(Disclaimer: I am a mathematician, not a physicist, and I hated the one thermodynamics course I took at university several decades ago. If anything I write here looks wrong, it probably is.)
Let’s try to do this from lower-level first principles.
The temperature is proportional to the average kinetic energy of the molecules.
Suppose you have particles with (mass,velocity) (m1,v1) and (m2,v2) and they collide perfectly elastically (which I believe is basically what happens for collisions between individual molecules). The centre-of-mass frame is moving with velocity m1v1+m2v2m1+m2 relative to the rest frame, and there the velocities are m2m1+m2(v1−v2) and m1m1+m+2(v2−v1). In this frame, the collision simply negates both velocities (since this preserves the net momentum of zero and the net kinetic energy) so now they are m2m1+m2(v2−v1) and m1m1+m2(v1−v2), so back in the rest frame the velocities are (m1−m2)v1+2m2v2m1+m2 and (m2−m1)v2+2m1v1m1+m2. So the first particle’s kinetic energy has changed by (scribble scribble) 2m1(m1+m2)2[m1(m1−m2)v1v2+2m22v22−4m1m2v21] .
(I should say explicitly that these velocities are vectors and when I multiply two of them I mean the scalar product.)
Now, the particles with which any given particle of our cooling object comes into contact will have randomly varying velocities. I think the following may be bogus but let’s suppose that the direction they’re moving in is uniformly random, so the distribution of v2 is spherically symmetrical. (It may be bogus because it seems like two particles are more likely to collide when their velocities are opposed than when they are in the same direction, and if that’s right then it will induce a bias in the distribution of v2.) In this case, the v1 v2 term averages out to zero and for any choice of |v2| we are left with constant . (KE2 - KE1). So as long as (1) our possibly-bogus assumption holds and (2) the rate at which object-particles and environment-particles interact isn’t changing, the rate of kinetic energy change should be proportional to the average value of KE2-KE1, which is to say proportional to the difference in termperatures.
This is the (unmodified) Newton cooling law.
So if my calculations are right then any wrongness in the Newton law is a consequence of assumptions 1 and 2 above failing in whatever ways they fail.
I think assumption 2 is OK provided we keep the environment at constant temperature. (Imagine an environment-molecule somewhere near the surface of our object. If it’s heading towards our object, then how long it takes before it collides with an object-molecule depends on how many object-molecules there are around, but not on how they’re moving.)
I am suspicious of this “Lambert’s law”. Suppose the environment is at absolute zero—nothing is moving at all. Then “Lambert’s law” says that the rate of cooling should be infinite: our object should itself instantly drop to absolute zero once placed in an absolute-zero environment. Can that be right?
I am suspicious of this “Lambert’s law”. Suppose the environment is at absolute zero—nothing is moving at all. Then “Lambert’s law” says that the rate of cooling should be infinite: our object should itself instantly drop to absolute zero once placed in an absolute-zero environment. Can that be right?
We’re assuming the environment carries away excess heat instantly. In practice the immediate environment will warm up a bit and the cooling rate will become finite right away.
But in the ideal case, yeah, I think instant cooling makes sense. The environment’s coldness is infinite!
Oh neat! Very interesting. I believe your argument is correct for head-on collisions. What about glancing blows, though?
Assume two rigid, spherical particles with the same mass and radius.
Pick a coordinate system (at rest) where the collision normal vector is aligned with the x-axis.
Then move the coordinate system along the x axis so that the particles have equal and opposite x-velocities. (The y-velocities will be whatever.) In this frame, the elastic collision will negate the x-velocities and leave the y-velocities untouched.
Back in the rest frame, this means that the collision swaps the x-velocities and keeps the y-velocities the same. Thus the energy transfer is half the difference of the squared x-velocities, 12((vx2)2−(vx1)2).
I’m not sure that’s proportional to T2−T1? The square of the x-velocity does increase with temperature, but I’m not sure it’s linear. If there’s a big temperature difference, the collisions are ~uniformly distributed on the cold particle’s surface, but not on the hot particle’s surface.
Hmm. You’re definitely right that my analysis (if it deserves so dignified a name) assumes all collisions are head-on, which is wrong. If “the x-axis” (i.e., the normal vector in the collision) is oriented randomly then everything still works out proportional to the kinetic energies, but as you say that might not be the case. I think this is basically the same issue as the possible bogosity of the “possibly-bogus assumption” in my original analysis.
Dealing with this all properly feels like more work than I want to do right now, though :-).
(Disclaimer: I am a mathematician, not a physicist, and I hated the one thermodynamics course I took at university several decades ago. If anything I write here looks wrong, it probably is.)
Let’s try to do this from lower-level first principles.
The temperature is proportional to the average kinetic energy of the molecules.
Suppose you have particles with (mass,velocity) (m1,v1) and (m2,v2) and they collide perfectly elastically (which I believe is basically what happens for collisions between individual molecules). The centre-of-mass frame is moving with velocity m1v1+m2v2m1+m2 relative to the rest frame, and there the velocities are m2m1+m2(v1−v2) and m1m1+m+2(v2−v1). In this frame, the collision simply negates both velocities (since this preserves the net momentum of zero and the net kinetic energy) so now they are m2m1+m2(v2−v1) and m1m1+m2(v1−v2), so back in the rest frame the velocities are (m1−m2)v1+2m2v2m1+m2 and (m2−m1)v2+2m1v1m1+m2. So the first particle’s kinetic energy has changed by (scribble scribble) 2m1(m1+m2)2[m1(m1−m2)v1v2+2m22v22−4m1m2v21] .
(I should say explicitly that these velocities are vectors and when I multiply two of them I mean the scalar product.)
Now, the particles with which any given particle of our cooling object comes into contact will have randomly varying velocities. I think the following may be bogus but let’s suppose that the direction they’re moving in is uniformly random, so the distribution of v2 is spherically symmetrical. (It may be bogus because it seems like two particles are more likely to collide when their velocities are opposed than when they are in the same direction, and if that’s right then it will induce a bias in the distribution of v2.) In this case, the v1 v2 term averages out to zero and for any choice of |v2| we are left with constant . (KE2 - KE1). So as long as (1) our possibly-bogus assumption holds and (2) the rate at which object-particles and environment-particles interact isn’t changing, the rate of kinetic energy change should be proportional to the average value of KE2-KE1, which is to say proportional to the difference in termperatures.
This is the (unmodified) Newton cooling law.
So if my calculations are right then any wrongness in the Newton law is a consequence of assumptions 1 and 2 above failing in whatever ways they fail.
I think assumption 2 is OK provided we keep the environment at constant temperature. (Imagine an environment-molecule somewhere near the surface of our object. If it’s heading towards our object, then how long it takes before it collides with an object-molecule depends on how many object-molecules there are around, but not on how they’re moving.)
I am suspicious of this “Lambert’s law”. Suppose the environment is at absolute zero—nothing is moving at all. Then “Lambert’s law” says that the rate of cooling should be infinite: our object should itself instantly drop to absolute zero once placed in an absolute-zero environment. Can that be right?
We’re assuming the environment carries away excess heat instantly. In practice the immediate environment will warm up a bit and the cooling rate will become finite right away.
But in the ideal case, yeah, I think instant cooling makes sense. The environment’s coldness is infinite!
Oh neat! Very interesting. I believe your argument is correct for head-on collisions. What about glancing blows, though?
Assume two rigid, spherical particles with the same mass and radius.
Pick a coordinate system (at rest) where the collision normal vector is aligned with the x-axis.
Then move the coordinate system along the x axis so that the particles have equal and opposite x-velocities. (The y-velocities will be whatever.) In this frame, the elastic collision will negate the x-velocities and leave the y-velocities untouched.
Back in the rest frame, this means that the collision swaps the x-velocities and keeps the y-velocities the same. Thus the energy transfer is half the difference of the squared x-velocities, 12((vx2)2−(vx1)2).
I’m not sure that’s proportional to T2−T1? The square of the x-velocity does increase with temperature, but I’m not sure it’s linear. If there’s a big temperature difference, the collisions are ~uniformly distributed on the cold particle’s surface, but not on the hot particle’s surface.
Hmm. You’re definitely right that my analysis (if it deserves so dignified a name) assumes all collisions are head-on, which is wrong. If “the x-axis” (i.e., the normal vector in the collision) is oriented randomly then everything still works out proportional to the kinetic energies, but as you say that might not be the case. I think this is basically the same issue as the possible bogosity of the “possibly-bogus assumption” in my original analysis.
Dealing with this all properly feels like more work than I want to do right now, though :-).