What's the effective difference between a helium molecule moving at 11.18 km/s and one moving at 11.2 km/s at the edge of the atmosphere? Is the idea that, with a particle moving just below the escape velocity directly away from the earth, (assuming it doesn't collide with another particle), gravity will slow it down and eventually cause it to reverse direction?

The problem I see with that is that it would require the kinetic energy of the particle to change, which would require the particle to lose thermal energy, since a gas' kinetic energy can only change through heat transfer. Given that temperature = kinetic energy in a gas particle, if the gas were to lose kinetic energy, where would that energy go in a vacuum?

Would the particle slow down until it reached 0 K and then reverse direction and heat up again by turning potential energy into kinetic energy? That doesn't make sense to me because, in the case of a gas particle, the energy would actually have to be stored somewhere, unlike gravitational potential energy, which isn't stored in a mass but is strictly a function of its position.

This problem wouldn't apply to a solid mass being thrown at escape velocity because the temperature of the solid mass would have no effect on its potential or kinetic energy under gravity. If you drop a red hot metal sphere or an extremely cold one, it will have no effect on its kinetic energy, in other words. So treating the particle as a solid mass will not solve the problem.

  • $\begingroup$ physics.stackexchange.com/q/158508 $\endgroup$ – BowlOfRed May 11 at 22:41
  • $\begingroup$ A comment from that thread: "The potential energy of the system can be converted to heat through various methods; but the GPE itself does not cause any change to the heat." So we're left with the same problem: given a gas particle's speed = kinetic energy = thermal energy, where does that energy go as it slows down due to gravitation? As its speed approaches zero before its velocity reverses by 180°, does its internal energy approach zero? That doesn't make sense. $\endgroup$ – numbynumb May 11 at 22:56
  • $\begingroup$ "The initial temperature increase comes from the transfer of GPE to KE as the particles condense. As the gas loses GPE, it gains KE (and therefore temperature)." The problem with this explanation is that kinetic energy = temperature in a gas and gravity can't change the thermodynamic state of a system because that would violate energy conservation. Gravity is not a source of energy for an isolated system because it's the result of the curvature of spacetime, not a force. $\endgroup$ – numbynumb May 11 at 23:17

The kinetic energy of a particle moving "up" a gravity well goes into its gravitational potential energy, whether that particle is part of a gas or part of a solid object. The energy of a gas particle is not somehow different from that of a solid in that it needs to be stored "somewhere".

Remember that for a gas, a change in internal energy can arise both from heat transfer or doing work. Thermal energy does not have to always remain thermal energy and it does not have to always come from thermal energy; it can turn into other forms of energy. A particle moving against gravity does work against gravity and therefore a gas expanding out of a gravity well should (thermodynamically speaking) get colder as it pushes against that pressure. Note that the reverse process—the heating of matter as it accretes from dust and gas clouds into large objects—is responsible for half the thermal energy of the Earth's interior and for igniting the Sun.

  • $\begingroup$ Unfortunately, this answer conflicts with what's written in standard physics texts: "In a statistical mechanical account of an ideal gas, in which the molecules move independently between instantaneous collisions, the internal energy is the sum total of the gas's independent particles' kinetic energies, and it is this kinetic motion that is the source and the effect of the transfer of heat across a system's boundary. For a gas that does not have particle interactions except for instantaneous collisions, the term 'thermal energy' is effectively synonymous with 'internal energy'." $\endgroup$ – numbynumb May 11 at 21:05
  • $\begingroup$ The problem remains that, if the particle gets colder doing work against gravity, that thermal energy would have to be accounted for due to conservation. In that respect, it's a completely different problem than dealing with the kinetic energy of a solid object subject to gravity, which is completely independent of the object's temperature. $\endgroup$ – numbynumb May 11 at 21:09
  • $\begingroup$ This is not contradictory at all. The lost thermal energy of a gas expanding against gravity is accounted for by the conservation of energy (assuming that's what you're referring to) by a compensatory increase in gravitational potential energy. It is the exact same as for the lost "collective" kinetic energy of a solid object moving up against gravity. The reason the gas gets colder and the solid doesn't is just the details of how exactly that transfer is happening. $\endgroup$ – HTNW May 11 at 21:28
  • $\begingroup$ And I also don't see what's wrong with your quote. All you have to do is define the gravitational potential energy of the gas to not be part of its internal energy (which makes sense; insofar as that energy has a location it's "between" the gas and the source of the gravitational field, not purely in either. Alternatively, removing the gravitational field's source changes the gas's gravitational potential energy without changing the gas, so obviously that energy isn't internal). The thermal energy of the gas is changed by gravity because gravity can affect the kinetic energy of the particles. $\endgroup$ – HTNW May 11 at 21:31
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    $\begingroup$ That's just... not true. It is a fact of the universe that neither kinetic nor potential energy are conserved alone, but their sum is. If you want to say "energy is conserved" then you must say "potential energy is a form of energy". And the various forms of potential energy can and do exchange with the various forms of internal energy an object may have. E.g. I can take a horizontal spring and hold it vertically, and gravity will stretch it so its center of mass drops (lowering GPE) and that energy is instead stored internally as stretching. That transfer simply does happen. $\endgroup$ – HTNW May 12 at 3:16

"Thermal energy" and "temperature" is a useful concept for collections, but not for a single particle. Thermal energy relate to the relative kinetic energy of the collection. Without a collection to come to equilibrium, it makes no sense to talk about the thermal energy of your particle.

If we imagine two identical collections of particles, one high in the gravity well and one lower. Then this particle can interact with either and the upper collection will come to equilibrium at a lower temperature than the other, because the particle gave it less energy than the other.

Would the particle slow down until it reached 0 K

Assuming it is a classical particle, you can find a frame where it is not moving and has 0 kinetic energy. But since it is not in equilibrium, we don't talk about a temperature.

The energy it has can be distributed among other particles and then examined to become a temperature.

That energy that "is the source and the effect of the transfer of heat across a system's boundary" must still be conserved in physical terms. Internal energy is leaving the system so where is it going?

As the other answer mentions, the energy comes from the change in potential energy of the system.

The sum of the mechanical energy (GPE + KE) is constant.

If the system equilibrates high in the gravitational field, you have a particle with greater GPE and the system has slightly lower internal energy.

If the system equilibrates low in the gravitational field, you have a particle with lower GPE and the system has slightly greater internal energy.

This should be no different than the classic idea of the GPE + KE of a ball in freefall remaining constant. There is an additional location for the energy (into thermal energy after equilibration), but the sum is still constant.

The motion of particles is what's known as thermal motion.

I would rephrase that and say that the aggregate motion of particles can be interpreted as thermal motion. Thermal motion is a means of coalescing multiple motions into a single statistical average. They aren't different things, they are different interpretations. And of course we require that the particles are in thermal equilibrium for this interpretation to be useful.

It differs greatly from the motion of solid objects. A key difference is that thermal motion is frame-independent.

I would say it's not a "difference". The thermal energy is the minimum kinetic energy of an object, the total KE measured in the frame where the center of mass is at rest. In any other frame, the total KE is higher. This sum over all particles is identical to the thermal energy measurement plus the KE of the mass of the system in that frame. We are free to interpret the system as a thermal collection or as independent objects and get the same total energy.

This explains why gravity can't change the thermodynamic state change in the thermal motion of particle.

We cannot interpret an isolated particle as having an aggregate temperature. It has a well-defined kinetic energy in any frame. Gravity can change its kinetic energy in the earth fixed frame. That well-defined kinetic energy can equilibrate with other particles once it joins the system.

  • $\begingroup$ If we put a single particle in a container that has a lower average thermal energy than the single particle's kinetic energy, the container will absorb energy in the form of heat. Likewise, if the particle is at rest in a container, it will absorb heat from the container and then possess kinetic energy. So perhaps the semantic distinction isn't of the essence here. That energy that "is the source and the effect of the transfer of heat across a system's boundary" must still be conserved in physical terms. Internal energy is leaving the system so where is it going? $\endgroup$ – numbynumb May 12 at 2:12
  • $\begingroup$ The motion of particles is what's known as thermal motion. It differs qualitatively from the motion of solid objects, a key difference being that thermal motion isnt relative. It remains constant in all frames of reference (or else ice might be forced to appear as water depending on the chosen frame). Conversely, the motion of solid objects can only be described in relative terms.This is why gravity cant affect the thermal motion of a system. It leads to paradox because thermal motion must appear the same from any frame of reference. **edited comment after noticing typing error $\endgroup$ – numbynumb May 13 at 16:06
  • $\begingroup$ Temperature is most definitely frame invariant. As I mentioned previously, paradoxes arise when considering the implications of non-invariant thermodynamic transformations. These 2 papers provide a number of relevant examples: zenodo.org/record/1334830#.YJ3HAlMXY0E inspirehep.net/literature/425456 I think you'll have to agree that there can be no reasonable doubt about the conclusion. And the implications are very interesting. $\endgroup$ – numbynumb May 14 at 0:55
  • $\begingroup$ I understand that 'temperature' is generally used to describe the measure of the aggregate motion of a large number of particles. I'm using 'thermal energy' here to describe the energy that's exchanged between particles during instantaneous collisions and, given that single particles can transfer this kind of energy, I don't think the terminology is adding confusion to the discussion. There are plenty of published papers that describe single particles along these lines. It's not unprecedented: journals.aps.org/pra/abstract/10.1103/PhysRevA.78.033425 $\endgroup$ – numbynumb May 14 at 1:25
  • $\begingroup$ I'm not sure what your comment is referring to. Nowhere do I say that temperature depends on the frame. $\endgroup$ – BowlOfRed May 14 at 1:25

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