{v} \right) = \frac{dx^i}{d\tau}\end{aligned} \hspace{\stretch{1}}(2.16)

The simplest action for a four vector field $A^i$ is then

\begin{aligned}S = - m c \int ds - \frac{e}{c} \int ds u^i A_i\end{aligned} \hspace{\stretch{1}}(2.17)

(Recall that $u^i A_i = u^i g_{ij} A^j$).

In this action $e$ is nothing but a Lorentz scalar, a property of the particle that describes how it “couples” (or “feels”) the electrodynamics field.

Similarily $mc$ is a Lorentz scalar which is a property of the particle (inertia).

It turns out that all the electric charges in nature are quantized, and there are some deep reasons (in magnetic monopoles exist) for this.

Another reason for charge quantitization apparently has to do with gauge invariance and associated compact groups. Poppitz is amusing himself a bit here, hinting at some stuff that we can eventually learn.

Returning to our discussion, we have

\begin{aligned}S = - m c \int ds - \frac{e}{c} \int ds u^i g_{ij} A^j\end{aligned} \hspace{\stretch{1}}(2.18)

with the electrodynamics four vector potential

\begin{aligned}A^i &= (\phi, \mathbf{A}) \\ u^i &= \left(\gamma, \gamma \frac{\mathbf{v}}{c} \right) \\ u^i g_{ij} A^j &= \gamma \phi - \gamma \frac{\mathbf{v} \cdot \mathbf{A}}{c}\end{aligned} \hspace{\stretch{1}}(2.19)

\begin{aligned}S &= - m c^2 \int dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - \frac{e}{c} \int c dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} \left( \gamma \phi - \gamma \frac{\mathbf{v}}{c} \cdot \mathbf{A} \right) \\ &= \int dt \left(- m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \phi(\mathbf{x}, t) + \frac{e}{c} \mathbf{v} \cdot \mathbf{A}(\mathbf{x}, t)\right) \\ \end{aligned}

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = \frac{m c^2}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} \frac{\mathbf{v}}{c^2} + \frac{e}{c} \mathbf{A}(\mathbf{x}, t)\end{aligned} \hspace{\stretch{1}}(2.22)

\begin{aligned}\frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = m \frac{d}{dt} (\gamma \mathbf{v}) + \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} + \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {x^\alpha}} v^\alpha\end{aligned} \hspace{\stretch{1}}(2.23)

Here $\alpha,\beta = 1,2,3$ and are summed over.

For the other half of the Euler-Lagrange equations we have

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {x^\alpha}} = - e \frac{\partial {\phi}}{\partial {x^\alpha}} + \frac{e}{c} v^\beta \frac{\partial {A^\beta}}{\partial {x^\alpha}}\end{aligned} \hspace{\stretch{1}}(2.24)

Equating these, and switching to coordinates for 2.23, we have

\begin{aligned}m \frac{d}{dt} (\gamma v^\alpha) + \frac{e}{c} \frac{\partial {A^\alpha}}{\partial {t}} + \frac{e}{c} \frac{\partial {A^\alpha}}{\partial {x^\beta}} v^\beta= - e \frac{\partial {\phi}}{\partial {x^\alpha}} + \frac{e}{c} v^\beta \frac{\partial {A^\beta}}{\partial {x^\alpha}}\end{aligned} \hspace{\stretch{1}}(2.25)

A final rearrangement yields

\begin{aligned}\frac{d}{dt} m \gamma v^\alpha = e \underbrace{\left( - \frac{1}{{c}} \frac{\partial {A^\alpha}}{\partial {t}} - \frac{\partial {\phi}}{\partial {x^\alpha}} \right)}_{E^\alpha} + \frac{e}{c} v^\beta \left( \frac{\partial {A^\beta}}{\partial {x^\alpha}} - \frac{\partial {A^\alpha}}{\partial {x^\beta}} \right)\end{aligned} \hspace{\stretch{1}}(2.26)

We can identity the second term with the magnetic field but first have to introduce antisymmetric matrices.

antisymmetric matrixes

\begin{aligned}M_{\mu\nu} &= \frac{\partial {A^\nu}}{\partial {x^\mu}} - \frac{\partial {A^\mu}}{\partial {x^\nu}} \\ &= \epsilon_{\mu\nu\lambda} B_\lambda,\end{aligned}

where

\begin{aligned}\epsilon_{\mu\nu\lambda} =\begin{array}{l l}0 & \quad \mbox{if any two indexes coincide} \\ 1 & \quad \mbox{for even permutations oflatex \mu\nu\lambda} \\ -1 & \quad \mbox{for odd permutations of $\mu\nu\lambda$}\end{array}\end{aligned} \hspace{\stretch{1}}(3.27)

Example:

\begin{aligned}\epsilon_{123} &= 1 \\ \epsilon_{213} &= -1 \\ \epsilon_{231} &= 1.\end{aligned}

We can show that

\begin{aligned}B_\lambda = \frac{1}{{2}} \epsilon_{\lambda\mu\nu} M_{\mu\nu}\end{aligned} \hspace{\stretch{1}}(3.28)

\begin{aligned}B_1 &= \frac{1}{{2}} ( \epsilon_{123} M_{23} + \epsilon_{132} M_{32}) \\ &= \frac{1}{{2}} ( M_{23} - M_{32}) \\ &= \partial_2 A_3 - \partial_3 A_2.\end{aligned}

Using

\begin{aligned}\epsilon_{\mu\nu\alpha} \epsilon_{\sigma\kappa\alpha} = \delta_{\mu\sigma} \delta_{\nu\kappa} - \delta_{\nu\sigma} \delta_{\mu\kappa},\end{aligned} \hspace{\stretch{1}}(3.29)

we can verify the identity 3.28 by expanding

\begin{aligned}\epsilon_{\mu\nu\lambda} B_\lambda&=\frac{1}{{2}} \epsilon_{\mu\nu\lambda} \epsilon_{\lambda\alpha\beta} M_{\alpha\beta} \\ &=\frac{1}{{2}} (\delta_{\mu\alpha} \delta_{\nu\beta} - \delta_{\nu\alpha} \delta_{\mu\beta})M_{\alpha\beta} \\ &=\frac{1}{{2}} (M_{\mu\nu} - M_{\nu\mu}) \\ &=M_{\mu\nu}\end{aligned}

Returning to the action evaluation we have

\begin{aligned}\frac{d}{dt} ( m \gamma v^\alpha ) = e E^\alpha + \frac{e}{c} \epsilon_{\alpha\beta\gamma} v^\beta B_\gamma,\end{aligned} \hspace{\stretch{1}}(3.30)

but

\begin{aligned}\epsilon_{\alpha\beta\gamma} B_\gamma = (\mathbf{v} \times \mathbf{B})_\alpha.\end{aligned} \hspace{\stretch{1}}(3.31)

So

\begin{aligned}\frac{d}{dt} ( m \gamma \mathbf{v} ) = e \mathbf{E} + \frac{e}{c} \mathbf{v} \times \mathbf{B}\end{aligned} \hspace{\stretch{1}}(3.32)

or

\begin{aligned}\frac{d}{dt} ( \mathbf{p} ) = e \left( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(3.33)

\paragraph{What is the energy component of the Lorentz force equation}

I asked this, not because I don’t know (I could answer this myself from $dp/d\tau = F \cdot v/c$, in the geometric algebra formalism, but I was curious if he had a way of determining this from what we’ve derived so far (intuitively I’d expect this to be possible). Answer was:

Observe that this is almost a relativisitic equation, but we aren’t going to get to the full equation yet. The energy component can be obtained from

\begin{aligned}\frac{du^0}{ds} = e F^{0j} u_j\end{aligned} \hspace{\stretch{1}}(3.34)

Since the full equation is

\begin{aligned}\frac{du^i}{ds} = e F^{ij} u_j\end{aligned} \hspace{\stretch{1}}(3.35)

“take with a grain of salt, may be off by sign, or factors of $c$”.

Also curious is that he claimed the energy component of this equation was not very important. Why would that be?

Gauge transformations.

Claim

\begin{aligned}S_{\text{interaction}} = - \frac{e}{c} \int ds u^i A_i\end{aligned} \hspace{\stretch{1}}(4.36)

changes by boundary terms only under

“gauge transformation” :

\begin{aligned}A_i = A_i' + \frac{\partial {\chi}}{\partial {x^i}}\end{aligned} \hspace{\stretch{1}}(4.37)

where $\chi$ is a Lorentz scalar. This ${\partial {}}/{\partial {x^i}}$ is the four gradient. Let’s see this

Therefore the equations of motion are the same in an external $A^i$ and ${A'}^i$.

Recall that the $\mathbf{E}$ and $\mathbf{B}$ fields do not change under such transformations. Let’s see how the action transforms

\begin{aligned}S &= - \frac{e}{c} \int ds u^i A_i \\ &= - \frac{e}{c} \int ds u^i \left( {A'}_i + \frac{\partial {\chi}}{\partial {x^i}} \right) \\ &= - \frac{e}{c} \int ds u^i {A'}_i - \frac{e}{c} \int ds \frac{dx^i}{ds} \frac{\partial {\chi}}{\partial {x^i}} \\ \end{aligned}

Observe that this last bit is just a chain rule expansion

\begin{aligned}\frac{d}{ds} \chi(x^0, x^1, x^2, x^3) &= \frac{\partial {\chi}}{\partial {x^0}}\frac{dx^0}{ds} + \frac{\partial {\chi}}{\partial {x^1}}\frac{dx^1}{ds} + \frac{\partial {\chi}}{\partial {x^2}}\frac{dx^2}{ds} + \frac{\partial {\chi}}{\partial {x^3}}\frac{dx^3}{ds} \\ &= \frac{\partial {\chi}}{\partial {x^i}} \frac{dx^i}{ds},\end{aligned}

so we have

\begin{aligned}S = - \frac{e}{c} \int ds u^i {A'}_i - \frac{e}{c} \int ds \frac{d \chi}{ds}.\end{aligned} \hspace{\stretch{1}}(4.38)

This allows the line integral to be evaluated, and we find that it only depends on the end points of the interval

\begin{aligned}S = - \frac{e}{c} \int ds u^i {A'}_i - \frac{e}{c} ( \chi(x_b) - \chi(x_a) ),\end{aligned} \hspace{\stretch{1}}(4.39)

which completes the proof of the claim that this gauge transformation results in an action difference that only depends on the end points of the interval.

\paragraph{What is the significance to this claim?}

References

[1] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.