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So I've been reading this answer.

The question boils down to how we calculate pressure on microscopic level in absence of a rigid boundary. In statistical physics pressure by a gas on a boundary is calculated from the change of momentum of atoms/molecules reflected by the boundary. In absence of a boundary one could suggest scattering against other molecules - but then, what is the difference between mascroscopic (bouyancy, convection) microscopic (diffusion, heat conduction) behavior of a gas?

It seems to me that one can start from BBGKY hierarchy and derive all properties pertaining to a gas?

How does one derive that the equation of motion in the continuum limit should be:

$$ \vec F = P \vec A$$

starting from BBGKY hierarchy?

More Anonymous
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  • I think, $\vec F = P\vec A$ is itself the definition of pressure. How can you derive it from any other theory? – SCh Jan 05 '23 at 05:27
  • @SCh I can define anything but then the question becomes why should my theory respect this definition. BBGKY hierarchy starts of from kinetic theory and aims to do this. – More Anonymous Jan 05 '23 at 06:57
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    I tend to agree, we would probably view F=PA as the definition of P. We then have to show that other definitions, for example P=dS/dV_{E,N} (thermodynamics), P=-T/V log Z (stat mech), P\delta_{ij}=T_{ij} (GR) are equivalent to it. This is discussed in various text books, but I don't quite see how it is related to BBGKY – Thomas Jan 05 '23 at 16:33
  • @thomas there are 2 ways to derive Euler's equations. 1. From using BBGKY hierarchy and 2. From using $F=PA$. I don't think then it's a stretch to think BBGKY can also derive it? – More Anonymous Jan 06 '23 at 00:24
  • Well, if BBGKY implies Euler, and Euler implies F=PA, then BBGKY implies F=PA, right? – Thomas Jan 06 '23 at 02:43
  • No $F=PA$ implies Euler not sure about the other way round. – More Anonymous Jan 06 '23 at 03:05
  • Seems pretty clear, just apply F=ma to a volume element with a free surface. But again, I don't think that BBGKY has anything to do with this. What is doing the "heavy lifting" here is the equivalence of different definitions of P. – Thomas Jan 06 '23 at 18:34
  • Would you recommend I rephrase the question and ask for this equivalence? – More Anonymous Jan 06 '23 at 21:51

1 Answers1

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According to your link $\mathbf{F}_{i}=-\frac{\partial \Phi_{i}^{ext}}{\partial q_{i}}$

the interaction potential $\Phi_{i,j} $ is assumed to be zero.

we know that: $F=-\frac{\partial E}{\partial r}\;\;\;(1)$

taking the average over this equality, we find: $\bar{F}=-\frac{\partial E}{\partial V}dS\;\;\;\;(2)$

so $\;\;P=-\frac{\partial E}{\partial V}\;\;\;\;(3)$, Pascal's law and: $F=PdS $

(1),(2),(3) L.Landau, E.Lifchitz statistical physics first part,volume V.

The Tiler
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  • I've awarded you the bounty but I am still dissatisfied, Because 1. This assumes the collision duration to be $0$. 2. Usually I would have to dot product $F$ with $dS$. I see no $\cos \theta$. $\bar F = \int F \cdot dS / dS = F \cos \theta$ (using Rolle's theorem) – More Anonymous Jan 09 '23 at 03:02
  • "We average over a set that satisfies the instantaneous external constraints", see page 108,109 https://www.physique.usherbrooke.ca/tremblay/cours/PHQ-340/stat-340.pdf, – The Tiler Jan 09 '23 at 09:59