 Right, so we had discussed last time the Boltzmann equation, very crucial equation and I would like to talk today about some consequences of it, specifically a very important consequence namely the equilibrium distribution. We have now taken this for granted that in equilibrium for a dilute gas, classical gas, the distribution is Maxwellian in the velocity but now we would like to see how that is rigorously derived from the Boltzmann equation itself. There are also a couple of points regarding the equation itself regarding the assumption which I would like to reiterate although I have mentioned this earlier. So if you recall just to refresh your memory in the Boltzmann equation for the one particle density, number density F of r v possibly time because it is due to collisions and external forces, this is evolving in time. This quantity F of r v t under the assumption that the gases dilute and only binary collisions are included. The details of the collision have not been specified, they have been kind of encompassed in cross section, a differential cross section for which there is a separate technology to compute that. But given a cross section, a non-zero cross section and some arbitrary external slowly varying field, this F satisfies the following equation Delta F over Delta t plus v dot a gradient with respect to r F plus any external force divided by m dot gradient with respect to, let us say v1. So v1 here, this quantity here, this is equal to on the right hand side by definition the collision integral is equal to this quantity here and for this Boltzmann derived a formula okay. And that formula was this is a integral dv2 d3v2 integral d sigma which depends on which angle you are looking at times modulus of v v2 minus v1 okay times this factor yes times F2 prime F1 prime minus F2 F1 in this fashion where these are the same function F but under the following conditions. F1 is short hand for F of r v1 and t, F2 is short hand for F r v2 and t where for any given v1, any v1 that you specify on the left hand side v2 is such that it undergoes scattering. So here is v1 coming in, here is v2 and then v1 goes out, this particle goes out with a velocity v1 prime and that goes out with the velocity v2 prime and the scattering occurs inside here. So these quantities v2 v1 v2 prime v1 prime etc they refer to what happens before and after the scattering. So F1 prime is F of r v1 prime t and F2 prime is F of r v2 prime and t. So once you fix v1 whatever be v1 then you integrate all over all v2 with these weight factors such that v1 and v2 after the scattering process go to v1 prime v2 prime with all the conservation that this implies namely that v1 plus v2 vector is equal to v1 prime plus v2 prime the squares of all the sums of these two fellows is equal to the sums of those two etc etc. This is the relative velocity and we also know that this relative velocity does not change before and after scattering. So we know that this is identically equal to by conservation this is v2 prime minus v1 prime. I call this u last time and this is u prime. So it is a nonlinear equation with different arguments here except for the position argument which is the same different velocity arguments for the 4 quantities. These 4 have different velocity arguments in there okay. Now a crucial assumption that went here was that the distribution of the position is independent the position of a part molecule is independent of it is not correlated to its velocity instantaneous velocity that was crucial and it is called the assumption of molecular chaos that was an Anzart's which Boltzmann used in order to derive this set on this set of this equation here. So the idea is that the velocity and position are not related to each other. So recall that f of r v1 t times d3r d3 v1 this quantity has the physical significance of being the number of particles with at position r in a cell located at the position r with a velocity v1. This quantity here is the number density multiplied by the volume element in phase space and since this is appearing always in this pair wise form the assumption of molecular chaos is equivalent to saying that this quantity multiplied by f of r v2 t d3r d3 v2 this gives you the number of pairs of particles in a cell centered about the point r such that the velocity of one of them is v1 and the other one is v2. Yes if this were not true if it said this is where the dilute gas approximation went in if this were not true and it was dense and there were reconsolidations being considered all the time then you cannot have this completely independent of this. These density functions would not factorize because whatever happens you would say how many pairs are there such that one of them has v1 the other has v2 if they are correlated then it is not just a product of the density here times the density there that could be correlations they could be memory of collisions and so on okay. It is because you are avoiding you are ignoring that you are neglecting that that you are able to derive this formula at all okay this is crucial and of course now the question is can one prove this can one prove this from dynamics etc. A lot of effort has been expended in trying to find out whether Boltzmann he called it molecular chaos this is not although that is the literal translation of the phrase he used this is not quite the same as chaos in the modern dynamical system sense although it is related closely related to each other the gas in this room for instance is certainly chaotic it is badly chaotic it is the Lyapunov exponent namely the number which governs how quickly the errors get amplified how quickly initial conditions separate out as a function of time that is of the order of the number of degrees of freedom of the gas in the room itself. So you have a Lyapunov exponent which is itself of the order of 10 to the 23 is incredibly chaotic but that is of course if you now are speaking about the exact dynamical state of all the molecules in this room then if you have two of them I mean if you have two configurations which are infinitesimally close to each other it would essentially double this error would essentially double in a time which is the Lyapunov time the inverse of the Lyapunov exponent so it is just incredibly chaotic even a few particles you put in this room and they have scatter of each other the dynamical chaos takes over almost immediately it is not integral. So although this is related to it this chaos is not the chaos of the then yeah yes indeed so you got to now ask what is the correction to it after all if you write the two particle density then the leading term is this quantity and then the question is what is the correction so one can make a systematic density expansion for instance in the simplest instance if you take a gas a real gas you do what is called a virial expansion that will immediately tell you what are the corrections to this the leading approximation here plus there are various other ways of doing this there are cluster expansions and so on when we talk about non-ideal gases one essentially uses these facts there is a rigorous way of asking what are the if this were not true then what would be what would appear here on the right hand side in the collision integral would be the two particle distribution function and then if you wrote an equation for that what would appear would be a three particle distribution function and so on so you would have an infinite hierarchy so in principle you have an infinite hierarchy but the point I am making here is that even in the simplest instance when you have got when you neglected all such correlations you still have a terrible equation to solve you have a non-linear integral differential equation to solve in general so that is something I wanted to call your attention to namely this assumption is crucial to do whatever you have done to do this derivation now the question is having got the Boltzmann equation can we say something about what happens at t equal to infinity can we say something about the equilibrium distribution by definition in equilibrium distribution would be one which does not depend on time explicitly so you would like to have Delta f over Delta t to be 0 and then let us see what this distribution is now if you have an equilibrium distribution Delta f over Delta t must be equal to 0 alternatively as we have seen from all our previous experience you could ask if I take f of r comma v comma t and let t tend to infinity then I expect to get if at all an equilibrium exists I would expect to get an equilibrium distribution so let us impose that and for simplicity let us put the external field to be 0 for the moment we will come back and see what happens if you apply an external force on the system so once this is gone the external force what breaks translation in variance in the system is an external force otherwise in the thermodynamic limit you have translation in variance so we can assume f if there is an equilibrium distribution to be independent not only of time in this fashion but also of r so I expect that the equilibrium distribution if such a distribution exists would be independent of r. So what happens then in this equation this term is not there f equilibrium let me call the solution to this equation Delta f not so the standard notation for an equilibrium distribution is f not or f equilibrium so f not is a function of v alone no t dependence no r dependence then this term goes away because there is gradient term on r acting on f is 0 of course and there is no external force so this is 0 so this would imply that f not satisfies this term equal to 0 the collision term to be equal to 0 so it must satisfy integral d3 v2 so f not of v1 must satisfy integral d3 v2 integral d sigma of omega mod v1 minus v2 times let us write this out explicitly so this is f not of v2 prime f not of v1 prime minus f not of v2 f not equal to 0 because that is Delta f not of v1 over Delta t and that is 0 so for every v1 this function must satisfy this equation here whatever be the scattering whatever be the scattering now of course we can write one solution down by inspection and that is to say if this quantity is 0 this bracket is 0 where v1 v2 v1 prime v2 primer related by scattering event nonzero cross section with a nonzero cross section otherwise this one would vanish so there is genuine scattering if the cross section is not 0 there is really collisions and some scattering then if this bracket vanishes then that is a solution and then we should now further examine to see what sort of solution it possibly is so let us write that out so the sufficiency condition for f not of v to be to be the equilibrium solution is the following f not of v2 prime f not of v1 prime minus f not of v2 f not of v1 equal to 0 where v1 v2 v1 prime v2 prime are related by a scattering nonzero scattering cross section so that is always at the back of our mind so let me draw that picture where these fellows are in this fashion and then we have to follow this up and see okay if this is so what is the consequences can we guess what the function or can we solve for the functional form of this f given all this information about scattering that is one possibility okay but notice the interesting thing if this is so if this is so there is no reference here to the actual details of the scattering it only says the scattering cross section should not be 0 so we do not care what sort of scattering cross section it is but you would still get that to be so so this of course has a very important consequence that whatever be this equilibrium distribution it cannot depend on the details of this actual scattering so that is the reason why different gases with different molecular interactions have exactly the same Maxwellian distribution we know that already by hand side but the reason for it is very clear here that this thing is independent of the scattering actual details of the scattering okay as long as the scattering is nonzero that is all that is required so that is one important lesson but it does not give us get us very far unless we find out if this is also a necessary condition we proved it sufficient that by inspection if this is so the integrand vanishes then delta f over delta t is automatically 0 that is the end of it but is it necessary or not to do that Boltzmann used a very clever trick okay and this leads us to the Boltzmann H theorem so let me write that so you define the quantity H of t to be integral d3 b1 f of b1 and t log f of b1 and t where f obeys the Boltzmann equation so define this where f of b1 t satisfied with satisfies and look at dh over dt it is integral d3 b1 the derivative of this quantity here but that is a function of v and t so the derivative would be delta over delta t plus v dot grad but this is independent of r this thing here does not have any r dependence okay therefore the b dot the convective derivative part is 0 and you just have delta f over delta t inside so delta f over delta t of v1 and t that is this term there is a log there the next term has 1 over f times delta f over delta t cancels this so you have this multiplied by 1 plus f of b1 log so that is dh over dt the next step that you took was to show that this dh over dt is not positive it cannot be positive it is less than equal to 0 okay but that proceeds in several steps to show this the first step is to note that if this fellow is 0 if this is an equilibrium distribution and this one vanishes because that is the equilibrium distribution then this is of course 0 so it is necessary that this be 0 in order that delta f by delta t be 0 it is a necessary condition if delta f over delta t is 0 identically it means that it cannot have explicit time dependence and is therefore an equilibrium distribution the statement is if this is 0 it implies dh over dt is 0 right so you cannot have this to be 0 without having dh by dt to be 0 therefore it is a necessary condition that dh by dt is 0 in order for this to be 0 okay so what I am trying to do now is to show sufficiency condition I am trying to show that this condition is sufficient for f to be an equilibrium distribution we know it is I am sorry we are trying to show it is necessary we have already shown that it is sufficient by inspection so you have to appreciate the subtlety of this argument you want to show that f0 of v is an equilibrium distribution if and only if this quantity is 0 so that is necessary and sufficient the sufficiency is by inspection we saw that the equation for delta f over delta t delta f not over delta t was a big integral the collision integral and if the integrand in this vanishes identically then of course the left hand side is 0 so that was a sufficient condition now we are trying to show that it is necessary and that proceeds by saying that define an h of t to be in this fashion this is time dependent by the way out here where this fellow satisfies Boltzmann equation with the collision integral now the statement is the time derivative of this is this quantity here and the observation is that you cannot have this to be 0 without having this to be 0 it is necessary because there is no way this can vanish without this being 0 if this is 0 then this vanishes yeah but there is no it is necessary without this being 0 how can that be 0 suppose this is not 0 there is no way this can be identically 0 why not it is necessary for this to be 0 it is necessary that dhp dt be 0 necessarily I had the same problem every time you look at the h theorem one has this little thing but it is exactly I mean look at the contrary suppose this is not 0 there is no way this can be identically 0 it is got to be 0 in order for this to be identically 0 it is necessary we are still not yet established that that condition is a necessary condition we have only established that dh by dt must be 0 is a necessary condition for this to vanish because if it is not 0 there is no way this can be identically 0 okay so now let us see what that implies we have already said that f of v1 t satisfies the Boltzmann equation so I am going to put that in here in this so this means let us write it out this means that for delta f by delta t I am going to put in the collision integral okay so it says integral d3 v1 integral d3 v2 this is already there this integral in the definition of dh over dt times delta f over delta t and that is equal to integral d sigma of omega mod v2 minus v1 times f2 prime f1 prime minus f1 prime f2 prime actually it is a little back sorry no prime f1 f2 f1 a little bit of a messed up notation you see once you put a 0 here there is nothing to do with 1 and 2 the velocities so I am stuck for notation these are all f0's of the f I should really write f superscript equilibrium or something like that okay it is too lazy to do that so this quantity that was this portion I hope they have not left out anything in it times 1 plus log f this is v1 so this is f1 equal to 0 this is now a necessary condition because this equal to 0 was a necessary condition I substitute from the Boltzmann equation for this quantity and that whole this double integral being equal to 0 is a necessary condition for this thing to be an equilibrium distribution with argument v1 okay it is the same function everywhere only the argument changes but now in any scattering event you can interchange v1 and v2 and no physics changes because what appears here is v2 minus v1 modulus and the scattering event does not change under the event v1 v2 the same scattering cross section appears so I do the same thing with an interchange of 1 and 2 v1 and v2 this remains the same this remains the same this remains the same because you got an integral over both right therefore this condition can be rewritten with v1 and v2 interchanged and it would look exactly the same except out here this would become f2 right so I add the 2 and take the average so it says 1 half and this becomes a 2 plus log f1 f2 equal to 0 1 plus log f1 plus 1 plus log f2 that is 2 plus log f1 f2 now let us interchange v1 v2 with v1 prime v2 prime because our scattering process said you have v1 v2 under scattering went to v1 prime but you know because this whole thing is time reversible you can interchange the final with the initial and you get exactly the same scattering cross section we saw that we asserted that the last time now what happens when you do that when you interchange the initial and final states this becomes 1 half integral v3 v2 prime v1 prime v2 prime integral d sigma prime of omega that is the reverse process and then v2 prime minus v1 prime v2 prime and what goes on here is now f2 f1 minus f2 prime f1 prime because this is the final state now and that is the initial state times 2 plus log f2 prime f1 prime and this must be equal to 0 but then we already saw that under the scattering process d3 v1 d3 v2 is equal to d3 v1 prime d3 v2 prime if you go to center of mass coordinates then this becomes d3 capital V d3 little u and that is the same as d3 capital V prime which is the same as v times d3 u prime which in turn is equal to this so I can actually erase this and write it in this fashion it is the same measure and the scattering is the same exactly the same scattering cross section so sigma prime is the same as sigma we also know the relative velocity does not change we know that v v1 we know that v2 minus v1 equal to u v2 prime minus v1 prime equal to u prime and we know that mod u prime u equal to mod u prime all that happens is it is rotated by the scattering angle so this 2 you can remove the prime so now this is 0 that is 0 we have done all the manipulations we need so the average of the 2 must be 0 right so that says it is one quarter there is a minus sign here there is a minus sign so I can get rid of that minus sign by making this f f2 prime f1 prime f1 f2 and then subtracting instead of adding so let me write it as this so this is the so what goes on inside here is f1 f2 minus 2 minus log f1 prime f2 prime equal to 0 so I kept the plus sign here but I subtracted this bracket and the 2 cancels out and I can rewrite this as right I can rewrite this as log f1 f2 divided by f1 prime f2 prime so if you permit write this as log f f2 f1 over f2 prime f1 times this quantity and this is equal to 0 the 1 4th is irrelevant this is non-negative this is non-negative the only way this thing can be 0 is for the integrand to vanish okay now call this part call it x and call it y then you got a term it says x minus y log y over x that is this integrand here that quantity cannot be positive where x and y are real this quantity cannot be positive because if x is bigger than y this is positive but this is log of a number less than 1 and that is negative if x is smaller than y this fellow is negative that is the number bigger than 1 that is positive but in any case this quantity cannot be negative it cannot be positive it is less than equal to 0 all equality if and only if x equal to 1 is no other way we have proved more than what we need we have actually shown that in general it is less than equal to 0 and we will see what the less than implies we will see what it does because when you are out of equilibrium that is going to play a crucial role yeah it is a convexity property definitely yeah I do not know what you call it what is the inequality called some name for it somewhere yeah it is a general property and this is yes there is more than one place where this inequality appears this thing here okay but whatever it is in our context it says this fellow must be equal to that x must be equal to y that is the only way this integral can vanish because everything else is positive this quantity this quantity etc so it is a necessary condition this is this thing or this quantity this equal to 0 implies this equal to 0 we have shown that we already saw that is a necessary sufficient condition now we are saying it is a necessary condition for equilibrium therefore the equilibrium density is given by this equation or rather equilibrium density is such that if and only if this condition is satisfied do you have equilibrium which is in yeah very much so very much so exactly exactly yeah h is like an entropy function f log f kind of structure okay but we are completely out of equilibrium in this situation because in general you define an h of t it is a time dependent quantity the only assumption we made is that there is no dependence on the position that was important now we got to put that back we got to put that back but as long as f is a function of v alone any equilibrium density has to satisfy this right so once we have that in place let us examine what this implies so this condition is needed for an equilibrium distribution here which means that log therefore for the equilibrium distribution f not of v you must have f not log f not of v 1 plus log f not of v 2 must be equal to log f not of v 1 prime plus log f not of v 2 where v 1 v 2 v 1 prime v 2 prime are related by a scattering event but these 2 things here are the initial state before scattering and this is final after the scattering but this is before scattering so it says in a scattering process arbitrary scattering process with velocity v 1 v 2 initially and v 1 prime v 2 prime initially f not which is happens to be the equilibrium distribution is such a function that the log of this f not is conserved for these 4 part velocities is conserved therefore log f not of v must be some constant of the motion which is conserved which is conserved depend on some quantity which is conserved under this what are the only quantities that are conserved momentum is conserved and energy is conserved so therefore this log f not of v can at best be a quadratic function of v so it immediately implies we are not home yet but getting there so it implies log f not of v and now let us forget all the subscripts must be equal to something of the form a v square plus b v possibly plus constant but it cannot be a thing like v it is clear that this is not a scalar because the energy is the only constant of the motion but some of the energies yeah some of the energies is constant kinetic energy is add up so no other function is possible it cannot stick out like this because this is a vector yeah so you can actually say this must depend on some arbitrary initial velocity so b v dot v not where capital A B C are constants independent of v depends on the mass of the molecules depends on whatever we do not know yet etc. Now this of course with the little effort one can show this will imply that the whole gas is moving with the velocity v not there is no other interpretation you can give this v not so this part of it I am slurring over but if you look at the elementary treatments on statistical mechanics and kinetic theory they will tell you that so this b goes away and you are left with this now of course f not is the exponential of this guy so the c becomes multiplicative it is the overall normalization constant and you can fix it by remembering that we needed this to start with we needed integral d 3 r f of r v t to be equal to be wanted d 3 r integral d 3 v f of r v t to be equal to the total number of particles and in the case where it did not depend on r you can do the r integration get a volume and therefore this is equal to n over equal to the number density n so you have a normalization which depends on the number density pardon me this is very important you want this to be normalizable right so if you do this this will imply it goes like e to the a v squared and if a is positive it cannot be normalized this would not be a finite integral in any case it is a Gaussian so if you try to normalize it you will get a 1 over square root of whatever it is and that is only valid if a is positive now you need 2 pieces of information you need to pardon me a is negative so a is called negative so there are 2 pieces of because you are going to use finally the fact that e to the minus a v squared integral this is going to exist and it is going to be some square root of pi over a or something like that now there are 2 pieces of information that you need to fix these constants and here is where the physics has to come in you have to say something about the average energy if you prescribe that for each particle then I find the moment of this thing multiplied by v squared that will be 1 and a normalization which is the density which will be the other so with these 2 fellows you get precisely the Maxwellian distribution but you have to put in 2 pieces of information after all we are doing kinetic theory and you have no other piece of information except what you know from thermodynamics so you say it is an ideal gas I know the equation of state I know the internal energy is 2 thirds the internal energy density is the pressure and I know P is equal to RT, P v equal to RT so I put that in if I put that piece of information in then purely from kinetic theory you get the Maxwellian distribution as the equilibrium distribution so it was a long route but the H theorem is a crucial one on the way now you could ask what happens if I put an external statement of the H theorem is that if f satisfies f of v, t satisfies the Boltzmann equation and you construct d3 v f of v, t log f of v, t you call that h of t then dh over dt is less than equal to 0 that is the statement of the theorem so this is the H theorem. Now finally what happens if you put in an external force on the system well the claim is the following not surprisingly and we can see this by verification the claim is in that case f0 of r, v if you put in an external force f of r so external force equal to minus gradient of some potential I do not want to use v because I used it for the relative velocity so for the total velocity so let us say it is derived from a scalar potential it is important that it should be this thing should be independent of the velocity this is crucial for a reason which you will see in a minute that this force should not depend on the velocity there is no velocity dependent force then this quantity equal to f0 of v where f0 of v is the distribution we just been talking about e to the minus phi of r over k Boltzmann is the equilibrium distribution okay how are we going to check this out because you are going to have delta f over delta t plus v dot gradient with respect to r f let us write the f outside plus f over m dot gradient with respect to v f of r comma v f0 this is equal to whatever the collision integral is and we went through this argument by saying that if it is an equilibrium distribution this guy is 0 and you had the collision integral on the right hand side now in that integral if instead of f0 of v you put in this fellow here then I assert that you can pull this factor out all the way out of the integral because there is no factor there that depends on the position at all so it comes right out and it is going to sit there and that collision part of it was 0 with f0 of v was 0 so that means this portion alone must be 0 otherwise it is not going to this answer is not going to satisfy it right. What I am saying is that when you have a force which is depend derived from a potential scalar potential of this kind the equilibrium density distribution is just the original one which we found without the external force multiplied by this factor there is some normalization suitable d3r of this should be equal to 1 okay. Now let us the way to argue is well if I put that in and this is equilibrium this term is 0 I put this whole business into this collision integral this factor comes right out of all the integrals because it does not scattering cross section does not depend on it etc etc but whatever is left is already been shown to be 0 because f0 of v satisfies that equation therefore this must be true you have to check this now with this solution but that is a trivial matter because if you pull this delta gradient with respect to r you are going to keep the same exponential but bring grad 5 downstairs that is going to be v dot f and this is going to act on a Gaussian in v so when this fellow acts on it you are going to get a v downstairs minus whatever and this f is grad 5 so this term will cancel that term and therefore this is really an equilibrium density suitably normalized check this piece of check this algebra no f is fine as it stands that is okay it is okay yeah but there is always this confusion as to what this phi is okay it is the potential energy per unit mass they are so used to saying it is the potential energy potential energy per unit mass okay this is the force per unit mass if you like so you have to put in some m factor somewhere so it is like putting in these 2 pi factors or whatever so I leave it to you to figure this out I mean the whole physics of the essential point is that to see how this is going to cancel out at all this is going to pull out gradient is going to pull out a grad 5 times an exponential but that is f it is already sitting here and this is going to act on a v square and pull out a grad v v dot f and they cancel we already know I mean now we know this is consistent with you know after all this fellow here was of the form this was of the form e to the minus one half m v square over k Boltzmann t so this whole thing is of the form e to the minus beta times the Hamilton you know that yeah because we said this is a necessary and sufficient condition that this vanish and once that vanishes it says this f not of v log f not of v can only depend on whatever is conserved in the collision process and that is it so of course once you put in correlations then matter becomes very different altogether so this is the first elementary consequence that this equilibrium density turns out to be what we know from the canonical ensemble but this of course now derived purely without making the fundamental postulate of statistical mechanics but in some sense it has been put in by saying there is the molecular chaos is on that but that is a different kind of statement altogether it is not immediately obvious these two are the same purely kinetic theory argument here this was Boltzmann's big contribution so let me stop here.