### 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…

### The Valentine's Day Magnetic Monopole

There's an assymetry to the form of the two Maxwell's equations shown in picture 1.  While the divergence of the electric field is proportional to the electric charge density at a given point, the divergence of the magnetic field is equal to zero.  This is typically explained in the following way.  While we know that electrons, the fundamental electric charge carriers exist, evidence seems to indicate that magnetic monopoles, the particles that would carry magnetic 'charge', either don't exist, or, the energies required to create them are so high that they are exceedingly rare.  That doesn't stop us from looking for them though!

Keeping with the theme of Fairbank[1] and his academic progeny over the semester break, today's post is about the discovery of a magnetic monopole candidate event by one of the Fairbank's graduate students, Blas Cabrera[2].  Cabrera was utilizing a loop type of magnetic monopole detector.  Its operation is in concept very simpl…

### Unschooling Math Jams: Squaring Numbers in their own Base

Some of the most fun I have working on math with seven year-old No. 1 is discovering new things about math myself.  Last week, we discovered that square of any number in its own base is 100!  Pretty cool!  As usual we figured it out by talking rather than by writing things down, and as usual it was sheer happenstance that we figured it out at all.  Here’s how it went.

I've really been looking forward to working through multiplication ala binary numbers with seven year-old No. 1.  She kind of beat me to the punch though: in the last few weeks she's been learning her multiplication tables in base 10 on her own.  This became apparent when five year-old No. 2 decided he wanted to do some 'schoolwork' a few days back.

"I can sing that song... about the letters? all by myself now!"  2 meant the alphabet song.  His attitude towards academics is the ultimate in not retaining unnecessary facts, not even the name of the song :)

After 2 had worked his way through the so…