There will be no Thermo Thursday this week because I’m in Washington, D.C. doing writing research. I got to be here for the last day of the government shutdown, and am now exploring how the city has changed since it reopened. Here’s a picture of DuPont Circle. Thermo Thursday returns next week.
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.
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.
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.
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.
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.
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.
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. ↩
The visual effects, action scenes, sets, and sound design are all brilliant, which is good because the dialog and score are largely terrible. The physics of an important scene where Dr. Stone has parachute ropes wrapped around her leg is very, very wrong. I still mostly loved it, and think it should be a lock for a VFX Oscar. Likely my favorite SF movie of 2013.
1.1: Thermal Equilibrium This chapter introduces the several basic definitions of temperature, starting with the operational (“temperature is what you measure with a thermometer”) and ending with a qualitative theoretical definition (“Temperature is a measure of the tendency of an object to spontaneously give up energy to its surroundings. When two objects are in thermal contact, the one that tends to spontaneously lose energy is at the higher temperature.”) Some time is spent on what exactly thermal contact entails–that the objects are in some way able to exchange energy, and that this energy is called heat. When the two objects have been in contact long enough, we learn that they are considered to be in thermal equilibrium, and that the time required to come to thermal equilibrium is the relaxation time. Curiously, Schroeder doesn’t say in the definition that the two objects are in thermal equilibrium when they’ve reached the same temperature. This is probably because of the way he defined relaxation time, which gets a footnote that some authors consider the relaxation time as the time required for the temperature difference between the two objects to decrease by a factor of e. The most interesting part of the section is an analogy of thermal equilibrium to other kinds of equilibrium. Schroeder gives the example that if you pour cold cream into a hot cup of coffee, the relaxation time of the cream/coffee system is very short, but the relaxation time for the coffee to come to thermal equilibrium with the room is still very long. But he notes that there are more kinds of equilibrium on display in the example; the cream in the coffee reaches diffusive equilibrium when it is blended with the coffee such that its molecules have no greater tendency to move in one direction than another. Also defined is mechanical equilibrium, when large-scale motions can take place, but no longer do. Whereas the exchanged quantity to reach thermal equilibrium is energy, the exchanged quantity for diffusive equilibrium is particles, and for mechanical equilibrium, volume. The rest of the section describes how thermometers work, and introduces the concepts of absolute zero and the kelvin scale. These things are sufficiently basic that I’m not going to synopsize them.
I don’t think I’ll always work every problem in a chapter, but to start out I’m going to.
Problem 1.1: The Fahrenheit temperature scale is defined so that ice melts at 32°F and water boils at 212°F. (a) Derive the formulas for converting from Fahrenheit to Celsius and back. (b) What is absolute zero on the Fahrenheit scale?
These are two linear scales, meaning that every degree F is the same size as every other degree F, and equivalently for every degree C. But F degrees and C degrees are different sizes, and the scales don’t start in the same place. So first I’ll figure out how do describe the size of a degree F in terms of a degree C, or the number of degrees F per degrees C. The magnitude of the temperature change between where water freezes and where it boils is the same regardless of scale, so let F be the size of a degree Fahrenheit and C be the size of a degree Celsius. Then,
( 212F - 32F ) = (100C - 0C).
Solving this equation gives us
F = \frac { 100 }{ 180 } C = \frac { 5 }{ 9 } C.
So one degree F is only five ninths the size of a degree C. So if we have a number of degrees Celsius, we have to divide by five ninths (same as multiplying by nine fifths) to get the equivalent number of degrees Fahrenheit. But there’s still the issue that the scales don’t start in the same place. When we’re at 0° C, we’re still at 32° F. So we need to add a constant factor of 32 to the equation above to adjust for the different zero-point, giving us a temperature conversion equation
{ T }_{ F }=\frac { 9 }{ 5 } { T }_{ C }+32 and { T }_{ C }=\frac { 5 }{ 9 } ({ T }_{ F }-32). (Solution to part a.)
For part b, we know from the chapter that absolute zero is -273°C, so
{ T }_{ F }=\frac { 9 }{ 5 } (-273)+32=-459. (Solution to part b.)
Problem 1.2: The Rankine temperature scale (abbreviated °R) uses the same size degrees as Fahrenheit, but measured up from absolute zero like kelvin (so Rankine is to Fahrenheit as kelvin is to Celsius). Find the conversion formula between Rankine and Fahrenheit, and also between Rankine and kelvin. What is room temperature on the Rankine scale?
Degrees Fahrenheit and Degrees Rankine are the same size, the scales just start in a different place. We know from the previous problem what absolute zero in °F is, so
{ T }_{ R }={ T }_{ F }+459.
Rankine and kelvin start in the same place, but have different degree sizes. The sizes are the same as for Fahrenheit and Celsius respectively, so we already know that one degree Rankine is five ninths the size of one kelvin. (I’ll note here that, for no reason I’m aware of, it’s considered improper to say “degree kelvin.”) So again we divide our kelvins by five ninths to get the number of degrees Rankine, giving us a temperature conversion equation
{ T }_{ R }=\frac { 9 }{ 5 } { T }_{ K }.
Room temperature is defined to be 300 K, so by the equation above, room temperature is also 540°R.
Problem 1.3: Determine the kelvin temperature for each of the following: (a) human body temperature; (b) the boiling point of water (at standard pressure of 1 atm); (c) the coldest day you can remember; (d) the boiling point of liquid nitrogen (-196°C); (e) the melting point of lead (327°C).
(a) 98.6°F = 37°C = 310 K. (b) 100°C = 373 K. (c) -10°F ≈ -23.3°C = 249.6 K. (d) -196°C = 77 K. (e) 327°C = 600 K.
Problem 1.4: Does it ever make sense to say that one object is “twice as hot” as another? Does it matter whether one is referring to Celsius or kelvin temperatures? Explain.
The book hasn’t explicitly defined “heat” yet, so I’m going to interpret this problem to be asking just about temperature scales. Without a defined scale, saying “twice as hot” is meaningless. Twice as many degrees Celsius is a much greater difference in temperature than twice as many degrees Fahrenheit. Also, suppose our first object is at 0°C. Within the Celsius scale, you can’t define what temperature an object “twice as hot” would be. On the kelvin scale, excepting the special case of absolute zero, the phrase “twice as hot” does always have a mathematically coherent definition. (I’m not sure that it has a physically coherent definition, but the book hasn’t gotten to specific enough definitions for me to make that argument.)
Problem 1.5: When you’re sick with a fever and you take your temperature with a thermometer, approximately what is the relaxation time?
I’m not sure what the intention with this problem is. Actually calculating this would both require information that hasn’t been introduced yet, and defining a specific type of thermometer. But I don’t see anything in the chapter to provide a basis for approximation. From anecdotal experience, the relaxation time (i.e. how long it takes for an oral thermometer to go from room temperature to the temperature of the inside of my mouth) is on the order of one minute.
Problem 1.6: Give an example to illustrate why you cannot accurately judge the temperature of an object by how hot or cold it feels to the touch.
The key here is judging accurately. Imagine that you touch an object that is at exactly your body temperature. Your body and the object are in thermal equilibrium, so there will be no transfer of energy, and you will feel no temperature difference. So you know that it’s the same temperature as you are, but what temperature is that? You can’t find out by touching yourself, as you are always in thermal equilibrium with yourself. You can estimate your own temperature, but that wouldn’t be accurate. Since body temperature is variable, it’s impossible to know where the “feels hot, feels cold” scale starts without recourse to a thermometer. (There’s also the issue that beyond certain temperature ranges the tissues of your body with do the sensing will be damaged, making you unable to feel temperature at all.)
Problem 1.7: When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume change increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:
\beta =\frac { { \Delta V }/{ V } }{ \Delta T }
(where V is volume, T is temperature, and the Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K^-1 = 1.81 x 10 ^-4 K^-1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)
(a) Get a mercury thermometer,
No.
estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.
Screw getting an thermometer. Let’s just assert that the bulb at the bottom has volume V of one cubic centimeter. Let’s also assert, because I’m making things up and I can, that the degree markings are a height h ofone millimeter apart. The amount of mercury that rises up the tube radius r with a 1 K increase in temperature, then, is
\beta (V)(\Delta T)=\pi { r }^{ 2 }hSolve for the radius of the cylinder,
r=\sqrt { \frac { \beta (V)(\Delta T)}{\pi h}} =0.02. So the radius is 0.002 cm. (Frustratingly, I haven’t been able to make the unit symbols parse in the LaTeX plugin, but the dimensional analysis checks out.
(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 x 10^-4 K^-1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of -0.68×10^-4 K^-1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.
Okay, this one is interesting. Imagine a lake getting colder and colder. The positive thermal expansion coefficient means that, above 4°C, the colder the water is the denser it is, so the coldest water will sink to the bottom. (This matches my intuition that the bottom of a swimming pool is colder than the surface.) Let this behavior continue until, hypothetically, we have an entire lake in which all the water is at 4°C. Then let some of the water in that lake get colder. Because the thermal expansion coefficient turns negative now, the colder water will be less dense, and will start rising to the top. That means that by the time the first water gets down to 0°C and freezes, it will be at the surface, and the water below will be warmer. So lakes freeze from the surface down. If the thermal expansion was always positive, I would expect lakes to freeze from the bottom up.
Problem 1.8: For a solid, we also define the linear thermal expansion coefficient, α, as the fractional increase in length per degree:
\alpha =\frac { { \Delta L }/{ L } }{ \Delta T } .
(a) For steel, α is 1.1 x 10^-5 K^-1. Estimate the total variation in the length of a 1-km steel bridge between a cold winter night and a hot summer day.
Let’s say that a hot summer day is 43°C, and a cold winter night is 0°C. For an L of 1km, we have
\alpha (L)(\Delta T)=\alpha (1)(43)=0.00047in units of kilometers, which is just under half a meter.
(b) The dial thermometer in Figure 1.2 uses a coiled metal strip made of two different metals laminated together. Explain how this works.
The two different metals have different thermal expansion coefficients. Assume that they are the same length when laminated together at some thermal equilibrium. When the temperature changes, the strips will not be able to expand linearly because they are attached to each other and changing by different lengths. As such, there will be a lateral displacement; the laminated strip will curl. This curling is proportional to the change in temperature.
(c) Prove that the volume thermal expansion coefficient of a solid is equal to the sum of its linear expansion coefficients in the three directions: \beta ={\alpha}_{ x }+{\alpha}_{y}+{\alpha}_{z}. (So for an isotropic solid, which expands the same in all directions, β=3α.)
Recall that \Delta V={ V }_{ final }-{ V }_{ initial }. If {V}_{initial}=xyz, then it follows that
{ V }_{ final }=(x+\Delta x)(y+\Delta y)(z+\Delta z)\\ =xyz+\Delta xyz+x\Delta yz+xy\Delta z+x\Delta y\Delta z+\Delta xy\Delta z+\Delta x\Delta yz+\Delta x\Delta y\Delta z\\ \approx xyz+\Delta xyz+x\Delta yz+xy\Delta zThat last step works because all of the terms with multiple deltas are so tiny compared to the other terms that they are physically negligible, so I can drop them. This gives me
{\Delta V=V }_{ final }-{ V }_{ initial }=xyz+\Delta xyz+x\Delta yz+xy\Delta z-xyz\\ =\Delta xyz+x\Delta yz+xy\Delta zWhich I can plug into the equation for β to get
\beta =\frac { { \Delta V }/{ V } }{ \Delta T } =\frac { { (\Delta xyz+x\Delta yz+xy\Delta z) }/{ (xyz) } }{ \Delta T } =\frac { ({ \Delta x }/{ x })+({ \Delta y }/{ y })+({ \Delta z }/{ z }) }{ \Delta T }\\ ={ \alpha}_{ x }+{ \alpha}_{ y }+{ \alpha}_{ z }thus proving the theorem.
Once upon a time, long before I ever moved to Iowa or started teaching creative writing courses, I thought I was going to be a physicist. But I was the kind of physics student who occasionally got tired of doing math all day and would take a week off to write a story or make a wireframe sculpture or do large scale papercraft. Anything creative rather than analytical. I eventually followed those impulses out of the physical sciences and into my current position.
Now, though, I find my situation reversed; I spend my days thinking about literature and aesthetics, and discover that I am the kind of creative writing professor who intermittently longs to dust off his calculus. From time to time I’ve been pulling some of my favorite physics and math textbooks off the shelf and recreationally doing a couple of problems, but I’ve noticed that just jumping in and solving things is getting harder. My physics and math is atrophying from disuse. So I’ve decided to get more systematic about this. I’m going to set aside a day of the week for brushing up on the subjects that most interest me, starting with thermodynamics. Welcome to Thermo Thursday.
I’ll be using the book An Introduction to Thermal Physics by Daniel V. Schroeder, which I remember thinking highly of back when it was assigned in college. Every Thursday I’ll work through some sections of the book, synopsize the key concepts and work some problems here. (This will hopefully also give me a chance to brush up my LaTeX skills.) My math may be rusty enough that I reach a point where I have to stop the physics and refresh my fundamentals on more basic concepts, but I’ll deal with that problem when I get to it. Let’s get started.
I’ve just discovered, via Nick Mamatas, the YouTube account AboutSF, which are all videos uploaded by the Center for the Study of Science Fiction at the University of Kansas. Some of these, like the one embedded below, are excerpts from lectures given back when James Gunn was bringing science fiction authors to KU in the ’60s and ’70s. This is interesting for me personally, as my parents met at KU during this time and attended the Gunn-organized readings and lectures together. Here’s one of a bunch writers, including Harlan Ellison, Poul Anderson, John Brunner, Fred Pohl, and Isaac Asimov discussing the value of science fiction. I particularly like Anderson’s opening comments.
Just putting these here so I can easily find them later.
I haven’t been this excited about an upcoming movie in a long time.