sethoflagos

2024-04-18T14:51:23Z

55 minutes ago, Prajna said:

Sorry, I'm not sure of the dynamics yet... So, at the starting position the magnets with a tab in the gap will be at minimum separation.

Pretty sure there's a x-post here with @exchemist so briefly:

If we're starting from your declared position of maximum attraction, we're moving against an attraction force for 90⁰; then with a weakened repulsive force (poles wide apart); then against the same repulsive force; then finally with the mirror image of the attraction of the initial power stroke.

In the absence of a proper mathematical analysis, by symmetry we have a nett zero sum.

And then there's cam friction and the hysteresis braking mentioned earlier.

Granted I've ignored secondary effects of the movement of the magnets themselves but frankly, that's beyond my pay scale.

Suffice to say, if there was anything to see here, Faraday would have found it back in the day I think.

39 minutes ago, exchemist said:

OK.

As I understand it, the idea is inserting the tab, or finger, causes the magnets to be attracted to it, instead of repelled from one another as they are in the previous phase of the motion.

If we describe the operation in terms of an engine cycle, there are 4 phases:-

1) magnets close together no tab inserted, high energy of the field

2) magnets have moved apart due to mutual repulsion, reduction in field energy. Work imparted to output shaft

3) tab or finger inserted into the gap, causing magnets to be now attracted towards it, with further lowering of field energy. More work output to the output shaft (and some work output to the input shaft as well, due to the attraction)

4) tab removed from the gap between the magnets, which are now close together. This replaces the force of attraction to the tab or finger by mutual repulsion of the magnets, which are now at close separation, i.e. back to (1). It is this step that requires the substantial work input which returns the stored energy in the field to its stating value. Failure to realise the work need to do this is what can lead the incautious designer to think he has an over-unity machine, as the other steps all involve extracting work from the magnetic field.

At least, that is my energy-based analysis of this machine.

Looks right enough, so you've got the 180⁰ phase shift covered. Shall we leave the +/-90⁰ phase shifts to the OP?

2024-04-18T13:41:21Z

23 minutes ago, Prajna said:

In the current design the tabs or fingers are arranged every 20 degrees around the rotor. There are nine tabs and the rotor axle is co-planar with the centre line of the magnets, so when there is a tab central to the gap between the magnets on one side there is a space on the opposite side. The magnets are 10mm x 2mm neodymium (N52?) and the tabs centres are at approximately a 35mm radius. The 'bulb' on the tabs that lies between the magnets is 5mm radius to match the area of the magnets. This may be more or less optimal as far as effectiveness in switching the flux and suffering eddy current drag, I don't know yet.

This is somewhat arbitrary and is just my first best guess of what might work.

You're not answering the question I asked.

When there is a tab directly between the magnets, are the magnets at minimum separation, maximum separation, or somewhere in between.

Just to be absolutely clear on where you want to push and when you want to pull. It makes a difference. Rather like the ignition timing on a combustion engine.

2024-04-18T12:39:30Z

1 hour ago, Prajna said:

Thanks for that, @sethoflagos, that is a very useful comparison.

In return, perhaps you could clarify something for me. What is the phase relationship between magnet pole separation and finger position?

If we define zero degrees for the disk when a finger is directly between the poles, and zero degrees for the poles as minimum pole separation, then what phase difference between the two should we consider for optimum performance?

And how is that optimal phase difference maintained?

2024-04-18T12:02:26Z

4 minutes ago, exchemist said:

One important difference, though, is that in the OP's machine the poles of the two magnets are opposed so that they repel. The region in which the fingers on the input disc move is in principle an area in which the field lines will be squashed outwards in the plane of the fingers of the disc.

Different words, same thing. It's the repulsion of like poles that causes the field rotation I took as a given for sake of brevity.

2024-04-18T10:43:32Z

1 hour ago, exchemist said:

... So there is something wrong with your assumption. Either you will find the magnets are not attracted together when the steel finger is interposed, or you will find the finger resists being inserted or removed, such that the operator has to do work against the field, thus supplying the required energy...

Perhaps one way of looking at this contraption is to compare it with a Faraday disk (aka homopolar generator).

In the latter, both motion and induced current are in the plane of the disk with the magnetic field perpendicular. The OP is rotating this so that motion and magnetic field lines are in the disk plane therefore forcing induced current into the perpendicular. However, different portions of the disk will see different current polarities depending on whether they are moving towards or away from the magnetic poles. In particular, the portion of the disk passing directly between the poles will see a sharp switch in polarity and consequent current flow component appearing in the disk plane. This will in turn deflect the magnetic field lines somewhat out of the disk plane as if attracted by a temporary opposite pole.

I don't know whether it's a good picture, but in my mind's eye, I'm seeing this induced temporary pole falling into a potential well only to climb back out as it departs with no nett overall energy change in and of itself. However these circulating currents are a different matter as they will add a time lag to the ideal case making ascent harder than descent, acting as a brake in exchange for simply heating up the disk.

Wednesday at 02:17 PM

31 minutes ago, exchemist said:

There was a minor outbreak of monkeypox recently, but I’m not aware it became a recognised epidemic. As I recall, there was no more than a few hundred cases.

Either of the two main smallpox vaccines can control it, so if it were perceived as important (eg by killing white people instead), it would be easy enough to deal with.

We get the odd case here from time to time. Nature's way of telling us not to mess around with rope squirrels (suspected wild reservoir).

April 6

14 hours ago, martillo said:

You are wrong in this. T is not defined as T = dU/dS. T is defined as T = dH/dS.
So you are also wrong in this:

Actually both are correct providing dU/dS is evaluated at constant volume and dH/dS is evaluted at constant chemical potential. Look at the wikipedia pages on thermodynamic potential and Maxwell's relations.

However, since 2019 the international community has agreed to settle on the former as the official definition of thermodynamic temperature:

Quote

In particular, when the body is described by stating its internal energy U, an extensive variable, as a function of its entropy S, also an extensive variable, and other state variables V, N, with U = U (S, V, N), then the temperature is equal to the partial derivative of the internal energy with respect to the entropy:^[29]^[30]^[31]

$T=\left({\frac {\partial U}{\partial S}}\right)_{V,N}.$

(2)

Likewise, when the body is described by stating its entropy S as a function of its internal energy U, and other state variables V, N, with S = S (U, V, N), then the reciprocal of the temperature is equal to the partial derivative of the entropy with respect to the internal energy:^[29]^[31]^[32]

${\frac {1}{T}}=\left({\frac {\partial S}{\partial U}}\right)_{V,N}.$

(3)

The above definition, equation (1), of the absolute temperature, is due to Kelvin. It refers to systems closed to the transfer of matter and has a special emphasis on directly experimental procedures.

One thing this equation tells us is that if an amount of energy dU is transferred from a warmer body to a cooler one, the TdS values for each must be equal in magnitude (though opposite in sign). This can only be so if the lower value of T for the cooler body is balanced by a proportionally higher value of dS. Therefore the total entropy change for the two bodies is positive,

14 hours ago, martillo said:

I'm not violating 2nd law at all.
You are applying a wrong definition.

Howl at the moon as long as you like, your absorbed photons have been dissipated (as phonons in context) and cannot be recovered intact.

April 5

2 hours ago, martillo said:

I was wrong in that statement and I admitted it well correcting the thing. I after said that conductivity is actually accomplished in solid metal conductors by the absorption with posterior emission of photons in a process that has a time delay involved and so on and we have a long discussion about. I'm wondering why you aren't you considering all that now. By the way, I know I make mistakes sometimes but always try to correct them.

Well there's a big mistake here.

The vast majority of absorbed photons originate from emitters at a higher temperature because of S-B's T⁴.

Therefore the vast majority of absorbed photons lead to an entropy increase in the emitter/absorber system because dU = TdS.

Therefore the reemission of photons you propose would cause an overall system entropy decrease in contravention of the 2nd Law. No amount of arm waving will rescue your hypothesis from this.

April 5

1 hour ago, martillo said:

Good question. Something I also thought about at some time. What I know is that's the way Enthaply is defined anywhere.

But what you don't know is what enthalpy is and where its use is appropriate. It isn't here. Temperature is defined as the inverse partial derivative of entropy with internal energy, not enthalpy.

2 hours ago, martillo said:

P = F/A so F = PA and W = F.distance then PV = W makes sense but at in a perfect equilibrium both P and V are constant so no work done isn't it?
I think the right anser could come considering the following: Thermal equilibrium is a dynamical equilibrium. In the case of the term PV you can think in the environment continuously doing an infinitesimal work on the system and the system continuously reacting with a correspondent infinitesimal work. You can think the environment winning in some infinitesimal areas of the surface while the system winning in the other infinitesimal ones. I didn't arrive to the right conclusion to finalyze the answer giving the not zero term PV in the equation. Something yet needs to be reviewed and solved in that subject.

This is just off-topic waffle.

2 hours ago, martillo said:

Well, as I already said I could agree on "atomic and molecular vibrations" present although we could differ in what we really mean by vibrations, how energy is stored in them and how they are provoked in atoms and molecules.

Familiarise yourself with Debye's theorem. And why that superseded Einstein's photoelectron model. That should help clarify.

April 5

54 minutes ago, martillo said:

I can feel the coffee radiation even with my face at some distance.

No you can't. That's convection currents.

April 5

4 hours ago, martillo said:

I did some quantification. I talked about the enthalpy equation H = U + PV balancing the energies in the system.

Enthalpy includes work previously performed on the environment by the system. Since that energy is no longer contained 'in the system' how can it be pertinent to current system temperature?

4 hours ago, martillo said:

Theory predicts the total energy for an ideal gas to be H = (3/2)NKT (K is Boltzmann constant)...

If you ignore all degrees of freedom bar translational, it is a fair approximation for the internal energy of the noble gases. It is simply the wrong expression for enthalpy.

4 hours ago, martillo said:

... while for solids the total energy is 3NKT. I think the Kinetic Theory is right here including the vibrations energy in the system among the kinetic energy of the particles (due to movement of their center of mass). Is missing then to consider the other half of the total energy in the case of ideal gases.

A couple of points here. For solids, N refers not to number of molecules, but number of atoms.

Secondly, solids support shearing vibrations that gases do not, and therefore have the corresponding degrees of freedom to accommodate this.

Of both counts, you are not comparing like with like and therefore your logic has no foundation.

5 hours ago, martillo said:

My theory predicts that for the ideal gases exactly the half equal to (3/2)NKT is missing and is related to the photons present in the system. In my model the total energy of the system is H = 3NKT...

Consideration of degrees of freedom explains almost all the differences in expressions for molar heat capacity.

5 hours ago, martillo said:

and is equal to the total heat that was absorbed by the system to reach the considered thermal equilibrium, H = Q.

Isothermally? Like your last equation. your understanding of thermodynamics needs work.

March 30

On 3/28/2024 at 10:39 PM, thidmir said:

Yeah it looks like it should be roughly cubic, because newton's law of cooling says that the rate of heat transfer from a fluid to another substance is proportional to the temperature...

The temperature difference between two adjacent regions aotbe typically decays per a lag function (1 - exp(-t / T_n)) where t is elapsed time and T_n is a characteristic time constant.

Series of multiple connected regions (like Earth's surface) therefore tend towards the product of multiple lags as they move towards a new equilibrium.

A second order lag produces a smooth 'S' shaped function between initial and final states. Higher order lags produce more of an initial delay followed by a more abrupt transition.

The principal pattern falls straight out of Fourier's Law. I think you might find it a little more useful than a polynomial fit.

March 30

1 hour ago, martillo said:

But the energy transfer is always linear in the difference of the temperatures. dQ = CdT.

As @swansont infers: This is not true for heat transfer by thermal radiation where dQ is proportional to d(T⁴). ref (https://en.wikipedia.org/wiki/Thermal_radiation)

Quote

The net radiative heat transfer from one surface to another is the radiation leaving the first surface for the other minus that arriving from the second surface.

For black bodies, the rate of energy transfer from surface 1 to surface 2 is:

${\dot {Q}}_{1\rightarrow 2}=A_{1}E_{b1}F_{1\rightarrow 2}-A_{2}E_{b2}F_{2\rightarrow 1}$

where $A$ is surface area, $E_{b}$ is energy flux (the rate of emission per unit surface area) and $F_{1\rightarrow 2}$ is the view factor from surface 1 to surface 2. Applying both the reciprocity rule for view factors, $A_{1}F_{1\rightarrow 2}=A_{2}F_{2\rightarrow 1}$ , and the Stefan–Boltzmann law, $E_{b}=\sigma T^{4}$ , yields:

${\dot {Q}}_{1\rightarrow 2}=\sigma A_{1}F_{1\rightarrow 2}\left(T_{1}^{4}-T_{2}^{4}\right)\!$

where $\sigma$ is the Stefan–Boltzmann constant and $T$ is temperature.^[24] A negative value for ˙ ${\dot {Q}}$ indicates that net radiation heat transfer is from surface 2 to surface 1.

For two grey-body surfaces forming an enclosure, the heat transfer rate is:

${\dot {Q}}={\frac {\sigma \left(T_{1}^{4}-T_{2}^{4}\right)}{\displaystyle {\frac {1-\epsilon _{1}}{A_{1}\epsilon _{1}}}+{\frac {1}{A_{1}F_{1\rightarrow 2}}}+{\frac {1-\epsilon _{2}}{A_{2}\epsilon _{2}}}}}$

where $\epsilon _{1}$ and $\epsilon _{2}$ are the emissivities of the surfaces

This point has now been made to you several times and you have failed to address it properly.

1 hour ago, martillo said:

The energy of a radiation at some place is given by S-B Law at a thermal equilibrium. Energy transfer involves difference in temperatures and so no thermal equilibrium. The energy transferred is directly proportional (linear) in the difference in temperature (assuming no loss in the transfer).

There is no nett energy transfer at thermal equilibrium. Is that your defence? Under some very special circumstance 0 = 0 whichever way you calculate it?.

Not much of a defence is it, really?

March 26

2 hours ago, KJW said:

Perhaps you can state precisely where I have made my error. I believe that what I have said has been misinterpreted, and I seem to be unable to get my interpretation across. I would find it instructive to know precisely what you think I am saying.

Rather than state them, I'll quote them:

On 3/15/2024 at 3:23 PM, KJW said:

One thing should be mentioned: Only the translational modes of molecular motion contribute to the temperature. Different substances have different heat capacities because they absorb energy in all their modes, but only the translational modes increase the temperature, thus the more modes that are available to the molecule, the more energy that is absorbed for a given increase in temperature.

If you'd stated that this observation was limited to the monatomica gases and low temperature hydrogen, then no one would have batted an eyelid. However, as presented it appeared that you intended this as definitive and universal.

When this idea was immediately blown out of the water by @exchemist's query about the temperature of solids, you might have considered retracting. But instead you shifted your ground to the following argument:

On 3/15/2024 at 11:42 PM, KJW said:

One other thing: In the ideal gas law, pressure, which is proportional to temperature, depends only on the translational motion of the gas molecules. Neither rotational motion, nor vibrational motion affects the pressure and therefore the temperature of an ideal gas.

With the sole exception of the theoretical ideal constant volume process, this isn't even true for the monatomic gases. So far from strengthening your case, it actually weakens it.

3 hours ago, KJW said:

I don't regard my statement as a "conjecture", and I am quite surprised by the controversy it has generated.

It certainly comes across as an article of faith. Hence the degree of kickback perhaps.

The only instance I can think of where thermodynamics treats translational and internal degrees of freedom differently is in the theoretical analysis of diffusivity coefficients (both thermal and mass). Which sort of makes sense since diffusion is hardly likely to happen without some translational motion. Not particularly relevant to my own areas of study, but if you're interested they get a mention in:

https://en.wikipedia.org/wiki/Thermal_conductivity_and_resistivity

https://en.wikipedia.org/wiki/Mass_diffusivity

March 25

On 3/22/2024 at 11:44 AM, KJW said:

On 3/19/2024 at 3:12 PM, sethoflagos said:

A bit of algebraic rearrangement gives:

C_VT = C_PT - PV which is equivalent to U = Q - W.

A bit of algebraic rearrangement from what?

Case under consideration (for illustrative purposes), constant pressure heating/cooling of a substance for which internal energy is a function of temperature only.

PV = RT (per kilomole basis)

By Mayer's relation R = C_P - C_V

PV = (C_P - C_V)T

Take partial derivatives and substitute appropriate values for constant pressure process,

VdP + PdV = C_PdT_P - C_VdT_V

0 + dW = dQ - dU

Hence:

dU = dQ - dW

Obviously, for a constant pressure process, P is anything but proportional to T.

On 3/22/2024 at 11:44 AM, KJW said:

What I did say is this:

On 3/15/2024 at 3:23 PM, KJW said:

One thing should be mentioned: Only the translational modes of molecular motion contribute to the temperature. Different substances have different heat capacities because they absorb energy in all their modes, but only the translational modes increase the temperature, thus the more modes that are available to the molecule, the more energy that is absorbed for a given increase in temperature.

That the same ideal gas law applies to argon, nitrogen, carbon dioxide, water, and ethane proves the point I was making. That all these gases have different heat capacities, particularly at higher temperatures, also proves the point I was making.

Per the above, your 'proof' seems merely a tedious repetition of the patently false.

On 3/22/2024 at 11:44 AM, KJW said:

On 3/19/2024 at 3:12 PM, sethoflagos said:

A truly constant volume operation would be rather difficult to achieve in practice.

We are talking about gases... simply place it in a sealed borosilicate glass flask.

Last time I checked, even borosilicate glass had a finite Young's modulus and non-zero thermal expansion coefficient.

On 3/22/2024 at 11:44 AM, KJW said:

On 3/19/2024 at 3:12 PM, sethoflagos said:

says nothing of the nature of particle collisions.

Why are you fixated on particle collisions? Particle collisions are irrelevant to the point I was making.

What do temperature and pressure even mean when there are no particle inteeractions to communicate them?

On 3/22/2024 at 11:44 AM, KJW said:

Why are you mentioning "processes"? Processes, are also not relevant to the point I was making.

In context, it's any defined path between different thermodynamic states. Such as changes in temperature and pressure. Or is your conjecture also confined to conditions of thermodynamic equilibrium only?

March 23

6 hours ago, martillo said:

Fine, if you can't manage my definition of the temperature for a single atom just forget it. My consideration that in the Kinetic Theory of Gases the internal thermal radiation is not taken into account still holds.

The internal energy of thermal radiation within a space occupied by a gas is accounted for by the internal energy of the photon gas that co-occupies that same space, Look at the wikipedia page on 'photon gas' for an explanation.

On 3/17/2024 at 6:08 AM, martillo said:

The thermal equilibrium I'm conceiving is something different. Is not related to the kinetic energy of the particle. I consider the particle is capable to absorb and emit EM radiation (photons). All atoms have their characteristic levels of energy in accordance to the levels of its electrons. All atoms have their characteristic spectra. When electrons jump to other levels they absorb or emit photons. In a thermal equilibrium with its environment the particle continuously absorbs and emits photons maintaining an average level of energy.

At everyday temperatures, black body photons capable of inducing electron orbital jumps are to all intents and purposes non-existent. The dominant process for generating black body radiation is via the acceleration of charged particles, and as @swansont has pointed out, this can be many orders of magnitude below the internal energy of the matter phase.

There is interchange between the matter phase and proton gas phase, and this can have effects during dynamic changes in thermal equilibrium. Off the top of my head, it's one reason we can never quite achieve the theoretical adiabatic combustion temperature in fuel burning processes. But if you only start getting a glimpse of a phenomenon at >1500 K, there really are no grounds whatsoever for claiming it to be a dominant process at normal, lower temperatures.

March 22

4 hours ago, joigus said:

Absolutely. Sorry I missed this very good argument for so long. It's only because of what you say that different molecules have different specific heats as a function of temperature. The internal degrees of freedom are totally relevant. This is exactly the reason why different molecular components have different specific heats. What other reason could there be for different gases to display different specific heats if only the CoM DOF were relevant?

Quite a different matter is how quantum mechanics introduces a cutoff for short-length degrees of freedom (independently of how poly-atomic a gas is), and how this played a crucial role in the dawning of quantum mechanics itself. (Birth of the old quantum theory as a mechanism to freeze the short-wavelength DOF.)

Glad someone was listening 🙂

2 hours ago, joigus said:

Recently, a key to why CO₂ (kinda mysterious, as it's just a boring non-polar linear molecule) is such an important agent in global warming has been found to have a root in resonances of such non-obvious normal modes due to Fermi resonances:

Worth a +1 just for the mention. Hope someone else is listening.

March 19

1 hour ago, KJW said:

Yes, pressure and temperature are statistical quantities. I never said that non-translational motion had no effect on each individual collision. But the universality of the ideal gas law is a testament to the statistical independence of pressure and temperature on non-translational motion.

This simply doesn't follow. The IDE is not universal, particularly for higher pressures, and says nothing of the nature of particla collisions. A bit of algebraic rearrangement gives:

C_VT = C_PT - PV which is equivalent to U = Q + W.

This demonstrates that it's simply a convenient reformulation of the First Law. Obviously, C_V and C_P include all available thermal degrees of freedom on an equivalent footing.

1 hour ago, KJW said:

Why are you saying that constant volume is a theoretical very special case?

All practical thermal processes involve bulk expansion or compression. A truly constant volume operation would be rather difficult to achieve in practice.

1 hour ago, KJW said:

Can you elaborate on this?

Gas processes operate within a spectrum that ranges between two idealised endpoints: the isothermal (PV = constant) and the isentropic (PV^k = constant) where k is the ratio of specific heats C_P/C_V. As a general rule, the faster a process occurs, the more nearly it approaches the isentropic endpoint. This is a strong indicator that all thermal degrees of freedom are immediately available to momentum exchange.

Everybody needs some kind of mental picture to help get their heads around such physical processes, and as you've pointed out, equipartition allows estimation of pressure and temperature from consideration of linear momentum alone. But to extrapolate from this an independence of rotational and vibrational modes is to confuse correlation with causation I think. At least it rattles the mental pictures others have which may be no bad thing, but you're going to get some kickback for that.

March 19

@KJW. I understand that you regard my last few posts as too much of a waste your precious time to be worth responding to. But I trust that in turn, you will not object to me regarding your silence, particularly wrt the non-conservation of linear momentum in general gas collisions, as concessions to the arguments presented. Live long and prosper.

March 18

2 hours ago, KJW said:

If pressure is independent of non-translational motion...

But it isn't. The 'boink' of each collision that creates the emergent property of pressure is in itself the vector sum of translational, rotational and vibrational momentum changes along the axis of point contact. The 'independence' you claim relies on the fact that statistically the rotational and vibrational transfers average out to zero over a sufficiently large number of collisions. It's a mathematical artefact.

Imagine being hit by a spinning dumbell. It makes a difference which end hits you.

March 15

21 minutes ago, KJW said:

One other thing: In the ideal gas law, pressure, which is proportional to temperature

Only in one very special case: that of a theoretical constant volume thermodynamic process.

More general cases show P, T dependence involving the ratio of specific heats, (~7/5 for diatomic gases) which per force requires consideration of all available thermal degrees of freedom, not just the three translational ones.

3 minutes ago, swansont said:

Seems to me there's no unambiguous argument in either direction, because the equipartition force the energy to be shared between the modes. I see nothing in literature making the distinction that it's only translational KE in systems with additional degrees of freedom. Nothing shows up in the equations, and IMO it's confusing to make that distinction when it doesn't show up in or matter to the calculation.

Exactly!

March 15

1 hour ago, KJW said:

But this conflicts with the equipartition theorem.

No it doesn't. It just conflicts with your rather unusual interpretation.

1 hour ago, KJW said:

The more degrees of freedom, the more energy. But this requires that only the ever-present three translational degrees of freedom contribute to the temperature.

As @exchemist, points out, your unusual interpretation collapses for cases where there are no translational degrees of freedom.

March 15

34 minutes ago, KJW said:

All degrees of freedom do contribute to the entropy.

Therefore all degrees of freedom contribute to equilibrium system temperature. This is the traditional thermodynamic definition of temperature.

March 15

49 minutes ago, KJW said:

I was actually thinking the same question. The answer I came up with was that vibrational motion within crystals is translational motion of individual molecules. The modes that do not correspond to temperature are rotations of individual molecules, vibrations...

T = dq_rev/dS (constant V, N)

What's going to your calculation of T if you decide to ignore some of the degrees of freedom that contribute to system entropy?

March 15

1 hour ago, martillo said:

Right. The temperature is related to the thermal energy of the system but is not a direct relation, they verify Stefan-Boltzmann Law.

And yet when we measure temperature gradients between systems at different temperatures, in non-extreme conditions they are generally linear in agreement with the dominant mechanism for transfer of heat being by momentum exchange.

If the dominant mechanism were EMR as you suggest, then the measured gradients would be highly non-linear (cubic in delta T I think).

1 hour ago, KJW said:

One thing should be mentioned: Only the translational modes of molecular motion contribute to the temperature.

I understand your POV but I think it's a misleading one. Particle collisions in gases that support rotational and vibrational modes only follow conservation of linear momentum on average. In the general case, some momentum is transferred via the other modes.

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