### EMII Notes: Special Relativity and Gamma via Initial Conditions

I'm working on a new writing project as I prepare for Electricity and Magnetism II this semester.  I'll be reviewing the class lecture notes each day leading up to the start of the semester.  After each half hour review, I intend to write up a summary of the notes.  (Yes, I took a reflective writing workshop this summer, and I'm trying it out.)

Pages covered today:  7 & 8

We consider light as a spherical wavefornt

$x^2 + y^2 + z^2 - c^2t^2 = 0$

$x^{\prime 2} + y^{\prime 2} + z^{\prime 2} - c^2t^{\prime 2} = 0$

We also look at the usual situation with two frames, $S$ and $S^{\prime}$ that are moving parallel to each other where $S^{\prime}$ has positive velocity $v$ in the x direction with respect to $S$.

We have the usual postulate that c, the speed of light is constan in all frames.

We need a way to transform the distance and time coordinates in the two frames.  Because we want to preserve invariance with respect to translation, (conservation of momentum), and invariance with respectc to translations in space-time, (conservation of energy), we're stuck with a linear transformation.  so

$x^{\prime} = Ax + Bt$

$t^{\prime} = Cx + Dt$

Question:  Is the above an example of a bilinear transform?

We won't really bother with C and D I don't think.

At time t = 0, we set the origins of both frames at the same position and get

$x^{\prime} = 0 => x = vt$

so

$0 = Avt + Bt$
$B = -Av$

Now, since we know where we're headed, go ahead and rename $A$ to be $\gamma$.

This all gives:

$x^{\prime} = \gamma x - \gamma v t$
$x^{\prime} = \gamma\left( x - v t\right)$

For the $S$ frame at $x = 0$, we have $x^{\prime} = -vt^\{prime}$

We can use this reversed velocity to do a nice little trick giving

$x = \gamma\left( x^{\prime} - v t^{\prime}\right)$

without doing any math to solve for $x$ in terms of $x^{\prime}$

Since the speed of light is constant in every frame.  That's our postulate.  we get

$x = ct <==> x^{\prime} = ct^{\prime}$

Plugging this into the $AB$ transform equation above we get

$ct^{\prime} = \gamma x - \gamma v t$
$ct^{\prime} = \gamma\left(ct - vt\right)$

and the following string of algebra from solving the two equations above for gamma and then multiplying them together.

$\gamma = \dfrac{ct}{t^{\prime}\left(c+v\right)}$
$\gamma = \dfrac{ct^{\prime}}{t\left(c - v\right)}$

multiply the two equations for gamma to get.  (Why?  Because I know the answer I'm headed for and this seems simplest).

$\gamma^2 = \dfrac{c^2 t t^{\prime}}{t t^{\prime} \left(c + v\right)\left(c - v\right)}$
$= \dfrac{c^2}{c^2 - v^2}$
$= \dfrac{1}{1-\dfrac{v^2}{c^2}}$
$\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

Question:
Since we never used the $BD$ transform equation, is there more information that can be gleaned from it?

### Cool Math Tricks: Deriving the Divergence, (Del or Nabla) into New (Cylindrical) Coordinate Systems

The following is a pretty lengthy procedure, but converting the divergence, (nabla, del) operator between coordinate systems comes up pretty often. While there are tables for converting between common coordinate systems, there seem to be fewer explanations of the procedure for deriving the conversion, so here goes!

What do we actually want?

To convert the Cartesian nabla

to the nabla for another coordinate system, say… cylindrical coordinates.

What we’ll need:

1. The Cartesian Nabla:

2. A set of equations relating the Cartesian coordinates to cylindrical coordinates:

3. A set of equations relating the Cartesian basis vectors to the basis vectors of the new coordinate system:

How to do it:

Use the chain rule for differentiation to convert the derivatives with respect to the Cartesian variables to derivatives with respect to the cylindrical variables.

The chain rule can be used to convert a differential operator in terms of one variable into a series of differential operators in terms of othe…

### Lost Phone

We were incredibly lucky to have both been in university settings when our kids were born.  When No. 1 arrived, we were both still grad students.  Not long after No. 2 arrived, (about 10 days to be exact), mom-person defended her dissertation and gained the appellation prependage Dr.

While there are lots of perks attendant to grad school, not the least of them phenomenal health insurance, that’s not the one that’s come to mind for me just now.  The one I’m most grateful for at the moment with respect to our kids was the opportunities for sheer independence.  Most days, we’d meet for lunch on the quad of whatever university we were hanging out at at the time, (physics research requires a bit of travel), to eat lunch.  During those lunches, the kids could crawl, toddle, or jog off into the distance.  There were no roads, and therefore no cars.  And, I realize now with a certain wistful bliss I had no knowledge of at the time, there were also very few people at hand that new what a baby…

### Lab Book 2014_07_10 More NaI Characterization

Summary: Much more plunking around with the NaI detector and sources today.  A Pb shield was built to eliminate cosmic ray muons as well as potassium 40 radiation from the concreted building.  The spectra are much cleaner, but still don't have the count rates or distinctive peaks that are expected.
New to the experiment?  Scroll to the bottom to see background and get caught up.
Lab Book Threshold for the QVT is currently set at -1.49 volts.  Remember to divide this by 100 to get the actual threshold voltage. A new spectrum recording the lines of all three sources, Cs 137, Co 60, and Sr 90, was started at approximately 10:55. Took data for about an hour.
Started the Cs 137 only spectrum at about 11:55 AM

Here’s the no-source background from yesterday
In comparison, here’s the 3 source spectrum from this morning.

The three source spectrum shows peak structure not exhibited by the background alone. I forgot to take scope pictures of the Cs137 run. I do however, have the printout, and…