The Virtuosi (Posts by Brian)

Tragedy of Great Power Politics? Modeling International War

Brian — Sun, 23 Jun 2013 23:48:00 GMT

Recently I finished reading John Mearsheimer's excellent political science book The Tragedy of Great Power Politics. In this book, Mearscheimer lays out his ``offensive realism'' theory of how countries interact with each other in the world. The book is quite readable and well-thought-out -- I'd recommend it to anyone who has an inkling for political history and geopolitics. However, as I was reading this book, I decided that there was a point of Mearsheimer's argument which could be improved by a little mathematical analysis.

The main tenant of the book is that states are rational actors who act to to maximize their standing in the international system. However, states don't seek to maximize their absolute power, but instead their relative power as compared to the other states in the system. In other words, according to this logic the United Kingdom position in the early 19th century -- when its army and navy could trounce most of the other countries on the globe -- was better than it is now -- when many other countries' armies and navies are comparable to that of the UK, despite the UK current army and navy being much better now than they were in the early 19th century. According to Mearsheimer, the main determinant of state's international actions is simply maximizing its relative power in its region. All other considerations -- capitalist or communist economy, democratic or totalitarian government, even desire for economic growth -- matter little in a state's choice of what actions it will take. (Perhaps it was this simplification of the problem which made the book really appeal to me as a physicist.)

Most of Mearsheimer's book is spent exploring the logical corollaries of his main tenant, along with some historical examples. He claims that his idea has three different predictions for three different possible systems. 1) A balanced bipolar system (one where two states have roughly the same amount of power and no other state has much to speak of) is the most stable. War will probably not break out since, according to Mearsheimer, each state has little to gain from a war. (His example is the Cold War, which didn't see any actual conflict between the US and the USSR.) 2) A balanced multipolar system ($N>2$ states each share roughly the same amount of power) is more prone to war than a bipolar system, since a) there is a higher chance that two states are mismatched in power, allowing the more powerful to push the less around, and b) there are more states to fight. (One of his examples is Europe between 1815 and 1900, when there were several great-power wars but nothing that involved the entire continent at once.) 3) An unbalanced multipolar system ($N>2$ states with power, but one that has more power than the rest) is the most prone to war of all. In this case, the biggest state on the block is almost able to push all the other states around. The other states don't want that, so two or more of them collude to stop the big state from becoming a hegemon -- i.e. they start a war. Likewise, the big state is also looking to make itself more relatively powerful, so it tries to start wars with the little states, one at a time, to reduce their power. (His examples here are Europe immediately before and leading up to the Napoleonic Wars, WWI, and WWII.) There is another case, which is unipolarity -- one state has all the power -- but there's nothing interesting there. The big state does what it wants.

While I liked Mearsheimer's argument in general, something irked me about the statement about bipolarity being stable. I didn't think that the stability of bipolarity (corollary 1 above) actually followed from his main hypothesis. After spending some extra time thinking in the shower, I decided how I could model Mearsheimer's main tenant quantitatively, and that it actually suggested that bipolarity was actually unstable!!

Let's see if we can't quantify Mearsheimer's ideas with a model. Each state in the system has some power, which we'll call $P_i$. Obviously in reality there are plenty of different definitions of power, but in accordance with Mearsheimer's definition, we'll define power simply in a way that if State 1 has power $P_1 > P_2$, the power of State 2, then State 1 can beat State 2 in a war^[1]. Each state does not seek to maximize their total power $P_i$, but instead their relative power $R_i$, relative to the total power of the rest of the states, So the relative power $R_i$ would be

$$ R_i = P_i / \left( \sum_{j=1}^N P_j \right) \qquad , $$

where we take the sum over the relevant players in the system. If there was some action that changed the power of some of the players in the system (say a war), then the relative power would also change with time $t$:

$$ \frac{dR_i}{dt} = \frac{dP_i}{dt} \times \left( \sum_{j=1}^N P_j \right)^{-1} - P_i \times \left( \sum_{j=1}^N P_j \right)^{-2} \times \left(\sum_{j=1}^N \frac{dP_j}{dt} \right) \qquad (1) $$

A state will pursue an action that increases its relative power $R_i$. So if we want to decide whether or not State A will go to war with State B, we need to know how war affects a state's individual powers. While this seems intractable, since we can't even precisely define power, a few observations will help us narrow down the allowed possibilities to make definitive statements on when war is beneficial to a state:

War always reduces a state's absolute power. This is simply a statement that in general, war is destructive. Many people die and buildings are bombed, neither of which is good for a state. Mathematically, this statement is that in wartime, $dP_i/dt < 0$ always. Note that this doesn't imply that that $dR_i/dt$ is always negative.

The change in power of two states A & B in a war should depend only on how much power A & B have. In addition, it should be independent of the labeling of states. Mathematically, $dP_a / dt = f(P_a, P_b)$, and $dP_b/dt = f(P_b, P_a)$ with the same function $f$^[2].
If State A has more absolute power than State B, and both states are in a war, then State B will lose power more rapidly than State A. This is almost a re-statement of our definition of power. We defined power such that if State A has more absolute power than State B, then State A will win a war against State B. So we'd expect that power translates to the ability to reduce another state's power, and more power means the ability to reduce another state's power more rapidly.
For simplicity, we'll also notice that the decrease of a State A's absolute power in wartime is largely dependent on the power of State B attacking it, and is not so much dependent on how much power State A has.

In general, I think that assumptions 1-3 are usually true, and assumption 4 is pretty reasonable. But to simplify the math a little more, I'm going to pick a definite form for the change of power. The simplest possible behavior that capture all 4 of the above assumptions is:

$$ \frac{dx}{dt} = -y \qquad \frac{dy}{dt} = -x \qquad (2) $$

where $x$ is the absolute power of State X and $y$ is the absolute power of State y. (I'm switching notation because I want to avoid using too many subscripts^[3]). Here I'm assuming that the rate of change of State X's power is directly proportional to State Y's power, and depends on nothing else (including how much power State Y actually has). We'll also call $r$ the relative power of State X, and $s$ the relative power of State Y^[4]. Now we're equipped to see when war is a good idea, according to our hypotheses.

Let's examine the case that was bothering me most -- a balanced bipolar system. Now we have only two states in the system, X and Y. For starters, let's address the case where both states start out with equal power $(x = y)$. If State X goes to war with State Y, how will the relative powers $r =x/(x+y)$ & $s=y/(x+y)$ change? Looking at Eq. (1), we see that by symmetry both states have to lose absolute power equally, so $x(t) = y(t)$ always, and thus $r(t) = s(t)$ always. In other words, from a relative power perspective it doesn't matter whether the states go to war! For our system to be stable against war, we'd expect that a state will get punished if it goes to war, which isn't what we have! So our system is a neutral equilibrium at best.

But it gets worse. For a real balanced bipolar system, both states won't have exactly the same power, but will instead be approximately equal. Let's say that the relative power between the two states differs by some small (positive) number $e$, such that $x(0) = x0$ and $y(0) = x0 + e$. Now what will happen? Looking at Eq. (2), we see that, at $t=0$,

$$ \frac{dr}{dt} = -(x_0 + e) / (2x_0 + e) + x_0(2x_0 + e) / (2x_0 + e)^2 = -e/(x_0 + e) $$

$$ \frac{ds}{dt} = -(x_0) / (2x_0 + e) + (x_0+e)(2x_0 + e) / (2x_0 + e)^2 = + e/(x_0 + e) \qquad . $$

In other words, if the power balance is slightly upset, even by an infinitesimal amount, then the more powerful state should go to war! For a balanced bipolar system, peace is unstable, and the two countries should always go to war according to this simple model of Mearsheimer's realist world.

Of course, we've just considered the simplest possible case -- only two states in the system (whereas even in a bipolar world there are other, smaller states around) who act with perfect information (i.e. both know the power of the other state) and can control when they go to war. Also, we've assumed that relative power can change only through a decrease of absolute power, and in a deterministic way (as opposed to something like economic growth). To really say whether bipolarity is stable against war, we'd need to address all of these in our model. A little thought should convince you which of these either a) makes a bipolar system stable against war, and b) makes a bipolar system more or less stable compared to a multipolar system. Maybe I'll address these, as well as balanced and unbalanced multipolar systems, in another blog post if people are interested.

1. ^ $P_i$ has some units (not watts). My definition of power is strictly comparative, so it might seem that any new scale of power $p_i = f(P_i)$ with an arbitrary monotonic function $f(x)$ would also be an appropriate definition. However, we would like a scale that facilitates power comparisons if multiple states gang up on another. So we would need a new scale such that

$$ p_{i+j} = f(P_i + P_j) = f(P_i) + f(P_j) = p_i + p_j $$

for all $P_i, P_j$ . The only function that behaves like this is a linear function of $P(p_i) = A \times P_i $, where A is some constant. So our definition of power is basically fixed up to what "units" we choose. Of course, defining $P_i$ in terms of tangibles (e.g. army size or GDP or population size or number of nuclear warheads) would be a difficult task. Incidentally, I've also implicitly assumed here that there is a power scale, such that if $P_1 > P_2$, and $P_2 > P_3$, then $P_1 > P_3$. But I think that's a fairly benign assumption.

2. ^ This implicity assumes that it doesn't matter which state attacked the other, or where the war is taking place, or other things like that.

3. ^ Incidentally this form for the rate-of-change of the power also has the advantage that it is scale-free, which we might expect since there is no intrinsic "power scale" to the problem. Of course there are other forms with this property that follow some or all of the assumptions above. For instance, something of the form $dx/dt = -xy = dy/dt$ would also be i) scale-invariant, and ii) in line with assumptions 1 & 2 and partially inline with assumption 3. However I didn't use this since a) it's nonlinear, and hence a little harder to solve the resulting differential equations analytically, and b) the rate of decrease of both state's power is the same, in contrast to my intuitive feeling that the state with less power should lose power more rapidly.

4. ^ Homework for those who are not satisfied with my assumptions: Show that any functional form for $dP_i/dt$ that follows assumptions 1-3 above does not change the stability of a balanced bipolar system.

When will the Earth fall into the Sun?

Brian — Thu, 29 Nov 2012 22:58:00 GMT

The time I spent making this poster could have been spent doing research

Since December 2012 is coming up, I thought I'd help the Mayans out with a look at a possible end of the world scenario. (I know, it's not Earth Day yet, but we at the Virtuosi can only go so long without fantasizing about wanton destruction.) As the Earth zips around the Sun, it moves through the heliosphere, which is a collection of charged particles emitted by the Sun. Like any other fluid, this will exert drag on the Earth, slowly causing it to spiral into the Sun. Eventually, it will burn in a blaze of glory, in a bad-action-movie combination of Sunshine meets Armageddon. Before I get started, let me preface this by saying that I have no idea what the hell I'm talking about. But, in the spirit of being an arrogant physicist, I'm going to go ahead and make some back-of-the-envelope calculations, and expect that this post will be accurate to within a few orders of magnitude. Well, how long will the Earth rotate around the Sun before drag from the heliosphere stops it? This seems like a problem for fluid dynamics. How do we calculate what the drag is on the Earth? Rather than solve the fluid dynamics equations, let's make some arguments based on dimensional analysis. What can the drag of the Earth depend on? It certainly depends on the speed of the Earth v -- if an object isn't moving, there can't be any drag. We also expect that a wider object feels more drag, so the drag force should depend on the radius of the Earth R. Finally, the density of the heliosphere might have something to do with it. If we fudge around with these, we see that there is only 1 combination that gives units of force:

$$ F_{drag} \sim \rho v^2 R^2 $$

Now that we have the force, the energy dissipated from the Earth to the heliosphere after moving a distance $d$ is $E_\textrm{lost} = F\times d$. If the Earth moves with speed v for time t, then we can write $E_\textrm{lost} = F v t$. So we can get an idea of the time scale over which the Earth starts to fall into the Sun by taking

$E_\textrm{lost} = E_\textrm{Earth} \sim 1/2 M_\textrm{Earth} v^2$. Rearranging and dropping factors of 1/2 gives

$$ T_\textrm{Earth burns} \sim M_{Earth} v^2 / (F_{drag}\times v) \ \qquad \sim M_{Earth} / (\rho R^2 v) $$

Using the velocity of the Earth as $2\pi \times 1 \mbox{Astronomical unit/year}$, Googlin' for some numbers, and taking the density of the heliosphere to be $10^{-23}$ g/cc we get...

$$ T \approx 10^{19} \textrm{ years} $$

Looks like this won't be the cause of the Mayan apocalypse. (By comparison, the Sun will burnout after only $\sim10^9$ years.)

Creating an Earth

Brian — Sat, 27 Oct 2012 19:07:00 GMT

GAAAAAAAAH

A while ago I decided I wanted to create something that looks like the surface of a planet, complete with continents & oceans and all. Since I've only been on a small handful of planets, I decided that I'd approximate this by creating something like the Earth on the computer (without cheating and just copying the real Earth). Where should I start? Well, let's see what the facts we know about the Earth tell us about how to create a new planet on the computer.

Observation 1: Looking at a map of the Earth, we only see the heights of the surface. So let's describe just the heights of the Earth's surface.

Observation 2: The Earth is a sphere. So (wait for it) we need to describe the height on a spherical surface. Now we can recast our problem of making an Earth more precisely mathematically. We want to know the heights of the planet's surface at each point on the Earth. So we're looking for field (the height of the planet) defined on the surface of a sphere (the different spots on the planet). Just like a function on the real line can be expanded in terms of its Fourier components, almost any function on the surface of a sphere can be expanded as a sum of spherical harmonics $Y_{lm}$. This means we can write the height $h$ of our planets surfaces as

$$ h(\theta, \phi) = \sum A_{lm}Y_l^m(\theta, \phi) \quad (1) $$

If we figure out what the coefficients $A$ of the sum should be, then we can start making some Earths! Let's see if we can use some other facts about the Earth's surface to get get a handle on what coefficients to use.

Observation 3: I don't know every detail of the Earth's surface, whose history is impossibly complicated. I'll capture this lack-of-knowledge by describing the surface of our imaginary planet as some sort of random variable. Equation (1) suggests that we can do this by making the coefficients $A$ random variables. At some point we need to make an executive decision on what type of random variable we'll use. For various reasons,^[1] I decided I'd use a Gaussian random variable with mean 0 and standard deviation $a_{lm}$:

$$ A_{lm} = a_{lm} N(0,1) $$

(Here I'm using the notation that $N(m,v)$ is a normal or Gaussian random variable with mean $m$ and variance $v$. If you multiply a Gaussian random variable by a constant $a$, it's the same as multiplying the variance by $a^2$, so $a N(0,1)$ and $N(0,a^2)$ are the same thing.)

Observation 4: The heights of the surface of the Earth are more-or-less independent of their position on the Earth. In keeping with this, I'll try to use coefficients $a_{lm}$ that will give me a random field that is is isotropic on average. This seems hard at first, so let's just make a hand-waving argument. Looking at some pretty pictures of spherical harmonics, we can see that each spherical harmonic of degree $l$ has about $l$ stripes on it, independent of $m$. So let's try using $a_{lm}$'s that depend only on $l$, and are constant if just $m$ changes^[2]. Just for convenience, we'll pick this constant to be $l$ to some power $-p$:

$$ a_{lm} = l^{-p} \quad \textrm{ or} $$

$$ h(\theta, \phi) = \sum_{l,m} N_{lm}(0,1) l^{-p} Y_l^m(\theta, \phi) \quad (2) $$

At this point I got bored & decided to see what a planet would look like if we didn't know what value of $p$ to use. So below is a movie of a randomly generated "planet" with a fixed choice of random numbers, but with the power $p$ changing.

As the movie starts ($p=0$), we see random uncorrelated heights on the surface.^[3] As the movie continues and $p$ increases, we see the surface smooth out rapidly. Eventually, after $p=2$ or so, the planet becomes very smooth and doesn't look at all like a planet. So the "correct" value for p is somewhere above 0 (too bumpy) and below 2 (too smooth). Can we use more observations about Earth to predict what a good value of $p$ should be?

Observation 5: The elevation of the Earth's surface exists everywhere on Earth (duh). So we're going to need our sum to exist. How the hell are we going to sum that series though! Not only is it random, but it also depends on where we are on the planet! Rather than try to evaluate that sum everywhere on the sphere, I decided that it would be easiest to evaluate the sum at the "North Pole" at $\theta=0$. Then, if we picked our coefficients right, this should be statistically the same as any other point on the planet. Why do we want to look at $\theta = 0$? Well, if we look back at the wikipedia entry for spherical harmonics, we see that

$$ Y_l^m = \sqrt{ \frac{2l +1}{4\pi}\frac{(l-m)!}{(l+m)!}} e^{im\phi}P^m_l(\cos\theta) \quad (3)$$

That doesn't look too helpful -- we've just picked up another special function $P_l^m$ that we need to worry about. But there is a trick with these special functions $P_l^m$: at $\theta = 0$, $P_l^m$ is 0 if $m$ isn't 0, and $P_l^0$ is 1. So at $\theta = 0$ this is simply:

$$ Y_l^m(\theta = 0) = \bigg { ^{\sqrt{(2l+1)/4\pi},\,m=0}_{0,\,m \ne 0} $$

Now we just have, from every equation we've written down:

$$ h(\theta = 0) = \sum_l \times l^{-p} \times \sqrt{(2l+1)/4\pi }\times N(0,1) $$

$$ \quad \qquad = \times \frac{1}{\sqrt{4\pi}} \times \sum_l N(0,l^{-2p}(2l+1)) $$

$$ \quad \qquad = \times \frac{1}{\sqrt{4\pi}} \times N(0,\sum_l l^{-2p}(2l+1) ) $$

$$ \quad \qquad = \times \frac{1}{\sqrt{4\pi}} \sqrt{\sum_l l^{-2p}(2l+1)} \times N(0,1) $$

$$ \quad \qquad \sim \sqrt{\sum_l l^{-2p+1}} N(0,1) \qquad (4) $$

So for the surface of our imaginary planet to exist, we had better have that sum converge, or $-2p+1 < -1 ~ (p > 1)$. And we've also learned something else!!! Our model always gives back a Gaussian height distribution on the surface. Changing the coefficients changes the variance of distribution of heights, but that's all it does to the distribution. Evidently if we want to get a non-Gaussian distribution of heights, we'd need to stretch our surface after evaluating the sum. Well, what does the height distribution look like from my simulated planets? Just for the hell of it, I went ahead and generated ${\sim}400$ independent surfaces at ${\sim}40$ different values for the exponent $p$, looking at the first 22,499 terms in the series. From these surfaces I reconstructed the measured distributions; I've combined them into a movie which you can see below.

As you can see from the movie, the distributions look like Gaussians. The fits from Eq. (4) are overlaid in black dotted lines. (Since I can't sum an infinite number of spherical harmonics with a computer, I've plotted the fit I'd expect from just the terms I've summed.) As you can see, they are all close to Gaussians. Not bad. Let's see what else we can get.

Observation 6: According to some famous people, the Earth's surface is probably a fractal whose coastlines are non-differentiable. This means that we want a value of $p$ that will make our surface rough enough so that its gradient doesn't exist (the derivative of the sum of Eq. (2) doesn't converge). At this point I'm getting bored with writing out calculations, so I'm just going to make some scaling arguments. From Eq. (3), we know that each of the spherical harmonics $Y_l^m$ is related to a polynomial of degree $l$ in $\cos \theta$. So if we take a derivative, I'd expect us to pick up another factor of $l$ each time. Following through all the steps of Eq. (4) we find

$$ \vec{\nabla}h \sim \sqrt{\sum_l l^{-2p+3}}\vec{N}(0,1) \quad , $$

which converges for $p > 2$. So for our planet to be "fractal," we want $1<p<2$. Looking at the first movie, this seems reasonable.

Observation 7: 70% of the Earth's surface is under water. On Earth, we can think of the points underwater as all those points below a certain threshold height. So let's threshold the heights on our sphere. If we want 70% of our generated planet's surface to be under water, Eq (4) and the cumulative distribution function of a Gaussian distribution tells us that we want to pick a critical height $H$ such that

$$ \frac{1}{2} \left[ 1 + \textrm{erf}(H \sqrt{2\sigma^2}) \right] = 0.7 \quad \textrm{or} $$

$$ H = \sqrt{2\sigma^2}\textrm{erf}^{-1}(0.4) $$

$$\sigma^2 = \frac 1 {4\pi} \sum_l l^{-2p}(2l+1) \quad (5)\, , $$

where $\textrm{erf}()$ is a special function called the error function, and $\textrm{erf}^{-1}$ is its inverse. We can evaluate these numerically (or by using some dirty tricks if we're feeling especially masochistic). So for our generated planet, let's call all the points with a height larger than $H$ "land," and all the points with a height less than $H$ "ocean." Here is what it looks like for a planet with $p=0$, $p=1$, and $p=2$, plotted with the same Sanson projection as before.

_{Top to bottom: p=0, p=1, and p=2. I've colored all the "water" (positions with heights < $H$ as given in Eq. (5) ) blue and all the land (heights > $H$) green.}

You can see that the the total amount of land area is roughly constant among the three images, but we haven't fixed how it's distributed. Looking at the map above for $p=0$, there are lots of small "islands" but no large contiguous land masses. For $p=2$, we see only one contiguous land mass (plus one 5-pixel island), and $p=1$ sits somewhere in between the two extremes. None of these look like the Earth, where there are a few large landmasses but many small islands. From our previous arguments, we'd expect something between $p=1$ and $p=2$ to look like the Earth, which is in line with the above picture. But how do we decide which value of p to use?

Observation 8: The Earth has 7 continents This one is more vague than the others, but I think it's the coolest of all the arguments. How do we compare our generated planets to the Earth? The Earth has 7 continents that comprise 4 different contiguous landmasses. In order, these are 1) Europe-Asia-Africa, 2) North- and South- America, 3) Antartica, and 4) Australia, with a 5th Greenland barely missing out. In terms of fractions of the Earth's surface, Google tells us that Australia covers 0.15% of the Earth's total surface area, and Greenland covers 0.04%. So let's define a "continent" as any contiguous landmass that accounts for more than 0.1% of the planet's total area. Then we can ask: What value of p gives us a planet with 4 continents? I have no idea how to calculate exactly what that number would be from our model, but I can certainly measure it from the simulated results. I went ahead and counted the number of continents in the generated planets.

The results are plotted above. The solid red line is the median values of the number of continents, as measured over 400 distinct worlds at 40 different values of $p$. The red shaded region around it is the band containing the upper and lower quartiles of the number of continents. For comparison, in black on the right y-axis I've also plotted the log of the total number of landmasses at the resolution I've used. The number of continents has a resonant value of $p$ -- if $p$ is too small, then there are many landmasses, but none are big enough to be continents. Conversely, if $p$ is too large, then there is only one huge landmass. Somewhere in the middle, around $p=0.5$, there are about 20 continents, at least when only the first ${\sim}23000$ terms in the series are summed. Looking at the curve, we see that there are roughly two places where there are 4 continents in the world -- at $p=0.1$ and at $p=1.3$. Since $p=0.1$ doesn't converge, and since $p=0.1$ will have way too many landmasses, it looks like a generated Earth will look the best if we use a value of $p=1.3$ And that's it. For your viewing pleasure, here is a video of three of these planets below, complete with water, continents, and mountains.^[4]

Notes

1. ^ Since I wanted a random surface, I wanted to make the mean of each coefficient 0. Otherwise we'd get a deterministic part of our surface heights. I picked a distribution that's symmetric about 0 because on Earth the bottom of the oceans seem roughly similar in terms of changes in elevation. I wanted to pick a stable distribution & independent coefficients because it makes the sums that come up easier to evalutate. Finally, I picked a Gaussian, as opposed to another stable distribution like a Lorentzian, because the tallest points on Earth are finite, and I wanted the variance of the planet's height to be defined.

2. ^ We could make this rigorous by showing that a rotated spherical harmonic is orthogonal to other spherical harmonics of a different degree $l$, but you don't want to see me try.

3. ^ Actually $p=0$ should correspond to completely uncorrelated delta-function noise. (You can convince yourself by looking at the spherical harmonic expansion for a delta-function.) The reason that the bumps have a finite width is that I only summed the first 22,499 terms in the series ($l=150$ and below). So the size of the bumps gives a rough idea of my resolution.

4. ^ For those of you keeping score at home, it took me more than 6 days to figure out how to make these planets.

A Homemade Viscometer I

Brian — Tue, 24 Jul 2012 22:32:00 GMT

Stirring a bowl of honey is much more difficult than stirring a bowl of water. But why? The mass density of the honey is about the same as that of water, so we aren't moving more material. If we were to write out Newton's equation, $ma$ would be about the same, but yet we still need to put in much more force. Why? And can we measure it?

The reason that honey is harder to stir is of course that the drag on our spoon depends on more than just the density of the fluid. The drag also depends on the viscosity of the fluid -- loosely speaking, how thick it is -- and the viscosity of honey is about 400 times that of water, depending on the conditions. In fact, a quick perusal of the Wikipedia article on viscosity shows that viscosities can vary by a fantastic amount -- some 13 orders of magnitude, from easy-to-move gases to thick pitch that behaves like a solid except on long time scales. The situation is even more complicated than this, as some fluids can have a viscosity that changes depending on the flow. I wanted to find a way to measure the viscosities of the stuff around me, so I made the viscometer pictured below for about $1.75 (the vending machines in Clark Hall are pretty expensive).

My homemade viscometer, taking data on the viscosity of water.

To do this, I

Enjoyed the crisp, refreshing taste of Diet Pepsi from a 20 oz bottle (come on, sponsorships).
Cut the top and bottom off the bottle, so all that was left was a straight tube.
Mounted the bottle with on top of a small piece of flat plastic.
Mounted a single-tubed coffee stirrer horizontally out of the bottle (I placed the end towards the middle of the bottle to avoid end effects).
Epoxied or glued the entire edge shut.
Marked evenly-spaced lines on the side of the bottle.

I can load my "sample" fluid in the top of the Pepsi bottle, and time how long it takes for the sample level to drop to a certain point. A more viscous fluid will take more time to leave the bottle, with the time directly proportional to the viscosity. (This is a consequence of Stokes flow and the equation for flow in a pipe. It will always be true, as long as my fluid is viscous enough and my apparatus isn't too big.)

So we're done! All we need to do is calibrate our viscometer with one sample, measure the times, and then we can go out and measure stuff in the world! No need to stick around for the boring calculations! We can do some fun science over the next few blog posts!

But this is a physics blog written by a bunch of grad students, so I'm assuming that a few of you want to see the details. (I won't judge you if you don't though.) If we think about the problem for a bit, we basically have flow of a liquid through a pipe (i.e. the coffee stirrer), plus a bunch of other crap which hopefully doesn't matter much.

We first need to think about how the fluid moves. We want to find the velocity of the fluid at every position. This is best cast in the language of vector calculus -- we have a (vector) velocity field $\vec{u}$ at a position $x$. There are two things we know: 1) We don't (globally) gain or lose any fluid, and 2) Newton's laws $F=ma$ hold. We can write these equations as the Navier-Stokes equations:

$$ \vec{\nabla}\cdot \vec{u} = 0 \quad (1) $$

$$ \rho \left( \frac {\partial \vec{u}} {\partial t} + (\vec{u}\cdot\vec{\nabla})\vec{u} \right) = - \vec{\nabla}p + \eta \nabla^2 \vec{u} \quad (2) $$

The first equation basically says that we don't have any fluid appearing or disappearing out of nowhere, and the second is basically $m \vec{a}=\vec{F}$, except written per unit volume. (The fluid's mass-per-unit-volume is $\rho$, the rate of change of our velocity is $\frac{d\vec{u}}{dt}$, and our force per unit volume is $\vec{\nabla}p$, plus a viscous term $\nabla^2 \vec{u}$. The only complication is that $\frac{d\vec{u}}{dt}$ is a total derivative, which we need to write as

$$ \frac{d\vec{u}}{dt} = \frac{\partial \vec{u}}{\partial t} + \frac{\partial \vec{u}}{\partial x} \frac{d x}{d t}$$

I won't drag you through the gory details, unless you want to see them, but it turns out that for my system the height of the fluid $h$ (measured from the coffee stirrer) versus time $t$ is

$$ h(t) = h(0)e^{-t/T}, \quad T= 60.7 \textrm{sec} \times [\eta / \textrm{1 mPa s}] \times [\textrm{ 1 g/cc} / \rho] $$

[For my viscometer, the coffee stirrer has length 13.34 cm and inside diameter 2.4 mm, and the Pepsi bottle has a cross-sectional area of 36.3 square centimeters (3.4 cm inner radius). You can see how the timescale scales with these properties in the gory details section.]

A run with measured heights vs times & error bars. The majority of the uncertainty turns out to come from not knowing the exact proportions of the viscometer. I don't know exactly why the heights are systematically deviating from the fit, but I suspect it's that my gridlines aren't perfectly lined up with the bottom of my viscometer (it looks like $\sim 5$ mm off would do it, which I can totally believe looking at the picture of my viscometer). However, because of the linearity of the equations for steady flow in a pipe, we know that the time scales linearly with the viscosity, so we should be able to accurately measure relative viscosities.

Well, how well does it work? Above is a plot of the height of water in my viscometer versus time, with a best-fit value from the equations above. To get a sense of my random errors (such as how good I am at timing the flow), I measured this curve 5 separate times. If I take into account the uncertainties in my apparatus setup as systematic errors, I find a value for my viscosity as

$$ \eta \approx 1.429 \textrm{mPa s} \pm 0.5 \% \textrm{Rand.} \pm 55\% \textrm{Syst.} $$

The actual value of the viscosity of water at room temperature (T=25 C) is about $0.86~\textrm{mPa s}$, which is more-or-less within my systematic errors. So it looks like I won't be able to measure absolute values of viscosity accurately without a more precise apparatus. But if I look at the variation of my measured viscosity, I see that I should probably be able to measure changes in viscosity to 0.5% !! That's pretty good! Hopefully over the next couple weeks I'll try to use my viscometer to measure some interesting physics in the viscosity of fluids.

How Cold is the Ground II

Brian — Sat, 26 May 2012 21:28:00 GMT

Images from Wikipedia

Last week (ok, it was a little more than a few days ago...) I used dimensional analysis to figure out how the ground's temperature changes with time. But although dimensional analysis can give us information about the length scales in the problem, it doesn't tell us what the solution looks like. From dimensional analysis, we don't even know what the solution does at large times and distances. (Although we can usually see the asymptotic behavior directly from the equation.) So let's go ahead and solve the the heat equation exactly:

$$ \frac {\partial T}{\partial t} = a \frac {\partial ^2 T}{\partial x^2} \quad (1)$$

Well, what type of solution do we want to this equation? We want the temperature at the Earth's surface $x=0$ to change with the days or the seasons. So let's start out modeling this with a sinusoidal dependence -- we'll look for a solution of the form

$$ T(x,t) = A(x)e^{i wt} $$

for some function $A(x)$, then we can take the real part for our solution. Plugging this into Eq. (1) gives $A^{\prime\prime} = i\omega/a \times A$, or

$$ A(x) = e^{ \pm \sqrt{w/2a } (1+i) x} $$

Since we have a second-order ordinary differential equation for $A$, we have two possible solutions, which are like $\exp(+x)$ or $\exp(-x)$. Which one do we choose?

Well, we want the temperature very far away from the surface of the ground to be constant, so we need the solution that decays with distance, $A\exp(-x)$. Taking the real part of this solution, we find^[1]

$$ T(x,t) = T_0 \cos (wt + \sqrt{w/2a}\times x ) e^{-\sqrt{w/2a}x} \quad (2) $$

Well, what does this solution say? As we expected from our scaling arguments last week, the distance scale depends on the square root of the time scale -- if we decrease our frequency by 4 (say, looking at changes over a season vs over a month), the ground gets cooler only $2{\times}$ deeper. We also see that the temperature oscillation drops off quite rapidly as we go deeper into the ground, and that there is a "lag" the farther you go into the ground. In particular, we see that at distances deep into the ground, the temperature drops to its average value at the surface. You can see this all in the pretty plot below (generated with Python):

Single frequency plot

Let's recap. To model the temperature of the ground, we looked for a solution to the heat equation which had a sinusoidally oscillating temperature at $x=0$, and decayed to 0 at large $x$. We found a solution such a solution, and it shows that the temperature decays rapidly as we go far into the ground. At this point, there are two questions that pop into mind:

1) Is the solution that we found unique? Or are there other possible solutions?

2) This is all well and good, but what if our days or seasons aren't perfect sines? Can we find a solution that describes this behavior?

I'll give one (1) VirtuosiPoint to the first commenter who can prove to what extent the above solution is unique^[2]. But how about the second point? Can we solve this for non-sinusoidal time variations? Well, at this point most of the readers are rolling their eyes and shouting "Use a Fourier series and move on." So I will. Briefly, it turns out that (more or less) any periodic function can be written as a sum of sines & cosines. So we can just add a bunch of sines and cosines together and construct our final solution. So just for fun, here is a plot of the temperature of the ground in Ithaca (data from Wikipedia) over a year. (I used a discrete Fourier transform to compute the coefficients.)

The temperature (colorbar) is in degrees C, assuming a=0.5 mm^2/s from before.

Looks pretty boring, but I swear that all the frequencies are in that plot. It just turns out that the seasons in Ithaca are pretty sinusoidal. So about 20 meters below Ithaca, the temperature is a pretty constant 8 C. While I was postponing writing this, I wondered what the temperature on Mercury's rocks would be. If we dig deep enough, can we find an area with habitable temperatures? Some quick Googlin' shows that the daytime and nighttime temperatures on Mercury are ${\sim}550-700~{\rm K}$ and ${\sim}110~{\rm K}$ at the "equator." While I don't think that Mercury's temperature varies symmetrically, let's assume so for lack of better data.^[3] Then we'd expect that deep into the surface, the temperature would be fairly constant in time, at the average of these two extremes. Plugging in the numbers (assuming $a\approx0.52~{\rm mm}^2 / s$ and using a Mercurial solar day as 176 days), we get

$T=94~{\rm C}$ at 2.75 meters into the surface.

1. ^ More precisely, since the heat equation is linear and real, if $T(x,t)$ is a solution to the equation, then so are $\frac{1}{2}(T+T^{})$ or $\frac{1}{2i}(T-T^{})$.

2. ^ Hint: It's not unique. For instance, here is another solution that satisfies the constraints, with no internal heat sources or sinks (I'll call it the "freshly buried" solution):

Freshly buried.

Can you prove that all the other solutions decay to the original solution? Or is there a second or even a spectrum of steady state solutions?

[3] ^ If someone provides me with better data of the time variation of Mercury's surface at some specific latitude, I'll update with a full plot of the temperature as a function of depth and time.

How Cold is the Ground?

Brian — Fri, 18 May 2012 00:20:00 GMT

It snowed in Ithaca a few weeks ago. Which sucked. But fortunately, it had been warm for the previous few days, and the ground was still warm so the snow melted fast. Aside from letting me enjoy the absurd arguments against global warming that snow in April birthed, this got me thinking: How cold is the ground throughout the year? At night vs. during the day? And the corollary: How cold is my basement? If I dig a deeper basement, can I save on heating and cooling? (I'm very cheap.)

Well, we want to know the temperature distribution $T$ of the ground as a function of time $t$ and position $x$. So some googlin' or previous knowledge shows that we need to solve the heat equation.

For our purposes, we can treat the Earth as flat (I don't plan on digging a basement deep enough to see the curvature of the Earth), so we can assume the temperature only changes with the depth into the ground $x$:

$$ \frac{\partial T}{\partial t} = a \frac{\partial^2 T}{\partial x^2}\qquad (1) $$

where $a$ is the thermal diffusivity of the material, in units of square meters per second. It looks like we're going to have to solve some partial differential equations! Or will we? We can get a very good estimate of how much the temperature changes with depth just by dimensional analysis.

Let's measure our time $t$ in terms of a characteristic time of our problem $w$ (it could be 1 year if we were trying to see the change in the ground's temperature from summer to winter, or 1 day if we were looking at the change from day to night). Then we can write:

$$ \frac{\partial T }{\partial t} = \frac{1}{w} \frac {\partial T} {\partial t/w} $$

plugging this in Eq. (1), rearranging, and calling $l= \sqrt{wa}$ gives....

$$ \frac{\partial T}{\partial (t/w)} = \frac{\partial ^2 T}{\partial (x/l )^2} $$

Now let's say we didn't know how to or didn't want to solve this equation. (Don't worry, we do & we will). From rearranging this equation, we see right away there is only one "length scale" in the problem, $l$. So if we had to guess, we could guess that the ground changes temperature a distance $l$ into the ground. A quick look at Wikipedia for thermal diffusivities gives us the following table, for materials we'd find in the ground:

Tables	Are	Cool
col 3 is	right-aligned	$1600
col 2 is	centered	$12
zebra stripes	are neat	$1

blloop

Material	$a$ (${\rm mm}^2 / {\rm s}$)	$l$ ($\rm{cm}$, $1~{\rm day}$)	$l$ ($\rm{m}$, $1~{\rm year}$)
Polycrystalline Silica (glass, sand)	0.83	27	5.1
Crystalline Silica (quartz)	1.4	35	6.6
Sandstone	1.15	32	6.0
Brick	0.52	21	4.0
Soil	0.3-1.25	16-33	3.1-6.3

So we would expect that the temperature of the ground doesn't change much on a daily basis a foot or so below the ground, and doesn't change ever about 15-20 feet into the ground. Just to pat ourselves on the back for our skills at dimensional analysis, a quick check shows that permafrost penetrates 14.6 feet into the ground after 1 year. So our dimensional estimates looks pretty good! In the next few days I'll solve this equation exactly and throw up a few pretty graphs, and maybe talk a little about PDE's and Fourier series in the process.

Earth Day 2012: Escape to the Moon

Brian — Sun, 22 Apr 2012 15:12:00 GMT

It is now Earth Day 2012, and, according to the Mayan predictions, The Virtuosi will destroy the earth. In a futile attempt to fight my own mortality, I decided to send something to the Moon. It seems, for a poor graduate student trying to get to the Moon, the most difficult part is the Earth holding me back. So first I'll focus on escaping the Earth's gravitational potential well, and if that's possible, then I'll worry about more technical problems, such as actually hitting the moon. Moreover, in honor of the destructive spirit of The Virtuosi near Earth Day, I'll try to do this in the most Wiley-Coyote-esque way possible. Preliminaries If we want to get to the Moon, we need to first figure out how much energy we'll need to escape the Earth's gravitational pull. "That's easy!" you say. "We need to escape a gravitational well, and we know from Newton's law that the potential from a spherical mass ME 's gravity for a test mass m is : \begin{equation} \Phi = - \frac {G M_{E} m}{r} \label{eqn:gravpot} \end{equation} We're currently sitting at the radius of the Earth RE, so we simply need to plug this value in and we'll find out how much energy we need." This is all well and good, but i) I can never remember what the gravitational constant G is, and ii) I have no idea what the mass of the Earth ME is. So let's see if we can recast this in a form that's easier to do mental arithmetic in. Well, we know that the force of gravity is the related to the potential by: $$ \vec{F}(r) = - \vec{\nabla} \Phi = - \frac {d\Phi}{dr} \hat{r} \ \vec{F} = - \frac {G m M_E } {r^2} \label{eqn:gravforce} $$ Moreover, we all know that the force of gravity at the Earth's surface is F(r=RE)=-mg. Substituting this in gives: $$ \frac {G m M_E} {R_E^2} = m g \quad \textrm{, or} $$ \begin{equation} \frac {G m M_E}{R_E} = m g {R_E} \quad . \label{eqn:betterDef} \end{equation} So the depth of the Earth's potential well at the Earth's surface is mgRE. If we use g = 9.8 m/s^2 \~ 10 m/s^2 and RE = 6378 km \~ 6x10^6 m, then we can write this as \begin{equation} \Delta \Phi = m g {R_E} \approx m \times 6 \times 10^7 \textrm{m}^2/\textrm{s}^2 \quad \textrm{(1)}, \end{equation} give or take. How fast do we need to go if we're going to make it to the Moon? Well, at the minimum, we need the kinetic energy of our object to be equal to the depth of the potential well [1], or $$ \frac 1 2 m v^2 = 6 m \times 10^7 \textrm{m}^2/\textrm{s}^2 \quad \textrm{or} \ v \approx 1.1 \times 10^4 \textrm{ m/s (2)} . $$ So we need to go pretty fast -- this is about Mach 33 (33 times the speed of sound in air). At this speed, we'd get from NYC to LA in under 7 minutes. Looks difficult, but let's see just how difficult it is. Attempt I: Shoot to the Moon What goes fast? Bullets go fast. Can we shoot our payload to the moon? Let's make some quick estimates. First, can we shoot a regular bullet to the moon? Well, we said that we need to go about Mach 33, and a fast bullet only goes about Mach 2, so we won't even get close. Since energy is proportional to velocity squared, we'll only have (2/33)^2 \~ 0.4 % of the necessary kinetic energy. [2] So let's make a bigger bullet. How big does it need to be? Well, loosely speaking, we have the chemical potential energy of the powder being converted into kinetic energy of the bullet. Let's assume that the kinetic energy transfer ratio of the powder is constant. If a bullet receives kinetic energy 1/2mbvb^2 from a mass mP of powder, then for our payload to have kinetic energy 1/2 M V^2, we need a mass of powder MP such that \begin{equation} \frac {M_P} {m_P} = \frac M {m_b} \times \frac {V^2}{v_b^2} \end{equation} A quick reference to Wikipedia for a 7.62x51mm NATO bullet shows that \~25 grams of powder propels a \~10 gram bullet at a speed of \~Mach 2.5. We need to get our payload moving at Mach 33, so (V/vb)^2 \~ 175. If we send a 10 kg payload to the Moon, we have M/mb \~ 1000. So we'll need about 1.75 x 10^5 the amount of powder of a bullet to get us to the Moon, or about 4400 kg, which is 4.8 tons (English) of powder. That's a lot of gunpowder to get us to the Moon. For comparison, if we are going to construct a tube-like "case" for our 10 kg bullet-to-the-Moon, it will have to be about half a meter in diameter and 17 feet tall. So I'm not going to be able to shoot anything to the Moon anytime soon. Attempt II: Charge to the Moon OK, shooting something to the Moon is out. Can we use an electric field to propel something to the Moon? Well, we would need to pass a charged object through a potential difference such that \begin{equation} q \Delta \Phi_E = m g R_E = 6 m \times 10^7 \textrm{m}^2/\textrm{s}^2 \quad . \label{eqn:chargepot} \end{equation} After the humiliation of the last section, let's start out small. Can we send an electron to the Moon? We could plug numbers into this equation, but I'm too lazy to look up all those values. Instead, we know that we need to get our electron (rest mass 511 keV) to a speed which is (Eq. 2) $$v \approx 1.1 \times 10^4 \textrm{m/s} \approx 4 \times 10^{-5} c. $$ So an electron moving at this velocity will have a kinetic energy of $$ \textrm{KE} = m c^2 \times \frac 1 2 \frac {v^2}{c^2} = 511 \textrm{ keV} \times \frac 1 2 \frac {v^2}{c^2} \ \qquad \approx 511 \textrm{ keV} \times 0.8 \times 10^{-9} \approx 0.4 \times 10^{-3} eV. $$ So we can give an electron enough kinetic energy to get to the moon with a voltage difference of 0.4 mV, assuming it doesn't hit anything on the way up (it will). We can send an electron to the Moon! How about a proton? Well, the mass of a proton is 1836x that of an electron, but with the same charge, so we'd need 1836 * 0.4 mV \~ 0.73 V to get a proton to the Moon -- again, pretty easy. Continuing this logic, we can send a singly-charged ion with mass 12 amu (i.e. C-) with a 9V battery, and a singly-charged ion with mass 150 amu (something like caprylic acid) using a 110V voltage drop. (Again, assuming these don't hit anything on the way up.) How about our 10 kg object? Let's say we can miraculously charge it with 0.01 C of charge. [3] Then from Eq. (1), we'd need $$ 0.01 C \times \Delta \Phi_E \approx 6 \times 10^8 \textrm{ J ,} $$ or a potential difference of $$ \Delta \Phi_E = 6 \times 10^{10} \textrm{ V. } $$ That is a HUGE potential drop. For comparison, if we have 2 parallel plates with a surface charge of 0.01 C/m^2 (again, a huge charge density), they'd have to be a distance $$ d = 6 \times 10^{10} \textrm{V} \times \epsilon_0 / (0.01 \textrm{C/m}^2) \approx 53 \textrm{ meters apart} $$ It looks like I won't be able to send something to the Moon using tools from my basement anytime soon. [1] We'll ignore both air resistance and the Moon's gravitational attraction for simplicity. [2] Since the potential U \~ - 1/r, if we increase our potential energy by 0.4%, this is (to 1st order) the same as increasing r by 0.4%. So we'll get 0.004 * 6378 km \~ 25 km above the Earth's surface. Of course air resistance slows it down a lot. [3] According to Wikipedia, this is 0.04% of the total charge of a thundercloud. And if our object is uniformly charged with a radius of 1 m, it will have an electrical self-energy of ** $$ U = \frac 1 2 \int \epsilon_0 E^2 dV \approx 36 \textrm{kJ} $$