Noticing confusion in physics

post by Jacob G-W (g-w1) · 2023-10-12T15:21:48.183Z · LW · GW · 27 comments

This is a link post for https://jacobgw.com/blog/observation/2023/10/12/noticing-confusion.html

Contents

27 comments

I recently had a Physics test, and the only thing I got wrong was the answer to this question: "What happens to the speed of sound in a gas when the gas is heated up?"

My answer was "The speed of sound decreases because heating up a gas decreases its density. Lower density means lower speed of sound because the particles have to travel further to bump into other particles." I got 1/2 points on this question because although I got the incorrect answer, my justification was "correct".

When I got the test back and saw what I got wrong and how I had gotten it wrong, I realized that my mental process went wrong. I was equally able to explain the truth as the opposite of the truth, and this is bad. I want to find the truth.

The reason that my teacher gave was that if a gas was hotter, the molecules moved faster, and thus bumped into each other faster and the speed of sound was faster.

This explanation is not that convincing to me, so I'm going to do an analysis of what went wrong in my thought process and how I can prevent it in the future.

  1. My mind immediatly jumped to the density explanation without considering how the opposite could be correct.
  2. My first thought now is to see how much molecular speed vs density would change with an increase of speed. Since , it would follow that if you double you increase by . But if you double , density should halve. This still leaves me confused.
  3. At this point, I'm thinking that a decrease in density does not decrease the speed of sound in gases, only in liquids and solids. I searched the answer up and found this explanation. This says that only. I kept searching and found this question, which cleared it up even more for me.
  4. Here is my synthesis of what happened: in solids and liquids, density does affect the speed of sound, but in gases, a third factor is pressure. If you increase the temperature, density can decrease, but pressure also increases by a portional amount since the molecules are hitting harder. These two changes cancel out. Thus, in a gas only.

So it seems the main thing that I missed was the extra factor of pressure, in addition to density. In the future, I should try to include all of the factors in a problem and give equal weight to the true or false answers. I'll admit thought that this might be hard to prevent in the general case since it didn't seem to be a meta-level confusion – just an object level one.


If you have a better explanation for this, please feel free to email me at jacoblevgw at gmail dot com and I'll put it here.

27 comments

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comment by AnthonyC · 2023-10-12T17:24:45.082Z · LW(p) · GW(p)

Before going into why the real answer is what it is, there is a flaw in your reasoning for your original answer. Your reasoning about density assumes the gas is free to expand and contract, which is not an assumption given in the problem. An ideal gas in a box of fixed size does not gain or lose mass or volume when heat or cooled, only T and P change. And the speed of sound in air does not change if you measure it inside with the doors and windows closed and sealed. So what variables can we rule out?

If I make the room bigger or smaller while holding T and P constant, v(sound) does not change. If it did, it would be very obvious in daily life.

If I increase pressure while holding V and T constant, I do it by adding more gas molecules. This increases density, but does not change the velocity of the gas molecules. In this scenario, your answer is correct. Speed of sound varies with P. However, this is only true because real gases are not ideal gases, and there are more complex interactions between molecules. In an ideal gas where all the interactions are perfectly elastic and instantaneous collisions, the density has no effect because mean free path is irrelevant to instantaneous or average speed of movement across the group of molecules as a whole, it just changes which molecules are moving. Side note: Remember sound is a pressure wave. So, if this were not true, then the speed of sound would depend on how loud it is. Also, it would be different at the peak and trough of each cycle of the pressure wave. Again, in real gases there are effects of this kind, and there is a maximum volume sound can be (in air at sea level, around 194 decibels the pressure at the trough hits zero; this is a shockwave and it can move faster than sound). But ideal gases assume all that away.

If we increase or decrease the amount of gas at fixed T, then either we're just adding volume again, or we're just increasing pressure again, and we've seen that those shouldn't have an effect on an ideal gas' speed of sound.

So now let's increase T. It doesn't matter what effect this has on P and V and n, as seen in the above. So what's left? Increasing T linearly increases the average kinetic energy of the gas molecules (PV and NkT both have units of energy, this is why), and velocity increases as the sqrt of kinetic energy. So if gas molecule velocity is what determines v(sound), then it has to be that v(sound) increases as sqrt(T).

I'm not sure what level physics class you're in, but I have two very general pieces of advice. 

The first is that you can often get very far by figuring out what doesn't or can't matter, and what has to be held constant. This is why conservation laws are so important! It's also why boundary conditions of a problem are important. They let you write down equations that constrain what the solution can possibly depend on. As an example, a string attached to fixed points on both ends can only vibrate at frequencies where those ends don't move. Air vibrating in a tube that's closed at one end can't have net motion at the closed end, but has to be moving at the other end for any sound to come out at all. These kinds of considerations let you figure out what resonance frequencies are possible in principle for idealized versions of different types of musical instruments. (If you get to more advanced physics, this is why there's so much discussion of symmetry. Noether's theorem tells us that every symmetry of the laws of physics implies a conservation law, and vice versa. "Energy is conserved" is equivalent to saying "The laws of physics don't change over time." "Linear and angular momentum are conserved" is equivalent to saying "the laws of physics don't depend on where you are or what direction you're facing." There's a reason Einstein originally wanted Relativity to be called "The Theory of Invariants": he arrived at it in large part by thinking about what couldn't matter and what couldn't change, then letting everything else vary however it needed to in order to accommodate that.)

The second is something a professor used to like to say: that if you really want an intuition for physics you need to feel it in your bones. Try things out. Look for places in your regular life where different concepts could apply, and try to reason out how they apply. It's a slow process that doesn't always match up to a semester-long class schedule, but it's the kind of thing where a year or two later you'll find yourself saying, "Oh yeah, that's what they meant!" Sometimes this takes a long time and a lot more physics than you'd expect! I was in grad school for materials science when one professor wrote down an equation at the end of a handful of math-heavy lectures and said, "And that's why metals are shiny." Other times it turns out you can explain seemingly complicated things with high school physics and very careful reasoning.

Replies from: gw
comment by gw · 2023-10-12T19:05:02.666Z · LW(p) · GW(p)

If I make the room bigger or smaller while holding T and P constant, v(sound) does not change. If it did, it would be very obvious in daily life.

This feels a bit too handwavy to me, I could say the same thing about temperature: if the speed of sound were affected by making a room hotter or colder, it would be very obvious in daily life, therefore the speed of sound doesn't depend on temperature. But it isn't obvious in daily life that the speed of sound changes based on temperature either.

So now let's increase T. It doesn't matter what effect this has on P and V and n, as seen in the above. So what's left? Increasing T linearly increases the average kinetic energy of the gas molecules (PV and NkT both have units of energy, this is why), and velocity increases as the sqrt of kinetic energy. So if gas molecule velocity is what determines v(sound), then it has to be that v(sound) increases as sqrt(T).

I think this also falls short of justifying that v(sound) increases as T increases.  Why does it have to be that v(sound) increases with gas molecule velocity and not decreases instead? Why is it the case that gas molecule velocity determines v(sound) at all?

Replies from: AnthonyC
comment by AnthonyC · 2023-10-12T20:12:32.489Z · LW(p) · GW(p)

To your first point: there is no scenario in daily life where we experience a change in absolute temperature spanning orders of magnitude (at most about 30%, from ~250K to ~325K, but we do experience room sizes that span multiple orders of magnitude (a closet vs. a concert hall vs outdoors in an open grassland). Similarly, our experiences of pressure variation almost never span more than about +/-30% from 1 atm. So I maintain that a dependence on volume would be much more obvious than a dependence on temperature or pressure unless it were something like log(V) or V^.1 (which would be hard to reconcile with ending up with units of m/s), and even then there would be scenarios that should be very odd, like a long, narrow cave with an open mouth where the speed of sound was several times faster along the short axis than the long axis, or vice versa. Or think about it this way. If you stand across a football field in a sealed, airtight indoor stadium bang cymbals together, I hear it about 1/4 of a second later. If we do the same thing without the stadium in a place where there's nothing blocking sound propagation for many miles, I hear it about 1/4 of a second later. This happens even though the volume of gas is at least 6 orders of magnitude larger (say, if we call the distance to the horizon the new relevant room size) or much more than that (if we count the whole atmosphere as the room size). 

 

To your second point - I agree this is not obvious. So, we have to dig deeper. What is sound? A pressure wave. What does that mean? Well it means you create a pattern of high and low pressure regions that propagates, for example by vibrating a membrane and imparting force to (initially randomly/thermally moving) molecules. Well why does it propagate? Because all the individual molecules move randomly, so the molecules in the high pressure regions tend to flow into the nearby low pressure regions more than the reverse (diffusion along concentration gradients). Now, could the initial forces creating the high and low pressure regions at the source impart some net kinetic energy in a particular direction or set of directions? After all, sound moves net away from the source (note: diffusion in 2D and 3D is also automatically net away from any point, but not in 1D, that's another fun derivation). It takes a bit more investigation to figure out why that doesn't let you make a sound wave that's faster or slower, but basically that has to do with boundary conditions again. You can totally get sound to a distant point sooner if your source has net velocity in that direction (aka I can theoretically hear a bullet that passes a foot away from me before the sound from the gun itself reaches me), or if your sound is loud enough to make a shockwave, but the former isn't propagation of the sound itself and the latter violates the ideal gas assumption. If you have a stationary source, though, then that boundary condition applies the constraint that each time you move right to provide rightward force must be balanced by moving left to apply leftward force (or less rightward force, if the source is the wall of your box). So you aren't applying net force to the gas (that would violate F=ma=0 for reaction forces on the source in the frame of reference where the source is stationary). Therefore the already-existing thermal motion of the gas is the only velocity we have to work with. Now it doesn't automatically have to be a simple relationship with that velocity. After all, there's a wide distribution of velocities of individual molecules, and depending on how the source is behaving you can even get into subtle distinctions like group vs phase velocity (it's the group velocity that matters for what we're discussing here). But that's the velocity distribution we've got. And somehow we have to end up with something measured in m/s. Those are pretty significant constraints on the form of the resulting equation.

You can also do all sorts of fancy things that make sound have different speeds based on direction and frequency, if you carefully combine materials with different densities and bulk moduli (see acoustic metamaterials). In this case you're interrupting the movement of molecules on distance scales shorter than the wavelength of sound, which isn't explicitly listed as an assumption of something forbidden in the problem, but it is one implicitly forbidden in the conventional understanding of what "sound" means at the level where teachers are asking questions about the behavior of ideal gases.

 

After writing all this I'm reminded of a story/joke where a professor says "It's easy to show that [something.]" A student asks, "Um... is that easy?" The professor starts writing and checking his notes. He leaves, comes back an hour later, says "Yes, it's easy," and then continues on with the lecture.

Replies from: Algon
comment by Algon · 2023-10-13T15:12:28.441Z · LW(p) · GW(p)

note: diffusion in 2D and 3D is also automatically net away from any point, but not in 1D, that's another fun derivation

 

What? The Green's function for the 1D heat equation has the same dependance on distance as the 2D and 3D case. 

"It's easy to show that [something.]" A student asks, "Um... is that easy?" The professor starts writing and checking his notes. He leaves, comes back an hour later, says "Yes, it's easy," and then continues on with the lecture.

The version I heard used 'obvious' instead of 'easy'. I wonder 'trivial' would be funnier yet.

 

Replies from: AnthonyC
comment by AnthonyC · 2023-10-13T20:00:24.315Z · LW(p) · GW(p)

Ah, you're right, sorry. Yes, diffusion rate works the same in 1D for concentration gradient over time. The difference I was thinking of is that in 1D for an individual molecule a random walk returns to the origin with probability 1, even though avg distance rises over time, while in higher dimensions that isn't true.

Replies from: Algon
comment by Algon · 2023-10-13T20:14:30.512Z · LW(p) · GW(p)

Simple random walks return to the origin in 2D as well, but not 3D or higher. I don't know if the continuous case is different, but I suspect not. 
 

Replies from: AnthonyC
comment by AnthonyC · 2023-10-13T20:54:46.398Z · LW(p) · GW(p)

Then I'm probably just misremembering, it's been about 15 years since I looked at that one. Thanks!

comment by gw · 2023-10-12T19:13:21.713Z · LW(p) · GW(p)

Here's a handwavy attempt from another angle:

Suppose you have a container of gas and you can somehow run time at 2x speed in that container. It would be obvious that from an external observer's point of view (where time is running at 1x speed) that sound would appear to travel 2x as fast from one end of the container to the other. But to the external observer, running time at 2x speed is indistinguishable from doubling the velocity of each gas molecule at 1x speed. So increasing the velocity of molecules (and therefore the temperature) should cause sound to travel faster.

(Also, for more questions like this, see this post on Thinking Physics) [LW · GW]

Replies from: AnthonyC
comment by AnthonyC · 2023-10-12T21:32:54.458Z · LW(p) · GW(p)

This seems like it should work at first glance, but doesn't. The initial intervention (double particle speed, which quadruples total kinetic energy) at first doubles pressure (each particle imparts twice as much force when it hits a wall and reverses direction), but the velocity distribution is no longer thermal. In a statistical mechanics sense, you've added energy without adding any entropy, and that means the colloquial concept of temperature for a gas doesn't really apply, just like accelerating a car by applying macroscopic kinetic energy doesn't mean you increased it's temperature (but over time, friction will thermalize the kinetic energy).  After that initial transformation, I think the gas will thermalize over a fairly short time but I'm not 100% sure. If it does, then in that case you've quadrupled total internal energy (1.5nRT or PV for a monatomic ideal gas), so I think it should stabilize at quadruple the T and P. Which, yes, will double v(sound), but doesn't tell you whether that's because of the T or the P or both. In any case this transformation put the system in an unnaturally low-entropy state, and so a lot of the usual assumptions about ideal gas behavior won't apply.

Replies from: gw
comment by gw · 2023-10-13T03:25:22.531Z · LW(p) · GW(p)

Interesting, thanks for the detailed responses here and above!

comment by Simon DeDeo (simon-dedeo) · 2023-10-12T17:01:55.303Z · LW(p) · GW(p)

We've been thinking about explanations in our research (see, e.g., https://arxiv.org/abs/2205.07938) and your example of explaining the wrong answer well is lovely.

I dislike these kinds of questions, because they're usually posed at a point well before the wave equations are presented. At this point, you are largely working with verbal explanations and, as you point out, verbal explanations are much harder to pin down. 

Mathematically, if A implies B, and you are working to the best of your ability, you can't derive ~B (you may not be able to derive B, of course!) Verbally, this is not so clear; a lot of philosophy is people arguing about whether A implies B or ~B.

If the underlying logic is (as it is in physics) mathematical, then a verbal account of mathematical fact A can be loose enough that you can derive ~B, because the verbal account is also consistent with A', which implies ~B.

In these explanations, there's another factor: a lot of the talk is relying on intuitions for "ordinary solids". One explanation I encountered when I googled referred to the "stiffness" of a hotter gas. While you might be able to cash this out in more formal terms, the temptation is to think about stiffness in one's normal experience. (It might have been possible to get the right answer by imagining heating up a bicycle tire that's already inflated; intuitively, the casing will be stiffer, "ring" at a higher frequency, etc.)

If you continue on in physics to relativity, quantum mechanics, etc, you end up dropping this kind of talk very quickly. This is why I don't like these questions; it's a bit useless to get good at them, because the more advanced you get the more you learn to rely on mathematical intuitions to get the right answer (and then perhaps informal folk talk afterwards, if you communicate to a popular audience.)

Replies from: ben-lang, AnthonyC, M. Y. Zuo
comment by Ben (ben-lang) · 2023-10-12T21:03:44.464Z · LW(p) · GW(p)

Personally i think the verbal kinds of explanations still have an important role in more advanced physics, as something you can present after the calculation or simulation. Things like 'it kind of makes sense that we see this because...' or 'this may seem surprising but it actually makes sense if you look at it as...'.

comment by AnthonyC · 2023-10-12T21:43:00.297Z · LW(p) · GW(p)

I dislike these kinds of questions, because they're usually posed at a point well before the wave equations are presented. At this point, you are largely working with verbal explanations and, as you point out, verbal explanations are much harder to pin down. 

I agree that they're not great test questions, but they can be excellent class discussion or homework problem set questions (as long as you encourage working together on homework, which can work in college but not usually in high school). If anything, using them well puts a much higher burden of understanding on the teacher to not only know the answer but also all the ways students are likely to go wrong in trying to reason about the answer and how to steer the discussion without just giving the answer.

In this case, yeah, I'm sure this question was posed at a point where the student doesn't really know what T and P mean at a fundamental level, what makes a gas "ideal," what the Maxwell-Boltzmann velocity distribution is and why, and a whole bunch of other relevant things. Given that, you should still be able to reason it out using dimensional analysis, the definition of kinetic energy, the idea that T is proportional to kinetic energy, and looking at some limiting cases and boundary conditions, but it isn't easy.

comment by M. Y. Zuo · 2023-10-12T17:06:57.616Z · LW(p) · GW(p)

I dislike these kinds of questions, because they're usually posed at a point well before the wave equations are presented. At this point, you are largely working with verbal explanations and, as you point out, verbal explanations are much harder to pin down

What would the same question look like when presented in mathematical form?

At a quick glance I can't see a concise way to express it.

Replies from: AnthonyC
comment by AnthonyC · 2023-10-12T21:51:53.065Z · LW(p) · GW(p)

That's the point - to actually be precise you need both wave mechanics and statistical mechanics as background to even define the "speed" of a "sound" "wave," the "temperature" and "pressure" of a gas, and how a gas gets "heated."

Replies from: M. Y. Zuo
comment by M. Y. Zuo · 2023-10-12T22:19:53.572Z · LW(p) · GW(p)

So how  instead should this category of questions be ideally presented?

Replies from: AnthonyC
comment by AnthonyC · 2023-10-12T23:53:53.765Z · LW(p) · GW(p)

At what appears to be this level of coursework, it should be something they've already discussed in class. Otherwise, it should be either asked with a lot more context, or as part of a discussion instead of on a test.

Replies from: M. Y. Zuo
comment by M. Y. Zuo · 2023-10-16T19:28:19.112Z · LW(p) · GW(p)

I'm not sure how to imagine 'a lot more context' for this question, can you provide an example?

Replies from: AnthonyC
comment by AnthonyC · 2023-10-17T23:47:10.328Z · LW(p) · GW(p)

I'm assuming this is something like a high school level physics class, and that this specific problem hasn't been discussed before the test. In that case I think it's a sufficiently hard problem that if you're going to ask it on a test, then it should be a multi-part series of questions that leads to towards the right mode of thinking. Maybe something like, "1) What happens to the average speed of molecules in an ideal gas when it gets heated up? 2) How, if at all, does this affect the speed of sound? 3) How, if at all, do changes to V and P affect the speed of sound in an ideal gas?"

comment by Soapspud · 2023-10-13T15:43:16.075Z · LW(p) · GW(p)

My thought after reading the first sentence of your post, and before reading any of the comments, was that gases become less compressible at higher temperatures, which should make it more responsive to a pressure-wave, raising the speed of sound in that medium.

Replies from: g-w1
comment by Jacob G-W (g-w1) · 2023-10-13T20:07:45.123Z · LW(p) · GW(p)

Wow, this is impressive intuition. Do you know what made you think of compressibility first? Or is it just intuition gained through hard work?

Replies from: Soapspud
comment by Soapspud · 2023-10-16T03:12:18.950Z · LW(p) · GW(p)

It's definitely intuition gained from a few years of doing those kinds of problems.

Also, there's an important point that makes my intuition a bit less impressive, and it's the fact that the problem-statement sounded like an intro-physics problem, so I assumed away many ambiguities that would make the solution particularly complicated to think through.

For example, thought it's not specified, it matters whether your gas is in a fixed volume or not, but if you assume the gas can expand, you're getting into solutions where you might need to know the boundary conditions at the edge of your gas, and you might need to figure out relative pressures and/or temperature gradients. Since the question doesn't specify any of that, I guessed that it's probably not that kind of problem.

comment by Daniel L (daniel-lyakovetsky) · 2023-10-13T15:05:40.512Z · LW(p) · GW(p)

I think the confusion is about unverified assumptions. The author assumed that the problem is about an open or a leaky space (perhaps thinking about the sound traveling across the quad in the summer vs winter), while the teacher meant some closed container (e.g., a pressure cooker). The teacher wins. 

comment by Eggrenade · 2023-10-12T17:34:14.794Z · LW(p) · GW(p)

You're making assumptions beyond the problem statement.  The problem never says pressure is constant, so a temperature change does not imply a change in volume.  (An inferred change in particle number could also change the density, but I'm guessing that wasn't your mistake.)

comment by simon · 2023-10-12T16:39:46.475Z · LW(p) · GW(p)

At this point, I'm thinking that a decrease in density does not decrease the speed of sound in gases, only in liquids and solids.

Actually, a decrease in density increases speed of sound in liquids and solids.  (intuition: less mass -> less inertia, faster movement for same forces). The same also does apply to gases (depending on what else is held constant, but consider, e.g. helium).

And, in the case of a gas, it's not clear to me that, other things being equal, moving farther before hitting another particle should be expected to decrease the speed of sound. After all, even if it takes a longer time for a particle to hit another particle, it's traveling farther in that time, so the total distance per time isn't dropping. 

Here is my synthesis of what happened: in solids and liquids, density does affect the speed of sound, but in gases, a third factor is pressure. If you increase the temperature, density can increase, but pressure also increases by a portional amount since the molecules are hitting harder. These two changes cancel out. Thus, in a gas  only.

It's confusing to me here what you are assumptions you are making about what else is happening as the temperature changes. E.g. mentioning both pressure and density changing rules out both constant pressure and constant volume. Also, you mention density increasing instead of  decreasing, which might be a typo?

One potential way to think about it, if you want to think in terms of pressure and density: (for actual math, see e.g. https://en.wikipedia.org/wiki/Bulk_modulus)

Constant volume case: as the temperature increases, the density stays the same but the forces increase due to the increase in pressure, so sound moves faster.

Constant pressure case: as the temperature increases, the forces stay the same (since we're holding the pressure constant), but the density (and hence inertia) decreases, so sound moves faster.

Good for you to be thinking about improving your understanding, best of luck in building up a strong physical intuition.

comment by cubefox · 2023-10-15T08:10:47.108Z · LW(p) · GW(p)

If you increase the temperature, density can increase, but pressure also increases by a portional amount since the molecules are hitting harder. These two changes cancel out.

I think you meant to say "density can decrease" not "increase"?

Replies from: g-w1
comment by Jacob G-W (g-w1) · 2023-10-15T12:35:58.123Z · LW(p) · GW(p)

Yes, this is what I meant to say, sorry: typo.