Monday, November 24, 2014

EM II Notes 2014_11_24: Leinard-Wiechert Potentials

There's sooo much going on today.  I'm back in the lab again, but I'm also studying for the last little bit of my EM II class.   Here are the EM notes for today.  Hopefully, I'll get a lab book up again in the morning.

Looking at the Leinard-Wiechert Potentials.  

We'll have a particel mofin along hte path $\vec{r} = \vec{r_o}\left(t\right)$.  There is a quite lengthy explanation of IRFs, but I'll skip that for now and keep careful track of whether or not this comes back to bite me in the butt.  We define $\vec{R}\left(t^\prime\right) = \vec{r} - \vec{r_0}\left(t\right)$ which is the vector from the point charge at time $t^\prime$ to the observatin poitn $\left(\vec{r}, t\right)$.  This gives us a retarded time, $t^\prime$ determined by $t - t^\prime = R\left(t^\prime\right)$, where $R\left(t^\prime\right) = |\vec{R}\left(t^\prime\right)|$.  This makes far more sense if you translate one of the ever present ever invisible $1$s to a c to get $c\left(t - t^\prime\right) = R\left(t^\prime\right)$

The potentials in the IRf can be written as

$\phi = \dfrac{e}{R\left(t^\prime\right)}$, $\vec{A} = 0$.

A charge at rest will have 4-veclocity $U^\mu = \left(1, 0, 0, 0\right)$.  

We can noew define the 4 potential to be $A^\mu = f U^\mu$.  We can also form a four vector version of $R^\mu$ as $R^\mu = \left(t - t^\prime, r - r_0\left(t^\prime\right)\right) = \left(t - t^\prime, \vec{R}\left(t^\prime\right)\right)$.  Looking at this, you should see a four space distance without the time axis turned negative.  In a sense, this fits because it isnt' squared yet.  In a sense it doesnt' because if it isn't squared, then the time componet shoudl have an $i$ out in front.  This is somewhat Wick rotated, to coin a somewhat fancy phrase.

In the special case where the charge truly isn't moving, then $f$ above shoudl be $e/R\left(t^\prime\right)$.  For the more general case where the charge is moving with four velocity $U^\mu$, we get

$f = \dfrac{e}{\left(-U^\nu R_\nu\right)}$, so $A^\mu = -\dfrac{eU^\mu}{U^\nu R^\nu}$

Here, we have rather mysteriously gotten our negative sign back in front of the time coordinate.  Ask the professor about this tomorrow.  The pertinent point is near equation 7.29.  

Now, on to problem number 2
For 2.a., and b, see the Wake Forest notes:

expression 17 and up give the appropriate curl.  If time allows take a look at Dr. Nevels article on graphically protraying E and B fields.  There's no reason everything shouldn't apply here since the L\&W potentials were derived classically. 

The strategy is just to bludgeon through the curl of $\vec{A}$ equation and get the final result.  Then, bludgeon through the cross product of $\vec{E}$ and show that the results are equivalient.

Sunday, November 23, 2014

EM II Notes 2014_11_23: Homework sketches

Just a few notes on how to proceed on the penultimate homework of the semester.

We're to show that the solutions for the 30/60/90 triangular waveguide given in the last homework set will also work for a waveguide that's formed from an equilateral traingle.  The three corners of the equilateral traingle are located at $\left(x,y\right) = \left(0, 0\right)$, $\left(x,y\right) = \left(a, a/\sqrt{3}\right)$, and $\left(x,y\right) = \left(a, -a/\sqrt{3}\right)$.

This falls out immediately from last week's homeowrk.  Because the sine function is peiodic in $\pi$ over the domain from $\left(-\infty, \infty\right)$, the solution given last week in terms of sines will still evaluate to zero on the wall that falls at negative $y$. coordites.  The positive $x$ coordinates of the functions will evaluate to 0 on the wall in the same manner they did before???  There's an issue here.  It's products of the $x$ and $y$ sinusoids that all sum to zero.  These will need to be checked to determine if they still evaluate to zero.  Since the y terms are converted to x terms using the coordinates of the walls, there's very little to do in checking that things are stil OK.  Everyting converts down to a difference of cosines expressed as

$cos\dfrac{m-n}{3a} - cos\dfrac{n-m}{3a}$

However, because the three $y$ sin functions are odd, they will merely wind up with negative signs out in front that can be factored out of the entire expression.  The result will be the same.

Saturday, November 15, 2014

YBCO Four Point Measurements: Lab Book 2014_11_15

I was back in the lab this morning.  I’m working on getting the four point measurement to work on the YBCO sample.  In the grand scheme of things, this is low priority, but it’s important to know that we can successfully make these measurements here before we have a large bucket of liquid helium evaporating with a sample inside.  Here’s what the four point probe measurement looked like:

There are still no conclusive results.  With any luck this is a consequence of me not being able to interpret the results more than a bad experimental setup.  The table below details the four point probe readings in ohms as the superconductor cools

Table of four point readings
Resistance kohm
Negative reading is probably from swapped sense wires.
Immediately after nitrogen pour
Near minimum
Minimum immediately after new pour
Magnet in place beside superconductor in reservoir

I had expected more conclusive readings, a zero would have been nice for example. 

Wednesday, October 29, 2014

Here's today's special relativistic EM question.  Can the Thomas precession be shown to be a special case of the perihelion advance of relativistic elliptical orbits?  Any ideas?  Here's what's going on:

We've been deriving the special relativistic  orbit of a charged particles around another fixed charged particle.  At the end of the day, you wind up with a perihelion advance which is a fancy way to say that major axis of the elliptical orbit won't stay put.  It swivels around, (orbits), the charged particle as well.  The advance angle of the major axis winds up being\\

$\delta\phi = 2\pi\left[\left(1 - \dfrac{\kappa^2}{l^2}\right)^{-1/2} - 1\right]$

Which is very, very, similar to the Thomas angle for the spin precession, or gyroscopic precession along a circular orbit at special relativistic speeds:\\

$\delta\phi = 2\pi\left[cosh\left(w\right) - 1\right]$

$= 2\pi\left[\left(1 - \dfrac{v^2}{c^2}\right)^{-1/2} - 1\right]$

In the expression for the perihelion advance, $\kappa = eQ$ where $e$ is the charge of the orbiting particle and $Q$ is the charge of the fixed particle.  $l$ is the angular momentum and is written as $l = mr^2\left(\dfrac{d\phi}{d\tau}\right)$

Just a little more massaging of the above before I'm done for the day.  For a circular orbit, $\dfrac{d\phi}{d\tau} = \omega$ is constant and we can write $l = m\omega r^2$.  If we plug this in as to the Perihelion advance equation we wind up with

$\dfrac{\kappa^2}{l^2} = \dfrac{Q^2 e^2}{m^2v^2r^2}$ where $v = \omega r$

Here's where I get into trouble for playing fast and loose with things that migh actually be rapidities instead of velocities like $v$ above.  However, if you carry the simplificatins a bit further out, I believe you wind up with something that looks like an units of potential energy over momentum squared.

$\dfrac{Q^2 e^2}{m^2v^2r^2} = \dfrac{F}{m} \dfrac{1}{mv^2} \sim \dfrac{a}{E}$

Notes du jour:

Tuesday, October 28, 2014

EM Notes Part I: The visual bit of relativistic EM fields pointing at the observer

This is kind of cool from yesterday's EM notes.  Our professor pointed out that if you calculate the field from a relativistically moving electric charge, you'll always find that it's pointed straight at the point of observation.  Anyone have any idea why?  The argument could certainly be made that if you measure the field from a static charge that it will also be pointing straight at you.  Then, there's also the realization that the Lorentz transformation only affects the E and B fields in a frame that are perpendicular to the frame's tangential velocity.  I'm not sure that's either here or there since the point of observation can be anywhere.  Here's the associated diagram for the curious.