### Proper Velocity!!! and Getting Index Notation Worked Out: EM II Notes 2014_09_09

Summary:  It looks like I'll finally get a good understanding of the gamma notation for moving proper velocities to lab velocities and back.  It'll be nice to know it inside and out, but a little irksome given all that can be done with the hyperbolic notation we're not using.  I want to maintain my fluency in both.

There may be a subtle second notation for inverted Lorentz transforms.  As it turns out, the subtle notation difference of moving around indices in the top and the bottom with spaces is meant to keep track of which index comes first when you go back to side by side notation.

First, we cover Lorentz transforms, (which are not in fact tensors), and contractions and arrive at the interesting result in equation 1.99:

$\Lambda^\mu_\rho \Lambda^\sigma_\mu T^\rho_\sigma = \delta^\sigma_\rho T^\rho_\sigma$

Which indicates the transpose of the Lorentz transform times itself follows a sort of orthogonality rule making use of contravariant indices.

Q:  Does this obviate the need for the $\eta$ metric?

A:  But wait!  There's so much more!  This is a way to write things without the $\eta$ cruising through everywhere, but it also explains the oddball spacing of the indices.  Maybe I just don't remember this from MacConnel?  First, the Lorentz transform is not a tensor.  Now, we hve that down.  The next bit is how to arrive at the above expression.

$\eta_{\mu\nu} \Lambda^\mu_{\;\rho} \Lambda^\nu_{\;\sigma} = \eta_{\rho\sigma}$

The next step is to plow the $\eta$ on the l.h.s. in and lower the $\nu$ on the second transform.

$\Lambda^\mu_{\;\rho} \Lambda_{\mu\sigma} = \eta_{\rho\sigma}$

We then raise the sigma

$\eta^{\sigma\lambda} \Lambda^\mu_{\;\rho} \Lambda_{\mu\sigma} = \eta^{\sigma\lambda}\eta_{\rho\sigma}$

$\eta^{\sigma\lambda} \Lambda^\mu_{\;\rho} \Lambda_{\mu\sigma} = \eta_\rho^\lambda = \delta_\rho^\lambda$

$\Lambda^\mu_{\;\rho} \Lambda^{\;\lambda}_\mu = \eta_\rho^\lambda = \delta_\rho^\lambda$

Now, since we can see that the first two indices are $\mu$s, we can call the above statement a transpose.  The index locations matter to get the transpose to be a transpose, making sure the rows and columns are handled properly.

NOTE:  This doe arise in MacConnel.  His notation is slightly different.  Instaed of $\Lambda^\mu_{\;\rho}$, he writes $\Lambda^\mu_{.\rho}$

We're going to need the delLambertian soon and it's important to note that it is

$\Box = -\partial_0\partial_0 + \partial_i\partial_i = \partial^\mu \partial_\mu$

A few notes follow on why the D'Alambertian is written with one index up and one down.  It has to do with the negative sign in the first entry of the Minkowski metric.  As it turns out, the index up version of the Kronecker delta is the same as the index down version, but not so for $\eta$ because of the $-1$ $00$ entry.  This is all much simpler if you never start writing your indices in the 'wrong' location in the first place.

OK, now for the interesting stuff.  First, with the choice of signature for the Minkowski metric here, we wind up having to write down $d\tau$ as \

$d\tau = -ds^2 = dt^2 - dx^2 - dy^2 - dz^2$

Given $d\tau$ we define the four velocity to be

$U^\mu = \dfrac{dx^\mu}{d\tau}$

Which is technically mixing frames, but then, that's what proper velocity does.  Intersting that we're starting with four velocity, proper velocity here.  It's really nice to get this out of the way early on and benefit from it.

Picture of the Day
Presenting the Takeno metric line element.  Hopefully one of the payoffs of all this will be understanding this more fully, and possibly using it to explain the circular Unruh effect.  However, a far more productive use is in explaining the Thomas Precession.

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

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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 differe…

### Division: Distributing the Work

Our unschooling math comes in bits and pieces.  The oldest kid here, seven year-old No. 1 loves math problems, so math moves along pretty fast for her.  Here’s how she arrived at the distributive property recently.  Tldr; it came about only because she needed it.
“Give me a math problem!” No. 1 asked Mom-person.

“OK, what’s 18 divided by 2?  But, you’re going to have to do it as you walk.  You and Dad need to head out.”

And so, No. 1 and I found ourselves headed out on our mini-adventure with a new math problem to discuss.

One looked at the ceiling of the library lost in thought as we walked.  She glanced down at her fingers for a moment.  “Is it six?”

“I don’t know, let’s see,” I hedged.  “What’s two times six?  Is it eighteen?”

One looked at me hopefully heading back into her mental math.

I needed to visit the restroom before we left, so I hurried her calculation along.  “What’s two times five?”

I got a grin, and another look indicating she was thinking about that one.

I flashed eac…