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I would want to call this question [science-based], except that we need to restrict ourselves to newtonian/"pre-relativity" physics.

(Clarification: for this question, by "rigid body", I mean an object which is completely incapable of flexing/bending, stretching, or being compressed. This is, of course, impossible under special/general relativity--such an object could be used to send information at faster-than-light speeds. But if we're looking at a physics "simulation" based simply on mass, momentum, and perfectly-elastic collisions, I'm hoping that such a concept ought to be treatable...)

Let's assume we have a rigid body. We'll black-box what it's made of (I have thoughts about it, but they would probably add unnecessary complication to the question), but it has all the inertial properties one would expect of solid matter (mass, density, momentum, moment of inertia, etc). Any molecules or atoms that would enter the space contained by the rigid body instead behave as though they had collided with another molecule or atom, and all the imparted momentum that would go with such a collision takes place. What I'm wondering is...if this is all we know about how a rigid body works, do we have enough to make an educated guess about its thermal conductivity?

To help us think about it, let's do an experiment!

Materials

I have a rigid body of this sort which is shaped like a really long dumbbell (say, two 1-foot-diameter spheres which are connected to each other by a 100-foot long bar). I've melted two batches of metal, and I dip each end of this rigid body into one of the batches and let it cool and set around it. So now I have two lumps of metal connected by a rigid, 100-foot bar.

I also have a vacuum chamber big enough to house this assembly, an electrical hot plate that's safe to run in a vacuum, a pair of thermometers, and a bunch of mirrors.

Procedure

  • I put the partially metal-encased rigid body in the vacuum chamber. One of the lumps of metal is sitting on the hot plate.
  • I sit a thermometer on top of each lump of metal.
  • I set up mirrors so that the lumps of metal have almost no line-of-sight on each other; and any black-body radiation they emit is mostly reflected back at them.
  • I empty the chamber of air, hoping to minimize heat transfer from conduction and convection.
  • I turn on the hotplate under the lump of metal at one end of the rigid body, and I start observing the thermometers.

What are my results, if we can reason about them at all?

Can we expect a rigid body to transfer heat efficiently, essentially via a sort of Brownian motion? Would vibration parallel with the bar transmit more efficiently than vibration perpendicular to the bar, because of the reduced angular motion at the other end?

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  • $\begingroup$ (Also...should this actually be migrated to physics.stackexchange , in spite of the pre-relativity condition?) $\endgroup$
    – Qami
    Commented Jul 13, 2021 at 14:14
  • $\begingroup$ @ARogueAnt. Thanks for your comment! I'll try to clarify a bit...my understanding is that thermal energy in a solid consists of vibrations among its atoms - i.e. its very capacity to be heated requires that it be non-rigid. If we're excluding internal vibrations from an object (as I imagined we must, if we have a truly rigid body), then it would not (I imagine) be able to carry thermal energy at all! But it seems to be that it should still be able to conduct it via the kinetic energy of the whole body as it moves. Does that help/makes any sense to you? $\endgroup$
    – Qami
    Commented Jul 13, 2021 at 14:42
  • $\begingroup$ What is your definition of "rigid body"? $\endgroup$
    – L.Dutch
    Commented Jul 13, 2021 at 15:06
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    $\begingroup$ Ah, but if it is not made of atoms and molecules, then it does not have a thermodynamic state in the usual sense. We cannot really say much about how it transfers heat, or even if it transfers heat at all. $\endgroup$
    – AlexP
    Commented Jul 13, 2021 at 15:16
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    $\begingroup$ I’m voting to close this question because it has "science-based" as a tag, but puts in conditions that explicitly deny known science, specifically, that contradict relativity. $\endgroup$
    – puppetsock
    Commented Jul 13, 2021 at 18:58

3 Answers 3

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We do NOT have enough information.

you have not defined any characteristics that would answer that question. the collision between molecules is affected by the properties so both molecules, so just saying it behaves like a collision with another molecule does not help. The question is does energy transfer happen during the collision and at what rate. Can infrared light be absorbed by and change energy state of the rigid body. In a pure kinetic sense it should act like a perfect thermal conductor, but if it can't absorb and emit thermal radiation it actually becomes an incredibly good insulator because that is how most heat is transferred, at least for useful purposes of heat transfer.

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  • $\begingroup$ Good point about the interactions with light! Since the point of this question is focused on kinematic sense, I would want to choose optical characteristics that as much as possible leave optical interaction out of the equation. So, like, making it a perfect mirror would probably be ideal. But when you say "thermal radiation...is how most heat is transferred", are you talking about even when you have two surfaces in contact with each other? $\endgroup$
    – Qami
    Commented Jul 13, 2021 at 19:27
  • $\begingroup$ @Qami It does depend on different the temperatures are, the larger the difference the larger impact radiation has, you have a weird situation where your material is almost better at transferring heat between similar temperature objects than drastically different ones. the bigger the difference the more like an insulator it will behave. of course. Oddly you have also created a material that can transmit sound faster than the speed of light. which I am pretty sure takes physics into a dark alley and pistol whips it. $\endgroup$
    – John
    Commented Jul 13, 2021 at 19:48
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Consider the mass as a single atom.

It does not internally stretch, bend etc except on the quantum scale. When it moves, the entire thing moves as a piece. That is how individual molecules work. Thermal energy is carried as kinetic energy.

Impacts of molecules impart kinetic energy to the thing as noted in OP. / Any molecules or atoms that would enter the space contained by the rigid body instead behave as though they had collided with another molecule or atom/

Normally an impact would distribute its energy such that the mass struck would travel in a direction according to the original vector and some of the force would go towards motion of constituent molecules, heating up the struck thing.

For your hypothetical nonvibrating thing, none of the force goes into heating it. All of the force goes into moving it. The vector force of the impact of a hot molecule is converted into motion.

Thus if you put such a mass by the fire, it would gradually move away from the fire with kinetic energy losses to friction on the substrate. Circumferential heating of one end would cause the thing to vibrate as it received many impacts from many sides cancelling each other out as regards the sum of all vectors.

As regards how it would feel if you put your hand on it, it would feel the same temperature as your hand. All thermal energy is perfectly reflected back to the source.

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  • $\begingroup$ (1) The sentence "the vector force of the impact of a hot molecule is converted into motion" has no meaning. (Force is ill-defined with the given constraints; we can use only conservation of momentum.) (2) A perfectly rigid body cannot vibrate, because it is perfectly rigid. $\endgroup$
    – AlexP
    Commented Jul 13, 2021 at 18:16
  • $\begingroup$ @AlexP In the context he was using it ("Circumferential heating of one end would cause the thing to vibrate as it received many impacts from many sides"), I was interpreting "vibrate" as meaning more bouncing-back-and-forth between impacts on different sides. Willk, is that what you meant? $\endgroup$
    – Qami
    Commented Jul 13, 2021 at 18:58
  • $\begingroup$ Qami is right - vibration happens because of changes in the vibrating solid. Many random impacts would produce a similar output although likely with more variety. $\endgroup$
    – Willk
    Commented Jul 13, 2021 at 20:54
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by "rigid body", I mean an object which is completely incapable of flexing/bending, stretching, or being compressed.

You are probably familiar with the model of a solid body being made by spheres (the atoms) connected by springs, allowing the spheres to oscillate around their equilibrium position. This model can help explain how heat is transmitted through a solid, via the propagation of the oscillation.

Your body, in this model, would have the spring replaced by infinitely stiff bars, preventing any oscillation.

Any atom hitting it would therefore bounce back with a perfectly elastic collision, transferring no energy to it: the object would behave like a sort of middle part of a Newton's cradle.

However the lack of vibration in the atoms of the body would be translated in it having a temperature of 0 K, which sort of violates the indetermination principle (you can't know both position and momentum of a particle with high precision) and thermodynamic law (you can't have a body that doesn't reach thermal equilibrium with its surrounding without any external intervention).

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  • $\begingroup$ I agree with your analysis up until the point where you say it transfers no energy--yes, the infinitely stiff bars prevent oscillation, but shouldn't the energy still be transferred into the whole assembly? $\endgroup$
    – Qami
    Commented Jul 13, 2021 at 17:33
  • $\begingroup$ Why wouldn't it transfer energy? It's an elastic collision. Conservation of momentum applies, conservation of kinetic energy applies, so that after a small elastic body of mass $m$ with velocity $v_{0, m}$ collides with the perfectly rigid and stationary body of mass $M$ I would expect that the perfectly rigid body moves with velocity $v_{1, M} = 2v_{0, m} m/(M + m)$ (and the colliding body recoils with velocity $v_{1, m} = v_{0, m} (M - m)/(M + m)$, of course). $\endgroup$
    – AlexP
    Commented Jul 13, 2021 at 18:24
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    $\begingroup$ @AlexP, there would be several collisions from multiple directions, averaging to 0 $\endgroup$
    – L.Dutch
    Commented Jul 13, 2021 at 18:58

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