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So in this answer I learned one can derive pressure simply as a proportionality constant by only considering the kinetic energy terms. However, in this answer I learned one cannot get the pressure we know and love by only considering the kinetic energy terms.

How can both answers be correct? (What am I missing)? Is it sufficient to only consider kinetic energies of molecules for pressure?

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  • $\begingroup$ You have summarized the answers a bit too succinctly here - the second answer isn't saying "you have to consider non-kinetic terms to get pressure", it's saying that in order to understand the behaviour of a gas of particles with collision you have to think differently from trying to understand the behaviour of a photon gas - note that it's not saying you should consider something other than the particles' movement to compute the pressure. I don't see a contradiction here. $\endgroup$
    – ACuriousMind
    Nov 12, 2021 at 12:03
  • $\begingroup$ @ACuriousMind Isn't the collision a consequence of a potential? en.wikipedia.org/wiki/Hard_spheres $\endgroup$ Nov 12, 2021 at 12:10
  • $\begingroup$ Yes, but neither of the answers uses that potential for anything more than saying "the particles collide and so spread everywhere through the container". The first answer isn't even about any box, it just computes a pressure term across (mathematical) surfaces. I don't know what you want an answer to say here. $\endgroup$
    – ACuriousMind
    Nov 12, 2021 at 12:12
  • $\begingroup$ I still think it's a fair question. To have any kind of collision one needs a potential energy term. The second answer says one cannot model pressure by thinking of them as collisionless partciles. The first answer does that. $\endgroup$ Nov 12, 2021 at 12:16
  • $\begingroup$ The two answers are considering two entirely different situations. The first has "pressure" as a component of the stress-energy tensor and does not consider any containers at all. The second has "pressure" as the actual pressure in the Newtonian sense of a real-world gas on the sides of a container. Words can mean different things in different contexts, this is not a contradiction. $\endgroup$
    – ACuriousMind
    Nov 12, 2021 at 12:20

2 Answers 2

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The question is interesting because it touches on issues related to the origin of pressure in a fluid.

In the case of the first answer, the context was about the stress-energy tensor of a perfect fluid in General Relativity (GR). One must remember that GR is a macroscopic theory, and the expression of the stress-energy tensor contains quantities relevant for the description of fluid at the Hydrodynamic level. For a perfect fluid, the stress-energy tensor field's independent variables depend on are the local hydrodynamical velocity and the local pressure. Notice that these quantities are not about individual molecules but a small fluid elements. It must be small enough to represent the local behavior, but large enough to justify neglecting microscopic fluctuations. At this level, the interaction between molecules does not appear explicitly but only through the equation of state connecting pressure to the local density and temperature. Interactions between different fluid particles are limited to the surface interaction associated with pressure differences.

A completely different picture is necessary at the microscopic level. When the relevant degrees of freedom are not lagrangian fluid particles but single atoms or molecules, it is possible to introduce a microscopic stress tensor (whose macroscopic average will result in its macroscopic counterpart), but its ingredients are different. There is a microscopic kinetic term, containing the sum of the tensor products, ${\bf v}_\alpha \otimes {\bf v}_\alpha$, of the velocity of each particle, and a term, containing the tensor products, ${\bf f}_\alpha \otimes {\bf r}_\alpha$, of the internal force on the particle $\alpha$ and its position. This last term contains information about the interactions that are not included in the kinetic term.

In the case of the second answer, implicitly, the level of discussion shifted to the microscopic/molecular level.

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  • $\begingroup$ Doesn't the dirac delta function represent "the stress-energy tensor of a discrete set of particles" in the first answer? (see the question therein) $\endgroup$ Nov 12, 2021 at 16:37
  • $\begingroup$ @MoreAnonymous It is a specific model deriving the macroscopic stress-energy tensor from the hypothesis of a perfect gas microscopic model. However, it should be taken as an example (to allow simple calculations). In general, the perfect fluid approximation is not directly connected to non-interacting Hamiltonians. It is more related to specific hydrodynamic regimes. $\endgroup$ Nov 12, 2021 at 21:55
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To add just little: the gas law that relates kinetic energy to gas pressure is a very good approximation for relatively low pressures and temperatures, when no phase changes are afoot. In practical terms this means that the higher-order corrections to those gas laws that take into account things like relativistic effects can safely be ignored in our everyday ("engineering") business.

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