Good effort. Almost had it.
Category: Science
Jorge Luis Borges’ story “The Library of Babel” has long been an obsession of mine. The 1941 short story1 posits a library that contains every possible book-length2 combination of words. It’s probably my second-favorite short story; I think about it all the time and teach it whenever I can. I once even wrote a program to output the digits of 251,312,000, the number of distinct books the Library of Babel contains, which produced a 2Mb text file of mostly zeros. So when my friend Tony Tulathimutte (about whom I’ve written before) asked me to consult on a “Library of Babel”-inspired essay he is writing on the algorithmic generation of literature, I was happy to help. Tony asked:
Even if 251,312,000 is beyond astronomically large, I’m interested in getting as close as possible to a non-theoretical implementation of the Library. Can we work on a Fermi estimate of what it would take to assemble the library? Like, if we distributed the workload to every computer on Earth, or used the world’s fastest supercomputer (China’s Tianhe-2, 33.86 petaflops), or even assembled a Douglas-Adams-style Deep Thought Computational Matrix made of human brains (the human brain runs at an estimated 36.8 petaflops)? Or if Moore’s law holds, at what point would the processing power on Earth suffice to create the Library within the lifespan of the universe.
This is a completely reasonable question, but one that illustrates just how unnatural it is to think about numbers that are “beyond astronomically large.” The number of books in the Library of Babel is so big, no set of adjectives can meaningfully capture its hugeness. After all, things like petaflops or the computational capacity of the human brain are also too big to really conceptualize. So it makes sense that one might treat them all as members in equal standing of the Numbers Too Big To Think About club. But they aren’t. Here are three illustrations of the absurd magnitude of the Library of Babel.
1. Time
First we’ll look at the initial question, how long would it take to generate the Library of Babel? Instead of addressing it the way Tony suggests, though, let’s approach the problem from the opposite direction: what is the fastest it’s possible to imagine generating the Library of Babel?
The Heisenberg uncertainty principle implies that there is a smallest possible size something can be, and a shortest possible time in which something can happen. These minimum quantities are built in to the basic workings of the universe, and are called the Planck units. The Planck time is equal to about 5.391 x 10-44 seconds. It isn’t physically possible for an event to occur in less time than that. Let’s imagine that we have computers capable of generating one Library of Babel Book (LoBB) per unit of Planck time. How many of these computers? Let’s be ambitious: through some impossible alchemy, we will now turn every single atom in the observable universe into a computer capable of generating one LoBB per unit of Planck time.
There are on the order of 1080 atoms in the observable universe. So let’s say we have that many computers… what’s that? Oh, you’re asking, “but what about dark matter?” It’s true. Scientists think there might be five times as much dark matter in the universe as there are atoms. So let’s be generous and bump it up ten times. We’ll say with have 1081 computers, each of which generates one LoBB per unit of Planck time. So, if we have 1081 computers generating about 1043 LoBBs per second, that means we generate 10124 LoBBs every second, 10131 LoBBs per year.
There are 251,312,000 possible LoBBs, which is on the order of 101,834,097. At a rate of 10131 LoBBs per year, it will take 101,833,966 years to finish making the whole Library, or on the order of 10106. Take a quick look at Wikipedia’s timeline of the far future. You’ll notice that the time when we finish making the Library at the fastest imaginable rate would be one of the last items on the list, coming well after the entire universe is a cold, dead, iron cinder.
So the answer to Tony’s question is: never.
2. World Enough
But maybe you noticed that I cheated a little. I said I would consider the fastest it’s possible to imagine generating LoBBs, but calculated based on the fastest it would be physically possible to make them. We can imagine things faster than that, though. We can imagine just snapping our fingers and–poof!–a complete Library of Babel made in an instant. So, why not? Let’s consider that case. We now have the power to instantly assemble a Library of Babel.
Assemble it… out of what? I mean, what are we going to make the literal books out of? Not out of atoms; we already said that there are, generously, 1081 atoms worth of matter in the observable universe. Even if we could somehow encode a LoBB in every atom, we wouldn’t come close to making 10106 of them. Not even if we could make a LoBB out of every subatomic particle.
The universe just doesn’t have enough stuff in it to make the Library of Babel.
3. Vaster Than Empires
So let’s add more stuff. We’ve already given ourselves the power to instantly reconfigure every atom in the universe. Why not give ourself the power to make new matter out of nothing while we’re at it? What happens then?
Turns out, even if we could conjure enough new matter to make the Library of Babel, the universe itself would be too small to hold it.
There’s a weird and fascinating result from black hole physics called the holographic principle, which says that all the information needed to describe a volume of space, down to the minutest quantum detail, only ever takes as much space to encode as the surface area of the volume.3 That is, if you wanted to write down all the information necessary to perfectly describe every detail of what’s inside a room, you would always be able to fit all the information on just the walls. In this way, the entire universe can be thought of as a three dimensional projection of what is, on the level of information, a strictly two dimensional system. Sort of like a hologram, which is 2D but looks 3D, a metaphor from which the principle gets its name.
In any normal region of the universe, the amount of information in a given volume will actually be much less than what you could encode on its surface area. For reasons having to do with thermodynamics that are too complicated to go into here, when you max out the amount of information a volume of space can contain, what you have is a black hole.4 Now, remember those Planck units from the beginning? Length was one of them; there’s a smallest possible size that the laws of nature will let something be, and we can use that length to define a new unit, the Planck area. The most efficient possible encoding of information, per the holographic principle, is one bit per unit of Planck area, which is on the order of 10-70 square meters.
The observable universe has a radius of around 4.4 x 1026 meters. That gives it a surface area on the order of 1053 square meters, which means it can hold 10123 bits of information. That’s just the observable universe though; the whole universe is much, much bigger. We aren’t sure exactly how much bigger, it isn’t observable. But inflationary universe theory, which just got some strong confirming evidence, provides an estimate that the whole universe is 3 x 1023 times larger than the part of the universe we can see. Carry out the same calculations, and the estimated size of the whole universe means that it can contain 10170 bits of information. As for the Library, if you assume that it takes a string of at least six bits to encode one of a set of 25 characters, then the whole Library of Babel would require a number of bits on the order, once again, of 10106. Even if we demiurgic librarians do violate the law of conservation of energy to bring the Library into being, the entire universe would collapse into a black hole long before we finished our project.
So: the Library of Babel is so large that the universe isn’t going to be around long enough to make it. And even if it was, there isn’t enough matter and energy to do it. And even if there was, before that point all of reality as we know it would be destroyed. That is how extreme things can get when you start dealing with “beyond astronomically large” numbers.
There is a version of the story online, but I much prefer the translation by Andrew Hurley in Collected Fictions. ↩
As described by Borges: 25 symbols, 40 symbols per line, 80 lines per page, 410 pages. ↩
I’ve previously posted a video to an excellent introduction to the holographic principle. You can find that here. ↩
This is because, physically speaking, information is the same thing as entropy. ↩
Thermo Thursday returns. But on a Tuesday! WHO COULD HAVE SUSPECTED?
This section introduces isothermal and adiabatic compression. Isothermal compression is compression that occurs so slowly that no heat is added to the gas. For isothermal compression of an ideal gas, the temperature remains constant, so you can use the ideal gas law, PV=NkT, and the integral equation for work done during compression, W=-\int _{{V}_{i}}^{{V}_{f}}{P(V)dV}, to derive that {W}_{isothermal}=NkT\ln {\frac {{V}_{i}}{{V}_{f}}}. Since work is being done but the temperature is not changing, heat must be flowing out of the gas, in an amount equal to the work being done. On a PV graph, an isothermal compression takes the shape of a concave-up hyperbola.
Adiabatic compression, in contrast, happens so fast that no heat escapes from the gas during the process. Thus, for adiabatic compression, \Delta U = W. The PV curve for this starts on a lower temperature isotherm and ends on a higher temperature isotherm. 
To find an equation for the shape of that curve, we look at the equipartition of energy theorem, U = \frac {f}{2}NkT, where f is the number of degrees of freedom per molecule. The infinitesimal change in energy along the curve is then given by dU=\frac {f}{2}NkdT. If we assume the compression is quasistatic, then from the equation for work we know that dU=-PdV. (We can say this because we previously established that in adiabatic compression the entire change in energy comes from work.) This gives \frac {f}{2}NkdT=-PdV. Now you can plug in the ideal gas law for P and do some canceling to get \frac {f}{2}\frac {dT}{T}=-\frac {dV}{V} \rightarrow \frac {f}{2}\ln {\frac {{T}_{f}}{{T}_{i}}}=-\ln{\frac{{V}_{f}}{{V}_{i}}}. This ends up simplifying down to V{T}^{{f}/{2}}=C for some constant C. If you are looking for pressure instead of temperature, you can use the ideal gas law to rewrite this {V}^{\gamma}P=D for some constant D, where \gamma = {(f+2)}/{f} and is called the adiabatic constant.
Problem 1.35: Derive {V}^{\gamma}P=constant from V{T}^{{f}/{2}}=constant.
I’m going to use a subscript on the constant term to show when that side of the equations changes. Let’s start with V{T}^{{f}/{2}}={C}_{0}. I want to get this equation in terms of pressure, so we use the ideal gas law to say that T=\frac {PV}{Nk}. This gives us
V{(\frac{PV}{Nk})}^{{f}/{2}}={C}_{0}.
Raising both sides of the equation to the power 2/f and moving the constant terms to the right side gives
{V}^{{2}/{f}}VP={C}_{1}.
I can then combine the volume terms,
{V}^{1+\frac{2}{f}}P={V}^{{(f+2)}/{f}}P={V}^{\gamma}P={C}_{1}.
Problem 1.36: In the course of pumping up a bicycle tire, a liter of air at atmospheric pressure is compressed adiabatically to a pressure of 7 atm. (Air is mostly diatomic nitrogen and oxygen.)
(a) What is the final volume of this air after compression?
I will treat the air as a diatomic ideal gas for these calculations, which means the air molecules have five degrees of freedom. This gives me \gamma =7/5. For adiabatic compression, {V}_{i}^{\gamma}{P}_{i}={V}_{f}^{\gamma}{P}_{f}, so
{V}_{f}={(\frac{{V}_{i}^{7/5}{P}_{i}}{{P}_{f}})}^{5/7}
which for an initial pressure of 1 atmosphere and volume of 1 liter, and final pressure of 7 atmospheres, gives a final volume of 0.25 liters.
(b) How much work is done in compressing the air?
The total change in energy is the heat added or lost plus the work done, \Delta U = Q + W. Since this is adiabatic compression, there is no heat lost to the environment, so the entire change in energy is due to work. From the equipartition of energy theorem and the ideal gas law I can write \Delta U= \frac {f}{2} \Delta (PV), which for this system gives
\Delta U = \frac {5}{2} ({P}_{f}{V}_{f}-{P}_{i}{V}_{i})
Plugging in {V}_{f}=0.25 L, {P}_{f}=7 atm, {V}_{i}=1.0 L, {P}_{i}=1 atm, converting to SI units, and calculating gives an energy added due to work of 189.5 J.
(c) If the temperature of the air is initially 300 K, what is the temperature after compression?
I can use the equation V{T}^{{f}/{2}}=constant to say that
{V}_{i}{T}_{i}^{\frac{5}{2}}={V}_{f}{T}_{f}^{\frac{5}{2}}
and so
{T}_{f}={(\frac {{V}_{i}{T}_{i}^{\frac{5}{2}}}{{V}_{f}})}^{\frac {2}{5}}.
Plugging in the values and calculating this out gives a final temperature of 522 K.
Problem 1.37: In a Diesel engine, atmospheric air is quickly compressed to about 1/20 of its original volume. Estimate the temperature of the air after compression, and explain why a Diesel engine does not require spark plugs.
Since the air is compressed “quickly,” I will assume adiabatic compression, and since it’s air that’s being compressed it has the same degrees of freedom as the previous problem. So I can use the last equation from part (c) above, plug in {V}_{f}=\frac {1}{20}{V}_{i}, and simplify to get
{T}_{f}={(20{T}_{i}^{\frac {5}{2}})}^{\frac {2}{5}} \approx 3.3{T}_{i}.
So if the initial temperature of the air is 300 K, then after compression the temperature will rise to around 990 K. Since the ignition temperature of Diesel fuel is 483 K, the air after compression is hot enough to ignite it without a spark.
Problem 1.38: Two identical bubbles of gas form at the bottom of a lake, then rise to the surface. Because the pressure is much lower at the surface than at the bottom, both bubbles expand as they rise. However, bubble A rises very quickly so not heat is exchanged between it and the water. Meanwhile, bubble B rises slowly (impeded by a tangle of seaweed), so that it always remains in thermal equilibrium with the water (which has the same temperature everywhere). Which of the two bubbles is larger by the time they reach the surface? Explain your reasoning fully.
Bubble A undergoes adiabatic compression (in this case, expansion), while bubble B undergoes isothermal compression. Initially, the bubbles are identical, so their pressures and volumes are equal. For isothermal compression, {P}_{B}{V}_{B}=C where C is a constant, and for adiabatic compression {P}_{A}{V}_{A}^{\gamma}=C. The constants C must be the same, because the bubbles are initially identical. Since \gamma > 1, {V}_{A}<{V}_{B}. Thus, the bubble that undergoes isothermal compression, bubble B, is larger when it reaches the surface. This can be seen by visual inspection of a graph depicting an isotherm and an adiabat. Notice that for the same initial conditions, the volume of the isotherm rises faster than that of the adiabat.
It’s the end of the semester and I’ve got other projects I need to be working on, so this week I’m just going to do the last problem from the compression work section and leave the adiabats and isotherms for next time.
Problem 1.34: An ideal diatomic gas, in a cylinder with a movable piston, undergoes the rectangular cyclic process shown in the figure to the right. Assume that the temperature is always such that rotational degrees of freedom are active, but vibrational modes are “frozen out.” Also assume that the only type of work done on the gas is quasistatic compression-expansion work. (a) For each of the four steps A through D, compute the work done on the gas, the heat added to the gas, and the change in the energy content of the gas. Express all answers in terms of {P}_{1}, {P}_{2}, {V}_{1}, and {V}_{2}. (Hint: Compute \Delta U before Q, using the ideal gas law and the equipartition theorem.)
Step A: This is a diatomic gas with no vibration modes, so for this system the equipartition theorem says that U = \frac {5}{2}PV. For this step, \Delta U = \frac{5}{2}({P}_{2}-{P}_{1}){V}_{1}. Since the volume of the gas doesn’t change in this step, there is no work being done on the gas. Recalling that \Delta U = W + Q, that means that Q = \frac{5}{2}({P}_{2}-{P}_{1}){V}_{1}.
Step B: here \Delta U = \frac{5}{2}{P}_{2}({V}_{2}-{V}_{1}). The volume is increasing, so the work done on the gas is negative, W=-{P}_{2}({V}_{2}-{V}_{1}). Since W is negative, that means that the heat added must be the total energy change plus the amount of energy subtracted by negative work. Q=\Delta U - W = \frac{7}{2}{P}_{2}({V}_{2}-{V}_{1}).
Step C: this time \Delta U = \frac{5}{2}({P}_{1}-{P}_{2}){V}_{2}. Note that this is a negative quantity, the system has lost energy. Again, there is no volume change, so no work is being done on the gas. Thus the entire energy change is due to the system losing heat energy, so Q=\frac{5}{2}({P}_{1}-{P}_{2}){V}_{2}.
Step D: in the last step, \Delta U = \frac{5}{2}{P}_{1}({V}_{1}-{V}_{2}). This is another negative quantity. But this time, the work being done on the gas is positive, W=-{P}_{1}({V}_{1}-{V}_{2}). Since the work done on the gas is positive, and the energy change is negative, it must be losing even more heat energy than it is gaining in work energy. Here Q= \Delta U - W = \frac{7}{2}{P}_{1}({V}_{1}-{V}_{2}).
(b) Describe in words what is physically being done during each of the four steps; for example, during step A, heat is added to the gas (from an external flame or something) while the piston is held fixed.
The author went ahead and did step A for me. In step B, the volume of the gas increases, so the piston is being drawn out (or, more likely, pushed out by internal pressure). But this doesn’t result in any decrease in pressure, which means that the gas is also being heated as the volume increases. In step C, the system just loses heat energy, so the system is being cooled while the piston is held fixed. In step D the piston is compressed, but the cylinder is still being cooled so the pressure doesn’t change.
(c) Compute the net work done on the gas, the net heat added to the gas, and the net change in energy of the gas during the entire cycle. Are the results as you expected? Explain briefly.
The net change in energy is zero, that’s what makes it a cycle. For the work over the whole cycle, that’s {W}_{cycle}=-({P}_{2}-{P}_{1})({V}_{2}-{V}_{1}). That’s a negative quantity, meaning that over the course of the whole cycle work was done on the environment. Since the energy change is zero, the heat added perfectly balances the work, so {Q}_{cycle}=({P}_{2}-{P}_{1})({V}_{2}-{V}_{1}). That’s a positive quantity. As mentioned at the end of the last Thermo Thursday, this cycle takes in heat energy and converts it to work done on the environment. So these results are as I expected.
I never got around to finishing the problem set last week, so instead of editing the last entry, I’ll just finish it for this week.
Problem 1.32: By applying a pressure of 200 atm, you can compress water to 99% of its usual volume. Sketch this process (not necessarily to scale) on a PV diagram, and estimate the work required to compress a liter of water by this amount. Does the result surprise you?
I’m just going to assume a linear function here and say that the work done is the area under the PV curve. Here that ends up being about 100 J to compress water by 1%.
Problem 1.33: An ideal gas is made to undergo a cyclic process shown in the figure to the right. For each of the steps A, B, C, determine whether each of the following is positive, negative, or zero: (a) the work done on the gas; (b) the change in the energy content added to the gas; (c) the heat added to the gas. Then determine the sign of each of these three quantities for the whole cycle. What does this process accomplish?
For A, the work being done on the gas is (a) negative, by the equation W = -P\Delta V. The energy added to the gas is (b) positive, which I could demonstrate with the equipartition of energy principle, but would rather show by observing that if the gas increases in volume but maintains the same pressure, that means that the frequency of collisions with the container must remain the same as when the gas was at lower volume. The only way for that to happen is if the molecules are moving faster, so energy must have been added to the system. Since the energy of the system is the heat plus the work, and the work is negative and the energy is positive, then the heat added must also be (c) positive (and greater than the absolute value of the work).
For B, the work done on the gas is (a) zero, since there is no change in volume. By equipartition of energy for an ideal gas, U = \frac{3}{2}PV, the energy increase is (b) positive. You can also get this from observing that increased pressure at the same volume means more collisions with the container per unit time, which means that the molecules of the gas must be moving faster, and so must have more energy. Since there is more energy and no work has been done, the heat added is (c) positive.
For C, the work done on the gas must be (a) positive, because the volume decreases. The energy added to the gas is (b) negative, because even though the volume is decreasing, the pressure is going down. Since the work done on the gas is positive and the overall energy change is negative, the heat added to the gas is also (c) negative.
For the whole cycle, the net work must be (a) positive, because the average pressure is higher during step C than during step A. Since the pressure and volume at the end of the cycle are the same as at the start, the net energy change must be (b) zero. And since the work is positive and the energy change is zero, the heat added must be (c) negative. So this cycle takes in energy as work and emits the energy as heat.
There’s another problem here, but it’s basically the same, except it’s a four-step rectangular cycle that goes clockwise, and turns heat added into work done by the gas. It’s late, so I’m not going to go through the details.
This section introduces the equations for the work done on a system by compression, such as when a piston is used to compress a volume of gas. Assuming nearly quasistatic compression (that is, compression slow enough that all of the gas can be said to be at a single pressure; in practice, usually any compression slower than the speed of sound in the gas), the work done per infinitesimal change in volume is W = -P\Delta V. When the pressure changes significantly during compression, you have to approximate the process as a series of small compressions, unless you have an equation for the pressure as a function of volume. If you do, then you can integrate, and the equation for work becomes W=-\int _{{V}_{i}}^{{V}_{f}}{P(V)dV} .
Problem 1.31: Imagine some helium in a cylinder with an initial volume of 1 liter and an initial pressure of 1 atm. Somehow the helium is made to expand to a final volume of 3 liters, in such a way that its pressure rises in direct proportion to its volume. (a) Sketch a graph of the pressure vs. volume for this process.
(b) Calculate the work done on the gas during this process, assuming that there are no “other” types of work being done.
Here I actually do have an equation for pressure with respect to volume. Since the relationship is linear, I can say that P = V (\frac {atm}{L}). Plugging into the equation for work gives W = -\frac {atm}{L} \int _{1 (L)}^{3 (L)}{V dV}. This ends up giving –4 atmosphere-liters, which converts to approximately –400 J. The negative sign indicates that, rather than work being done on the gas, it is the gas that is doing work on the environment (presumably by displacing whatever was in the volume it expands to occupy).
(c) Calculate the change in the helium’s energy content during this process.
Helium is a monatomic gas, so it has three quadratic degrees of freedom. By the equipartition of energy theorem, U = \frac {3}{2}NkT, which by the ideal gas law is equivalent to U = \frac {3}{2}PV. The change in energy, then, is \Delta U = \frac {3}{2} (P_{final}V_{final} - P_{initial}V_{initial}). This gives 12 atmosphere-liters, or approximately 1200 J.
(d) Calculate the amount of heat added to or removed from the helium during this process.
The first law of thermodynamics is \Delta U = Q + W, which I can rewrite Q = \Delta U - W. In this case, the work done is -400 J, and the change in energy is 1200 J, so the heat added to the system is 1600 J.
(e) Describe what you might do to cause the pressure to rise as the helium expands.
Normally as volume increased pressure would decrease. (Solve the ideal gas law for pressure and volume ends up in the denominator). To counter that, you would have to increase the temperature. So you could cause the pressure to rise as the helium expands by heating it.
I’ll have to finish these problems another time, as I need to go to the reception for Marina Warner, who is receiving the Truman Capote award. I’ll edit this post later. (Tumblr, you won’t see the edit, you’ll have to click through.)
This section defines heat, “the spontaneous flow of energy from one object to another caused by the difference in temperature between the objects,” and work, any non-heat transfer of energy into or out of a system. It also introduces the first law of thermodynamics, \Delta U = Q + W, where U is the total energy of the system, Q is the energy that enters the system as heat, and W is the energy that enters the system as work. Schroeder points out that these definitions of heat and work are counter to the way we use the terms in common speech. For example, if you rub your cold hands together to warm them up, by these definitions there is no heating being done to the system, because the change in temperature isn’t due to spontaneous energy flow. Rather, your hands are being warmed by work.
Schroeder goes on to introduce the units Joule, calorie (equal to 4.186 J), and kilocalorie (also known as the food calorie, 4186 J); and the three types of heat transfer, conduction (transfer by molecular contact), convection (transfer by the bulk motion of a gas or liquid), and radiation (emission of electromagnetic waves.)
Problem 1.26: A battery is connected in series to a resistor, which is immersed in water (to prepare a hot cup of tea). Would you classify the flow of energy from the batter to the resistor as “heat” or “work”? What about the flow of energy from the resistor to the water?
The flow of energy from the battery to the resistor is work, because the energy transferred by the electrons moving through the resistor is driven by the voltage difference, it doesn’t occur spontaneously due to a temperature difference. The flow of energy from the resistors to the water, though, is heat.
Problem 1.27: Give an example of a process in which no heat is added to a system, but its temperature increases. Then give an example of the opposite: a process in which heat is added to a system but its temperature does not change.
Imagine you have a cylinder of compressed air that is at room temperature. If you open the valve and start letting air out, the pressure within the cylinder will drop. By the ideal gas law, the temperature of the air inside the cylinder will drop proportionally with the pressure. (This is why the propane tanks attached to gas grills occasionally ice up while the grill is in use.) The opposite condition would hold if you pumped air from the room into the cylinder. Then the temperature of the gas in the cylinder would rise proportionally with the pressure, but no heat would be added.
A system where heat is added but the temperature doesn’t change would be one where either the heat energy is being compensated for by work, or where the heat energy is causing a non-temperature change, like a phase change. So for example if you boil water, energy is constantly flowing into the water due to heat, but the temperature stays at 100°C. The energy goes into changing the water from liquid phase to gas phase.
Problem 1.28: Estimate how long it should take to bring a cup of water to boiling temperature in a typical 600-watt microwave oven, assuming that all the energy ends up in the water. (Assume any reasonable initial temperature for the water.) Explain why no heat is involved in this process.
A calorie is the amount of energy needed to raise one gram of water by 1° C. A watt is one joule per second. One cup of water is 236.6 ml, which is also 236.6 grams. If we assume the water is at 21° C, then the the amount of energy needed to bring the water to boiling is 236.6 x 79 = 18691.4 calories, which is 78,242 joule. Dividing by 600, we get 130 seconds to heat the water to boiling. No heat is involved because the microwaves that are physically vibrating the polar water molecules are being generated, not emitted spontaneously. This is work, just like when you rub your hands together so that the friction warms them. This isn’t a spontaneous transfer of energy due to temperature, it’s a driven energy transfer.
Problem 1.29: A cup containing 200 g of water is sitting on your dining room table. After carefully measuring its temperature to be 20° C, you leave the room. Returning ten minutes later, you measure the temperature again and find that it is now 25° C. What can you conclude about the amount of heat added to the water? (Hint: This is a trick question)
You can infer that the net heat added to the water was less than or equal to 1000 calories. It’s possible that no heat was added, and the temperature increase was due entirely to work. It’s possible that some work was done, and also some heat added. It’s possible that more than 1000 calories of heat were added, and then removed again. You can’t know for sure anything other than that there are 1000 new calories of energy, some or all or none of which could have come from heat.
More student meetings today, so I’m not sure how far I’ll get with this (I type these up as I work the problems), but I want to finish the last problem from two weeks ago because it ends up in a nice equation. I don’t know how to solve it, though, without invoking something that, weirdly, the book doesn’t introduce until after the problem: the equipartition of energy theorem. So let’s start there.
The equipartition of energy theorem states that the average energy of any quadratic degree of freedom, at temperature T, is given by \frac {1}{2}kT. A quadratic degree of freedom is any way a particle can move whose equation of motion is quadratic. So examples of this would translational kinetic energy in each individual spatial dimension (e.g. \frac {1}{2}m{v}_{x}^{2}), rotational kinetic energy (e.g. \frac {1}{2}I{\omega}_{x}^{2}), elastic potential energy (\frac {1}{2}{k}_{s}{x}^{2}), etc. So a system of N molecules in which each molecule has f degrees of freedom and no other (non-quadratic) temperature-dependent forms of energy will have a total average thermal energy given by {U}_{thermal}=Nf(\frac {1}{2}kT). For a very large N, as will usually be the case, deviations away from the average will be negligible.
It’s important to note that {U}_{thermal} doesn’t include other sources of energy, such as relativistic rest energy, or energy stored in chemical bonds. So the equipartition theorem is best used to look at changes in energy due to temperature variation, and not things like phase transformations.
So now that I have the equipartition of energy theorem, I can go back and finish problem 1.22 from last time.
(b) It’s not easy to calculate \bar {{v}_{x}}, but a good enough approximation is \sqrt {\bar {{v}_{x}^{2}}}, where the bar now represents an average over all molecules in the gas. Show that \sqrt {\bar {{v}_{x}^{2}}} = \sqrt{\frac {kT}{m}}.
This comes directly from the equipartition of energy theorem, which gives me \frac {1}{2}m\bar {{v}_{x}^{2}} = \frac {1}{2}kT so \bar {{v}_{x}^{2}} = \frac {kT}{m} and \sqrt {\bar{{v}_{x}^{2}}} = \sqrt {\frac {kT}{m}} .
(c) If we now take away this small part of the wall of the container, the molecules that would have collided with it will instead escape through the hole. Assuming that nothing enters through the hole, show that the number N of molecules inside the container as a function of time is governed by the differential equation \frac {dN}{dt} = -\frac {A}{2V} \sqrt {\frac {kT}{m}}N. Solve this equation (assuming constant temperature) to obtain a formula of the form N(t)=N(0){e}^{{-t}/{\tau}} where \tau is the “characteristic time” for N (and P) to drop by a factor of e.
In my answer to part (a) I got the expression for the number of molecules colliding with the area of the hole over a given time interval as PA\Delta t/(2m\bar { { v }_{ x } }). Now I will let the time interval become infinitesimal, and rewrite this as \frac {PAdt}{2m\bar {{v}_{x}}}. Now once you poke a hole in the container, what was previously a number of collisions becomes a number of molecules leaving the container, which turns N into a decreasing function of time, with the infinitesimal change in N(t) given by dN = -\frac {PAdt}{2m\bar {{v}_{x}}}.
By using the ideal gas law and bringing the dt to the lefthand side, I can rewrite the above as \frac {dN}{dt} = - \frac {N(t)kTA}{2Vm\bar {{v}_{x}}} (notice that N from the ideal gas law is now N(t), a value that is varying with time). As shown before, kT=m\bar {{v}_{x}^{2}}, so I can rewrite this \frac {dN}{dt} = - \frac {N(t)m\bar {{v}_{x}^{2}}A}{2Vm\bar {{v}_{x}}}.
Using the approximation \sqrt {\bar {{v}_{x}^{2}}}=\bar {{v}_{x}} and the result from part (b), I can express the above as \frac {dN}{dt} = - \frac {N(t)\bar {{v}_{x}}A}{2V} = -\frac {N(t)A}{2V}\sqrt {\bar{{v}_{x}^{2}}} =-\frac {N(t)A}{2V}\sqrt {\frac {kT}{m}}. Reorganizing the terms, I get the first sought expression, \frac {dN}{dt} = -\frac {A}{2V}\sqrt {\frac {kT}{m}}N(t). (The textbook expresses that final term as N instead of N(t), but I’m pretty sure that it has to be a function of time since we have a derivative of N with respect to t on the other side, so for the sake of clarity I’m expressing it as a function in my answer.)
The above is a separable differential equation, so \int {\frac {dN}{N(t)}} = \int {-\frac {A}{2V}\sqrt {\frac{kT}{m}}dt} \rightarrow \ln {N(t)} = -\frac {A}{2V}\sqrt {\frac{kT}{m}}t + C. Exponentiating both sides gives N(t) = {e}^{c}{e}^{-\frac {A}{2V}\sqrt {\frac{kT}{m}}t} = B{e}^{-\frac {A}{2V}\sqrt {\frac{kT}{m}}t} for some constant B. Setting t=0 gives N(0)=B, so if we plug that in and we let \tau = \frac {2V}{A\sqrt{\frac{kt}{m}}}, then we get the sought equation, N(t)=N(0){e}^{-\frac {t}{\tau}}.
The rest of the parts of this problem are just using the equations for N(t) and \tau to solve how long it takes air to escape through a given hole in a given container, or figure out how big a hole was given how long it takes a tire to deflate. Things like that. The most interesting one is part (f):
(f) In Jules Verne’s Round the Moon, the space travelers dispose of a dog’s corpse by quickly opening a window, tossing it out, and closing the window. Do you think they can do this quickly enough to prevent a significant amount of air from escaping? Justify your answer with some rough estimates and calculations.
In the preliminary chapter of Round the Moon, the spacecraft is described as a shell with an interior diameter of 84 inches (108 inch diameter with 12 inch thick walls), giving a radius of approximately one meter. (Given that the work was originally in French, I would bet that in the original the diameter is two meters and a conversion was made for the English translation.) Since there are many compartments in the shell, one above another, and they have to hunt “a long time” to find the dog in the upper compartment, I will say the shell is 90 feet long with 10 compartments, which would make it 1/10 as long as the gun it as fired from. Since they had to hunt for the dog, I assume all the compartments are open to each other, and since it makes things simpler I declare the shell to be cylindrical (even though it isn’t). If you plug in these values you get a volume of approximately 86 cubic meters. Assume the area of the window is 0.2 cubic meters and the shell is filled with oxygen, giving a mass of 5 x 10^-26 kg. Recalling that \tau = \frac {2V}{A\sqrt{\frac{kt}{m}}}, using these numbers and room temperature gives a characteristic time of about 3 seconds. That’s the amount of time for the air in the capsule to reduce by a factor of e, so after 3 seconds more than half the oxygen is gone. Throwing the body of Satellite the dog into space is a bad idea.
For chapter 1.3, the problems are all plug-and-play scenarios with the equipartition of energy theorem, except for
Problem 1.25: List all the degrees of freedom, or as many as you can, for a molecule of water vapor. (Think carefully about the various ways in which the molecule can vibrate.)
A water molecule is shaped like this, so to my mind it has nine quadratic degrees of freedom. There are the three translational degrees in the x, y, and z axes, and the three rotational degrees as well. (If it were radially symmetric there would be fewer rotational degrees, but it’s not.) Then there is the possibility of vibration in those bonds. One vibration mode would have the bond distance between the oxygen and the hydrogens oscillating (imagine holding the oxygen and pulling the hydrogens away and letting them bounce back). One would have the bond angles oscillating (imagine holding the oxygen and waving it up and down so the hydrogens flap around it). And one would be a torsional oscillation (imagine holding onto the hydrogens and twisting the oxygen back and forth). Those are the ones I can think of, and Googling tells me I have the number right. There’s a chance my physical descriptions are off, though. Chemistry, bonds; these were never my strong suit. I’m just imagining the molecule as a toy made of balls and springs and trying to figure it out from that.
Section opens with the ideal gas law in the form that I was taught it in high school, PV=nRT where P is pressure, V is volume, T is temperature in kelvins, n is the number of moles of gas, and R is the ideal gas constant, R = 8.31 J/(mol K). Also from high school, a mole of molecules is Avogadro’s number of them, 6.02 x 10^23.
For physics it is often more useful to talk about the number of molecules, rather than the number of moles, so we multiply n by Avogadro’s number to get N, the number of molecules of gas. This transforms the ideal gas law into PV=NkT where k is Boltzmann’s constant, k = 1.381 x 10^-23 J/K. The section then shows how if you consider an ideal gas one molecule at a time, you can eventually conclude that this constant k is essentially a conversion factor between temperature and molecular energy. From there it introduces the electron-volt (eV) as a more convenient unit than the Joule to talk about the tiny kinetic energies of molecules, and derives the equation for the root-mean-square velocity of a molecule in an ideal gas, {v}_{rms} = \sqrt {\frac {3kT}{m}}.
Interesting problems:
Problem 1.11: Rooms A and B are the same size, and are connected by an open door. Room A, however, is warmer (perhaps because its windows face the sun). Which room has a greater mass of air? Explain carefully.
The rooms are the same size, so the volume of the two rooms is the same. The rooms are connected by an open door, so the pressure of the two rooms is the same. The only variable term in the ideal gas law between the two rooms is temperature. Expressing the ideal gas law in terms of the number of molecules in each room, I get N=\frac {PV} {kT}. Since the temperature term is in the denominator, the room with the higher temperature is going to contain a lower number of gas molecules, and therefore a lower mass of air. So the colder room, room B, has the greater mass of air.
Problem 1.12: Calculate the average volume per molecule for an ideal gas at room temperature and atmospheric pressure. Then take the cube root to get an estimate of the average distance between molecules. How does this distance compare to the size of a molecule like { N }_{ 2 } or { H }_{ 2 }O?
From \frac { V }{ N } =\frac { kT }{ P } we get an average volume per molecule of 4.1 x 10^-26 cubic meters. The cube root gives us an average distance between molecules of 3.4 x 10^-9 meters, or 3.4 nm. The size of the nitrogen gas molecule is on the order of tens of picometers, so 100 times smaller. The size of a water molecule is on the order of hundreds of picometers, so 10 times smaller.
Problem 1.16: The exponential atmosphere.
(a) Consider a horizontal slab of air whose thickness (height) is dz. if this slab is at rest, the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for dP/dz, the variation of pressure with altitude, in terms of the density of air.
It’s embarrassing how rusty my math has gotten. I had to look back at my old work to recall how to start solving this problem. Density is mass per unit volume, and usually denoted by \rho (the Greek letter rho). If you assume a fictitious “air molecule” with mass m, then for a slab of air the density would be given by \rho = \frac{Nm}{V}. Since I’m assuming that the air is uniform horizontally and only varies in the z direction, let the area of the top of the slab go to zero, which gives \rho = \frac {Nm}{dz}. Then the mass of the air in the slab is just \rho dz.
If the pressure below the slab is equal to the pressure above plus the weight of the slab, then we have
P(z) = P(z+dz) + (\rho dz)gwhat can be rewritten
P(z) - P(z+dz) = \rho g dzIf the change in pressure dP = P(z + dz) - P(z), then the left side of the equation above is -dP. So I have
-dP = \rho g dzwhich gives me
\frac {dP}{dz} = -\rho g.
(b) Use the ideal gas law to write the density of air in terms of pressure, temperature, and the average mass of air molecules. Show, then, that the pressure obeys the differential equation \frac {dP}{dz}= -\frac {mg}{kT}P, called the barometric equation.
Let’s consider my fictitious air molecule from the previous part to have a mass equal to the average molecular mass of air. (If I were doing this for real, I’d need to make use of the fact that air is 78% molecular nitrogen, 21% molecular oxygen, and 1% argon. But that seems tedious.) Going back to the definition of density, I can rewrite my last result
\frac {dP}{dz} = -\rho g = -\frac {Nm}{V}g.
Then I can use the ideal gas law PV=NkT \rightarrow \frac {N}{V}=\frac {P}{kT} to show
\frac {dP}{dz} = -\frac {P}{kT}mg = -\frac {mg}{kT}P.
(c) Assuming that the temperature of the atmosphere is independent of height (not a great assumption but not terrible either), solve the barometric equation to obtain the pressure as a function of height: P(z) = P(0){e}^{{-mgz}/{kT}}. Show also that the density obeys a similar equation.
Finally, it’s time to pull out my hideously rusty calculus skills. I only really remember how to solve this for a separable differential equation. Fortunately, it looks like this is one; it fits the form \frac {dy}{dt}=g(t)h(y). So I can take my differential equation from part b and do this
\frac {dP}{dz} = -\frac {mg}{kT}P \rightarrow \frac {dP}{P} = -\frac {mg}{kT} dz.
Now I integrate both sides of the separated equation
\int {\frac {dP}{P}} = \int{-\frac {mg}{kT} dz} \Rightarrow \ln {P} = -\frac {mgz}{kT}+C.
To solve for pressure I exponentiate both sides
P(z) = {e}^{{-mgz}/{kT}+C}={e}^{C}{e}^{{-mgz}/{kT}}.
Notices that when z=0, P(z)=P(0)=e^C. Thus I can rewrite the constant term and get the sought equation,
P(z) = P(0){e}^{{-mgz}/{kT}}.
For density, I recall that \rho = \frac {Nm}{V}, which lets me rewrite the ideal gas law as P=\frac {NkT}{V}=\frac {\rho kT}{m}, and thus \rho = \frac {Pm}{kT}. Plugging that into my equation for pressure, I get
\rho (z) = \frac {m}{kT} P(z) = \frac {m}{kT}P(0){e}^{{-mgz}/{kT}}.
From the ideal gas law above, though, we can say that \frac {P(0)m}{kT}=\rho (0), so I can rewrite the above as
\rho (z) = \rho (0){e}^{{-mgz}/{kT}}.
Problem 1.22: If you poke a hole in a container full of gas, that gas will start leaking out. In this problem you will make a rough estimate of the rate at which gas escapes through a hole. (This process is called effusion, at least when the hole is sufficiently small.)
(a) Consider a small portion (area = A) of the inside wall of a container full of gas. Show that the number of molecules colliding with this surface in a time interval \Delta t is PA\Delta t/(2m\bar { { v }_{ x } }), where P is the pressure, m is the average molecular mass, and \bar { { v }_{ x } } is the average x velocity of those molecules that collide with the wall.
First, we need to consider only those molecules that are able to reach the area A in the time allowed. Those would be only those molecules in the volume given by A\bar {{v}_{x}}\Delta t. The number of molecules in this volume would be (\frac {N}{V})A\bar {{v}_{x}}\Delta t, which I can use the ideal gas equation to rewrite as \frac {P}{kT}A\bar {{v}_{x}}\Delta t. That’s the number of molecules in the volume we’re concerned with, but I wouldn’t expect all of them to be moving toward the wall of the container. At any given time I’d expect half of them, on average, to me moving toward the wall of the container, and the other half to be moving away. So let’s express the molecules in the relevant volume that are actually going to hit the wall as \frac {1}{2}\frac {P}{kT}A\bar {{v}_{x}}\Delta t.
A result from earlier in the chapter had that kT=m\bar {{v}_{x}^{2}}, so I’ll invoke that now to rewrite the previous expression as \frac {1}{2}\frac {P}{m\bar {{v}_{x}^{2}}}A\bar {{v}_{x}}\Delta t. Then I cancel to get the sought expression, PA\Delta t/(2m\bar { { v }_{ x } }).
There’s more to this problem, including developing a differential equation for the number of molecules that escape from the container as a function of time once you punch a hole in it. But I had to interrupt Thermo Thursday this week for a student meeting, so I’m not going to get to them here.
Recently I was having a fun and interesting conversation with someone I’d just met, a clearly very bright person, and during our talk I commented that I’m a compatibilist determinist – someone who believes that free will and determinism are compatible concepts. My clearly very bright new friend dismissed the idea as obviously false. In fact, he trivially dismissed it; his immediate response was that claiming that free will and determinism are compatible didn’t even merit the weight of consideration he’d given the rest of the conversation.
I get that reaction a lot. Narratives about the future are almost always about how it is an unfixed fog of infinite possibility, crystalizing onto history and becoming solid in the flash of the present moment. Almost all time travel stories include the instance or threat of someone going back in the past and changing the future, positing an assumed indeterminist model of reality. In those very few (and usually very good) time travel stories that do take place in a deterministic universe, the discovery of the fixed nature of the future is usually treated as a tragic diminishment of possibility, an attack on free will. But outside of academic writing, free will is very rarely defined. My impression is that most people, even irreligious people, have a sort of vague sense of what they mean by “free will” that is inherently numinous, that free will is some ineffable quality by which we are masters of our own fate. When I’ve asked people to try to explicate to me what they mean when they say “free will,” they usually say something about the existence of choice, and free will being the thing that makes choices possible, and that choice is what gives life meaning. The notion that a fixed future annuls choice seems intuitively obvious.
It isn’t. Here I’d like to offer a fairly straightforward argument for why not.
Prologue: The Predictor
Consider the following hypothetical scenario conceived by Ted Chiang.1 Imagine that I gave you a toy, a little box the size of a remote for unlocking a car, called a Predictor. The Predictor has only a button, a small LED light, and some internal circuitry that I tell you has a built-in negative time delay. When you play with the Predictor, you find that the light always illuminates exactly one second before you push the button.
I give you a Predictor to play with, and at first if feels like a game, like the goal is to push the button right after you see the light flash. But if you try to break the rules, you find you can’t. If you try to press the button without having seen the light, no matter how fast you move it always flashes exactly one second before your finger gets there. If you try to wait for the light to flash, intending to not push the button after, then the light never turns on. There’s no way to fool the Predictor.
So, I’ve given you a toy that conclusively demonstrates that your future actions are already determined. Having played with this toy, the question is, do you stop feeding yourself? Or paying your bills? Or caring for your family? You’ve been shown a demonstration that the events of the future are, in principle, precisely determinable from the present. It either is or is not true, already, that you will feed yourself, get dressed, pay your bills, bathe your children, etc. So do you think that you, the person reading this article, would stop doing those things after I gift you a small toy? If you don’t believe that you would stop, then you have some intuition that a fatalist attitude of defeated meaninglessness is not a necessary consequence of determinism. It remains for me to motivate exactly why that might be, and to convince you that your choice to continue doing things like feeding yourself is, in fact, a choice, despite the revealed deterministic nature of the world. I will try to do that now, then return to this hypothetical example.
1. Definitions
Let’s get the terms out of the way.2 As used here, determinism is the notion that at any given moment the physics of the universe admits only a single possible future.
Choice is a little more complicated. When I use the terms “choice” or “decision,” I am referring to categories of actions undertaken by specific agents. For these choices to be free, they must be caused strictly by processes internal to the agent, but that alone is not a sufficient definition. Taking people to be the agents in question, there are plenty of actions that people perform for reasons that are strictly internal, but are not volitional. (Sneezing, for example, or breathing while asleep.) So let us say that an action is a choice or decision if and only if it is caused by the agent’s beliefs and desires.
This definition still admits some quibbling. How are we to consider, for example, addiction? It motivates action based on beliefs and desires, but we feel in some sense that the desires have been warped by an external factor. Similarly with compulsion, where we feel that desire has been warped by a non-volitional internal factor. So let us even more narrowly define a free choice: a choice is free if and only if (1) the agent would have acted differently had it so chosen, (2) the action was voluntary (unaffected by internal restraint), (3) the action was uncoerced (unaffected by external restraint).
It’s worth noting that if we consider things like addiction and compulsion to be, in some fashion, an adulteration of free will, then seeing such things exist in human beings lends credence to the mechanistic description of free will that I am going to develop. But what do I, specifically, mean by “free will?” Let us describe an agent as having free will if and only if the agent possesses the ability to perform deliberative processes that result in choice.
2. A Brief Hearing for Indeterminism
Let’s look quickly at the notion of free will in the absence of determinism. Imagine, instead, that the future is completely independent of the present, that there is no causal relationship between this present moment and the one to come. What, then, does choice consist of? If there is no causal relationship between the present and the future, then, as David Hume noted, any attempt to make an informed choice as to how you should act in pursuit of a goal is futile. In the purely indeterminist case, past experience is not a logical guide for future behavior. So rather than free will being obviously incompatible with determinism, it is in fact pure indeterminism that trivially excludes the existence of free will. Choice can not exist in a universe where the future is random. Choice requires some degree of causal relationship between past, present, and future.
“Some degree,” though, is a quibble phrase. Perhaps, one might argue, the universe is determinist-ish. It’s predictable enough for choice to exist, but not perfectly predictable. Perfect predictability, one might wish to argue, also excludes free will. If one wishes to take this perspective, then one has to answer the question: at what point does the somewhat predictable universe become too predictable for free will to exist? I will now argue that choice can exist even in a purely deterministic universe, sliding that boundary point right off the scale.
3. Deliberative Process as Physical Event
Posit that we exist in a perfectly deterministic universe, and consider the case where you throw a ball at my face. If I am asleep–eyes closed, largely insensitive to my environment–then you are very likely to hit me. If I am awake–eyes open, watching your throw–then you are unlikely to hit me. (What do I mean though, in this deterministic universe, by “likely” and “unlikely?” I’ll address that more later. For now, let’s say that if we repeated many trials with slightly varying contextual conditions, in most of the cases where you throw the ball at my face while I’m asleep you hit me, and in most of the cases where you do so when I’m awake, you miss.) I am, when awake, able to avoid the ball, using faculties unavailable to me when I am asleep. I am able to avoid the ball, but I don’t have to do so. If, say, you’ve thrown a ball at my face because we are playing baseball, I might perceive some advantage in allowing it to hit me. That is to say, I might avoid avoiding the ball. I’m able to do this because our species has evolved sensory apparatus (sight, proprioception) that allows me to know when projectiles are coming at my head and move out of the way. When those apparatus are nonfunctional, such as when I’m asleep, I can’t move out of the way.
Events that I have the capacity to perceive and avoid are what Daniel Dennett likes to call evitable3 (to distinguish from those that are inevitable). Now, you may be objecting, “the universe was posited to be deterministic. Whether or not the ball hits your face was already determined the moment it leaves my hand. That makes it inevitable.” To which I would respond: inevitable to whom? It is clearly not inevitable to me; as the baseball/non-baseball example shows, whether or not the ball hits me is influenced by my beliefs and desires. Perhaps you mean inevitable to the universe. But that is not a meaningful notion, the universe is not a volitional agent. It does not make sense to speak of the universe avoiding, or avoiding avoiding.
The motions I make with my body after you throw the ball at me result from a cascade of electrochemical events in my brain, which correspond to my weighing the desirability of the ball hitting my face, the position of my body, how I would have to move to avoid the ball, etc. This electrochemical cascade, this physical event, is itself a deliberative process, one that results me choosing to dodge or not dodge the ball. All of my initial definitions for free will have now been met: I am an agent possessed of the ability to engage in deliberative processes that result in the perpetration of a choice: decision to dodge or not dodge the ball. That decision is definitionally a choice, in that it is an action caused by my beliefs and desires, where my beliefs are my sensory/conceptual model of the world, and my desires are my internal preferences. It even meets the more restrictive definition of being a free choice. The fact that the dodge or the hit was already extant in the future when the ball left your hand is irrelevant. The free choice was already extant, too.
I know of no logically coherent way to define freedom of choice that is incompatible with choice as an event that can occur within a deterministic universe. As long as choice is a behavior arising from a deliberative process, it is compatible with determinism. Thus free will, as I have defined it, is also compatible with determinism.
4. The Issue of Counterfactuals
One reason that it seems (incorrectly) to many people that determinism is incompatible with choice is that our deliberative processes which result in choice involve the consideration of counterfactuals. We think to ourselves, “What would be true if I did X? What would be true if I did Y? How likely is it that any action I take will result in Z?” This notion of likeliness seems to be challenged by determinism. How can one meaningfully think of things being likely or unlikely if the future is already determined? But in truth there is no contradiction. When we utilize counterfactuals in our deliberative processes, we are conducting mental simulations based on our beliefs and our understanding of past experiences. Our ability to judge how likely we think something is does not depend on what actually later occurs. The mental events of simulation and prediction are just part of the deliberative process that results in choice.
We talk about counterfactuals in a confusing way, though. If I am standing at the free throw line on a basketball court, and I shoot a free throw that bounces off the rim, I might be heard to say, “I could have made that.” What does that statement mean? I’m not actually saying that if everything about the state of myself and the world were somehow exactly the same down to the minutest detail, and the situation were to recur, I would make the shot. Rather, I’m making a claim about counterfactuals. I’m saying that among the family of possible worlds admitting minute variations of the air, moisture on the ball, potential gradients along the ion channels of the cells in my muscles; in many of those possible worlds I make the shot. Here again, the fact that when I took the shot there was only one physically possible future does not invalidate my counterfactual analysis. Just as it is meaningful to say that it is, was, and always will be the case that I missed the shot, it is also meaningful to say that I (counterfactually) could have made it. It is simply semantic ambiguity that makes these two notions seem to be in conflict.
Epilogue: The Predictor, Again
So let’s go back to the case of the Predictor. If I were to gift to you a tiny toy that happened, by implication, to demonstrate that the future already existed, of course you wouldn’t stop feeding yourself, or paying your bills, or acting in the interest of others you care for. You do these things because you believe they matter, and make choices motivated by that belief. The only reason for your choices to change would be if your beliefs fundamentally changed. Maybe you’ve been previously convinced, for no good reason, that determinism would mean that nothing matters. Then playing with the Predictor might be dangerous. You might then, as some people in Ted’s story do, choose to abdicate all personal responsibilities and never do anything again. But that’s not the Predictor’s fault, nor the universe’s; it’s the fault of you having “determinism = meaninglessness” in your head as a disabling, destructive narrative . That would be a tragedy if it were to happen–but why should it? The Predictor is just a small piece of plastic that you can throw in the trash if you want4, and anyway, reading this has taken a long time and you’re hungry. Might as well go eat something.
In his story “What’s Expected of Us,” Nature, 2005 ↩
I’m indebted for portions of this section to Curtis Brown, my symbolic logic professor at Trinity University. ↩
Dennett discusses evitable and inevitable events at length in his book Freedom Evolves, which is a much more learned and thorough explanation of compatibilist determinism than this article. ↩
All of the discussion up to this point has been about a hypothetical universe that I simply posited at the start was deterministic. I haven’t made any claims about the actual reality you and I inhabit, nor have I placed a tiny plastic Predictor in your hand. And, of course, I can’t. They don’t really exist. But I think I can give you something that is very close to the same.
Special relativity has as one of its basic results that simultaneity does not have any meaning across reference frames. The math for this isn’t too complicated, but instead of writing out equations, here’s a two minute video that clearly demonstrates the phenomenon.
There are many versions of this thought experiment, but the one in the video is the one Einstein proposed. As you can see in the video, the man on the platform sees the bolts of lightning strike both ends of the train simultaneously, and the woman riding in the train sees lighting first strike the front of the train, and then strike the back of the train. And neither person is wrong. In the man’s frame of reference, the strikes were simultaneous. In the woman’s, one came before the other.
Special relativity has been experimentally confirmed time after time. As theories go, it’s one we are as sure of as we are sure of anything at all. Consequently it is already widely accepted that simultaneity as a concept has no meaning across reference frames. But lets think through the implications: the man on the platform observes an event that, at the time he observes it, is still in the woman on the train’s future. That is to say, there exist an event–the lightning striking the back of the train–that is in the man’s past, and in the woman’s future. Thus it is possible for a physical event that is already in my past to still be in your future. But the past is unchangeable. Anything that is in the past, for anyone, is necessarily a thing that happened in the universe. But if my unchangeable past can be your future, that means that an event in your future is determined. And this reasoning can, in principle, apply to any arbitrary event. Therefore the future already exists, and the events of the future are already determined.
Conclusion: special relativity implies that the universe in which we live is, in fact, deterministic. ↩