# Is it sufficient to only consider kinetic energies of molecules for pressure?

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?

• 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. Nov 12, 2021 at 12:03
• @ACuriousMind Isn't the collision a consequence of a potential? en.wikipedia.org/wiki/Hard_spheres Nov 12, 2021 at 12:10
• 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. Nov 12, 2021 at 12:12
• 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. Nov 12, 2021 at 12:16
• 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. Nov 12, 2021 at 12:20

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.