Suppose you had a gas at a particular temperature but the gas did not have a container. As such the particles could have nothing to collide with except themselves. Moving freely through space they lose no energy. Collisions with themselves are elastic and so no energy is lost. Without a container to give energy to, they would stay at the same temperature.
Is this a correct assumption?
If they collide with themselves and the collision is inelastic, where does the energy go?
To start off, I think you are confusing energy with temperature. The “temperature” of a gas is a thermodynamic property that is proportional to the average kinetic energy of the molecules making up the gas.
With that said, you’re right, the energy doesn’t go anywhere, it’s just spread out over a larger volume as the gas expands (assuming a free expansion in which the gas performs no work on its surroundings). In a free expansion of an gas that is thermally isolated from its surroundings, the change in internal energy of the gas is zero. See here. For an ideal gas, the internal kinetic energy would therefore remain constant, and the temperature would remain constant. For real gases, there would be some change in internal potential energy (due to intermolecular forces), so the temperature would drop.
However, the system that you are describing is not steady-state, so this makes it even more difficult to meaningfully describe the temperature of the gas in question. Note that the concept of temperature for a single atom or molecule also does not have any real meaning.
At which level do you want your answer at, macroscopic or quantum? As tobby mentioned the concept of heat is really not defined at the quantum level.
As you mentioned “gas” I will error on the side of macroscopic, so the energy will be “lost” through the emission of EM radiation as all black bodies do. Even in the case of a gas in a container the energy transfer to the container through the photon and the “empty” space will reach an equilibrium point with that EM radiation.
As for the transfer of energy with colliding particles, remember even under classic mechanics the transfer of kinetic energy is frame dependent.
Consider two frames of reference, the paddle and the player, with a ball hit in a way that it doesn’t impart any spin.
In a frame where the paddle is inertial and not moving it sees no energy transfer but the player (and the ball) would claim there was a transfer of inertial energy.
You need to consider the conservation of momentum, angular momentum, and energy. Where the energy mediated by photons ends up really depends on having better defined conditions. If you were in free space they would simply just radiate away as an example.
The Ideal Gas Equation if PV=nRT. Where R is a constant, and n is just the amount. Apart from the inconvenient fact that the system you have described, with zero pressure and infinite volume, doesn’t give a meaningful answer, Yes. If you let the volume expand adiabaticly, say double the volume and half the pressure, then an ideal gas stays at the same temperature.
We normally don’t want gas to expand adiabitcally. We allow it to interact with the environment so that it gets colder as the pressure drops and/or the volume increases, for example, a refrgerator, or a gasoline engine. And we often are doing all kinds of other things at the same time: phase change or chemical reaction, which are completely non-ideal-gas.
Note if you just want an answer the context of the basic thermodynamics, you don’t have two systems and typically the ideal gas is considered to be a the same temperature so you would in theory be at equilibrium.
In more realistic models the individual gas particles never have the exact same temperature, but if you are working on pure classical thermodynamics this is the answer you want to find.
In that context you have no differential and a single system, thus nothing to transfer the energy to go. If you want to expand on that particular model please let us know which rabbit hole you prefer to go down.
In addition to the points previously mentioned (or possibly, another way of looking at something someone else has mentioned), the cloud of atoms will have a distribution of kinetic energies. The atoms with high kinetic energy are the atoms that tend to be the ones that will head off into the universe, never to interact with the cloud again. This lowers the average kinetic energy of the remaining atoms (the remaining atoms can’t collide with the atoms that have escaped to get that energy back) so the portion of the cloud near where the cloud was initially placed will lower in temperature.
It should be noted that astronomers do observe gases that are not contained. Or only contained by the gas cloud’s own gravity, not a container. And I’m not talking about stars, I mean interstellar or intra-cluster gas. However, since they observe them, the gases are emitting radiation, which is a way that they cool off. Why they are emitting the radiation is a significant part of the science of astronomy.
While this works for a theoretical thought experiment, in actual reality it won’t. Your “nothing to collide with” part presupposes an absolutely empty space. But in reality, there ain’t no such place. Everywhere we know of contains interstellar dust, light & electro-magnetic waves, and gravity moving through it. Way ot in the emptiest regions of space, the amount of matter may be very small, but it’s still there. (But it may be small enough to be negligible for your experiment.)
“Adiabatically” usually means that the temperature does change. Specifically, the gas doesn’t absorb any heat from its environment, which means that if it expands (and does work on its surroundings), that energy must come from the internal thermal energy of the gas, and thus the temperature decreases.
If gas molecules collide inelastically, there are two possibilities: One, the collision might result in the emission of a photon, in which case the temperature of the gas would decrease (but you’d have some extra energy in light to make up for it). Or two (especially if the molecules are very complicated), you could put one of the molecules into some sort of excited state. This could result in the molecule eventually relaxing on its own and (again) emitting a photon, or it could be released the next time that molecule collides with another one, producing the opposite sort of inelastic collision, like a pinball bumper.
Ultimately, would they not interact with the ‘quantum vacuum’, which AFAIK is not a vacuum at all in the classical sense. If so, would such interactions affect the particles’ ‘temperature’, i.e. their energy? Maybe a sort of quantum friction?
Avoid the word friction like the plague, as it was really the intuition tool that lead to the idea of Luminiferous Ether. It is probably safer to remember that mass is simply the emergent property of bounded energy.
For a boson or monotonic gas the thermal capacity will mostly be limited to translational kinetic energy or momentum in the up-down,forward-back,right-left degrees of freedom. In gasses made of molecules rotational kinetic energy becomes a significant portion of the heat capacity.
Nitrogen in the form N[sub]2[/sub] as an example has the three translational degrees of freedom as well as two internal rotational degrees of freedom. Note that this 5 degrees is actually less than the 6 total that the individual atoms would have.
Counter intuitively, at higher temperatures where rotational degrees of freedom become accessible N[sub]2[/sub] will have a higher capacity than the atomic version.
Temperature is an ensemble property or a macroscopic property, if you get too diffuse the kinetic theory of gases will no longer hold and actually talking about temperature becomes problematic.
The Wikipedia article for gas in a box will link to a few overviews of the various forms of quantum ideal gasses but all of these thought experiments will fail if the gas is too spread out.
This is particularly relevant to the OP’s scenario, since some components of the interstellar medium are technically extremely hot, even though they could not warm an object in space. A very low density of particles with high velocity.
The quantum vacuum is, so far as we can tell, completely frame-invariant. If there were some sort of friction with it, that would tend to slow down particles, that would give it a preferred reference frame, and relativity goes out the window.
As an aside, I get really annoyed with people (even people who should know better) who describe mass due to the Higgs effect as being due to a sort of friction or viscosity with the Higgs field. Friction and inertia do not behave at all in the same way, and even if they did, most mass that we know of is unrelated to the Higgs field. And a lot of the people who repeat that explanation know this, and just really suck at analogies.
To try and help people get away from the intuitive yet incorrect friction analogy.
The Higgs field is where the property of inertia comes from. Remember that mass is the sum of all energies of a system divided by the speed of light squared.
M[sub]rest[/sub] = E/c[sup]2[/sup]
Is really:
M[sub]rest[/sub] = {E[sub]inertial[/sub] + E[sub]potential[/sub] + E[sub]kinetic[/sub] + E[sub]rotational[/sub] + …}/c[sup]2[/sup]
If you could make a mass-less box of perfect mirrors and filled it with photons it would have a M[sub]rest[/sub] that would include the energy of the photons. The mass a H[sub]2[/sub] is less than that of two separated hydrogen atoms. In the proton, the 3 quarks it is made of are only ~1/100 of it’s M[sub]rest[/sub] because the binding and kinetic energy are so huge.
To avoid the issues with the term “friction” in analogies simply think of the Higgs mechanism as adding another term for Energy to the above.
The Higgs mechanism results in electroweak symmetry breaking, resulting in an effect that limits the ways that particles can freely move. This energy confinement works better as an analogy for what causes the property of inertia to rise. It also has the benefit of working as just another E[sub]higgs[/sub] term to the sum for most needs.
Thus:
M[sub]rest[/sub] = {E[sub]inertial[/sub] + E[sub]potential[/sub] + E[sub]kinetic[/sub] + E[sub]rotational[/sub] + E[sub]higgs[/sub] + …}/c[sup]2[/sup]
Really this is just hard to explain because lots of our very useful models and theories depend on the law of conservation for mass which really only holds in reductive cases. The amount of matter changes all the time as energy is released or added to a confined system, but the outdated concept of mass invariance has more practical applications and is maintained despite the increased confusion it causes in cases like this.
Fortunately in the cases where conservation of mass rule is critical, the still flawed analogy of added energy works and is better than the “friction” version. Just don’t try to relate it to how mass-less particles interact with the Higgs field etc…but at least you don’t have to explain away a “drag”.
I should add,
H[sub]2[/sub] has less M[sub]inertial[/sub] because E[sub]potential[/sub] is negative. Individual Hydrogen atoms do not have an intrinsic positive E[sub]potential[/sub].
(No energy was created or destroyed during the creation of this post. I swear I was just borrowing it!)
And right there, you’re already wrong.
The Higgs boson is responsible for rest mass of elementary particles that couple to it. While the boson is not responsible for the majority of what adds up to be an object’s total rest mass, without the Higgs boson those elementary particles would be massless and traveling the speed of light. The fact that this is a small part of mass of an object is irrelevant.
The Brout-Englert-Higgs mechanism allows weak particles, electrons, quarks, and other fundamental particles a way to interact with the Higgs field, and yet this interaction is proportional to how massive an object is.
Perhaps you though I was saying that the “Higgs boson gave other particles mass”? Or perhaps you could share how you think I am wrong so I don’t have to guess why you think I am.
Because if you are going to argue that “The Higgs field is the stuff that gives all other particles a mass.” I need a cite, and you will need to argue with Firmilab’s analogy too.
http://www.fnal.gov/pub/science/inquiring/questions/higgs_boson.html
If you are ignoring modern parlance in that mass is the rest mass please call that out. Please do the same with inertia if you do not accept it as what mass does when there is no external force acting on it.
I don’t understand this. If there were some sort of friction, wouldn’t that just red shift a photon, sort of like expanding spacetime does? Why would that give it a preferred reference frame?
There is no friction, that was disproved by experiment and was the basis of the superseded luminous ether theory of light.
Once emitted and until absorbed, photons don’t really change energy from their frame (or a co-moving frame). The geometry of space-time is changing with red shift, not the photon.
If there was “friction” the speed of light wouldn’t be the same for all observers, which would mean that there is some frame that is more privileged than others.