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 :-).
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 :-).