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# Archive for April, 2011

## Why is the speed of light the limiting velocity that a particle can travel. Attempt to answer with (almost) no math.

Posted by peeterjoot on April 30, 2011

” I’ve googled some data about “c”, the speed of light, but found really no data about why this is the speed limit that a particle can travel. It is just stated to be so but no allusions ( except Einstein’s smoke and mirrors) were indicated as to why this was the case.”

Here was my attempt at answering.  I liked it and share it here.

For light itself, this is an observed quality.  Light in vacuum hasn’t been observed travelling at any other velocity.  There have been lots of experiments attempting to show otherwise, and none of them worked.  Where people get hung up is that they think of light as a thrown object, like a ball tossed out of a car window.  If you are (the passenger) in a convertible (let’s say this is a Porsche for fun) that is zipping down the street at 90mph, and toss a baseball at 50mph somebody on the sidewalk will see that ball travelling at 140mph.  If you through it backwards, an observer will see it moving at just 40mph.  We have lots of experience with “addition of velocities like this”.

With light, something that’s already tricky to measure because it is so fast, we don’t have a lot of experience.  However, when an experiment like this is done carefully, from a fast moving object, it is carries a light pointing in its direction of motion, an observer will see the light emitted go at the speed of light, regardless of the speed of the object.  If you imagine a rocket that’s moving at 1/2 the speed of light itself, it’s headlight emits light that seen to travel at c (not 1.5c), and it’s taillight is seen to emit light that travels at c (not 0.5c).  This is very different from the ball and the car example.

The way that I think of this is that the light was not, then is.  At that point of creation, it moves at speed c, but this motion is never associated with the speed of any nearby objects (ie.  this light at the point of creation was never carried by the object that’s travelling in its vicinity).

Since we have this phenomena that always appears to have a fixed velocity, we can use it as both a measuring stick and as a clock.  If light is bounced off of a mirror and returns (assuming that the delay for the interaction with the mirror is negligible), then if we’ve counted out the distance to the mirror (say d), then we would know that an interval of time 2d/c has passed.  Similarly, if clocks are synchronized at two points in space, with you standing by one clock and your buddy standing by another with an agreement to turn on his light at 12:00, and you observe that his light signal gets to you at 12:00:000003, then the distance between you is .0003 seconds x c = 90km.  The basic idea is that you can use light signals as a mechanism for both time and space depending on what you know.

What Einstein did was argue very carefully that there is no such concept as simultaneity.  Events that are simultaneous can be observer dependent and we have to alter the way that we describe time and space to account for this.  Light ends up having a place as a measuring stick for this new way of describing time and space.  When space and time and motions in space and time are treated carefully, to ensure that notions of time and distance are related to the observer and the observers own motion, there are some interesting consequences.

One of these consequences is that velocities don’t add like the ball and car example.  There is a very very small correction required, and if one were to able to measure the speed of the ball thrown forward over the front windshield of the Porsche, you would find that it is actually a tiny bit less than 140mph.  That correction gets bigger and bigger, as the speeds of the objects are increased.

If you had a rocket turbocharger on the Porche and was able to launch it into space at .75 c, and then tossed the ball into a turbocharged baseball pitching device that could propel it at 0.5 c, then you’ll see the ball moving at 0.5 c, but an observer will NOT see it moving at 0.75c+0.5c=1.25c.  It will actually be observed to travel at less than 1.0 c.  There is a lot of math involved and this can be thought of as Smoke and Mirrors if you like (and perhaps justifiably so since it’s not terribly easy math to learn), but again this is something that is backed up by experiment.  It takes a lot of care to measure things when interactions happen at such high velocities (velocities that are significant factions of the speed of light), but when you do velocities are never seen to add to more than the speed of light, regardless of the motion of the observed interactions.  This is why in a very real way: it has never been observed to be otherwise.

## My first arxiv submission. Change of basis and Gram-Schmidt orthonormalization in special relativity

Posted by peeterjoot on April 29, 2011

Now that I have an academic email address I was able to make an arxiv submission (I’d tried previously and been auto-rejected) :

Change of basis and Gram-Schmidt orthonormalization in special relativity

This is based on a tutorial from our relativistic electrodynamics class, which covered non-internal relativistic systems. Combining what I learned from that with some concepts I learned from ‘Geometric Algebra for Physicists’ (particularly reciprocal frames) I was able to write up some notes that took those ideas plus basic linear algebra (the Graham-Schmidt procedure) and apply them to relativity and/or non-orthonormal Euclidean bases. How to do projections onto non-orthonormal Euclidean bases isn’t taught in Algebra I, but once you figure out that the same thing works for SR.

Will anybody read it? I don’t know … but I had fun writing it.

## Just Energy cancellation. Finally!

Posted by peeterjoot on April 29, 2011

I’d blogged previously about my objection to the Just Energy contract that I’d unknowningly ended up with:

http://peeterjoot.wordpress.com/2011/01/04/just-energy-contract-complaints-going-nowhere/

http://peeterjoot.wordpress.com/2010/12/30/1693/

Finally my complaint to the Ontario Energy Board about this paid off, and I’ve got the cancellation that I should have just been given straight up in the first place (in my opinion).  Both Just Energy and Enbridge gas sent me notifications that Just Energy will no longer be my “supplier”.

It was interesting to see that the attempts of this company to perpetuate the scam did not end with the initial Ontario Energy Board complaint.  They contacted me once they received the complaint, basically told me that I had no other option than to take their inflated rate “market rate program” (where they charge a bit more than market value for the supply cost and add surcharges to that).  This was better than their extortion rate cancellation fees, and better than  their extortion rate fixed rate fees, so I accepted that, but told them straight up that I was being coerced into doing so, and that the ethical and reasonable thing to do would be to just cancel the contract that I considered invalid.  I also told them that I’d be following up with the OEB about this.

When I did talk to somebody in the OEB it turned out that Just Energy claimed that they’d settled the matter with me, and that I was happy with the resolution!  The complaint went to a higher level OEB representitive for the conflict resolution with gas suppliers.  I hadn’t heard anything from them about this since, but they quietly worked away on this in the background and it’s now deal with.  My thanks to the OEB!

## PHY450HS1: Relativistic electrodynamics: some exam reflection.

Posted by peeterjoot on April 28, 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.)]

# Charged particle in a circle.

From the 2008 PHY353 exam, given a particle of charge $q$ moving in a circle of radius $a$ at constant angular frequency $\omega$.

\begin{itemize}
\item Find the Lienard-Wiechert potentials for points on the z-axis.
\item Find the electric and magnetic fields at the center.
\end{itemize}

When I tried this I did it for points not just on the z-axis. It turns out that we also got this question on the exam (but stated slightly differently). Since I’ll not get to see my exam solution again, let’s work through this at a leisurely rate, and see if things look right. The problem as stated in this old practice exam is easier since it doesn’t say to calculate the fields from the four potentials, so there was nothing preventing one from just grinding away and plugging stuff into the Lienard-Wiechert equations for the fields (as I did when I tried it for practice).

## The potentials.

Let’s set up our coordinate system in cylindrical coordinates. For the charged particle and the point that we measure the field, with $i = \mathbf{e}_1 \mathbf{e}_2$

\begin{aligned}\mathbf{x}(t) &= a \mathbf{e}_1 e^{i \omega t} \\ \mathbf{r} &= z \mathbf{e}_3 + \rho \mathbf{e}_1 e^{i \phi}\end{aligned} \hspace{\stretch{1}}(1.1)

Here I’m using the geometric product of vectors (if that’s unfamiliar then just substitute

\begin{aligned}\{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\} \rightarrow \{\sigma_1, \sigma_2, \sigma_3\}\end{aligned} \hspace{\stretch{1}}(1.3)

We can do that since the Pauli matrices also have the same semantics (with a small difference since the geometric square of a unit vector is defined as the unit scalar, whereas the Pauli matrix square is the identity matrix). The semantics we require of this vector product are just $\mathbf{e}_\alpha^2 = 1$ and $\mathbf{e}_\alpha \mathbf{e}_\beta = - \mathbf{e}_\beta \mathbf{e}_\alpha$ for any $\alpha \ne \beta$.

I’ll also be loose with notation and use $\text{Real}(X) = \left\langle{{X}}\right\rangle$ to select the scalar part of a multivector (or with the Pauli matrices, the portion proportional to the identity matrix).

Our task is to compute the Lienard-Wiechert potentials. Those are

\begin{aligned}A^0 &= \frac{q}{R^{*}} \\ \mathbf{A} &= A^0 \frac{\mathbf{v}}{c},\end{aligned} \hspace{\stretch{1}}(1.4)

where

\begin{aligned}\mathbf{R} &= \mathbf{r} - \mathbf{x}(t_r) \\ R = {\left\lvert{\mathbf{R}}\right\rvert} &= c (t - t_r) \\ R^{*} &= R - \frac{\mathbf{v}}{c} \cdot \mathbf{R} \\ \mathbf{v} &= \frac{d\mathbf{x}}{dt_r}.\end{aligned} \hspace{\stretch{1}}(1.6)

We’ll need (eventually)

\begin{aligned}\mathbf{v} &= a \omega \mathbf{e}_2 e^{i \omega t_r} = a \omega ( -\sin \omega t_r, \cos\omega t_r, 0) \\ \dot{\mathbf{v}} &= -a \omega^2 \mathbf{e}_1 e^{i \omega t_r} = -a \omega^2 (\cos\omega t_r, \sin\omega t_r, 0)\end{aligned} \hspace{\stretch{1}}(1.10)

and also need our retarded distance vector

\begin{aligned}\mathbf{R} = z \mathbf{e}_3 + \mathbf{e}_1 (\rho e^{i \phi} - a e^{i \omega t_r} ),\end{aligned} \hspace{\stretch{1}}(1.12)

From this we have

\begin{aligned}R^2 &= z^2 + {\left\lvert{\mathbf{e}_1 (\rho e^{i \phi} - a e^{i \omega t_r} )}\right\rvert}^2 \\ &= z^2 + \rho^2 + a^2 - 2 \rho a (\mathbf{e}_1 \rho e^{i \phi}) \cdot (\mathbf{e}_1 e^{i \omega t_r}) \\ &= z^2 + \rho^2 + a^2 - 2 \rho a \text{Real}( e^{ i(\phi - \omega t_r) } ) \\ &= z^2 + \rho^2 + a^2 - 2 \rho a \cos(\phi - \omega t_r)\end{aligned}

So

\begin{aligned}R = \sqrt{z^2 + \rho^2 + a^2 - 2 \rho a \cos( \phi - \omega t_r ) }.\end{aligned} \hspace{\stretch{1}}(1.13)

Next we need

\begin{aligned}\mathbf{R} \cdot \mathbf{v}/c&= (z \mathbf{e}_3 + \mathbf{e}_1 (\rho e^{i \phi} - a e^{i \omega t_r} )) \cdot \left(a \frac{\omega}{c} \mathbf{e}_2 e^{i \omega t_r} \right) \\ &=a \frac{\omega }{c}\text{Real}(i (\rho e^{-i \phi} - a e^{-i \omega t_r} ) e^{i \omega t_r} ) \\ &=a \frac{\omega }{c}\rho \text{Real}( i e^{-i \phi + i \omega t_r} ) \\ &=a \frac{\omega }{c}\rho \sin(\phi - \omega t_r)\end{aligned}

So we have

\begin{aligned}R^{*} = \sqrt{z^2 + \rho^2 + a^2 - 2 \rho a \cos( \phi - \omega t_r ) }-a \frac{\omega }{c} \rho \sin(\phi - \omega t_r)\end{aligned} \hspace{\stretch{1}}(1.14)

Writing $k = \omega/c$, and having a peek back at 1.4, our potentials are now solved for

\begin{aligned}\boxed{\begin{aligned}A^0 &= \frac{q}{\sqrt{z^2 + \rho^2 + a^2 - 2 \rho a \cos( \phi - k c t_r ) }} \\ \mathbf{A} &= A^0 a k ( -\sin k c t_r, \cos k c t_r, 0).\end{aligned}}\end{aligned} \hspace{\stretch{1}}(1.24)

The caveat is that $t_r$ is only specified implicitly, according to

\begin{aligned}\boxed{c t_r = c t - \sqrt{z^2 + \rho^2 + a^2 - 2 \rho a \cos( \phi - k c t_r ) }.}\end{aligned} \hspace{\stretch{1}}(1.16)

There doesn’t appear to be much hope of solving for $t_r$ explicitly in closed form.

## General fields for this system.

With

\begin{aligned}\mathbf{R}^{*} = \mathbf{R} - \frac{\mathbf{v}}{c} R,\end{aligned} \hspace{\stretch{1}}(1.17)

the fields are

\begin{aligned}\boxed{\begin{aligned}\mathbf{E} &= q (1 - \mathbf{v}^2/c^2) \frac{\mathbf{R}^{*}}{{R^{*}}^3} + \frac{q}{{R^{*}}^3} \mathbf{R} \times (\mathbf{R}^{*} \times \dot{\mathbf{v}}/c^2) \\ \mathbf{B} &= \frac{\mathbf{R}}{R} \times \mathbf{E}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(1.18)

In there we have

\begin{aligned}1 - \mathbf{v}^2/c^2 = 1 - a^2 \frac{\omega^2}{c^2} = 1 - a^2 k^2\end{aligned} \hspace{\stretch{1}}(1.19)

and

\begin{aligned}\mathbf{R}^{*} &= z \mathbf{e}_3 + \mathbf{e}_1 (\rho e^{i \phi} - a e^{i k c t_r} )-a k \mathbf{e}_2 e^{i k c t_r} R \\ &= z \mathbf{e}_3 + \mathbf{e}_1 (\rho e^{i \phi} - a (1 - k R i) e^{i k c t_r} )\end{aligned}

Writing this out in coordinates isn’t particularly illuminating, but can be done for completeness without too much trouble

\begin{aligned}\mathbf{R}^{*} = ( \rho \cos\phi - a \cos t_r + a k R \sin t_r, \rho \sin\phi - a \sin t_r - a k R \cos t_r, z )\end{aligned} \hspace{\stretch{1}}(1.20)

In one sense the problem could be considered solved, since we have all the pieces of the puzzle. The outstanding question is whether or not the resulting mess can be simplified at all. Let’s see if the cross product reduces at all. Using

\begin{aligned}\mathbf{R} \times (\mathbf{R}^{*} \times \dot{\mathbf{v}}/c^2) =\mathbf{R}^{*} (\mathbf{R} \cdot \dot{\mathbf{v}}/c^2) - \frac{\dot{\mathbf{v}}}{c^2}(\mathbf{R} \cdot \mathbf{R}^{*})\end{aligned} \hspace{\stretch{1}}(1.21)

Perhaps one or more of these dot products can be simplified? One of them does reduce nicely

\begin{aligned}\mathbf{R}^{*} \cdot \mathbf{R} &= ( \mathbf{R} - R \mathbf{v}/c ) \cdot \mathbf{R} \\ &= R^2 - (\mathbf{R} \cdot \mathbf{v}/c) R \\ &= R^2 - R a k \rho \sin(\phi - k c t_r) \\ &= R(R - a k \rho \sin(\phi - k c t_r))\end{aligned}

\begin{aligned}\mathbf{R} \cdot \dot{\mathbf{v}}/c^2&=\Bigl(z \mathbf{e}_3 + \mathbf{e}_1 (\rho e^{i \phi} - a e^{i \omega t_r} ) \Bigr) \cdot(-a k^2 \mathbf{e}_1 e^{i \omega t_r} ) \\ &=- a k^2 \left\langle{{\mathbf{e}_1 (\rho e^{i \phi} - a e^{i \omega t_r} ) \mathbf{e}_1 e^{i \omega t_r} ) }}\right\rangle \\ &=- a k^2 \left\langle{{(\rho e^{i \phi} - a e^{i \omega t_r} ) e^{-i \omega t_r} ) }}\right\rangle \\ &=- a k^2 \left\langle{{\rho e^{i \phi - i \omega t_r} - a }}\right\rangle \\ &=- a k^2 ( \rho \cos(\phi - k c t_r) - a )\end{aligned}

Putting this cross product back together we have

\begin{aligned}\mathbf{R} \times (\mathbf{R}^{*} \times \dot{\mathbf{v}}/c^2)&=a k^2 ( a -\rho \cos(\phi - k c t_r) ) \mathbf{R}^{*} +a k^2 \mathbf{e}_1 e^{i k c t_r} R(R - a k \rho \sin(\phi - k c t_r)) \\ &=a k^2 ( a -\rho \cos(\phi - k c t_r) ) \Bigl(z \mathbf{e}_3 + \mathbf{e}_1 (\rho e^{i \phi} - a (1 - k R i) e^{i k c t_r} )\Bigr) \\ &\qquad +a k^2 R \mathbf{e}_1 e^{i k c t_r} (R - a k \rho \sin(\phi - k c t_r)) \end{aligned}

Writing

\begin{aligned}\phi_r = \phi - k c t_r,\end{aligned} \hspace{\stretch{1}}(1.22)

this can be grouped into similar terms

\begin{aligned}\begin{aligned}\mathbf{R} \times (\mathbf{R}^{*} \times \dot{\mathbf{v}}/c^2)&=a k^2 (a - \rho \cos\phi_r) z \mathbf{e}_3 \\ &+ a k^2 \mathbf{e}_1(a - \rho \cos\phi_r) \rho e^{i\phi} \\ &+ a k^2 \mathbf{e}_1\left(-a (a - \rho \cos\phi_r) (1 - k R i)+ R(R - a k \rho \sin \phi_r)\right) e^{i k c t_r}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.23)

The electric field pieces can now be collected. Not expanding out the $R^{*}$ from 1.14, this is

\begin{aligned}\begin{aligned}\mathbf{E} &= \frac{q}{(R^{*})^3} z \mathbf{e}_3\Bigl( 1 - a \rho k^2 \cos\phi_r \Bigr) \\ &+\frac{q}{(R^{*})^3} \rho\mathbf{e}_1 \Bigl(1 - a \rho k^2 \cos\phi_r \Bigr) e^{i\phi} \\ &+\frac{q}{(R^{*})^3} a \mathbf{e}_1\left(-\Bigl( 1 + a k^2 (a - \rho \cos\phi_r) \Bigr) (1 - k R i)(1 - a^2 k^2)+ k^2 R(R - a k \rho \sin \phi_r)\right) e^{i k c t_r}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.24)

Along the z-axis where $\rho = 0$ what do we have?

\begin{aligned}R = \sqrt{z^2 + a^2 } \end{aligned} \hspace{\stretch{1}}(1.25)

\begin{aligned}A^0 = \frac{q}{R} \end{aligned} \hspace{\stretch{1}}(1.26)

\begin{aligned}\mathbf{A} = A^0 a k \mathbf{e}_2 e^{i k c t_r } \end{aligned} \hspace{\stretch{1}}(1.27)

\begin{aligned}c t_r = c t - \sqrt{z^2 + a^2 } \end{aligned} \hspace{\stretch{1}}(1.28)

\begin{aligned}\begin{aligned}\mathbf{E} &= \frac{q}{R^3} z \mathbf{e}_3 \\ &+\frac{q}{R^3} a \mathbf{e}_1\left(-( 1 - a^4 k^4 ) (1 - k R i)+ k^2 R^2 \right) e^{i k c t_r} \end{aligned}\end{aligned} \hspace{\stretch{1}}(1.29)

\begin{aligned}\mathbf{B} = \frac{ z \mathbf{e}_3 - a \mathbf{e}_1 e^{i k c t_r}}{R} \times \mathbf{E}\end{aligned} \hspace{\stretch{1}}(1.30)

The magnetic term here looks like it can be reduced a bit.

## An approximation near the center.

Unlike the old exam I did, where it didn’t specify that the potentials had to be used to calculate the fields, and the problem was reduced to one of algebraic manipulation, our exam explicitly asked for the potentials to be used to calculate the fields.

There was also the restriction to compute them near the center. Setting $\rho = 0$ so that we are looking only near the z-axis, we have

\begin{aligned}A^0 &= \frac{q}{\sqrt{z^2 + a^2}} \\ \mathbf{A} &= \frac{q a k \mathbf{e}_2 e^{i k c t_r} }{\sqrt{z^2 + a^2}} = \frac{q a k (-\sin k c t_r, \cos k c t_r, 0)}{\sqrt{z^2 + a^2}} \\ t_r &= t - R/c = t - \sqrt{z^2 + a^2}/c\end{aligned} \hspace{\stretch{1}}(1.31)

Now we are set to calculate the electric and magnetic fields directly from these. Observe that we have a spatial dependence in due to the $t_r$ quantities and that will have an effect when we operate with the gradient.

In the exam I’d asked Simon (our TA) if this question was asking for the fields at the origin (ie: in the plane of the charge’s motion in the center) or along the z-axis. He said in the plane. That would simplify things, but perhaps too much since $A^0$ becomes constant (in my exam attempt I somehow fudged this to get what I wanted for the $v = 0$ case, but that must have been wrong, and was the result of rushed work).

Let’s now proceed with the field calculation from these potentials

\begin{aligned}\mathbf{E} &= - \boldsymbol{\nabla} A^0 - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \\ \mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(1.34)

For the electric field we need

\begin{aligned}\boldsymbol{\nabla} A^0 &= q \mathbf{e}_3 \partial_z (z^2 + a^2)^{-1/2} \\ &= -q \mathbf{e}_3 \frac{z}{(\sqrt{z^2 + a^2})^3},\end{aligned}

and

\begin{aligned}\frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} =\frac{q a k^2 \mathbf{e}_2 \mathbf{e}_1 \mathbf{e}_2 e^{i k c t_r} }{\sqrt{z^2 + a^2}}.\end{aligned} \hspace{\stretch{1}}(1.36)

Putting these together, our electric field near the z-axis is

\begin{aligned}\mathbf{E} = q \mathbf{e}_3 \frac{z}{(\sqrt{z^2 + a^2})^3}+\frac{q a k^2 \mathbf{e}_1 e^{i k c t_r} }{\sqrt{z^2 + a^2}}.\end{aligned} \hspace{\stretch{1}}(1.37)

(another mistake I made on the exam, since I somehow fooled myself into forcing what I knew had to be in the gradient term, despite having essentially a constant scalar potential (having taken $z = 0$)).

What do we get for the magnetic field. In that case we have

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{A}(z)&=\mathbf{e}_\alpha \times \partial_\alpha \mathbf{A} \\ &=\mathbf{e}_3 \times \partial_z \frac{q a k \mathbf{e}_2 e^{i k c t_r} }{\sqrt{z^2 + a^2}} \\ &=\mathbf{e}_3 \times (\mathbf{e}_2 e^{i k c t_r} ) q a k \frac{\partial {}}{\partial {z}} \frac{1}{{\sqrt{z^2 + a^2}}} +q a k \frac{1}{{\sqrt{z^2 + a^2}}} \mathbf{e}_3 \times (\mathbf{e}_2 \partial_z e^{i k c t_r} ) \\ &=-\mathbf{e}_3 \times (\mathbf{e}_2 e^{i k c t_r} ) q a k \frac{z}{(\sqrt{z^2 + a^2})^3} +q a k \frac{1}{{\sqrt{z^2 + a^2}}} \mathbf{e}_3 \times \left( \mathbf{e}_2 \mathbf{e}_1 \mathbf{e}_2 k c e^{i k c t_r} \partial_z ( t - \sqrt{z^a + a^2}/c ) \right) \\ &=-\mathbf{e}_3 \times (\mathbf{e}_2 e^{i k c t_r} ) q a k \frac{z}{(\sqrt{z^2 + a^2})^3} -q a k^2 \frac{z}{z^2 + a^2} \mathbf{e}_3 \times \left( \mathbf{e}_1 k e^{i k c t_r} \right) \\ &=-\frac{q a k z \mathbf{e}_3}{z^2 + a^2} \times \left( \frac{ \mathbf{e}_2 e^{i k c t_r} }{\sqrt{z^2 + a^2}} + k \mathbf{e}_1 e^{i k c t_r} \right)\end{aligned}

For the direction vectors in the cross products above we have

\begin{aligned}\mathbf{e}_3 \times (\mathbf{e}_2 e^{i \mu})&=\mathbf{e}_3 \times (\mathbf{e}_2 \cos\mu - \mathbf{e}_1 \sin\mu) \\ &=-\mathbf{e}_1 \cos\mu - \mathbf{e}_2 \sin\mu \\ &=-\mathbf{e}_1 e^{i \mu}\end{aligned}

and

\begin{aligned}\mathbf{e}_3 \times (\mathbf{e}_1 e^{i \mu})&=\mathbf{e}_3 \times (\mathbf{e}_1 \cos\mu + \mathbf{e}_2 \sin\mu) \\ &=\mathbf{e}_2 \cos\mu - \mathbf{e}_1 \sin\mu \\ &=\mathbf{e}_2 e^{i \mu}\end{aligned}

Putting everything, and summarizing results for the fields, we have

\begin{aligned}\mathbf{E} &= q \mathbf{e}_3 \frac{z}{(\sqrt{z^2 + a^2})^3}+\frac{q a k^2 \mathbf{e}_1 e^{i \omega t_r} }{\sqrt{z^2 + a^2}} \\ \mathbf{B} &= \frac{q a k z}{ z^2 + a^2} \left( \frac{\mathbf{e}_1}{\sqrt{z^2 + a^2}} - k \mathbf{e}_2 \right) e^{i \omega t_r}\end{aligned} \hspace{\stretch{1}}(1.38)

The electric field expression above compares well to 1.29. We have the Coulomb term and the radiation term. It is harder to compare the magnetic field to the exact result 1.30 since I did not expand that out.

FIXME: A question to consider. If all this worked should we not also get

\begin{aligned}\mathbf{B} \stackrel{?}{=}\frac{z \mathbf{e}_3 - \mathbf{e}_1 a e^{i \omega t_r}}{\sqrt{z^2 + a^2}} \times \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(1.40)

However, if I do this check I get

\begin{aligned}\mathbf{B} =\frac{q a z}{z^2 + a^2} \left( \frac{1}{{z^2 + a^2}} + k^2 \right) \mathbf{e}_2 e^{i \omega t_r}.\end{aligned} \hspace{\stretch{1}}(1.41)

# Collision of photon and electron.

I made a dumb error on the exam on this one. I setup the four momentum conservation statement, but then didn’t multiply out the cross terms properly. This led me to incorrectly assume that I had to try doing this the hard way (something akin to what I did on the midterm). Simon later told us in the tutorial the simple way, and that’s all we needed here too. Here’s the setup.

An electron at rest initially has four momentum

\begin{aligned}(m c, 0)\end{aligned} \hspace{\stretch{1}}(2.42)

where the incoming photon has four momentum

\begin{aligned}\left(\hbar \frac{\omega}{c}, \hbar \mathbf{k} \right)\end{aligned} \hspace{\stretch{1}}(2.43)

After the collision our electron has some velocity so its four momentum becomes (say)

\begin{aligned}\gamma (m c, m \mathbf{v}),\end{aligned} \hspace{\stretch{1}}(2.44)

and our new photon, going off on an angle $\theta$ relative to $\mathbf{k}$ has four momentum

\begin{aligned}\left(\hbar \frac{\omega'}{c}, \hbar \mathbf{k}' \right)\end{aligned} \hspace{\stretch{1}}(2.45)

Our conservation relationship is thus

\begin{aligned}(m c, 0) + \left(\hbar \frac{\omega}{c}, \hbar \mathbf{k} \right)=\gamma (m c, m \mathbf{v})+\left(\hbar \frac{\omega'}{c}, \hbar \mathbf{k}' \right)\end{aligned} \hspace{\stretch{1}}(2.46)

I squared both sides, but dropped my cross terms, which was just plain wrong, and costly for both time and effort on the exam. What I should have done was just

\begin{aligned}\gamma (m c, m \mathbf{v}) =(m c, 0) + \left(\hbar \frac{\omega}{c}, \hbar \mathbf{k} \right)-\left(\hbar \frac{\omega'}{c}, \hbar \mathbf{k}' \right),\end{aligned} \hspace{\stretch{1}}(2.47)

and then square this (really making contractions of the form $p_i p^i$). That gives (and this time keeping my cross terms)

\begin{aligned}(\gamma (m c, m \mathbf{v}) )^2 &= \gamma^2 m^2 (c^2 - \mathbf{v}^2) \\ &= m^2 c^2 \\ &=m^2 c^2 + 0 + 0+ 2 (m c, 0) \cdot \left(\hbar \frac{\omega}{c}, \hbar \mathbf{k} \right)- 2 (m c, 0) \cdot \left(\hbar \frac{\omega'}{c}, \hbar \mathbf{k}' \right)- 2 \cdot \left(\hbar \frac{\omega}{c}, \hbar \mathbf{k} \right)\cdot \left(\hbar \frac{\omega'}{c}, \hbar \mathbf{k}' \right) \\ &=m^2 c^2 + 2 m c \hbar \frac{\omega}{c} - 2 m c \hbar \frac{\omega'}{c}- 2\hbar^2 \left(\frac{\omega}{c} \frac{\omega'}{c}- \mathbf{k} \cdot \mathbf{k}'\right) \\ &=m^2 c^2 + 2 m c \hbar \frac{\omega}{c} - 2 m c \hbar \frac{\omega'}{c}- 2\hbar^2 \frac{\omega}{c} \frac{\omega'}{c} (1 - \cos\theta)\end{aligned}

Rearranging a bit we have

\begin{aligned}\omega' \left( m + \frac{\hbar \omega}{c^2} ( 1 - \cos\theta ) \right) = m \omega,\end{aligned} \hspace{\stretch{1}}(2.48)

or

\begin{aligned}\omega' = \frac{\omega}{1 + \frac{\hbar \omega}{m c^2} ( 1 - \cos\theta ) }\end{aligned} \hspace{\stretch{1}}(2.49)

# Pion decay.

The problem above is very much like a midterm problem we had, so there was no justifiable excuse for messing up on it. That midterm problem was to consider the split of a pion at rest into a neutrino (massless) and a muon, and to calculate the energy of the muon. That one also follows the same pattern, a calculation of four momentum conservation, say

\begin{aligned}(m_\pi c, 0) = \hbar \frac{\omega}{c}(1, \hat{\mathbf{k}}) + ( \mathcal{E}_\mu/c, \mathbf{p}_\mu ).\end{aligned} \hspace{\stretch{1}}(3.50)

Here $\omega$ is the frequency of the massless neutrino. The massless nature is encoded by a four momentum that squares to zero, which follows from $(1, \hat{\mathbf{k}}) \cdot (1, \hat{\mathbf{k}}) = 1^2 - \hat{\mathbf{k}} \cdot \hat{\mathbf{k}} = 0$.

When I did this problem on the midterm, I perversely put in a scattering angle, instead of recognizing that the particles must scatter at 180 degree directions since spatial momentum components must also be preserved. This and the combination of trying to work in spatial quantities led to a mess and I didn’t get the end result in anything that could be considered tidy.

The simple way to do this is to just rearrange to put the null vector on one side, and then square. This gives us

\begin{aligned}0 &=\left(\hbar \frac{\omega}{c}(1, \hat{\mathbf{k}}) \right) \cdot\left(\hbar \frac{\omega}{c}(1, \hat{\mathbf{k}}) \right) \\ &=\left( (m_\pi c, 0) - ( \mathcal{E}_\mu/c, \mathbf{p}_\mu ) \right) \cdot \left( (m_\pi c, 0) - ( \mathcal{E}_\mu/c, \mathbf{p}_\mu ) \right) \\ &={m_\pi}^2 c^2 + {m_\nu}^2 c^2 - 2 (m_\pi c, 0) \cdot ( \mathcal{E}_\mu/c, \mathbf{p}_\mu ) \\ &={m_\pi}^2 c^2 + {m_\nu}^2 c^2 - 2 m_\pi \mathcal{E}_\mu\end{aligned}

A final re-arrangement gives us the muon energy

\begin{aligned}\mathcal{E}_\mu = \frac{1}{{2}} \frac{ {m_\pi}^2 + {m_\nu}^2 }{m_\pi} c^2\end{aligned} \hspace{\stretch{1}}(3.51)

## PHY450H1S. Relativistic Electrodynamics Lecture 19 (Taught by Prof. Erich Poppitz). Lienard-Wiechert potentials.

Posted by peeterjoot on April 26, 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 8 material from the text [1].

Covering lecture notes pp. 136-146: the Lienard-Wiechert potentials (143-146) [Wednesday, Mar. 9...]

# Fields from the Lienard-Wiechert potentials

(We finished off with the scalar and vector potentials in class, but I’ve put those notes with the previous lecture).

To find $\mathbf{E}$ and $\mathbf{B}$ need

$\frac{\partial {t_r}}{\partial {t}}$, and $\boldsymbol{\nabla} t_r(\mathbf{x}, t)$

where

\begin{aligned}t - t_r = {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r) }\right\rvert}\end{aligned} \hspace{\stretch{1}}(2.1)

implicit definition of $t_r(\mathbf{x}, t)$

In HW5 you’ll show

\begin{aligned}\frac{\partial {t_r}}{\partial {t}} = \frac{{\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r) }\right\rvert}}{{\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r) }\right\rvert} - \frac{\mathbf{v}_c }{c} \cdot (\mathbf{x} - \mathbf{x}_c(t_r))}\end{aligned} \hspace{\stretch{1}}(2.2)

\begin{aligned}\boldsymbol{\nabla} t_r = \frac{1}{{c}} \frac{\mathbf{x} - \mathbf{x}_c(t_r) }{{\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r) }\right\rvert} - \frac{\mathbf{v}_c }{c} \cdot (\mathbf{x} - \mathbf{x}_c(t_r))}\end{aligned} \hspace{\stretch{1}}(2.3)

and then use this to show that the electric and magnetic fields due to a moving charge are

\begin{aligned}\mathbf{E}(\mathbf{x}, t) &= \frac{e R}{ (\mathbf{R} \cdot \mathbf{u})^3 } \left( (c^2 - \mathbf{v}_c^2) \mathbf{u} + \mathbf{R} \times (\mathbf{u} \times \mathbf{a}_c) \right) \\ &= \frac{\mathbf{R}}{R} \times \mathbf{E} \\ \mathbf{u} &= c \frac{\mathbf{R}}{R} - \mathbf{v}_c,\end{aligned} \hspace{\stretch{1}}(2.4)

where everything is evaluated at the retarded time $t_r = t - {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert}/c$.

This looks quite a bit different than what we find in section 63 (63.8) in the text, but a little bit of expansion shows they are the same.

# Check. Particle at rest.

With

\begin{aligned}\mathbf{x}_c &= \mathbf{x}_0 \\ X_c^k &= (ct, \mathbf{x}_0) \\ {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert} &= c(t - t_r)\end{aligned}

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{particleAtRestTrCalc}
\caption{Retarded time for particle at rest.}

\end{figure}

As illustrated in figure (\ref{fig:particleAtRestTrCalc}) the retarded position is

\begin{aligned}\mathbf{x}_c(t_r) = \mathbf{x}_0,\end{aligned} \hspace{\stretch{1}}(3.7)

for

\begin{aligned}\mathbf{u} = \frac{\mathbf{x} - \mathbf{x}_0}{{\left\lvert{\mathbf{x} - \mathbf{x}_0}\right\rvert}} c,\end{aligned} \hspace{\stretch{1}}(3.8)

and

\begin{aligned}\mathbf{E} = e \frac{{ {\left\lvert{\mathbf{x} - \mathbf{x}_0}\right\rvert}} }{ (c {\left\lvert{\mathbf{x} - \mathbf{x}_0}\right\rvert})^3 } c^3 \frac{\mathbf{x} - \mathbf{x}_0}{{{\left\lvert{\mathbf{x} - \mathbf{x}_0}\right\rvert}}},\end{aligned} \hspace{\stretch{1}}(3.9)

which is Coulomb’s law

\begin{aligned}\mathbf{E} = e \frac{\mathbf{x} - \mathbf{x}_0}{{\left\lvert{\mathbf{x} - \mathbf{x}_0}\right\rvert}^3}\end{aligned} \hspace{\stretch{1}}(3.10)

# Check. Particle moving with constant velocity.

This was also computed in full in homework 5. The end result was

\begin{aligned}\mathbf{E} =e \frac{\mathbf{x} - \mathbf{v} t}{{\left\lvert{\mathbf{x} - \mathbf{v} t}\right\rvert}^3}\frac{1 -\boldsymbol{\beta}^2}{ \left(1 - \frac{(\mathbf{x} \times \boldsymbol{\beta})^2}{{\left\lvert{\mathbf{x} - \mathbf{v} t}\right\rvert}^2} \right)^{3/2} }\end{aligned} \hspace{\stretch{1}}(4.11)

Writing

\begin{aligned}\frac{\mathbf{x} \times \boldsymbol{\beta}}{{\left\lvert{\mathbf{x} - \mathbf{v} t}\right\rvert}}&=\frac{1}{{c}} \frac{(\mathbf{x} - \mathbf{v} t) \times \mathbf{v}}{{\left\lvert{\mathbf{x} - \mathbf{v} t}\right\rvert}} \\ &=\frac{{\left\lvert{\mathbf{v}}\right\rvert}}{c} \frac{(\mathbf{x} - \mathbf{v} t) \times \mathbf{v}}{{\left\lvert{\mathbf{x} - \mathbf{v} t}\right\rvert} {\left\lvert{\mathbf{v}}\right\rvert}} \end{aligned}

We can introduce an angular dependence between the charge’s translated position and its velocity

\begin{aligned}\sin^2 \theta = {\left\lvert{ \frac{\mathbf{v} \times (\mathbf{x} - \mathbf{v} t)}{\Abs{\mathbf{v}} \Abs{\mathbf{x} - \mathbf{v} t}} }\right\rvert}^2,\end{aligned} \hspace{\stretch{1}}(4.12)

and write the field as

\begin{aligned}\mathbf{E} =\underbrace{e \frac{\mathbf{x} - \mathbf{v} t}{{\left\lvert{\mathbf{x} - \mathbf{v} t}\right\rvert}^3}}_{{*}}\frac{1 -\boldsymbol{\beta}^2}{ \left(1 - \frac{\mathbf{v}^2}{c^2} \sin^2 \theta \right)^{3/2} }\end{aligned} \hspace{\stretch{1}}(4.13)

Observe that ${*} = \text{Coulomb's law measured from the instantaneous position of the charge}$.

The electric field $\mathbf{E}$ has a time dependence, strongest when perpendicular to the instantaneous position when $\theta = \pi/2$, since the denominator is smallest ($\mathbf{E}$ largest) when $\mathbf{v}/c$ is not small. This is strongly $\theta$ dependent.

Compare

\begin{aligned}\frac{{\left\lvert{\mathbf{E}(\theta = \pi/2)}\right\rvert} - {\left\lvert{\mathbf{E}(\theta = \pi/2 + \Delta \theta)}\right\rvert} }{{\left\lvert{\mathbf{E}(\theta = \pi/2)}\right\rvert}}&\approx\frac{\frac{1}{{(1 - \mathbf{v}^2/c^2)^{3/2}}} - \frac{1}{{(1 - \mathbf{v}^2/c^2(1 - (\Delta \theta)^2))^{3/2}}}}{\frac{1}{{(1 - \mathbf{v}^2/c^2)^{3/2}}}} \\ &=1 - \left(\frac{1 - \mathbf{v}^2/c^2}{1 - \mathbf{v}^2/c^2 + \mathbf{v}^2/c^2(\Delta \theta)^2}\right)^{3/2} \\ &=1 - \left(\frac{1}{1 + \mathbf{v}^2/c^2 \frac{(\Delta \theta)^2}{1 - \mathbf{v}^2/c^2}}\right)^{3/2} \\ \end{aligned}

Here we used

\begin{aligned}\sin(\theta + \pi/2) = \frac{e^{i (\theta + \pi/2)} - e^{-i(\theta + \pi/2)}}{2i} = \cos\theta \end{aligned} \hspace{\stretch{1}}(4.14)

and

\begin{aligned}\cos^2 \Delta \theta \approx \left( 1 - \frac{(\Delta \theta)^2}{2} \right)^2 \approx 1 - (\Delta \theta)^2\end{aligned} \hspace{\stretch{1}}(4.15)

FIXME: he writes:

\begin{aligned}\Delta \theta \le \sqrt{1 - \frac{\mathbf{v}^2}{c^2}}\end{aligned} \hspace{\stretch{1}}(4.16)

I don’t see where that comes from.

FIXME: PICTURE: Various $\mathbf{E}$‘s up, and $\mathbf{v}$ perpendicular to that, strongest when charge is moving fast.

# Back to extracting physics from the Lienard-Wiechert field equations

Imagine that we have a localized particle motion with

\begin{aligned}{\left\lvert{\mathbf{x}_c(t_r)}\right\rvert} < l\end{aligned} \hspace{\stretch{1}}(5.17)

The velocity vector

\begin{aligned}\mathbf{u} = c \frac{\mathbf{x} - \mathbf{x}_c(t_r)}{{\left\lvert{\mathbf{x} - \mathbf{x}_c}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(5.18)

doesn’t grow as distance from the source, so from 2.4, we have for ${\left\lvert{\mathbf{x}}\right\rvert} \gg l$

\begin{aligned}\mathbf{B}, \mathbf{E} \sim \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}^2}}(\cdots) + \frac{1}{\mathbf{x}}(\text{acceleration term})\end{aligned} \hspace{\stretch{1}}(5.19)

The acceleration term will dominate at large distances from the source. Our Poynting magnitude is

\begin{aligned}{\left\lvert{\mathbf{S}}\right\rvert} \sim {\left\lvert{\mathbf{E} \times \mathbf{B}}\right\rvert} \sim \frac{1}{{\mathbf{x}^2}} (\text{acceleration})^2.\end{aligned} \hspace{\stretch{1}}(5.20)

\begin{aligned}\oint d^2 \boldsymbol{\sigma} \cdot \mathbf{S} \sim R^2 \frac{1}{{R^2}} (\text{acceleration})^2 \sim (\text{acceleration})^2 \end{aligned} \hspace{\stretch{1}}(5.21)

In the limit, for the radiation of EM waves

\begin{aligned}\lim_{R\rightarrow \infty} \oint d^2 \boldsymbol{\sigma} \cdot \mathbf{S} \ne 0\end{aligned} \hspace{\stretch{1}}(5.22)

The energy flux through a sphere of radius $R$ is called the radiated power.

# References

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

## Playing with complex notation for relativistic applications in a plane

Posted by peeterjoot on April 19, 2011

# Motivation.

In the electrodynamics midterm we had a question on circular motion. This screamed for use of complex numbers to describe the spatial parts of the spacetime trajectories.

Let’s play with this a bit.

# Our invariant.

Suppose we describe our spacetime point as a paired time and complex number

\begin{aligned}X = (ct, z).\end{aligned} \hspace{\stretch{1}}(2.1)

Our spacetime invariant interval in this form is thus

\begin{aligned}X^2 \equiv (ct)^2 - {\left\lvert{z}\right\rvert}^2.\end{aligned} \hspace{\stretch{1}}(2.2)

Not much different than the usual coordinate representation of the spatial coordinates, except that we have a ${\left\lvert{z}\right\rvert}^2$ replacing the usual $\mathbf{x}^2$.

Taking the spacetime distance between $X$ and another point, say $\tilde{X} = ( c \tilde{t}, \tilde{z})$ motivates the inner product between two points in this representation

\begin{aligned}(X - \tilde{X})^2 &= (ct - c \tilde{t})^2 - {\left\lvert{z - \tilde{z}}\right\rvert}^2 \\ &= (ct - c \tilde{t})^2 - (z - \tilde{z})(z^{*} - \tilde{z}^{*}) \\ &= (ct)^2 - 2 (ct) (c \tilde{t}) + (c \tilde{t})^2 - {\left\lvert{z}\right\rvert}^2 - {\left\lvert{\tilde{z}}\right\rvert}^2 + (z \tilde{z}^{*} + z^{*} \tilde{z}) \\ &= X^2 + \tilde{X}^2 - 2 \left( (ct) (c \tilde{t}) - \frac{1}{{2}}(z \tilde{z}^{*} + z^{*} \tilde{z}) \right) \\ \end{aligned}

It’s clear that it makes sense to define

\begin{aligned}X \cdot \tilde{X} = (ct) (c \tilde{t}) - \text{Real} (z \tilde{z}^{*}),\end{aligned} \hspace{\stretch{1}}(2.3)

consistent with our original starting point

\begin{aligned}X^2 = X \cdot X.\end{aligned} \hspace{\stretch{1}}(2.4)

Let’s also introduce a complex inner product

\begin{aligned}{\langle{{z}}, {{\tilde{z}}}\rangle} \equiv \frac{1}{{2}} \left( z \tilde{z}^{*} + z^{*} \tilde{z}) \right) = \text{Real} (z \tilde{z}^{*}).\end{aligned} \hspace{\stretch{1}}(2.5)

Our dot product can now be written

\begin{aligned}X \cdot \tilde{X} = (ct) (c \tilde{t}) - {\langle{{z}}, {{\tilde{z}}}\rangle}.\end{aligned} \hspace{\stretch{1}}(2.6)

# Change of basis.

Our standard basis for our spatial components is $\{1, i\}$, but we are free to pick any other basis should we choose. In particular, if we rotate our basis counterclockwise by $\phi$, our new basis, still orthonormal, is $\{ e^{i\phi}, i e^{i\phi} \}$.

In any orthonormal basis the coordinates of a point with respect to that basis are real, so just as we can write

\begin{aligned}z = {\langle{{1}}, {{z}}\rangle} + i {\langle{{i}}, {{z}}\rangle},\end{aligned} \hspace{\stretch{1}}(3.7)

we can extract the coordinates in the rotated frame, also simply by taking inner products

\begin{aligned}z = e^{i \phi} {\langle{{e^{i \phi}}}, {{z}}\rangle} + i e^{i\phi} {\langle{{i e^{i\phi} }}, {{z}}\rangle}.\end{aligned} \hspace{\stretch{1}}(3.8)

The values ${\langle{{e^{i \phi}}}, {{z}}\rangle}$, and ${\langle{{i e^{i\phi} }}, {{z}}\rangle}$ are the (real) coordinates of the point $z$ in this rotated basis.

This is enough that we can write the Lorentz boost immediately for a velocity $\vec{v} = c \beta e^{i\phi}$ at an arbitrary angle $\phi$ in the plane

\begin{aligned}\begin{bmatrix}ct' \\ {\langle{{e^{i\phi}}}, {{z'}}\rangle} \\ {\langle{{i e^{i\phi}}}, {{z'}}\rangle} \end{bmatrix}=\begin{bmatrix}\gamma & -\gamma \beta & 0 \\ -\gamma \beta & \gamma & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}ct \\ {\langle{{e^{i\phi}}}, {{z}}\rangle} \\ {\langle{{i e^{i\phi}}}, {{z}}\rangle} \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.9)

Let’s translate this to $ct, x, y$ coordinates as a check. For the spatial component parallel to the boost direction we have

\begin{aligned}{\langle{{e^{i\phi}}}, {{x + iy}}\rangle} &= \text{Real} ( e^{-i\phi} (x + i y) ) \\ &= \text{Real} ( (\cos\phi - i \sin\phi)(x + i y) ) \\ &= x \cos\phi + y \sin\phi,\end{aligned}

and the perpendicular components are

\begin{aligned}{\langle{{ i e^{i\phi}}}, {{x + iy}}\rangle} &= \text{Real} ( -i e^{-i\phi} (x + i y) ) \\ &= \text{Real} ( (-i \cos\phi - \sin\phi)(x + i y) ) \\ &= -x \sin\phi + y \cos\phi.\end{aligned}

Grouping the two gives

\begin{aligned}\begin{bmatrix}{\langle{{e^{i\phi}}}, {{x + iy}}\rangle} \\ {\langle{{i e^{i\phi}}}, {{x + iy}}\rangle} \end{bmatrix}=\begin{bmatrix}\cos\phi & \sin\phi \\ -\sin\phi & \cos\phi\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= R_{-\phi}\begin{bmatrix}x \\ y\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.10)

The boost equation in terms of the cartesian coordinates is thus

\begin{aligned}\begin{bmatrix}1 & 0 \\ 0 & R_{-\phi}\end{bmatrix}\begin{bmatrix}c t' \\ x' \\ y'\end{bmatrix}=\begin{bmatrix}\gamma & -\gamma \beta & 0 \\ -\gamma \beta & \gamma & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & R_{-\phi}\end{bmatrix}\begin{bmatrix}c t \\ x \\ y\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.11)

Writing

\begin{aligned}\begin{bmatrix}c t' \\ x' \\ y'\end{bmatrix}={\left\lVert{{\wedge^\mu}_\nu}\right\rVert} \begin{bmatrix}c t \\ x \\ y\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(3.12)

the boost matrix ${\left\lVert{{\wedge^\mu}_\nu}\right\rVert}$ is found to be (after a bit of work)

\begin{aligned}{\left\lVert{{\wedge^\mu}_\nu}\right\rVert} &=\begin{bmatrix}1 & 0 \\ 0 & R_{\phi}\end{bmatrix}\begin{bmatrix}\gamma & -\gamma \beta & 0 \\ -\gamma \beta & \gamma & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & R_{-\phi}\end{bmatrix} \\ &=\begin{bmatrix}\gamma & - \gamma \beta \cos\phi & -\gamma \beta \sin\phi \\ -\gamma \beta \cos\phi & \gamma \cos^2\phi + \sin^2 \phi & (\gamma -1) \sin\phi \cos\phi \\ -\gamma \beta \sin\phi & (\gamma -1) \sin\phi \cos\phi & \gamma \sin^2\phi + \cos^2\phi \\ \end{bmatrix} \\ \end{aligned}

A final bit of regrouping gives

\begin{aligned}{\left\lVert{{\wedge^\mu}_\nu}\right\rVert} =\begin{bmatrix}\gamma & - \gamma \beta \cos\phi & -\gamma \beta \sin\phi \\ -\gamma \beta \cos\phi & 1 + ( \gamma -1) \cos^2\phi & (\gamma -1) \sin\phi \cos\phi \\ -\gamma \beta \sin\phi & (\gamma -1) \sin\phi \cos\phi & 1 + (\gamma -1) \sin^2\phi \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.13)

This is consistent with the result stated in [1], finishing the game for the day.

# References

[1] Wikipedia. Lorentz transformation — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 20-April-2011]. http://en.wikipedia.org/w/index.php?title=Lorentz_transformation&oldid=424507400.

## Experiment to see what the default SIGCHLD handling was

Posted by peeterjoot on April 15, 2011

For some multi-process code I’d see that our waitpid() wrapper function blocked SIGCHLD before calling waitpid(). That didn’t quite make sense to me. If SIGCHLD had to be blocked, wouldn’t that have to be done before the fork?

I didn’t even want SIGCHLD to be delivered, since I was calling waitpid(). Is the default handler for SIGCHLD SIG_IGN? To experiment (non-portably), I wrote the following mini-test:

#include <signal.h>
#include <unistd.h>
#include <stdio.h>
#include <sys/wait.h>

int main()
{
pid_t pid = fork() ;

if ( pid )
{
int status ;
printf( "forked: %d\n", (int)pid ) ;

kill( pid, SIGKILL ) ; // does this generate SIGCHLD?  Is the default for that signal to be blocked?

int rc = waitpid(0, &status, 0) ;
printf( "waitpid(): rc, status = %d, %X\n", rc, status ) ;
}
else
{
sleep( 10 ) ;
_exit( 1 ) ;
}

return 0 ;
}


I see:

forked: 16787
waitpid(): rc, status = 16787, 9


and don’t appear to get taken out due to no handler for SIGCHLD. The default does appear to be SIG_IGN (on Linux and AIX). This makes sense (man 7 signal on linux also appears to confirm).

I wonder if there is a way to query the handler (like calling sigaction for read only). If I wanted to be super safe, putting in an assert that the handler was SIG_IGN would then be possible.

## PHY450H1S. Relativistic Electrodynamics Lecture 26 (Taught by Prof. Erich Poppitz). Radiation reaction force for a dipole system.

Posted by peeterjoot on April 11, 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 8 section 65 material from the text [1].

Covering pp. 181-195: (182-189) [Tuesday, Mar. 29]; the EM potentials to order $(v/c)^2$ (190-193); the “Darwin Lagrangian. and Hamiltonian for a system of nonrelativistic charged particles to order $(v/c)^2$ and its many uses in physics (194-195) [Wednesday, Mar. 30]

Next week (last topic): attempt to go to the next order $(v/c)^3$ – radiation damping, the limitations of classical electrodynamics, and the relevant time/length/energy scales.

# Recap.

A system of N charged particles $m_a, q_a ; a \in [1, N]$ closed system and nonrelativistic, $v_a/c \ll 1$. In this case we can incorporate EM effects in a Largrangian ONLY involving particles (EM field not a dynamical DOF). In general case, this works to $O((v/c)^2)$, because at $O((v/c))$ system radiation effects occur.

In a specific case, when

\begin{aligned}\frac{m_1}{q_1} = \frac{m_2}{q_2} = \frac{m_3}{q_3} = \cdots\end{aligned} \hspace{\stretch{1}}(2.1)

we can do that (meaning use a Lagrangian with particles only) to $O((v/c)^4)$ because of specific symmetries in such a system.

The Lagrangian for our particle after the gauge transformation is

\begin{aligned}\mathcal{L}_a = \frac{1}{{2}} m_a \mathbf{v}_a^2 + \frac{m_a}{8} \frac{\mathbf{v}_a^4}{c^2} -\sum_{b \ne a} \frac{q_a q_b}{{\left\lvert{\mathbf{x}_a(t) - \mathbf{x}_b(t)}\right\rvert}}+\sum_b q_a q_b \frac{\mathbf{v}_a \cdot \mathbf{v}_b + (\mathbf{n} \cdot \mathbf{v}_a) (\mathbf{n} \cdot \mathbf{v}_b)}{2 c^2 {\left\lvert{\mathbf{x} - \mathbf{x}_b}\right\rvert} }.\end{aligned} \hspace{\stretch{1}}(2.2)

Next time we’ll probably get to the Lagrangian for the entire system. It was hinted that this is called the Darwin Lagrangian (after Charles Darwin’s grandson).

We find for whole system

\begin{aligned}\mathcal{L} = \sum_a \mathcal{L}_a + \frac{1}{{2}} \sum_a \mathcal{L}_a (interaction)\end{aligned} \hspace{\stretch{1}}(2.3)

\begin{aligned}\mathcal{L} = \frac{1}{{2}} \sum_a m_a \mathbf{v}_a^2 + \sum_a \frac{m_a}{8} \frac{\mathbf{v}_a^4}{c^2} -\sum_{ a < b} \frac{q_a q_b}{{\left\lvert{\mathbf{x}_a(t) - \mathbf{x}_b(t)}\right\rvert}}+\sum_b q_a q_b \frac{\mathbf{v}_a \cdot \mathbf{v}_b + (\mathbf{n} \cdot \mathbf{v}_a) (\mathbf{n} \cdot \mathbf{v}_b)}{2 c^2 {\left\lvert{\mathbf{x} - \mathbf{x}_b}\right\rvert} }.\end{aligned} \hspace{\stretch{1}}(2.4)

This is the Darwin Lagrangian (also Charles). The Darwin Hamiltonian is then

\begin{aligned}H = \sum_a \frac{p_a}{2 m_a} \mathbf{v}_a^2 + \sum_a \frac{p_a^4}{8 m_a^3 c^2} +\sum_{ a < b} \frac{q_a q_b}{{\left\lvert{\mathbf{x}_a(t) - \mathbf{x}_b(t)}\right\rvert}}- \sum_{ a < b } \frac{q_a q_b}{ 2 c^2 m_a m_b } \frac{\mathbf{p}_a \cdot \mathbf{p}_b + (\mathbf{n}_{a b} \cdot \mathbf{p}_a) (\mathbf{n}_{ a b } \cdot \mathbf{p}_b)}{{\left\lvert{\mathbf{x} - \mathbf{x}_b}\right\rvert} }.\end{aligned} \hspace{\stretch{1}}(2.5)

# Incorporating radiation effects as a friction term.

To $O((v/c)^3)$ obvious problem due to radiation (system not closed). We’ll incorporate radiation via a function term in the EOM

Again consider the dipole system

\begin{aligned}m \dot{d}{z} &= -k z \\ \omega^2 &= \frac{k}{m}\end{aligned} \hspace{\stretch{1}}(3.6)

or

\begin{aligned}m \dot{d}{z} = -\omega^2 m z\end{aligned} \hspace{\stretch{1}}(3.8)

gives

\begin{aligned}\frac{d{{}}}{dt}\left( \frac{m}{2} \dot{z}^2 + \frac{ m \omega^2 }{2} z^2 \right) = 0 \end{aligned} \hspace{\stretch{1}}(3.9)

The energy radiated per unit time averaged per period is

\begin{aligned}P = \frac{ 2 e^2 }{ 3 c^3} \left\langle{{ \dot{d}{z}^2 }}\right\rangle\end{aligned} \hspace{\stretch{1}}(3.10)

We’ll modify the EOM

\begin{aligned}m \dot{d}{z} = -\omega^2 m z + f_{\text{radiation}}\end{aligned} \hspace{\stretch{1}}(3.11)

Employing an integration factor $\dot{z}$ we have

\begin{aligned}m \dot{d}{z} \dot{z} = -\omega^2 m z \dot{z} + f_{\text{radiation}} \dot{z}\end{aligned} \hspace{\stretch{1}}(3.12)

or

\begin{aligned}\frac{d{{}}}{dt} \left( m \dot{z}^2 + \omega^2 m z^2 \right) = f_{\text{radiation}} \dot{z}\end{aligned} \hspace{\stretch{1}}(3.13)

Observe that the last expression, force times velocity, has the form of power

\begin{aligned}m \frac{d^2 z}{dt^2} \frac{dz}{dt} = \frac{d{{}}}{dt} \left( \frac{m}{2} \left( \frac{dz}{dt} \right)^2 \right)\end{aligned} \hspace{\stretch{1}}(3.14)

So we can make an identification with the time rate of energy lost by the system due to radiation

\begin{aligned}\frac{d{{}}}{dt} \left( m \dot{z}^2 + \omega^2 m z^2 \right) \equiv \frac{d{{\mathcal{E}}}}{dt}.\end{aligned} \hspace{\stretch{1}}(3.15)

Average over period both sides

\begin{aligned}\left\langle{{ \frac{d{{\mathcal{E}}}}{dt} }}\right\rangle = \left\langle{{ f_{\text{radiation}} \dot{z} }}\right\rangle=- \frac{2 e^2 }{3 c^3} \left\langle{{\dot{d}{z}^2}}\right\rangle\end{aligned} \hspace{\stretch{1}}(3.16)

We demand this last equality, by requiring the energy change rate to equal that of the dipole power (but negative since it is a loss) that we previously calculated.

Claim:

\begin{aligned}f_{\text{radiation}} = \frac{2 e^2 }{3 c^3} \dddot{z}\end{aligned} \hspace{\stretch{1}}(3.17)

Proof:

We need to show

\begin{aligned}\left\langle{{ f_{\text{radiation}} }}\right\rangle= - \frac{2 e^2 }{3 c^3} \left\langle{{\dot{d}{z}^2}}\right\rangle\end{aligned} \hspace{\stretch{1}}(3.18)

We have

\begin{aligned}\frac{2 e^2}{3 c^3} \left\langle{{ \dddot{z} \dot{z} }}\right\rangle &= \frac{2 e^2}{3 c^3} \frac{1}{{T}} \int_0^T dt \dddot{z} \dot{z} \\ &= \frac{2 e^2}{3 c^3} \frac{1}{{T}} \int_0^T dt {\frac{d{{}}}{dt} (\dot{d}{z} \dot{z}) }-\frac{2 e^2}{3 c^3} \frac{1}{{T}} \int_0^T dt (\dot{d}{z})^2\end{aligned}

We first used $(\dot{d}{z} \dot{z})' = \dddot{z} \dot{z} + (\dot{d}{z})^2$. The first integral above is zero since the derivative of $\dot{d}{z} \dot{z} = (-\omega^2 z_0 \sin\omega t)(\omega z_0 \cos\omega t) = -\omega^3 z_0^2 \sin(2 \omega t)/2$ is also periodic, and vanishes when integrated over the interval.

\begin{aligned}\frac{2 e^2}{3 c^3} \left\langle{{ \dddot{z} \dot{z} }}\right\rangle =-\frac{2 e^2}{3 c^3} \left\langle{{ (\dot{d}{z})^2 }}\right\rangle\end{aligned} \hspace{\stretch{1}}(3.19)

We can therefore write

\begin{aligned}m \dot{d}{z} = -m \omega^2 z + \frac{2 e^2}{3 c^3} \dddot{z}\end{aligned} \hspace{\stretch{1}}(3.20)

Our “frictional” correction is the radiation reaction force proportional to the third derivative of the position.

Rearranging slightly, this is

\begin{aligned}\dot{d}{z} = - \omega^2 z + \frac{2}{3 c} \left( \frac{e^2}{m c^2} \right) \dddot{z} = - \omega^2 z + \frac{2}{3 c} \frac{r_e}{c} \dddot{z},\end{aligned} \hspace{\stretch{1}}(3.21)

where $r_e \sim 10^{-13} \text{cm}$ is the “classical radius” of the electron. In our frictional term we have $r_e/c$, the time for light to cross the classical radius of the electron.

There are lots of problems with this. One of the easiest is with $\omega = 0$. Then we have

\begin{aligned}\dot{d}{z} = \frac{2}{3} \frac{r_e}{c} \dddot{z}\end{aligned} \hspace{\stretch{1}}(3.22)

with solution

\begin{aligned}z \sim e^{\alpha t},\end{aligned} \hspace{\stretch{1}}(3.23)

where

\begin{aligned}\alpha \sim \frac{c}{r_e} \sim \frac{1}{{\tau_e}}.\end{aligned} \hspace{\stretch{1}}(3.24)

This is a self accelerating system! Note that we can also get into this trouble with $\omega \ne 0$, but those examples are harder to find (see: [2]).

FIXME: borrow this text again to give that section a read.

The sensible point of view is that this third term ($f_{\text{rad}}$) should be taken seriously only if it is small compared to the first two terms.

# References

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

[2] D.J. Griffith. Introduction to Electrodynamics. Prentice-Hall, 1981.

## PHY450H1S. Relativistic Electrodynamics Tutorial 9 (TA: Simon Freedman). Some worked problems. EM reflection. Stress energy tensor for simple configurations.

Posted by peeterjoot on April 11, 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.)]

# HW6. Question 3. (Non subtle hints about how important this is (i.e. for the exam)

## Motivation.

This is problem 1 from section 47 of the text [1].

Determine the force exerted on a wall from which an incident plane EM wave is reflected (w/ reflection coefficient $R$) and incident angle $\theta$.

Solution from the book

\begin{aligned}f_\alpha = - \sigma_{\alpha \beta} n_\beta - {\sigma'}_{\alpha \beta} n_\beta\end{aligned} \hspace{\stretch{1}}(1.1)

Here $\sigma_{\alpha \beta}$ is the Maxwell stress tensor for the incident wave, and ${\sigma'}_{\alpha \beta}$ is the Maxwell stress tensor for the reflected wave, and $n_\beta$ is normal to the wall.

## On the signs of the force per unit area

The signs in 1.1 require a bit of thought. We have for the rate of change of the $\alpha$ component of the field momentum

\begin{aligned}\frac{d{{}}}{dt} \int d^3 \mathbf{x} \left( \frac{S^\alpha}{c^2} \right) = - \int d^2 \sigma^\beta T^{\beta \alpha}\end{aligned} \hspace{\stretch{1}}(1.2)

where $d^2 \sigma^\beta = d^2 \sigma \mathbf{n} \cdot \mathbf{e}_\beta$, and $\mathbf{n}$ is the outwards unit normal to the surface. This is the rate of change of momentum for the field, the force on the field. For the force on the wall per unit area, we wish to invert this, giving

\begin{aligned}df^\alpha_{\text{on the wall, per unit area}}= (\mathbf{n} \cdot \mathbf{e}_\beta) T^{\beta \alpha}= -(\mathbf{n} \cdot \mathbf{e}_\beta) \sigma_{\beta \alpha}\end{aligned} \hspace{\stretch{1}}(1.3)

## Returning to the tutorial notes

Simon writes

\begin{aligned}f_\perp &= - \sigma_{\perp \perp} - {\sigma'}_{\perp \perp} \\ f_\parallel &= - \sigma_{\parallel \perp} - {\sigma'}_{\parallel \perp} \end{aligned} \hspace{\stretch{1}}(1.4)

and then says stating this solution is very non-trivial, because $\sigma_{\alpha \beta}$ is non-linear in $\mathbf{E}$ and $\mathbf{B}$. This non-triviality is a good point. Without calculating it, I find the results above to be pulled out of a magic hat. The point of the tutorial discussion was to work through this in detail.

# Working out the tensor.

FIXME: PICTURE:

The Reflection coefficient can be defined in this case as

\begin{aligned}R = \frac{ {\left\lvert{\mathbf{E}'}\right\rvert}^2 }{ {\left\lvert{\mathbf{E}}\right\rvert}^2 },\end{aligned} \hspace{\stretch{1}}(2.6)

a ratio of the powers of the reflected wave power to the incident wave power (which are proportional to ${\mathbf{E}'}^2$ and ${\mathbf{E}}^2$ respectively.

Suppose we pick the following orientation for the incident fields

\begin{aligned}E_x &= E \sin\theta \\ E_y &= -E \cos\theta \\ B_z &= E ,\end{aligned} \hspace{\stretch{1}}(2.7)

With the reflected assumed to be in some still perpendicular orientation (with this orientation picked for convienence)

\begin{aligned}E_x' &= E' \sin\theta \\ E_y' &= E' \cos\theta \\ B_z' &= E'.\end{aligned} \hspace{\stretch{1}}(2.10)

Here

\begin{aligned}E &= E_0 \cos(\mathbf{p} \cdot \mathbf{x} - \omega t) \\ E' &= \sqrt{R} E_0 \cos(\mathbf{p}' \cdot \mathbf{x} - \omega t)\end{aligned} \hspace{\stretch{1}}(2.13)

FIXME: there are assumptions below that $\mathbf{p}' \cdot \mathbf{x} = \mathbf{p} \cdot \mathbf{x}$. I don’t see where that comes from, since the propagation directions are difference for the incident and the reflected waves.

\begin{aligned}\sigma_{\alpha\beta} = -T^{\alpha\beta} = \frac{1}{{4\pi}} \left(\mathcal{E}^\alpha\mathcal{E}^\beta+\mathcal{B}^\alpha\mathcal{B}^\beta- \frac{1}{{2}} \delta^{\alpha\beta} ( \vec{\mathcal{E}}^2 + \vec{\mathcal{B}}^2 )\right)\end{aligned} \hspace{\stretch{1}}(2.15)

## Aside: On the geometry, and the angle of incidence.

According to wikipedia [2] the angle of incidence is measured from the normal.

Let’s use complex numbers to get the orientation of the electric and propagation direction fields right. We have for the incident propagation direction

\begin{aligned}-\hat{\mathbf{p}} \sim e^{i (\pi + \theta) }\end{aligned} \hspace{\stretch{1}}(2.16)

or

\begin{aligned}\hat{\mathbf{p}} \sim e^{i\theta}\end{aligned} \hspace{\stretch{1}}(2.17)

If we pick the electric field rotated negatively from that direction, we have

\begin{aligned}\hat{\mathbf{E}} &\sim -i e^{i \theta} \\ &= -i (\cos\theta + i \sin\theta) \\ &= -i \cos\theta + \sin\theta\end{aligned}

Or

\begin{aligned}E_x &\sim \sin\theta \\ E_y &\sim -\cos\theta\end{aligned} \hspace{\stretch{1}}(2.18)

For the reflected direction we have

\begin{aligned}\hat{\mathbf{p}}' \sim e^{i (\pi - \theta)} = - e^{-i \theta}\end{aligned} \hspace{\stretch{1}}(2.20)

rotating negatively for the electric field direction, we have

\begin{aligned}\hat{\mathbf{E}'} &\sim -i (- e^{-i\theta} ) \\ &= i(cos\theta - i\sin\theta) \\ &= i cos\theta + \sin\theta\end{aligned}

Or

\begin{aligned}E_x' &\sim \sin\theta \\ E_y' &\sim \cos\theta \end{aligned} \hspace{\stretch{1}}(2.21)

## Back to the problem (again).

Where $\vec{\mathcal{E}}$ and $\vec{\mathcal{B}}$ are the total EM fields.

\paragraph{Aside:} Why the fields are added in this fashion wasn’t clear to me, but I guess this makes sense. Even if the propagation directions differ, the total field at any point is still just a superposition.

\begin{aligned}\vec{\mathcal{E}} &= \mathbf{E} + \mathbf{E}' \\ \vec{\mathcal{B}} &= \mathbf{B} + \mathbf{B}'\end{aligned} \hspace{\stretch{1}}(2.23)

Get

\begin{aligned}\sigma_{3 3} &= \frac{1}{{4 \pi}} \left( \underbrace{B_z B_z}_{=\vec{\mathcal{B}}^2} - \frac{1}{{2}} (\vec{\mathcal{E}}^2 + \vec{\mathcal{B}}^2) \right) = 0 \\ \sigma_{3 1} &= 0 = \sigma_{3 2} \\ \sigma_{1 1} &= \frac{1}{{4 \pi}} \left( (\mathcal{E}^1)^2 - \frac{1}{{2}} (\vec{\mathcal{E}}^2 + \vec{\mathcal{B}}^2) \right) \end{aligned} \hspace{\stretch{1}}(2.25)

\begin{aligned}\vec{\mathcal{B}}^2 &= (B_z + B_z')^2 = (E + E')^2 \\ \vec{\mathcal{E}}^2 &= (\mathbf{E} + \mathbf{E}')^2\end{aligned} \hspace{\stretch{1}}(2.28)

so

\begin{aligned}\sigma_{1 1} &= \frac{1}{{4 \pi}} \left( (\mathcal{E}^1)^2 - \frac{1}{{2}} ((\mathcal{E}^1)^2 + (\mathcal{E}^2)^2 + (E + E')^2 \right) \\ &= \frac{1}{{8 \pi}} \left( (\mathcal{E}^1)^2 - (\mathcal{E}^2)^2 - (E + E')^2 \right) \\ &= \frac{1}{{8 \pi}} \left( (E + E')^2 \sin^2\theta -(E' - E)^2 \cos^2\theta -(E + E')^2\right) \\ &= \frac{1}{{8 \pi}} \left( E^2( \sin^2\theta - \cos^2\theta - 1)+(E')^2( \sin^2\theta - \cos^2\theta - 1)+ 2 E E' (\sin^2\theta + \cos^2\theta -1 )\right) \\ &= \frac{1}{{8 \pi}} \left( - 2 E^2 \cos^2\theta - 2 (E')^2 \cos^2\theta \right) \\ &= -\frac{1}{{4 \pi}} (E^2 + (E')^2) \cos^2\theta \\ &= \sigma_\parallel + {\sigma'}_\parallel\end{aligned}

This last bit I didn’t get. What is $\sigma_\parallel$ and ${\sigma'}_\parallel$. Are these parallel to the wall or parallel to the normal to the wall. It turns out that this appears to mean parallel to the normal. We can see this by direct calculation

\begin{aligned}\sigma_{x x}^{\text{incident}} &= \frac{1}{{4 \pi}} \left( E_x^2 - \frac{1}{{2}} (\mathbf{E}^2 + \mathbf{B}^2)\right) \\ &= \frac{1}{{4 \pi}} \left( E^2 \sin^2 \theta - \frac{1}{{2}} 2 E^2 \right) \\ &= -\frac{1}{{4 \pi}} E^2 \cos^2\theta\end{aligned}

\begin{aligned}{\sigma}_{x x}^{\text{reflected}} &= \frac{1}{{4 \pi}} \left( {E_x'}^2 - \frac{1}{{2}} ({\mathbf{E}'}^2 + {\mathbf{B}'}^2)\right) \\ &= \frac{1}{{4 \pi}} \left( {E'}^2 \sin^2 \theta - \frac{1}{{2}} 2 {E'}^2 \right) \\ &= -\frac{1}{{4 \pi}} {E'}^2 \cos^2\theta\end{aligned}

So by comparison we see that we have

\begin{aligned}\sigma_{1 1} = {\sigma}_{x x}^{\text{incident}} +{\sigma}_{x x}^{\text{reflected}} \end{aligned} \hspace{\stretch{1}}(2.30)

Moving on, for our other component on the $x,y$ place $\sigma_{12}$ we have

\begin{aligned}\sigma_{12} &= \frac{1}{{4 \pi}} \mathcal{E}^1 \mathcal{E}^2 \\ &= \frac{1}{{4 \pi}} (E + E') \sin\theta (-E + E') \cos\theta \\ &= \frac{1}{{4 \pi}} ((E')^2 - E^2) \sin\theta \cos\theta \end{aligned}

Again we can compare to the sums of the reflected and incident tensors for this $x,y$ component. Those are

\begin{aligned}\sigma_{12}^{\text{incident}} &= \frac{1}{{4 \pi}} ( E^1 E^2 ) \\ &= -\frac{1}{{4 \pi}} E^2 \sin\theta \cos\theta,\end{aligned}

and

\begin{aligned}\sigma_{12}^{\text{reflected}} &= \frac{1}{{4 \pi}} ( {E'}^1 {E'}^2 ) \\ &= \frac{1}{{4 \pi}} {E'}^2 \sin\theta \cos\theta\end{aligned}

Which demonstrates that we have

\begin{aligned}\sigma_{12} = \sigma_{12}^{\text{incident}} + \sigma_{12}^{\text{reflected}} \end{aligned} \hspace{\stretch{1}}(2.31)

Summarizing, for the components in the $x,y$ plane we have found that we have

\begin{aligned}\sigma_{\alpha\beta}^{\text{total}} n_\beta = \sigma_{\alpha 1 }^{\text{total}} = \sigma_{\alpha 1} + {\sigma'}_{\alpha 1}\end{aligned} \hspace{\stretch{1}}(2.32)

(where $n_\beta = \delta^{\beta 1}$)

This result, assumed in the text, was non-trivial to derive. It is also not generally true. We have

\begin{aligned}\sigma_{2 2} &= \frac{1}{{4 \pi}} \left( (\mathcal{E}^y)^2 - \frac{1}{{2}} ( \vec{\mathcal{E}}^2 + \vec{\mathcal{B}}^2 ) \right) \\ &= \frac{1}{{8 \pi}} \left( (\mathcal{E}^y)^2 - (\mathcal{E}^x)^2 - \vec{\mathcal{B}}^2 ) \right) \\ &= \frac{1}{{8 \pi}} \left( (E' - E)^2 \cos^2\theta - (E + E')^2 \sin^2\theta - (E + E')^2\right) \\ &= \frac{1}{{8 \pi}} \left( E^2 ( -1 + \cos^2 \theta - \sin^2\theta )+{E'}^2 ( -1 + \cos^2 \theta - \sin^2\theta )+ 2 E E' ( -\cos^2\theta - \sin^2\theta -1 ) \right) \\ &= -\frac{1}{{4 \pi}} \left( E^2 \sin^2 \theta + (E')^2 \sin^2 \theta + 2 E E' \right)\end{aligned}

If we compare to the incident and reflected tensors we have

\begin{aligned}\sigma_{y y}^{\text{incident}} &= \frac{1}{{4 \pi}} \left( (E^y)^2 -\frac{1}{{2}} E^2 \right) \\ &= \frac{1}{{4 \pi}} E^2 ( \cos^2\theta - 1 ) \\ &= -\frac{1}{{4 \pi}} E^2 \sin^2\theta \end{aligned}

and

\begin{aligned}\sigma_{y y}^{\text{reflected}} &= \frac{1}{{4 \pi}} \left( ({E'}^y)^2 -\frac{1}{{2}} {E'}^2 \right) \\ &= \frac{1}{{4 \pi}} {E'}^2 ( \cos^2\theta - 1 ) \\ &= -\frac{1}{{4 \pi}} {E'}^2 \sin^2\theta \end{aligned}

There’s a cross term that we can’t have summing the two, so we have, in general

\begin{aligned}\sigma_{2 2}^{\text{total}} \ne \sigma_{y y}^{\text{incident}} +\sigma_{y y}^{\text{reflected}} \end{aligned} \hspace{\stretch{1}}(2.33)

## Force per unit area?

\begin{aligned}f_\alpha = n^x \sigma_{x \alpha} \end{aligned} \hspace{\stretch{1}}(2.34)

Averaged

\begin{aligned}\left\langle{{\sigma_{xx}}}\right\rangle &= -\frac{1}{{8 \pi}} E_0^2 ( 1 + R) \cos^2\theta \\ \left\langle{{\sigma_{xy}}}\right\rangle &= -\frac{1}{{8 \pi}} E_0^2 ( 1 - R) \sin\theta \cos\theta\end{aligned} \hspace{\stretch{1}}(2.35)

\begin{aligned}\left\langle{\mathbf{S}}\right\rangle &= -\frac{c}{8 \pi} E_0^2 \hat{\mathbf{n}} \\ \left\langle{{\mathbf{S}'}}\right\rangle &= -\frac{c}{8 \pi} E_0^2 \hat{\mathbf{n}}'\end{aligned} \hspace{\stretch{1}}(2.37)

\begin{aligned}\left\langle{{{\left\lvert{\mathbf{S}}\right\rvert}}}\right\rangle = \text{Work} = W\end{aligned} \hspace{\stretch{1}}(2.39)

\begin{aligned}f_x &= n^x \sigma_{x x} = W (1 + R) \cos^2\theta \\ f_y &= n^y \sigma_{x y} = W (1 - R) \sin\theta \cos\theta \\ f_z &= 0\end{aligned} \hspace{\stretch{1}}(2.40)

# A problem from Griffiths.

FIXME: try this.

Two charges $q+$, $q-$ reflected in a plane, separated by distance $a$. Work out the stress energy tensor from the Coulomb fields of the charges on the plane.

Will get the Coulomb force:

\begin{aligned}\mathbf{F} = k \frac{q^2}{2 a^2}.\end{aligned} \hspace{\stretch{1}}(3.43)

# Infinite parallel plate capacitor

Write $\sigma_{\alpha\beta}$.

\begin{aligned}\mathbf{B} &= 0 \\ \mathbf{E} &= - \frac{\sigma}{\epsilon_0} \mathbf{e}_z\end{aligned} \hspace{\stretch{1}}(4.44)

FIXME: derive this. Observe that we have no distance dependence in the field because it is an infinite plate.

\begin{aligned}\sigma_{1 1} &= \left( - \frac{1}{{2}} \delta^{1 1} \left( \frac{-\sigma}{\epsilon_0} \right)^2 \right) = - \frac{ \sigma^2}{ 2 \epsilon_0^2 } = \sigma_{22} \\ \sigma_{3 3} &= \left( (E^3)^2 - \frac{1}{{2}} \mathbf{E}^2 \right) = - \frac{1}{{2}} \mathbf{E}^2 = - \sigma_{2 2}\end{aligned} \hspace{\stretch{1}}(4.46)

Force per unit area is then

\begin{aligned}f_\alpha &= n_\beta \sigma_{\alpha \beta} \\ &= n_3 \sigma_{\alpha 3}\end{aligned}

So

\begin{aligned}f_1 &= 0 = f_2 \\ f_3 &= \sigma_{3 3} = -\frac{\sigma^2}{2 \epsilon_0^2}\end{aligned} \hspace{\stretch{1}}(4.48)

\begin{aligned}\mathbf{f} = -\frac{\sigma^2}{2 \epsilon_0^2} \mathbf{e}_z\end{aligned} \hspace{\stretch{1}}(4.50)

REMEMBER: EXAM WEDNESDAY!

# References

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

[2] Wikipedia. Angle of incidence — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 11-April-2011]. http://en.wikipedia.org/w/index.php?title=Angle_of_incidence&oldid=421647114.