### Accelerating Frames: Cosmology Homework

Our cosmology course is well under way and it's  a lot of fun so far!  The class direction overall is towards describing the inflationary universe by means of quantum field theory, but this week we're focused on relativity.  We're allowed to work on our homework together, however, I'm spending most of my time in the lab this semester, so I'll be posting my homework notes here.  If you'd like to grab bits and pieces, make suggestions, or contribute, the whole shooting match will also be archived on github.

Our first homework contains a problem that involves accelerating reference frames.  The question is, given the transformation between the lab and the accelerating reference frame, figure out if the line element $ds^2$ is preserved.  There are a few interesting aspects to this problem.  First, while the transform looks similar to Rindler coordinates, it's not, (as ar as I can tell.)  Second, looking into Rindler coordinates a bit, they seem to maintain an invariant $ds^2$, which kind of makes sense, because the calculation is made moving form one instantaneous inertial rest frame to another in succession.  This also kind of doesn't make sense because mentions of acceleration fairly scream, non-inertial.

3.  We need to find out if $ds^2$, the four-space line element is invariant under a given transformation.
The given transformation is

$t' = \dfrac{1}{g}\left(e^{gz}\right)sinh\left(gt\right)$

$z' = \dfrac{1}{g}\left(e^{gz}\right)cosh\left(gt\right)$

So,

$dt' = \dfrac{1}{g}g\left(e^{gz}\right)sinh\left(gt\right)dz + \dfrac{1}{g}g\left(e^{gz}\right)cosh\left(gt\right)dt$

$dz' = \dfrac{1}{g}g\left(e^{gz}\right)cosh\left(gt\right)dz + \dfrac{1}{g}g\left(e^{gz}\right)sinh\left(gt\right)dt$

$dt^{\prime 2} = e^{2gz}sinh^2\left(gt\right)dz^2 + e^{2gz}cosh^2\left(gt\right)dt^2$

Where the mixed differentials have been omitted because they will cancel with the mixed terms from $dz^{\prime 2}$.

$dz^{\prime 2} = e^{2gz}cosh^2\left(gt\right)dz^2 + e^{2gz}sinh^2\left(gt\right)dt^2$

$ds^{\prime 2} = dt^{\prime 2} - dz^{\prime 2} = e^{2gz}sinh^2\left(gt\right)dz^2 + e^{2gz}cosh^2\left(gt\right)dt^2 -$
$\left(e^{2gz}cosh^2\left(gt\right)dz^2 + e^{2gz}sinh^2\left(gt\right)dt^2\right)$

$= e^{2gz}\left(sinh^2\left(gt\right) - cosh^2\left(gt\right)\right)dz^2 + e^{2gz}\left(cosh^2\left(gt\right) - sinh^2\left(gt\right)\right)dt^2$

$ds^{\prime 2} = e^{2gz}\left(dt^2 - dz^2\right)$

Consequently, $ds^{\prime 2}$ is not invariant in this metric.

Just as a quick check, what if we had factored the $gz$ term back in as a phase to the $sinh$ and $cosh$ functions right away?  How would that have changed things?  Can the $e^{gz}$ even be factored back in as a phase?

$sinh$ can be written as

$sinh\left(x\right) = \dfrac{e^x +e^{-x}}{2}$

Factoring the $e^{gz}$ back in gives us

$\dfrac{e^{gt + gz} +e^{-gt + gz}}{2}$

So, it looks like life is good.  We can't get back to a single $sinh$ or $cosh$ when you take the $gz$ back into the expression.

Questions
The metric given for this problem looks like the Davies and Birrell form of the Rindler coordinates.  However, if we write down Rindler coordinates from Rindler's "Special Relativity" book, (1960, p. 41), they are:

$t = \dfrac{1}{g}sinh\left(g \tau\right)$

$z = \dfrac{1}{g}cosh\left(g \tau\right) - \dfrac{1}{g}$

While the missing leading exponential term, (since it's not there), will not cause the final result to not be invariant in $ds^2$, other issues may.  Playing the same $dt^2 - dz^2$ trick as above, (even though we're not looking at the primed, but the unprimed frame), results in

$ds^2 = dt^2 - dz^2 = cosh^2\left(g\tau\right)d\tau^2 - sinh^2\left(g\tau\right)d\tau^2$

or $ds^2 = d\tau^2$

This seems to indicate that Rindler coordinates maintain an invariance in $ds$ even though they incorporate acceleration and a non-inertial frame.  It would also seem to indicate that the metric given in this problem is not in fact the same as Rindler coordinates.

Picture of the Day:
East Texas Sunset

Associated .tex file:
https://github.com/hcarter333/CosmologyHWork/blob/master/CosmologyHWI.tex

### 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…