### 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…

### 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…

### Kids R Kapable

Just a little note to concerned ‘grownups’ everywhere.  If you look at a kid—and I mean really look—I don’t mean notice a person shorter than you, I mean make eye contact, notice their facial expression and observe their body language—If you look at a kid, don’t assume they need your help unless they’re obviously distressed, or ask for it.  You might think this is difficult call to make.  You might think, not having kids of your own, that you’re unable to make this determination.  You are.  You do in fact, already have the skills even if you’ve never been around kids  It’s a remarkably simple call to make, just use the exact same criteria you would for determining if an adult was in distress.  Because, guess what, kids and adults are in fact the same species of animal and communicate in the same way.  Honest.  If someone—adult or child—doesn’t need your help, feel free to say hello, give a wave, give a smile, but don’t—do not—try to force help on anyone that doesn’t want or need it.

Y…