### The Handy Contour Integral

Suppose you have an integral that looks like the following seen during yesterday's quantum lecture:

Your professor turns, looks at you and says, "Who can do this integral?"  After a bit, no one answers and he grins and writes down that the answer is pi.  Then he waves it away as being easy as a contour integral.  Well yeah, but how?  Here's how...

We already said we're going to a contour integral, so that mystery is solved.  We're going to move the integral into the complex plane, and choose a contour that skips the pole at u equal to zero.  Something like this

The portion lableled B skips around the pole at u equals 0, the portion labeled A is on the real axis and stretches from negative infinity to infinity.  The slightly off-screen semi-circle labled C is the return path that in this case integrates out to zero at infinity, (Jordan's lemma and whatnot if you're into the details).  We kept the pole outside of the contour, so we know at the end of the day, the entire integral has to wind up being zero, (Cauchy theorem?).  That means that A is equal to negative B.  A is pretty much our original integral that we didn't know how to to do in the first place, so maybe B will be easier and it is (sort of).

First, we re-write our integral as

We've just re-written cosine as the real part of e to the i times u power, which is true, (check out Euler's formulas).  This still isn't exactly tidy, so the next step is to remember that our little jig around the pole was small. Then, we can approximate the exponential as it's Taylor series.

Things are getting better.  Now, we're down to an integral that just has powers of the variable.  Since we're integrating around a semi-circle, the next step will be to represent u as an angular variable.

We've re-written u as e to the i times phi.  I know, I know, I just put the exponential back in that I used my Taylor Series to get rid of.  Don't worry, though, it will go away almost immediately as it gets cancelled by the new differential for the new variable of integration phi.

Finally, we're just left with the very simple integral in the last step and we get that our original integral along A which is equal to negative B, (the integral we just did), is in fact pi.

What to Look For
The two keys that should have pointed me iand everyone else in the direction of contours are that the integral extended from positive to negative infinity and that the integral involved a trigonometric function to begin with.

The Preachy Bit
Physics professors, (at least mine), spend a lot of time saying "Don't memorize the formulas.  Look them up."  They only kind of mean it though.  The big kids, (professors), have all the tools used above memorized including the Taylor series for the exponential, Euler's formula, and the fact that they're headed into the complex plane in the first place.

A Silly Question
I've been trying to perfect getting paper to look like a chalk board.  It's not much, but you gotta have a hobby.  In today's post, I thought I'd nailed it with pictures 3 - 5.  Elaine said not so much.  She thought picture 2 was the way to go.  Any thoughts?

### 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…

### 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…

### 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…