# Monthly Archives: February 2013

## Möbius Transformations

Möbius transformations (also called linear fractional transformations) are maps from the complex plane to itself of the form
$f(z) = \frac{a z + b}{c z + d}$
where $$a, b, c, d \in \mathbf{C}$$ and $$a d – b c \neq 0$$. This definition extends to functions on the Riemann sphere, where the geometry of such transformations becomes more apparent. The following short video shows off a remarkable correspondence between symmetries of the sphere and Möbius transformations.

teaching

## Explanation of 17.5 #48

In this problem, you are asked to compute the gravitational potential from uniform density spherical shell of total mass $$m$$ and radius $$R$$. In general, the gravitational potential of a massive surface $$S$$ is given by
$V(a, b, c) = – G \iint_S \frac{\rho\, dS}{\sqrt{(x – a)^2 + (y – b)^2 + (z – c)^2}}$
where $$\rho$$ is the density (mass per area) of the surface. In this case, $$\rho = m / (4 \pi R^2)$$.

The first part of the question asks you to reduce the problem to the case where $$(a, b, c) = (0, 0, r)$$. We can do this by spherical symmetry: given any $$(a,b,c)$$ we can rotate the entire picture so that the point $$(a, b, c)$$ lies on the positive $$z$$ axis. That is, $$(a, b, c) = (0, 0, r)$$ with $$r = \sqrt{x^2 + y^2 + z^2}$$. Applying this rotation doesn’t change the gravitational potential because the rotated sphere is indistinguishable from the original sphere.

The second part of the question asks you to set up the integral for $$V(0, 0, r)$$ in spherical coordinates. In spherical coordinates (with radius $$R$$) we have
$x = R \cos \theta \sin \varphi, \quad y = R \sin \theta \sin \varphi, \quad z = R \cos \varphi.$
You should verify that the normal vector for this parametrization satisfies $$\|n\| = R^2 \sin \varphi$$. Notice that the denominator of the integral defining the gravitational potential is just the distance from $$(a, b, c)$$ to $$(x, y, z)$$ (where the latter is a point on the surface). In the case where $$(a, b, c) = (0, 0, r)$$ and $$S$$ is the sphere of radius $$R$$ centered at the origin, we can compute this distance using the law of cosines as indicated in the following figure:

In the figure, $$d$$ is the distance from the point $$P = (0,0,r)$$ to a point $$Q = (x, y, z)$$ lying on the sphere — that is, $$d$$ is the denominator of the integral we’re trying to compute. By the law of cosines,
$d^2 = r^2 + R^2 – 2 R r \cos \varphi.$
Therefore, the potential is given by
$V(0, 0, r) = \frac{- G m}{4 \pi} \int_0^{\pi} \int_{0}^{2 \pi} \frac{\sin \varphi \, d\theta\, d\varphi}{\sqrt{R^2 + r^2 – 2 R r \cos \varphi}}.$
The book suggests using the substitution $$u = R^2 + r^2 – 2 R r \cos \varphi$$ to compute this integral.

teaching

## Math 32B/H Midterm 2 Review

I just finished writing up some review materials for the second midterm for Math 32B/H. There are two documents: the problems (here) and solutions (here). If you find any typos or points that need clarification, feel free to let me know in the comments below.

teaching

## Problem 17.2 #31

In this problem, we are asked to compute the line integral
$\int_C \frac{-y\, dx + x\, dy}{x^2 + y^2}$
where $$C$$ is the line segment from $$(1,0)$$ to $$(0,1)$$. It turns out that this integral is particularly easy to compute in polar coordinates (although the work that I did in class probably made it seem hopelessly complicated!).

In polar coorinates, we have $$x = r \cos \theta$$ and $$y = r \sin \theta$$. To get $$dx$$ and $$dy$$ in terms of $$r$$, $$\theta$$ and $$d\theta$$ we have to do a little work differentiating. Since we are integrating over the curve $$C$$, we will eventually have $$r = r(\theta)$$, a function of theta. We compute
$dx = d(r \cos \theta) = \left(\frac{dr}{d\theta} \cos \theta – r \sin \theta \right)\, d\theta$
and similarly
$dy = d(r \sin \theta) = \left(\frac{dr}{d\theta} sin \theta + r \cos \theta \right)\, d\theta.$
Using these formulas, the numerator of the integrand simplifies immensely:
$– y\, dx + x\, dy = r^2 \, d\theta.$
Therefore, the integral becomes (in polar coordinates)
$\int_C \frac{-y\, dx + x\, dy}{x^2 + y^2} = \int_C \, d\theta.$
The curve $$C$$ can be written in polar coordinates as
$r(\theta) = \frac{1}{\cos \theta + \sin \theta}$
where $$0 \leq \theta \leq \pi / 2$$. To see this, notice the line satisfies $$y = 1 – x$$, or equivalently $$x + y = 1$$. In polar coordinates, this equation is $$r \cos \theta + r \sin \theta = 1$$, and solving for $$r$$ gives the equation for $$C$$. Therefore, we can compute the integral
$\int_C\, d\theta = \int_0^{\pi / 2} \, d\theta = \frac{\pi}{2}.$
So despite the imposing setup to the problem, the final integral turns out to be very simple.

## Getty Center Textures

Last weekend my sister visited from Boston. During her well-timed visit, we went to the Getty Center. For me, the highlight of the Getty is usually the well-manicured grounds more than the exhibitions. During this visit I took pictures that focused on the textures I noticed around the gardens. Here are a few that I thought turned out reasonably well:

brewing

## Poseidon Double IPA

A couple months ago, a friend of mine drove down from Portland to visit. In exchange for putting him up for a few nights, I asked that he bring down some of those delicious Portland microbrews to share. He arrived with 3 growlers of ale from Hair of the Dog, which is probably my favorite brewery in Portland, if not the country.

One of the growlers contained Blue Dot Double IPA. Since leaving Portland, I had forgotten what a fantastic and unique beer it is. After doing some research, I decided that it was time to attempt to craft a similar beer in my kitchen. As one would expect from a West Coast double IPA, Blue dot contains a boat load of hops. What I think makes this beer special though is the grain bill: it only contains pilsner malt and flaked rye. Not an ounce of pale ale malt to be found. After sifting through some home brew forums, I decided on this recipe for a 5 gallon batch:

Grain Bill

• 15 lb German Pilsner malt
• 2.5 lb flaked rye

Hops

• 3.5 oz Warrior (15% AA) 75 minutes
• 3.5 oz Magnum (German) (14% AA) 30 minutes
• 3.5 oz Columbus (15% AA) 5 minutes
• 3 oz Centennial dry hop
• 3 oz Chinook dry hop

Yeast

For the mash, I performed a single temperature infusion at 154 F for 90 minutes, followed by 2 batch sparges at 165 F. The result was about 7.5 gallons of wort at a gravity of 1.070. After the boil, I was left with around 7 gallons of wort at 1.074. I lost probably another gallon of liquid which the insane amount of hops absorbed during the boil, so around 5.5 – 6 gallons went into the fermentor.

I finally got a chance to brew a batch of this beer on Sunday. I came home from campus on Monday to find the airlock of my fermentor clogged with hops and the bucket about to burst. I quickly pulled the airlock out of the lid, which resulted in a geyser of hops and beer foam in my dining room. After cleaning up the mess, I rigged a makeshift blow-off tube from the fermentor to relieve the pressure of a decidedly vigorous fermentation. In light of the beer’s tempestuous beginnings, I’ve decided to dub it Poseidon Double IPA. I am expecting a final gravity of around 1.020 for an ABV of 6.8%.

Update

I just racked Poseidon to the secondary fermenter. The final gravity is lower than I had anticipated, about 1.012 as I should have expected from the vigorous primary fermentation. So the ABV should be around 8%. I was going to rack on Friday, but I dropped my glass carboy while I was sanitizing it. Pro tip: buy a carboy handle so you don’t drop your carboy. They are expensive, heavy and when they shatter, they fill your kitchen with about 10 pounds of razor sharp shrapnel.

In case you’ve ever wondered what six ounces of whole leaf hops look like in a 5 gallon batch of beer, here you go:

teaching

## Explanation of 17.1 #34

This problem asks you to argue that the depicted vector field is not conservative. Recall that a vector field $$\mathbf{F}(x, y)$$ is conservative if we can write
$\mathbf{F}(x, y) = \nabla V(x, y)$
where $$V(x, y)$$ is a scalar function. That is,
$\mathbf{F} = \left\langle \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right\rangle.$
The geometric interpretation of the gradient $$\nabla V$$ is that

1. $$\nabla V$$ is perpendicular to the level curves given by $$V(x, y) = c$$
2. $$\nabla V$$ points in the direction of greatest increase of $$V$$
3. $$\|\nabla V\|$$ is the slope of $$V$$ in the direction of greatest increase.

From the figure, the vector field $$\mathbf{F}$$ always points horizontally to the right. Therefore, by property 1 of the gradient, the level curves of $$V$$ must be vertical lines.

The figure also depicts the lengths of the vectors $$\mathbf{F}(x, y)$$ as decreasing as $$y$$ increases. By properties 2 and 3, this tells us that the supposed potential function $$V$$ is increasing as $$x$$ increases, but that this increase is slower for larger values $$y$$. However, this implies that the level curves are further apart for larger values $$y$$. This contradicts the previous observation that the level curves of $$V$$ must be vertical lines as vertical lines are a constant $$x$$–width apart!

We can also think about this problem analytically (as opposed to geometrically). Notice that if
$\mathbf{F} = \langle F_1, F_2 \rangle = \left\langle \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right\rangle,$
then by Clairaut’s theorem, we must have
$\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}.$
In the depicted vector field, $$F_1$$ decreases as $$y$$ increases, but $$F_2$$ is constant (in fact, $$0$$). Therefore, $$\frac{\partial F_1}{\partial y}\neq 0$$, but $$\frac{\partial F_2}{\partial x} = 0$$. Thus the vector field cannot possibly be conservative, as these two partial derivatives would be equal for a conservative vector field.