Among the assumptions required for the applicability of Maxwell-Boltzmann statistics, there is one that assumes the particles to be evenly distributed in the volume $V$ they occupy. Suppose now that they have within the same volume $V$ a density distribution $\rho(\mathbf{x}, t)$ instead. Is there a way to adapt the Maxwell-Boltzmann distribution of velocities $$ {\displaystyle f(v)~d^{3}v=\left({\frac {m}{2\pi kT}}\right)^{3/2}\,e^{-{\frac {mv^{2}}{2kT}}}~d^{3}v}? $$ to this specific case?
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$\begingroup$ I think we can not use Maxwell-Boltzmann distribution of velocities to this specific case for any density distribution $\rho(x,t)$. There must be some reason (interactions or constraints) to make the system have the density distribution $\rho(x,t)$ rather than uniform distribution. So we should deriv the new distribution from the principle of maximum entropy with the constraints of density distribution $\rho(x,t)$. It may be helpful to assume one concrete form of density distribution and get the corresponding distribution of velocities. $\endgroup$– lbyshareCommented Jan 1, 2023 at 11:32
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3 Answers
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Maxwell-Boltzmann distribution (and a big part of what is studied in introductory statistical mechanics) refers to a gas in equilibrium. Maxwell distribution specifically deals with the case where no external potential is present, in which case the equilibrium distribution of gas has uniform density. Boltzmann generalizes this to the case when external potential is present, and the density might be non-uniform in this case, $$W(\mathbf{p},\mathbf{x})\propto e^{-\frac{H(\mathbf{p},\mathbf{x})}{k_BT}}.$$
However, general non-unform density is not necessarily in thermal equilibrium and thus not necessarily obeys MB. In this case one needs to use non-equilibrium statistical mechanics (kinetic equation, etc.) - although this one also often uses an assumption of local equilibrium.
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$\begingroup$ What If I have a fixed volume $V$ made of an (initial) unevenly distributed liquid substance in water at a constant temperature $T$? $\endgroup$– ric.sanCommented Jan 3, 2023 at 14:34
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$\begingroup$ @ric.san Generally, the water will flow, the entropy will grow, until it settles in equilibrium. The fact that (local) temperature is the same everywhere does not mean that the system is in equilibrium. $\endgroup$– Roger V.Commented Jan 3, 2023 at 14:46
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Yes, you can adapt the Maxwell-Boltzmann distribution to a case of non uniform spatial distribution of particles as long as the process is isothermal. This means that mixing should have no effect on temperature. This condition is satisfied for ideal gases and also for ideal solutions. In both cases if components start with arbitrary spatial distribution but at the same temperature, temperature will be maintained during mixing and both components will satisfy the same MB distribution.
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$\begingroup$ Thanks, that is what I was hoping to hear. Can you provide some references that could support what you said? $\endgroup$– ric.sanCommented Jan 3, 2023 at 16:47
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$\begingroup$ I don't have a reference but the proof is simple: imagine that the particles are "black" and "white" but otherwise identical. To a color-blind observer the system is in in equilibrium and obeys the MB distribution. To a color-able observer the particles are distributed non uniformly in space and still obey the MB distribution. $\endgroup$– ThemisCommented Jan 4, 2023 at 11:52
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What is the point of evenly distributed or not, when result is in statistical form then there is variation. This distribution although assume thermal equilibirium to have only one variable as speed or mechanical energy, otherwise this distribution itself meant there is no thermal equilibirium. At equilibirium there is no distribution. But it is hard to have equilibirium if density is more or source has low power.
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$\begingroup$ You should edit your answer because in its present form it makes no grammatical or physical sense. $\endgroup$– ThemisCommented Jan 4, 2023 at 11:56
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$\begingroup$ @Themis I think your theory must first have idea of equilibirium. What is equilibirium of particles in closed system, when all particles have that energy. Can you compute total energy with above distribution, what is it. Is it $NkT$, now where is distribution when all particles gained that energy. $\endgroup$ Commented Jan 4, 2023 at 15:23