## Archive for the ‘physics’ Category

### Unwinding: Physics of a spool of string

June 1, 2012

It’s been a long day. Let’s unwind with a physics problem.

This problem was on the pre-entrance exam I took before arriving at Caltech for my freshman year. I’ve seen it from time to time since, and here I hope to find an intuitive solution.

Here it is in side view. The dashed circle is the inside of the spool and the green line is the thread. Take a minute to see if you can tell how it works. Does the spool go right or left?

The usual method is to work it out with torques. The forces you must account for are the force of tension from the string and the force of friction from the table.

Torques are actually a pretty easy way to solve this problem, especially if you calculate the torque around the point of contact between the spool and the table (since in that case friction has no moment arm and exerts no torque).

This method is direct, but it’s useful to find another viewpoint if you can.

Let’s first examine a different case where the string is pulled up rather than sideways.

In this case, even if the first situation was unclear, you probably know that the spool will roll off to the left. To see why, let’s imagine that the thread isn’t being pulled by your hand, but by a weight connected to a pulley.

I put a red dot on the string to help visualize its motion.

The physics idea is simply that the weight must fall, so the red dot must come closer to the pulley. Which way can the spool roll so the red dot moves upward?

When the spool rolls (we assume without slipping), the point at the very bottom, where it touches the table, is stationary. The spool’s motion can be described, at least instantaneously, as rotation around that contact point.

Googling, I found a nice description of this by Sunil Kumar Singh at Connexions. This image summarizes the point:

If the spool rolls to the right, as above, the point where the string leaves the spool (near point B), will have a somewhat downward motion. This will pull the red dot down and raise the weight. That’s the opposite of what we want, so what really happens is that the spool rolls to the left, the string rises, and the weight falls.

With this scenario wrapped-up (or unwrapped, I suppose), let’s return to the horizontal string segment.

Again, the weight must fall and so the red dot must go towards the pulley.

If we check out Mr. Singh’s graphic, we’re now concerned with the motion of a point somewhere near the bottom-middle, between points A and C. As the spool rolls to the right, this point also moves to the right. This is indeed what happens as the weight falls.

Notice that the red point actually moves more slowly than the spool as a whole. This means the spool catches up to the string as we move along – the spool winds itself up. If the inside of the spool is 3/4 as large as the outside (like it is in my picture), the spool rolls 4 times as fast the string moves, and so for every centimeter the weight falls, the spool rolls four centimeters.

Here’s a short video demonstration:

### Electric Fields in a Wire

March 8, 2012

A student asked me the following insightful question:

Suppose we have a situation like this:

The lines are wires. The circle is a light bulb. Current runs from left to right. Because the wire has negligible resistance, essentially all the current runs through the top section of wire, skipping around the light bulb, and the bulb doesn’t light.

But what if we look only at this red region?

At the junction, we see current coming in from the left, and deciding to go up rather than continue on to the right. But inside the box, both regions are just made of wire. How does the current know to go up rather than continue straight through?

The answer is that, at the junction, the electric field points mostly up, and very little to the right. But why is that?

Consider two resistors in series. Each one has a voltage drop proportional to its resistance. If the resistors are the same length, the electric field strength inside the resistors is then proportional to the resistance. If one resistor has very high resistance compared to the other, the electric field is much stronger in that resistor.

Go back to the bottom path of our picture. We essentially have three resistors – a wire, a light bulb, and another wire – in series. The light bulb has significant resistance, while the wires do not. Therefore the electric field is much stronger in the light bulb than it is in the wires. It’s weak everywhere, since the bulb isn’t lit. But still, as weak as it is in the bulb, it’s much weaker still in the surrounding wire, by an amount that is proportional to the ratio of the bulb’s resistance to the wire’s resistance.

On the other hand, in the top path, the electric field is the same strength everywhere because it is all just a wire.

Both paths have the same voltage drop because they are in parallel. If they are the same length (they aren’t drawn the same length, but it is easy to imagine), the average electric field strength in them must be the same.

So the average electric field strength is the same in bottom and top. But the electric field in the bottom is localized almost entirely in the light bulb. That means that right at the junction, the electric field is much weaker in the wire heading right than in the wire heading up. Hence, the current almost entirely goes up, making a decision about where to go based only on the local electric field.

### Collecting the Harmonic Numbers

January 1, 2011

When I was a kid, I collected baseball cards. I’ve stopped long since. I can no longer recite streams of players’ yearly RBI totals, and I think my parents threw the cards out when I was away at college. Still, one problem related to those baseball cards still interests me.

Cards were sold in packs of twelve, each containing a random sample selected from the complete set of seven hundred. With my one-dollar allowance and the fifty-cent price of a pack of a cards, how long should I have expected to invest my income before getting a complete set?

To get a complete set of 700 cards, I first need to get a card that’s not a repeat, then get another one that’s not a repeat, and so on 700 times. The first card I pull out of the first pack can’t be a repeat. After that, the second card has a 699/700 chance of being new. Once I have c different cards, the probability of the next one being distinct is (700-c)/700.

We need to know how many cards I expect to draw before getting a new one, when the probability of the next card being new is p. This is similar to the classic problem of how many times one expects to roll a die before getting a 4. The answer is just what you’d think – six times. One argument for this is that if I roll the die many times, then one sixth of the rolls will be 4′s, so the average number of rolls from one 4 to the next is six. That means the expected number of rolls to get a 4 is six. The general answer is 1/p.

So once I have c different cards, the expected number of cards I have to look at to get another new one is 700/(700-c). This makes my expected number of cards purchased to complete the set

$\frac{700}{700} + \frac{700}{699} + \frac{700}{698} + \ldots + \frac{700}{2} + \frac{700}{1}$

Factoring out 700 and writing the terms backwards, this is

$700 \left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{700}\right)$

In other words, if there are n possible equally-likely outcomes to a random event, the expected value of the number of trials to get every value to occur is n times the nth harmonic number.

An image from Wikipedia shows that the harmonic numbers are approximated by the integral of 1/x, so that the answer scales with n as n*log(n).

For my baseball card problem, n=700 and I’d need about 4500 cards, or around 4 years’ allowance, to complete the set (I never made it).

The integral picture immediately suggests an improvement. The curve drawn is an underestimate – the rectangles always pop up higher than the curve. What is the approximate area left over above the curve?

A simple guess comes from taking each of those left-over areas to be a triangle. These triangles all have a base of 1. Their heights vary, but the nice thing is that the top of one triangle begins at the bottom of the next. Imagine sliding all the triangles over so they’re stacked on top each other. This would give a total height of 1-1/(n+1) for the triangles out to the nth harmonic number. So a better guess for the for the nth harmonic number $H_n$ is

$H_n = \ln(n) + \frac{1 - 1/(n+1)}{2}$

This is also an underestimate because the area above the curve doesn’t really consist of triangles, as you can see. $H_n$ truly converges to $\ln(n) + \gamma$. $\gamma$ is the Euler-Mascheroni constant. It’s now know to 30 billion digits, but the most important ones are 0.577…

Browsing Wikipedia reveals a trove of information about these mathematical entities. For example, if you have ever tried to sum a geometric series, you saw that

If you integrate $x^n$ from zero to one, you get 1/n, so

This lets us define fractional harmonic numbers, and for simple fractions they involve things like $\pi$ and $\sqrt{2}$. Further down that page we learn that harmonic numbers are related to the Riemann zeta function and the Riemann hypothesis and a bunch of other things I don’t know about.

Why might I find this stuff interesting? If you have a closed physical system with many particles, there are many possible physical states of those particles. The ergodic hypothesis assumes that as the system evolves in time, it passes through all those states, and all states are equally likely to be accessed at some far future time (assuming we restrict attention to those states that obey conservation laws, like conservation of energy). So the harmonic numbers might give some insight into how long I have to wait to observe particular configuration of the system. Would the harmonic numbers be discovered by a Boltzmann brain?

### The Hanging Chain

December 20, 2010

One of the many simple, charming exhibits in San Francisco’s Exploratorium is a tall metal chain that hangs over a spinning pulley. That’s the whole exhibit. It looks like this:

That video conveys the size and scope of the exhibit, but makes it hard to see what’s going on in detail. Fortunately, YouTube user grahamj21 created this simulation:

The chain, though spinning past, appears to writhe, stationary, in mid-air. This only happens at the bottom of the Exploratorium’s chain. If you tap it on the side, the side itself quickly goes back to being straight, while a moment later the bottom of the chain begins its strange contortions.

When we hit the chain, we send two traveling waves out in either direction from our tap. (The first video incorrectly identifies them as standing waves.) One wave travels in the same direction as the chain. It moves away quickly and we don’t see it. The second wave moves against the chain. This second wave would appear to move fairly slowly to us; the chain might be pulling it up while it is trying to propagate down.

Let’s further suppose that the wave wins this race and swims all the way down the up-flowing chain. When it reaches the bottom, something interesting happens. The further down the chain we go, the less tension. As the tension goes down, the wave speed goes down, too. Finally, at the bottom of the chain, the wave speed and the chain speed match. The wave gets stuck there, and we enjoy watching its pained death throes.

This is the story I’ve told about the chain since the first time I saw it, years ago. I’ve never stopped to check if it makes sense, though. Would the wave speed at the bottom of the chain be a meter or two per second, as the observed speed of the chain is? For simplicity, let’s imagine the chain is stationary and estimate the wave speed at the bottom.

The chain has a mass per unit length $\lambda$, is subject to a gravitational acceleration $g$, is a height $h$, and has a wheel with radius $r$. The wave speed $v$ must be some function of these variables.

Dimensionally, speed is just a length divided by a time, and does not include mass at all. $\lambda$ cannot come into our formula for $v$ – there is nothing else able to cancel its mass dimension. Of the remaining quantities, only $g$ has units of time, so the velocity must depend on $\sqrt{g}$. We can conclude

$v = \sqrt{gl}f(r/h)$,

with $f$ some unknown function of the dimensionless quantity $r/h$. When we look at the chain, we see that $h >> r$, so $r/h$ is a small number, and it’s reasonable to assume $f$ is well-approximated by a first-order Taylor series, $f \approx \alpha r/h$. This gives

$v = \alpha r\sqrt{g/h}$.

I have no particular way to estimate $\alpha$, but we can hope that it’s not too far from $1$. This would mean

$v \approx r\sqrt{g/h}$.

Setting $g = 10 m/s^2$, $h = 3m$, $r = 0.2m$, we get that, roughly speaking

$v \approx 1m/s$.

This is a pretty reasonable speed for the chain to be moving, based on the video. The estimate shows that our hypothesis about why the chain writhes is a reasonable one, without having to go through a complicated calculation. The problem is far from solved. The moving chain accelerates, especially at the bottom, so the tension and wave speeds are higher than for a stationary chain. We know very little about why the chain writhes in quite the way it does, and can’t predict its motion accurately. But we know that our rough understanding makes sense.

The first time I did this problem, I tried calculating the shape of the hanging chain, getting the tension from that, and evaluating the phase velocity $v = \sqrt{T/\lambda}$. This was kind of ugly because it involved solving a transcendental equation on the computer, and it gave me almost the same answer. Such a calculation is hardly superior, since I’m guessing at the dimensions of the chain anyway.

### Leakier, Slower, and No Rain

December 7, 2010

A while ago, I asked a standard freshman physics problem about a cart that has rain fall into it, then opens a hole and rain leaks out. Then I gave an answer saying that as rain falls vertically into an open cart running on a frictionless track, the cart slows down, but as rain leaks out it shows no change in speed.

That was mostly correct, given the picture I drew of the hole:

Water leaks out a hole in the bottom of the cart as it slides to the right.

The key is that the hole is in the center. Yesterday, Martin Gales posed a question on Physics.StackExchange pointing out that this makes a difference, because if we imagine a stationary cart with a hole all the way to the left, then as it drains, the water moves left, and so the cart will have to move right a little to conserve momentum. But then once the cart starts moving the water leaking out of the cart is moving…

I spent three hours last night trying to solve this seemingly-trivial problem. (My answer is at the original question.) It’s simple enough to pose to first-term freshmen, and yet I went through dozens of slightly-wrong ideas and calculations before hitting on the surprising answer. Further, once I knew what happens, it didn’t seem very complicated any more, leaving me to wonder what the hell is wrong with my overclocked simian brain. The feeling you get when thinking about such a problem is an asymmetric oscillation of healthy frustration and premature joy unparalleled in other pursuits. I want to be mind-fucked like this every night.

### Thermal energy of a gas

November 11, 2010

If we have a gas with some heat, the atoms are all bouncing around and stuff. Their kinetic energy is

$KE = 1/2 m v^2$.

But momentum is given by $mv = p$, so we can rewrite the energy as

$KE = 1/2 pv$.

Everyone knows $pv = nRT$, so

$KE = 1/2 nRT$.

The kinetic energy of the motion of molecules is just the temperature of the gas!

(PS – this is a joke, but it actually gets the right answer for a 1-dimensional ideal gas. See wikipedia Thermal Energy)

October 25, 2010

### A Good High School Science Fair Idea

Suppose you drop a hamster off a tall building. It will fall, initially with an acceleration of $g$, about $10 m/s^2$. But after the hamster has swept out a volume of air whose mass equals the hamster’s own, the effects of air resistance dominate and the acceleration is small.

A hamster is about as dense as water, which is about a thousand times as dense as air. So air resistance becomes important when a hamster falls 1000 times its own thickness. The thickness of a spread-eagled hamster is 5 cm, so air resistance dominates falls over 50m.

(Compare this to a sheet of paper. A sheet of paper is as dense as a hamster, but 1000 sheets of paper are only as thick as a 1000-page book, so for the sheet of paper, air resistance is important even if you fall just a few centimeters.)

50m is about 20 stories, so if we want the speed of the hamster when dropping it off a building at least that tall, we had better include air resistance.

A reasonable model for air resistance is that its force is proportional to the square of the hamster’s velocity, because if the hamster falls twice as fast, it strikes the air twice as hard, but also strikes twice as much air per second. Thus

$F_{air} = -kv^2$

for some constant $k$. To use Newton’s second law, include the force of gravity and set the force equal to $ma$.

$F_{air} + F_{grav} = m a$

$-kv^2 + gm = m a$

Let $k/m = b$ and you get

$-bv^2 + g = a = \frac{\textrm{d}v}{\textrm{d}t}$

This is a separable equation. We get

$\textrm{d}t = \frac{\textrm{d}v}{-b v^2 + g}$

which integrates to

$t+c = \frac{1}{\sqrt{bg}}\textrm{arctanh}\left(\sqrt{\frac{b}{g}} v\right)$.

Solving for $v$ yields

$v = \sqrt{\frac{g}{b}}\tanh\left(\sqrt{bg}(t+c)\right)$

If the hamster is dropped from rest, the initial condition gives $c = 0$ so

$v = \sqrt{\frac{g}{b}}\tanh\left(\sqrt{bg}t\right)$

The hamster speeds up, approaching its terminal velocity

$v_{ter} = \sqrt{\frac{g}{b}}$

To estimate the terminal velocity, go back to the heuristic that air resistance dominates when the hamster sweeps out its own mass worth of air. That happened at 50m, and in a 50m free-fall you accelerate to about 30m/s, or 70mph (the hamster should survive, I’m told).

If we measure time in units of $\sqrt{gb}$ and then choose units of length so that the velocity $\sqrt{g/b} = 1$, the hamster’s velocity during free fall is just

$v = \tanh(t)$

### An Impractical High School Science Fair Idea

Suppose the hamster survives, so we put it in a rocket blasting into deep space. We stay behind on Earth. The hamster has constant acceleration $g$, as viewed from its own reference frame.

The effects of relativity will be important when $gt \approx c$, which happens after about a year. If the hamster’s trip will be longer than that, we better include the effects of relativity.

At a given moment, the hamster is going along at some speed $v$. To transform from the reference frame of Earth to the hamster, we use the Lorentz transformation

$\left( \begin{array}{cc} \cosh\theta & \sinh\theta\\ \sinh\theta & \cosh\theta \end{array} \right) \left(\begin{array}{c} x \\ t \end{array} \right) = \left( \begin{array}{c} x' \\ t' \end{array}\right)$

where $\theta$ is the rapidity, defined by

$\tanh\theta = v/c$. We’ll choose the proportion of units of length and time such that $c = 1$.

The hamster accelerates, so this transformation depends on time. Let’s try to find $\theta(\tau)$, the rapidity as a function of the proper time experienced by the hamster.

In a small proper time $\textrm{d}\tau$, the hamster accelerates by $g \textrm{d}\tau$. The rapidity of this transformation is given by

$\tanh\theta' = g \textrm{d}\tau$.

For small $\theta'$, $\tanh\theta' = \theta'$, so

$\theta' = g \textrm{d}\tau$.

We can measure time in units of $1/g$. Then the transformation from the hamster’s comoving frame from one moment to the next is

$\left( \begin{array}{cc} \cosh \textrm{d}\tau & \sinh \textrm{d}\tau \\ \sinh \textrm{d}\tau & \cosh \textrm{d}\tau \end{array} \right) \left(\begin{array}{c} x' \\ t' \end{array} \right) = \left( \begin{array}{c} x'' \\ t'' \end{array}\right)$

To find the hamster’s new velocity in Earth’s frame, we simply compose the transformations

$\left( \begin{array}{cc} \cosh \textrm{d}\tau & \sinh \textrm{d}\tau \\ \sinh \textrm{d}\tau & \cosh \textrm{d}\tau \end{array} \right) \left( \begin{array}{cc} \cosh\theta & \sinh\theta\\ \sinh\theta & \cosh\theta \end{array} \right) \left(\begin{array}{c} x \\ t \end{array} \right) = \left( \begin{array}{c} x'' \\ t'' \end{array}\right)$

That’s an easy matrix multiplication – it’s almost a rotation matrix. We get

$\left( \begin{array}{cc} \cosh(\theta + \textrm{d}\tau)& \sinh(\theta + \textrm{d}\tau)\\ \sinh(\theta + \textrm{d}\tau) & \cosh(\theta + \textrm{d}\tau) \end{array} \right) \left(\begin{array}{c} x \\ t \end{array} \right) = \left( \begin{array}{c} x'' \\ t'' \end{array}\right)$

We see that $\theta(\tau + \textrm{d}\tau) = \theta(\tau) + \textrm{d}\tau$

Since $\theta(0) = 0$ (i.e. the hamster has 0 speed when $t = 0$), this integrates to

$\theta(\tau) = \tau$.

Returning to the velocity $v = \tanh(\theta)$ gives

$v = \tanh(\tau)$.

The velocity of the hamster dropped from the tall building, as a function of time, follows the same equation as the velocity of the hamster accelerating in the spaceship, as a function of proper time. All we have to do is substitute $c = v_{ter}$ and, choosing the correct units of length (depending on the constant in the equation for the hamster’s drag), the analogy is complete.

### Three-Way

October 21, 2010

I’ve been told that when you and your sweetheart get accepted to different grad schools, you’ve encountered the “Two-Body Problem”. It’s an inapt analogy, because the two-body problem can be solved elegantly. What’s really difficult is the three-body problem.

If we have three massive bodies interacting through Newton’s gravitational law, with arbitrary initial positions and velocities, it turns out that exact analytic solutions are extremely difficult to find. That’s not to say there is no solution – the massive bodies have no trouble finding it. But if we want to predict their motion we’ll need to do some numerical integration.

However, we might wonder if there are some particularly simple or intriguing solutions, perhaps in very symmetrical situations. There are. Quite a few of them are cited in the Scholarpedia article. Here are some:

I’m not sure how these solutions were discovered. But if we consider the problem a moment it’s clear the center of mass cannot accelerate, and that energy and angular momentum must be preserved. An especially simple way to do this is to have the angular momentum and energy of each individual mass be constant by making them all rotate around the center of mass at a fixed frequency.

To see if this works, measure the positions of the three masses by vectors $\vec{r_1}, \vec{r_2}, \vec{r_3}$ from the axis of rotation, which is through the center of mass and perpendicular to the plane of the bodies. Let the masses of the bodies be $m_1, m_2, m_3$. Then the equation for the center of mass states

$\vec{r_1}m_1 + \vec{r_2}m_2 + \vec{r_3}m_3 = 0$.

If we look at a reference frame comoving with the center of mass and corotating with the masses, the masses will be stationary, so there must be no force on them. Let’s take $m_1$ as an example. The centrifugal force on it is

$F_c = \omega^2 m_1 \vec{r_1}$.

The gravitational force on it is

$F_g = G m_1 \left(m_2\frac{\vec{r_2} - \vec{r_1}}{\left|\vec{r_1} - \vec{r_2}\right|^3} + m_3\frac{\vec{r_3} - \vec{r_1}}{\left|\vec{r_3} - \vec{r_1}\right|^3}\right)$

If those forces are going to add to zero to give $m_1$ zero acceleration, they better at least point the same direction. It’s not clear that they do. The centrifugal force points only in the direction of $\vec{r_1}$, while the gravitational force has all three position vectors in there. So we need the amounts of $\vec{r_2}$ and $\vec{r_3}$ to be such that they add up to point towards $\vec{r_1}$.

From the equation for the center of mass we get,

$-m_1\vec{r_1} = m_2\vec{r_2} + m_3\vec{r_3}$

whence we know the proportions in which the $\vec{r_2}$ and $\vec{r_3}$ must appear in the gravitational force. We conclude

$m_2\vec{r_2} + m_3\vec{r_3} \propto \left(\frac{m_2\vec{r_2}}{\left|\vec{r_1} - \vec{r_2}\right|^3} + \frac{m_3 \vec{r_3}}{\left|\vec{r_3} - \vec{r_1}\right|^3}\right)$.

This will work perfectly iff the denominators on the right hand side of that equation are equal, meaning mass $m_1$ must be equidistant from masses $m_2$ and $m_3$. Repeating the argument while writing out forces on $m_2$ this time, the masses must be in an equilateral triangle.

We still don’t know if such a solution exists, only that it’s possible to get the centrifugal force to point opposite the gravitational force for arbitrary masses, as long as we put them in an equilateral triangle. Let’s continue, setting the centrifugal and gravitational forces equal for $m_1$ and see what the resulting rotation rate is. Also let’s let the sides of the triangle be length $l$

$F_c = \omega^2 m_1 \vec{r_1} = F_g = \frac{G m_1}{l^3} \left( m_2(\vec{r_2} - \vec{r_1}) + m_3 (\vec{r_3} - \vec{r_1}) \right)$

Applying the center of mass equation

$-m_1\vec{r_1} = m_2\vec{r_2} + m_3\vec{r_3}$

one more time and setting the total mass equal to $M$, this simplifies to

$\omega^2 = \frac{G M}{l^3}$.

A clean result that is the same for all three masses, meaning these orbits indeed solve the equations of motion.

### Fucking Geomagnetism, How Does That Work?

October 14, 2010

Here’s a tidbit from the lore of physics history. It’s from Einstein.

A wonder of such nature I experienced as a child of 4 or 5 years, when my father showed me a compass… I can still remember – or at least believe I can remember – that this experience made a deep and lasting impression upon me. Something deeply hidden had to be behind things.

Einstein’s scientific work led us directly to a deeper understanding of magnets (among other things). Specifically, his seminal paper on special relativity is about electromagnetism.

With the later addition of quantum mechanics, we now pretty much understand magnets. We know how they work, meaning that we know what the rules are for magnets, and roughly how to go from those basic rules to making levitating trains.

I learned some of those rules in college. The fundamentals don’t require relativity or quantum mechanics and are not abstruse. They’re just things like this: you run current through a wire. It deflects a compass needle. You set up another wire next to the first, with current running the same way. The wires attract each other. You make the currents run the opposite way. The wires repel. This is all fairly simple. It really works, and it’s been known for 200 years.

Shortly after Facebook introduced its “Questions” app, someone asked, “Fucking magnets, how do they work?”. Innocently, I thought it was someone who wanted an answer, but not a very serious one, so I replied,

Some magnets hate each other because their electrons are spinning the wrong way, and this makes them very, very angry. They repel. On the other hand if you turn one around the electrons are spinning the opposite way and then they get along fine and attract.

Which is wrong for a few reasons. One is the anthropomorphizing; another is treating quantum mechanical “spin” as colloquial “spinning”, third is that it doesn’t really explain much, and fourth is that the person wasn’t asking a question, but quoting an internet meme I didn’t know about.

After I learned about the meme and the band, Insane Clown Posse, which created it, I didn’t think much more of it. Then yesterday, Sean at Cosmic Variance compared “Miracles” to a famous Whitman poem.

Whitman says, “When I heard the learn’d astronomer; … How soon, unaccountable, I became tired and sick;”. ICP says, “And I don’t wanna talk to a scientist / Y’all motherfuckers lying, and getting me pissed.” Juxtaposed, the sentiments sound similar (and even form a slant rhyme), but there’s a major difference of intention. Whitman is speaking about disillusionment with scientific thinking – the familiar sentiment that mathematical or scientific explanations kill the beauty of what they describe. ICP’s objection is not directly to science, but to scientists themselves, as a group. Whitman’s objection is essentially intellectual, ICP’s essentially cultural.

Whitman was much less likely to object to Y’All Motherfucker Scientists because science wasn’t yet a firm American establishment in 1855. There was certainly a culture of science, but there were not that many scientists, and Science, as a group of people representing a certain general worldview, was not prominent.

Today there is a huge culture of science, and most of the reaction I’ve read to “Miracles” is defensive; people identify with that culture and view “Miracles” as an affront to what they believe and value. They want to get in a room with ICP and tell them that, for example, their attitude is perverse, it’s hypocritical to impugn science while taking advantage of the benefits of scientific research, and there’s a perfectly good explanation for magnets that only adds to their wonder and beauty.

I’ve allowed this essay to wander slightly off course. I’d like to bring it back to compasses. I’m not a scientist but I do like to party with them. I remember one time I had just met a bunch of geology grad students, and I wanted to know how compasses work, so I asked them.

The problem here is that the more I learned about electricity and magnetism, the less sense the Earth’s magnetic field made to me. As a kid, I knew the basic story about where it comes from, exemplified by an article on the popular website How Stuff Works:

The “big bar magnet buried in the core” analogy works to explain why the Earth has a magnetic field, but obviously that is not what is really happening. So what is really happening?

No one knows for sure, but there is a working theory currently making the rounds. As seen on the above, the Earth’s core is thought to consist largely of molten iron (red). But at the very core, the pressure is so great that this superhot iron crystallizes into a solid. Convection caused by heat radiating from the core, along with the rotation of the Earth, causes the liquid iron to move in a rotational pattern. It is believed that these rotational forces in the liquid iron layer lead to weak magnetic forces around the axis of spin.

It kind of makes sense. Why does Earth have a magnetic field? Because there’s convection of molten iron in it. Why does convection of iron create a magnetic field? Because iron is magnet stuff. You know, like, when you want to make an electromagnet you need some iron, or you can trace out the field lines of a bar magnet with iron filings. It’s just… um.. iron! magnets! boom!

This is pretty much the answer I’ve gotten most times when I ask someone about Earth’s magnetic field. But it’s a pretty bad candidate for the next “rainbows are made by refracting and reflecting light in spherical raindrops” (which is pretty accurate). What I learned from physics was that if you make things hotter, they’ll lose their magnetic properties (magnetized iron has aligned domains, and hence lower entropy). The molten iron is obviously very hot and not very magnetic. And even so, just moving normal iron around doesn’t create a magnet anyway. If I take an iron crowbar and start waving it around like a madman, it doesn’t give me the powers of Magneto. It does attract things like police officers with tasers, but not by magnetism.

A few times, earlier on when the topic of geomagnetism came up, I got the feeling that no one was willing to acknowledge that the Cartoon Guide to Geomagnetism made no sense. I further felt like no one was acknowledging it because no one realized it. They had bought into Science hook line and sinker, forgetting about science, which holds that you should keep good track of how little you actually know about things.

My conversation with the geology grad students at that party went better than previous conversations. They were better informed, and more willing to admit that it’s a hella-complicated problem involving magnetohydrodynamics, and basically said it wasn’t their specialty and they didn’t know. That was more satisfying to me.

It’s not satisfying enough though, so reading Sean’s essay and being reminded of the fucking mysterious magnets, my internal compass brought me back to wondering about geomagnetism. I’ve spent the last day or so looking for a decent explanation of the origin of the Earth’s magnetic field. Here is the simplest, most direct answer I found in that time that meets the criterion of not sounding like an uncomfortable circumlocution, a lie, or the insane rambling of a misinformed member of the science posse.

That actually makes sense to me. I know what those equations mean, and I know where they come from (pretty much. Nobody’s perfect). I know basically what numerically solving means, too. It means that it’s all very complicated and you shouldn’t wrack your brain trying to understand it, because you won’t.

When I hear ignoramuses profess a love of ignorance, it’s probably either insincere or incorrigible. But when I hear scientists or science appreciators declaiming that science can only ever make our picture grander, only makes the night sky more beautiful, and then also hear the same people expounding scientific doctrines they don’t understand, I can only believe that their avowal of the beauty of science is, like their stories about how things work, a regurgitated ideology.

I don’t feel childish wonder when confronted with mysteries. I don’t feel naive awe. I don’t feel sophisticated, informed appreciation. I feel an intense desire to understand better – to be a little less stupid about the thing.

I’d like there to be a better story about compasses. I realize I don’t understand them well enough to write a good story. It might not even be possible to write a really good story. But I think I at least understand it a little better than I did before.

So maybe I will write a story. I want there to be a story that’s bad, but at least a little less bad than the one I heard over and over. Based on what I read so far, the knowledge about this phenomenon is impressive, though incomplete. But experience tells me that what’s been discovered is not broadly known.

There are a lot of basic things about geomagnetism I don’t understand, like why the convection currents should be planet-sized, or why solutions should exist that persist for thousands of years with very little change, or why the strength of the field is a Gauss to order of magnitude. I’m not even sure why the field aligns roughly with the axis of rotation. If there were a totally new geomagnetic theory ten years from now, I probably wouldn’t have too hard a time letting go of the old one. I want to learn the stuff I’m missing and fill in as much as I can. There are thousands of things I want to learn, though, and there won’t be time for them all. So for now, even though it’s incomplete and inaccurate and confusing and sloppy, here’s a story.

The Earth’s magnetic field is generated by the complex interplay of the motion of molten metal in its interior, the motion of electric charges through that metal, and the dynamic electromagnetic fields that affect those motions. The inside of the Earth is hot, melting the metals that constitute it. The very center core is solid due to the high pressure there, despite its intense heat. The solid core and the temperature gradient from the deep interior of Earth to its crust combine to create hydrodynamic forces that push on the molten metals. It turns out that metals respond by moving in huge, continent-to-planet scale loops. The metals are too hot to be magnetized, but they do have electrical conductivity. That means that if we did, for some reason, have a dynamic magnetic field in the Earth, it would induce electromagnetic current in these metals. Once the metals have electromagnetic current, the posited magnetic field exerts forces on them, influencing their motion. Further, since the metals are carrying current while being pushed around by heat gradients and things, they go on to create their own complicated magnetic fields. The magnetic fields they generate then feed back to change the forces on the currents, etc. In this way the originally hypothetical dynamic magnetic field becomes a plausible alternative, needing only a small seed field to grow into some approximately-steady-state field like the one we observe. The result is a huge, complicated mess. Because of that, it could conceivably do absolutely wild things, like completely change directions every few score millenia, or get stronger or weaker, or wind and curl in strange loops, or even go out.

### Viete’s Formula and Spinning Pizza

September 17, 2010

Have you seen Viete’s formula?:

It’s a special case of a trig identity found by Euler:

If you plug in $\pi/2$ to the trig identity and use the half-angle formula for cosine over and over, you get Viete’s formula.

But why would you want to consecutively cut angles in half and multiply their cosines? Well, you might be eating pizza.

You have a slice of pizza that is too hot to hold, so you want to balance it on your fork and gnaw at it instead. There’s a precise spot on the underside of the slice where the fork should go.

This point is called the slice’s center of mass, and we’re going to find it. By symmetry, it must be on the line cutting the slice in half lengthwise, but we don’t yet know how far down. It depends on the shape of the slice, which we measure by $\theta$, the angle its edges make.

The center of mass of the slice of pizza is a green dot. It lies on the line cutting the slice in half vertically.

A bigger slice will have its center of mass closer to the tip. We would like to know $r(\theta)$, the distance from the tip to the center of mass as a function of $\theta$.

We want to know the distance to the center of mass.

There are two limiting cases – a very skinny slice and a whole pie. A very skinny slice is basically an isosceles triangle. Its center of mass is 2/3 the way from the tip to the edge1, so

$\lim_{\theta \to 0} r(\theta) = \frac{2}{3}R$.

Let’s choose the radius of the pizza as unit of length, so $R = 1$ from here on.

In the other limiting case, an entire pizza has its center of mass right at the tip (i.e. center), so

$r(2\pi) = 0$.

To investigate intermediate cases, we start with a slice of angle $\theta$ and imagine cutting it in half lengthwise, creating two skinny pieces of angle $\theta/2$. These have their own centers of mass at $r(\theta/2)$.

The center of mass of the big piece is on the line connecting the smaller pieces’ centers of mass.

The center of mass of the entire slice lies on the line connecting the centers of mass of the half-slices. This creates a geometric relation between $r(\theta)$ and $r(\theta/2)$.

A bit of trigonometry tells us

$r(\theta) = r(\theta/2)\cos(\theta/4)$

If we take this formula and divide all angles by $2$, we get a formula for $r(\theta/2)$. We substitute this for where $r(\theta/2)$ appeared in the original. We obtain

$r(\theta) = \left[r(\theta/4)\cos(\theta/8)\right]\cos(\theta/4)$

Repeat the process ad infinitum. Rearranging the order of the terms and substituting the limiting value of $r$ for small $\theta$, we get

$r(\theta) =\frac{2}{3} \cos(\theta/4)\cos(\theta/8)\cos(\theta/16)\ldots$

It involves one half of Euler’s trig identity. If we find $r(\theta)$ by a different method and get a different expression for it, we can set our two expressions for $r(\theta)$ equal to each other, and prove Euler’s identity. We’ll do this by invoking some physics ideas.

Suppose you’re spinning some pizza dough in the air. You know, like this:

If the pizza is spinning, each little bit of dough undergoes centripetal acceleration. Where there’s acceleration, there’s force. The pizza isn’t touching anything, so the force on any one piece of pizza must be coming from the rest of the pizza.

Let’s again examine a slice of size $\theta$, this time still attached to the spinning pizza. It has two forces of size $F$ acting on it; one force is exerted by the slice to its left and one by the slice to its right.

There are two forces on the slice - one from the pizza to the left and one from the pizza to the right. They're both drawn originating from the center of mass. The slice is accelerating towards its tip (red arrow).

The sum of these forces is the mass of the slice times the acceleration of its center of mass. That acceleration is $\omega^2 r(\theta)$. Hence, if we determine the forces we can deduce $r(\theta)$.

Some trigonometry shows that the net force is $2F\sin(\theta/2)$.

Equating this to mass times acceleration, we get

$2F\sin(\theta/2) = \frac{m \theta}{2\pi} \omega^2 r(\theta)$

We might as well let $m = \omega = 1$ and solve for $r$ to get

$r(\theta) = 2F\sin(\theta/2)\frac{2 \pi}{\theta}$.

We still need to determine $F$, but we can do that because we know $r(\theta) \to 2/3$ as $\theta \to 0$. After a little algebra, we get

$r(\theta) = \frac{4}{3} \frac{sin(\theta/2)}{\theta}$

This gives us the sought two expressions for $r$. We can now equate them and simplify to

$\frac{sin(\theta)}{\theta} = \cos(\theta/2)\cos(\theta/4)\cos(\theta/8)\ldots$

1) To see why an isosceles triangle’s center of mass is 2/3 up the altitude, first show it’s true for an equilateral triangle. Then explain why all isosceles triangles have their center of mass the same fraction of the way down the altitude.