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Archive for February 17th, 2011

PHY450H1S. Relativistic Electrodynamics Lecture 14 (Taught by Simon Freedman). Wave equation in Coulomb and Lorentz gauges.

Posted by peeterjoot on February 17, 2011

[Click here for a PDF of this post with nicer formatting]

Covering chapter 4 material from the text [1].

Covering lecture notes pp.103-114: the wave equation in the relativistic Lorentz gauge (114-114) [Tuesday, Feb. 15; Wednesday, Feb.16]…

Covering lecture notes pp. 114-127: reminder on wave equations (114); reminder on Fourier series and integral (115-117); Fourier expansion of the EM potential in Coulomb gauge and equation of motion for the spatial Fourier components (118-119); the general solution of Maxwell’s equations in vacuum (120-121) [Tuesday, Mar. 1]; properties of monochromatic plane EM waves (122-124); energy and energy flux of the EM field and energy conservation from the equations of motion (125-127) [Wednesday, Mar. 2]

Trying to understand “c”

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} &= 0 \\ \boldsymbol{\nabla} \times \mathbf{B} &= \frac{1}{{c}} \frac{\partial {\mathbf{E}}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(2.1)

Maxwell’s equations in a vacuum were

\begin{aligned}\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) &= \boldsymbol{\nabla}^2 \mathbf{A} -\frac{1}{{c}} \frac{\partial {}}{\partial {t}} \boldsymbol{\nabla} \phi - \frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}}{\partial t^2} \\ \boldsymbol{\nabla} \cdot \mathbf{E} &= - \boldsymbol{\nabla}^2 \phi - \frac{1}{{c}} \frac{\partial {\boldsymbol{\nabla} \cdot \mathbf{A}}}{\partial {t}} \end{aligned} \hspace{\stretch{1}}(2.3)

There’s a redundancy here since we can change $\phi$ and $\mathbf{A}$ without changing the EOM

\begin{aligned}(\phi, \mathbf{A}) \rightarrow (\phi', \mathbf{A}')\end{aligned} \hspace{\stretch{1}}(2.5)

with

\begin{aligned}\phi &= \phi' + \frac{1}{{c}} \frac{\partial {\chi}}{\partial {t}} \\ \mathbf{A} &= \mathbf{A}' - \boldsymbol{\nabla} \chi\end{aligned} \hspace{\stretch{1}}(2.6)

\begin{aligned}\chi(\mathbf{x}, t) = c \int dt \phi(\mathbf{x}, t)\end{aligned} \hspace{\stretch{1}}(2.8)

which gives

\begin{aligned}\phi' = 0\end{aligned} \hspace{\stretch{1}}(2.9)

\begin{aligned}(\phi, \mathbf{A}) \sim (\phi = 0, \mathbf{A}')\end{aligned} \hspace{\stretch{1}}(2.10)

Maxwell’s equations are now

\begin{aligned}\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}') &= \boldsymbol{\nabla}^2 \mathbf{A}' - \frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}'}{\partial t^2} \\ \frac{\partial {\boldsymbol{\nabla} \cdot \mathbf{A}'}}{\partial {t}} &= 0\end{aligned}

Can we make $\boldsymbol{\nabla} \cdot \mathbf{A}'' = 0$, while $\phi'' = 0$.

\begin{aligned}\underbrace{\phi}_{=0} &= \underbrace{\phi'}_{=0} + \frac{1}{{c}} \frac{\partial {\chi'}}{\partial {t}} \\ \end{aligned} \hspace{\stretch{1}}(2.11)

We need

\begin{aligned}\frac{\partial {\chi'}}{\partial {t}} = 0\end{aligned} \hspace{\stretch{1}}(2.13)

How about $\mathbf{A}'$

\begin{aligned}\mathbf{A}' = \mathbf{A}'' - \boldsymbol{\nabla} \chi'\end{aligned} \hspace{\stretch{1}}(2.14)

We want the divergence of $\mathbf{A}'$ to be zero, which means

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{A}' = \underbrace{\boldsymbol{\nabla} \cdot \mathbf{A}''}_{=0} - \boldsymbol{\nabla}^2 \chi'\end{aligned} \hspace{\stretch{1}}(2.15)

So we want

\begin{aligned}\boldsymbol{\nabla}^2 \chi' = \boldsymbol{\nabla} \cdot \mathbf{A}'\end{aligned} \hspace{\stretch{1}}(2.16)

Can we solve this?

Recall that in electrostatics we have

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = 4 \pi \rho\end{aligned} \hspace{\stretch{1}}(2.17)

and

\begin{aligned}\mathbf{E} = -\boldsymbol{\nabla} \phi\end{aligned} \hspace{\stretch{1}}(2.18)

which meant that we had

\begin{aligned}\boldsymbol{\nabla}^2 \phi = 4 \pi \rho\end{aligned} \hspace{\stretch{1}}(2.19)

This has the identical form (with $\phi \sim \chi$, and $4 \pi \rho \sim \boldsymbol{\nabla} \cdot \mathbf{A}'$).

While we aren’t trying to actually solve this (just show that it can be solved). One way to look at this problem is that it is just a Laplace equation, and we could utilize a Green’s function solution if desired.

On the Green’s function.

Recall that the Green’s function for the Laplacian was

\begin{aligned}G(\mathbf{x}, \mathbf{x}') = \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}}\end{aligned} \hspace{\stretch{1}}(2.20)

with the property

\begin{aligned}\boldsymbol{\nabla}^2 G(\mathbf{x}, \mathbf{x}') = \delta(\mathbf{x} - \mathbf{x}')\end{aligned} \hspace{\stretch{1}}(2.21)

Our LDE to solve by Green’s method is

\begin{aligned}\boldsymbol{\nabla}^2 \phi = 4 \pi \rho,\end{aligned} \hspace{\stretch{1}}(2.22)

We let this equation (after switching to primed coordinates) operate on the Green’s function

\begin{aligned}\int d^3 \mathbf{x}' {\boldsymbol{\nabla}'}^2 \phi(\mathbf{x}') G(\mathbf{x}, \mathbf{x}') =\int d^3 \mathbf{x}' 4 \pi \phi(\mathbf{x}') G(\mathbf{x}, \mathbf{x}').\end{aligned} \hspace{\stretch{1}}(2.23)

Assuming that the left action of the Green’s function on the test function $\phi(\mathbf{x}')$ is the same as the right action (i.e. $\phi(\mathbf{x}')$ and $G(\mathbf{x}, \mathbf{x}')$ commute), we have for the LHS

\begin{aligned}\int d^3 \mathbf{x}' {\boldsymbol{\nabla}'}^2 \phi(\mathbf{x}') G(\mathbf{x}, \mathbf{x}') &=\int d^3 \mathbf{x}' {\boldsymbol{\nabla}'}^2 G(\mathbf{x}, \mathbf{x}') \phi(\mathbf{x}') \\ &=\int d^3 \mathbf{x}' \delta(\mathbf{x} - \mathbf{x}') \phi(\mathbf{x}') \\ &=\phi(\mathbf{x}).\end{aligned}

Substitution of $G(\mathbf{x}, \mathbf{x}') = 1/{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}$ on the RHS then gives us the general solution

\begin{aligned}\phi(\mathbf{x}) = 4 \pi \int d^3 \mathbf{x}' \frac{\rho(\mathbf{x}') }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(2.24)

Back to Maxwell’s equations in vacuum.

What are the Maxwell’s vacuum equations now?

With the second gauge substitution we have

\begin{aligned}\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}'') &= \boldsymbol{\nabla}^2 \mathbf{A}'' - \frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}''}{\partial t^2} \\ \frac{\partial {\boldsymbol{\nabla} \cdot \mathbf{A}''}}{\partial {t}} &= 0\end{aligned}

but we can utilize

\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{A}) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) - \boldsymbol{\nabla}^2 \mathbf{A},\end{aligned} \hspace{\stretch{1}}(2.25)

to reduce Maxwell’s equations (after dropping primes) to just

\begin{aligned}\frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}''}{\partial t^2} - \Delta \mathbf{A} = 0\end{aligned} \hspace{\stretch{1}}(2.26)

where

\begin{aligned}\Delta = \boldsymbol{\nabla}^2 = \boldsymbol{\nabla} \cdot \boldsymbol{\nabla} = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial y^2}\end{aligned} \hspace{\stretch{1}}(2.27)

Note that for this to be correct we have to also explicitly include the gauge condition used. This particular gauge is called the \underline{Coulomb gauge}.

\begin{aligned}\phi &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{A}'' &= 0 \end{aligned} \hspace{\stretch{1}}(2.28)

Claim: EM waves propagate with speed $c$ and are transverse.

\paragraph{Note:} Is the Coulomb gauge Lorentz invariant?
\paragraph{No.} We can boost which will introduce a non-zero $\phi$.

The gauge that is Lorentz Invariant is the “Lorentz gauge”. This one uses

\begin{aligned}\partial_i A^i = 0\end{aligned} \hspace{\stretch{1}}(3.30)

Recall that Maxwell’s equations are

\begin{aligned}\partial_i F^{ij} = j^j = 0\end{aligned} \hspace{\stretch{1}}(3.31)

where

\begin{aligned}\partial_i &= \frac{\partial {}}{\partial {x^i}} \\ \partial^i &= \frac{\partial {}}{\partial {x_i}}\end{aligned} \hspace{\stretch{1}}(3.32)

Writing out the equations in terms of potentials we have

\begin{aligned}0 &= \partial_i (\partial^i A^j - \partial^j A^i) \\ &= \partial_i \partial^i A^j - \partial_i \partial^j A^i \\ &= \partial_i \partial^i A^j - \partial^j \partial_i A^i \\ \end{aligned}

So, if we pick the gauge condition $\partial_i A^i = 0$, we are left with just

\begin{aligned}0 = \partial_i \partial^i A^j\end{aligned} \hspace{\stretch{1}}(3.34)

Can we choose ${A'}^i$ such that $\partial_i A^i = 0$?

Our gauge condition is

\begin{aligned}A^i = {A'}^i + \partial^i \chi\end{aligned} \hspace{\stretch{1}}(3.35)

Hit it with a derivative for

\begin{aligned}\partial_i A^i = \partial_i {A'}^i + \partial_i \partial^i \chi\end{aligned} \hspace{\stretch{1}}(3.36)

If we want $\partial_i A^i = 0$, then we have

\begin{aligned}-\partial_i {A'}^i = \partial_i \partial^i \chi = \left( \frac{1}{{c^2}} \frac{\partial^2}{\partial t^2} - \Delta \right) \chi\end{aligned} \hspace{\stretch{1}}(3.37)

This is the physicist proof. Yes, it can be solved. To really solve this, we’d want to use Green’s functions. I seem to recall the Green’s function is a retarded time version of the Laplacian Green’s function, and we can figure that exact form out by switching to a Fourier frequency domain representation.

Anyways. Returning to Maxwell’s equations we have

\begin{aligned}0 &= \partial_i \partial^i A^j \\ 0 &= \partial_i A^i ,\end{aligned} \hspace{\stretch{1}}(3.38)

where the first is Maxwell’s equation, and the second is our gauge condition.

Observe that the gauge condition is now a Lorentz scalar.

\begin{aligned}\partial^i A_i \rightarrow \partial^j {O_j}^i {O_i}^k A_k\end{aligned} \hspace{\stretch{1}}(3.40)

But the Lorentz transform matrices multiply out to identity, in the same way that they do for the transformation of a plain old four vector dot product $x^i y_i$.

What happens with a Massive vector field?

\begin{aligned}S = \int d^4 x \left( \frac{1}{{4}} F^{ij} F_{ij} + \frac{m^2}{2} A^i A_i \right)\end{aligned} \hspace{\stretch{1}}(4.41)

An aside on units

“Note that this action is expressed in dimensions where $\hbar = c = 1$, making the action is unit-less (energy and time are inverse units of each other). The $d^4x$ has units of $m^{-4}$ (since $[x] = \hbar/mc$), so $F$ has units of $m^2$, and then $A$ has units of mass. Therefore $d^4x A A$ has units of $m^{-2}$ and therefore you need something that has units of $m^2$ to make the action unit-less. When you don’t take $c=1$, then you’ve got to worry about those factors, but I think you’ll see it works out fine.”

For what it’s worth, I can adjust the units of this action to those that we’ve used in class with,

\begin{aligned}S = \int d^4 x \left( -\frac{1}{{16 \pi c}} F^{ij} F_{ij} - \frac{m^2 c^2}{8 \hbar^2} A^i A_i \right)\end{aligned} \hspace{\stretch{1}}(4.42)

Back to the problem.

The variation of the field invariant is

\begin{aligned}\delta (F_{ij} F^{ij})&=2 (\delta F_{ij}) F^{ij}) \\ &=2 (\delta(\partial_i A_j -\partial_j A_i)) F^{ij}) \\ &=2 (\partial_i \delta(A_j) -\partial_j \delta(A_i)) F^{ij}) \\ &=4 F^{ij} \partial_i \delta(A_j) \\ &=4 \partial_i (F^{ij} \delta(A_j)) - 4 (\partial_i F^{ij}) \delta(A_j).\end{aligned}

Variation of the $A^2$ term gives us

\begin{aligned}\delta (A^j A_j) = 2 A^j \delta(A_j),\end{aligned} \hspace{\stretch{1}}(4.43)

so we have

\begin{aligned}0 &= \delta S \\ &= \int d^4 x \delta(A_j) \left( -\partial_i F^{ij} + m^2 A^j \right)+ \int d^4 x \partial_i (F^{ij} \delta(A_j))\end{aligned}

The last integral vanishes on the boundary with the assumption that $\delta(A_j) = 0$ on that boundary.

Since this must be true for all variations, this leaves us with

\begin{aligned}\partial_i F^{ij} = m^2 A^j\end{aligned} \hspace{\stretch{1}}(4.44)

The RHS can be expanded into wave equation and divergence parts

\begin{aligned}\partial_i F^{ij}&=\partial_i (\partial^i A^j - \partial^j A^i) \\ &=(\partial_i \partial^i) A^j - \partial^j (\partial_i A^i) \\ \end{aligned}

With $\square$ for the wave equation operator

\begin{aligned}\square = \partial_i \partial^i = \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta,\end{aligned} \hspace{\stretch{1}}(4.45)

we can manipulate the EOM to pull out an $A_i$ factor

\begin{aligned}0 &= \left( \square -m^2 \right) A^j - \partial^j (\partial_i A^i) \\ &= \left( \square -m^2 \right) g^{ij} A_i - \partial^j (\partial^i A_i) \\ &= \left( \left( \square -m^2 \right) g^{ij} - \partial^j \partial^i \right) A_i.\end{aligned}

If we hit this with a derivative we get

\begin{aligned}0 &= \partial_j \left( \left( \square -m^2 \right) g^{ij} - \partial^j \partial^i \right) A_i \\ &= \left( \left( \square -m^2 \right) \partial^i - \partial_j \partial^j \partial^i \right) A_i \\ &= \left( \left( \square -m^2 \right) \partial^i - \square \partial^i \right) A_i \\ &= \left( \square -m^2 - \square \right) \partial^i A_i \\ &= -m^2 \partial^i A_i \\ \end{aligned}

Since $m$ is presumed to be non-zero here, this means that the Lorentz gauge is already chosen for us by the equations of motion.

References

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