 We are going to take a look today at some consequences of the Boltzmann transport equation. If you recall we made one fundamental assumption in writing the Boltzmann equation down and that was the assumption of molecular chaos. In other words, we assumed that in a cell in mu space, in single particle phase space, the probability that you have a particle at the position r with a velocity v at time t with some velocity v1 at time t, simultaneously with the probability that you have another particle in the same cell with a different velocity v2 was simply the product of the density. So this was multiplied by the corresponding volume elements in mu space d mu1 and d mu2. So this was the assumption of molecular chaos assumption that together with binary collisions led to the Boltzmann equation, the transport equation. So this plus binary collisions, elastic collisions in which you did not lose a particle, nothing got absorbed. This led to the Boltzmann equation, Boltzmann transport equation. In kinetic theory it is usual to call the Boltzmann equation the Boltzmann transport equation because as we will see very briefly today you can extract transport coefficients from the Boltzmann equation such as the diffusion coefficient, the coefficient of thermal conductivity, electrical conductivity and so on, viscosity etc. So they can all be computed for whatever system you are looking at in this case a dilute gas from the Boltzmann equation with a very systematic procedure. The catch was that the Boltzmann equation was a nonlinear integral differential equation in this function F. So that is what made it practically impossible to solve analytically and then there is a huge literature which over the last 100 plus years has been devoted to trying to develop systematic approximation procedures for going further and further towards the exact solution of the Boltzmann equation. Now we are going to look at a very very very oversimplified version of it today very briefly but I would like to go over again since it is so important to the foundations of statistical physics I would like to go over again some of the key steps that we went through in deriving the Boltzmann equation. I am not going to give the derivation again, I am just going to call attention to what are the features of this equation. So if you recall the equation itself said that delta F over delta t for a particle with velocity v1 say some arbitrary velocity v1 so by F I mean F of r v1 and t plus v1 dot del in position space del r times this F plus if you had an external force of some kind then that force F over m dot gradient in velocity space on F. This is a differential operator acting on the v dependence of F. This was equal to the famous collision integral which is delta F over delta t. I use delta F over delta t just to define this collision integral here through this equation here. This has got the dimensions of F divided by time which is why I call it this but very often in textbooks you would see it is called the collision integral that is it and the separate symbol is used for it all together. And this collision integral under the assumption of binary collisions that was equal to this equation here was equal to an integral d3 v2 for our scattering geometry integral d sigma is dependent on omega and now let me explicitly write its dependence also on the relative velocity the magnitude of the relative velocity u times u times F we have a strange notation F2 prime F1 prime minus F2 F1 the standard notation where the scattering geometry just to recall to you once again was v1 and v2 coming in the initial states and then going out with velocities v1 prime and v2 prime. And for instance F2 prime was identically F of r v2 prime t etc. So the 4 velocities involved in the collision v1 v2 v3 and oh yes and of course this quantity u was mod v2 minus v1 which is also equal to mod v2 prime minus v1 prime. This scattering cross section is dependent on where you are looking with respect to the incident direction v1 and this quantity here u which is the relative velocity. So this equation as we can see here is a non-linear integral differential equation for this unknown function F that is what creates the problem. Of course we know that certain quantities are conserved in this collision we know that the total momentum of the system is conserved. We know that the total energy kinetic energy is conserved and those will have to be put in. We also know that the total particle number is conserved. So no particles are going out anywhere in this gas they are all in there. So at any given time if you integrate this F the normalization was integral d3r integral d3v which is what I call d mu in mu space this multiplied by F of r v t at any given time this is equal to the number density the number of particles per unit volume. Now this F has two interchangeable meanings under the assumptions we have made large numbers and so on. The F multiplied by the volume element in phase space is either the total number of particles in that per unit volume in that space or it is also the probability that the given particle has the position r in the cell whatever is concerned the center at the point r with the velocity v at time t these two are interchangeably used. Now this equation is satisfied by those F's for which they are some for those states of the gas for which the assumption of molecular chaos is valid. It is not saying that this assumption is valid at all times. It is not saying that at all. It is just saying if the gas instantaneously happens to be a state in a state in which this assumption is valid then the gas obeys the Boltzmann equation okay and we found out that Boltzmann introduced and he proved this necessary and sufficient condition for equilibrium by very ingenious argument he introduced the H function and we discovered that DH over DT was not positive was either equal to 0 or less than 0 at any given time. What it actually means is that if at some instant of time the gas obeys the Boltzmann equation and obeys therefore the assumption of molecular chaos then an infinitesimal instant later so if at time t it obeys the equation there is molecular chaos then at t plus epsilon where epsilon goes to 0 from the positive side you are guaranteed DH over DT is negative. The instantaneous slope will be negative at that point that is what you are guaranteed okay. You are not guaranteed that the gas for any means satisfies the assumption of molecular chaos at all times there are fluctuations of course. Now the H function also tells you something about those fluctuations so we will discuss that but before that let me remind you of what equilibrium meant what was the meaning of equilibrium we discovered that equilibrium distribution of course means that this quantity must be 0 there is no explicit time dependence so that was achieved by saying that if F of R V2 prime so let me write it out if this quantity V2 and it is an equilibrium distribution so the time derivative is 0 that means this whole thing on the right hand side if it is independent of R or even if it depends on it even if there is an external force we do not care it put a specific condition on when the gas is in equilibrium and the condition was that in the state of thermal equilibrium the distribution was such that this bracket vanished identically. So essentially we had a statement that F2 prime F1 prime was equal to F1 F2 F1 does not matter what order where 1 2 and 1 prime 2 prime are related by this scattering geometry. Now what does it remind you of this thing if you like for the 2 particle system is like an initial state just like a final state so it is really saying that if you give me a V1 then all those V2's such that this condition is satisfied on this side the a priori probability of having V1 is represented by F1 this is what causes the transition to the final state because it comes on scatters on the right hand side this would have been the a priori probability and that is the scattering what causes the scattering. So it is very reminiscent of what is happening in the theory of Markov processes where you have the assumption of detailed balance in a Markov chain. So if you recall when you write a master equation for a Markov chain if you have states which are labeled by i, j, k, etc, etc in a Markov chain and you say that p sub i is the a priori probability that the system is in the state i then p i multiplied by the transition probability for i to go to some j and that is equal to w j i if this is equal to on this side w i j t j then you have the condition of detailed balance and you have an equilibrium distribution of course this is a sufficient condition for you to have equilibrium distribution because Markov master equation says d p over d t if I write all these p's as a column vector is given on the right hand side by some capital W some matrix transition matrix times p and the elements of it are given by these transition rates. So since this appears inside the summation if each term in the sum is 0 then of course the whole thing is the sum is 0 and you have a stationary distribution okay. So this assumption is called detailed balance because each pair of states satisfies this condition here and there is no reason why that should be the case for equilibrium but if it is so then this state then you have an equilibrium distribution. So that is what is happening here this is like a detailed balance condition and now in this situation in this physical system this dilute gas what the Boltzmann equation what Boltzmann's proof of necessary and sufficient condition for equilibrium being this says is that you automatically have detailed balance that as t tends to infinity any distribution you start with will tend to unique equilibrium distribution which satisfies detail this detailed balance condition. There is no Markov chain here there is no such thing here but it is detailed balance all the same. So it says so it is what is trying telling us of course is that why did this come about where this factor these factors come from they came from the scattering process. So as long as the scattering is time reversal invariant as long as the dynamics is time reversal invariant you are going to have this situation. So that is really what it is telling us saying something about time reversal that it is a time reversal invariant system for these binary and of course the binary collision assumption. So if you recall how did what Boltzmann do to prove a necessary and sufficient condition well we went through this proof we said first of all if it is this distribution in equilibrium is independent there is no external force this part is 0 it is uniform distribution this part is 0 and then the fact that you have an equilibrium distribution means that this must vanish and that vanishes implies that this integral must vanish here and a sufficient condition for the integral to vanish is for the integrand to vanish that was the first part of the proof. The second part of the proof showed that it was also necessary by saying that what appears in the H function was the Boltzmann H function by the way the H is not the Hamiltonian it is closely related to the entropy Boltzmann actually use the letter E and then later on somebody else change it to H so it should not be confused with the Hamiltonian this function was for any distribution whatsoever does not have to be a distribution which satisfies the Boltzmann equation at all you define an H of t to be equal to integral d 3 v f of v t log f of v t that is the definition of the H function pardon me but it did not put the minus sign it is left like this now we have to see where the minus sign comes in when we write the entropy of an ideal gas entropy of a gas in equilibrium so this immediately led to the conclusion that d H over trivially that this is equal to integral d 3 v delta f over delta t times 1 plus log f of v and now the statement was that if this vanishes if you have an equilibrium distribution there is no time dependence then this vanishes and the vanishing of this is a necessary condition because if this is not 0 there is no way this can be 0 it have to be finite clearly so the necessary condition for an equilibrium distribution is that this should vanish and now put in for delta f over delta t if you put in this whole thing here then using the inequality about y minus x log x over y is always negative non positive we discovered that you had to have this condition the detail balance condition for an equilibrium so it was necessary and sufficient in that case so that much was the argument okay so the part that is ingenious was the fact that Boltzmann used the invariances of this object that if you interchange v 1 and v 2 after you integrate over v 1 if you interchange v 1 v 2 nothing happens if you interchange this does not change at all if you interchange the initial and final states nothing happens except a change of sign here so that gave us what was needed to show that the detail balance condition uniquely specifies the equilibrium distribution okay now given that much we could ask what is this H function actually look like is it always is if I plot this H function what happens to it after all the system even if it is momentarily in equilibrium in other words even if the distribution is let me call that f equilibrium of v the equilibrium time independent distribution which satisfies the detail balance condition if you say that this system is momentarily in this state it does not remain so because collisions knock it out of that state instantaneously practically right so the system is always getting knocked out of equilibrium but it is being restored the equilibrium is in some sense overall there is an equilibrium right so this is how statistical mechanics the same molecular fluctuations or collisions that knock you out momentarily out of a state of equilibrium also help to restore this equilibrium there are connections between fluctuation and dissipation which help you to restore this right so there is a interesting proof which tells you that if momentarily the system is if you plot this H versus T and momentarily it is at a value here then if the system obeys the assumption of molecular chaos if f of v, t at this instant of time satisfies the assumption of molecular chaos then you are guaranteed that the H function will decrease the local slope will be negative that is all you are guaranteed okay but there is an interesting argument due to Francis law which also says that when the system is in when in this state molecular chaos assumption is valid in this state momentarily it comes down right but the argument is that if you now reverse all the velocities in this state at this point you are still in a state of molecular chaos nothing changes that has something to do with correlation between pairs of particles and that is independent of where the particles have positive velocities or negative or whatever have velocities completely reversed or not it does not matter but then it also implies for that gas too with all the velocities reversed you still have this negative slope but the future of that reversed gas is the past of the present gas so that in a few lines you can show that at this point if in the next instant you are going to have molecular chaos the slope would have had to be positive okay there is a delicate argument here it involves actually implementing the time reversal operation but in loose words what it means is exactly what I said that if you have in a state of molecular chaos at any instant of time then at that instant of time the H function is at a local peak but the function itself is fluctuating so it is going zigzag zigzag in this fashion what you are guaranteed and there is nothing which says that this must function must have a continuous slope does not say that at all so the slope can and does jump discontinuously over the function is continuous it all it says is these points at local peaks you have molecular chaos it does not even say that all local peaks must be states of molecular chaos it only says if you have a state of molecular chaos you have to be at a local peak that is all it says but there are fluctuations about it and what statistical mechanics does is precisely to calculate what these fluctuations do at this point. Now the other point is suppose you are in a rare state you prepare the system in a rare state it will relax back to the state and what it will do is if you are in a state like this it is far away and H is itself actually fluctuating like this but then it finds itself there then it will come down it will relax back and go back but it will do so in this fashion and it will fluctuate back into this little range in which the system is approximately got an equilibrium distribution differs from it infinitesimally but there are always local fluctuations. So it is not that the system is always in a state of thermal equilibrium even if you are in thermal equilibrium on the average. So this H function the Boltzmann equation is that solid curve is a continuous curve there is the smooth curve here so it is giving you something which is an average already and there are fluctuations about it and you need these fluctuations for instance you know that the sky is blue because of molecular scattering it is happening because of local density fluctuations if you did not have those density fluctuations you would not have this sky blue at all. So this scattering especially during and this becomes very very these fluctuations become very pronounced near critical points which is what we are going to study next and that is where you get phenomena like critical scattering critical or palisthenics etc. So the scattering can become enormous in those situations but even in the ordinary atmosphere the fact that the sky is blue is due to density scattering of light due to density fluctuations so they are always happening all around you and you can measure them there are measurable consequences etc but the fact is that equilibrium is a statistical state it is not as if the system is always in a state. Now the H theorem says a little more also one could ask what is the connection between that and thermal equilibrium entropy in the thermal equilibrium state well when f is equal to f equilibrium of v in that state then this H is equal to minus the entropy divided by the units we have used v times k bolts we define this as f log f without putting in a Boltzmann constant there but entropy has measured in units of Boltzmann constant so this is the relation and the v because of the way we define this our normalisation is integral d3 v f is equal to n the number density number per unit volume so this is what the connection is and the minus sign as Suresh pointed out because Boltzmann originally defined this H function with f log so I hope this kind of tells you what the role of this H function is because we are now going to go on and see what we can do with this Boltzmann equation whether we can try to solve it or we can try to extract something from it etc first let us look at how constants come out how various conservation laws come out you know that in mechanics in dynamical systems valid not just in particle mechanics but also in theory of fields as well as quantum mechanics and quantum fields etc when you have continuous symmetries you have associated with them certain in variances of the equations of motion and then you have conservation principles which are like equations of continuity and when you integrate them you get conservation laws now where does that kind of thing appear here well look at the following keep that equation since we need it go back to the way we derived the sufficient necessary condition from the Boltzmann equation for the equilibrium distribution the maneuvering was that you integrate this over v1 and then you interchange v1 and v2 interchange initial state with the final state and interchange v2 prime with v1 prime and you get several averages and we argued that all these fellows are the same quantity and we took one forth the whole lot etc similar kind of thing goes on here because now let us suppose that you have the Boltzmann equation by the way one small side remark this quantity here when we have an external potential when this f was minus the gradient of some potential I do not know what symbol I used here phi phi of r when you have a conservative force then this was the term that appeared in the Boltzmann equation in the derivative term on the left hand side. Now you could ask what happens if I have a magnetic field then you have a velocity dependent potential and can we now do this can we extend this formalism to it the answer is yes for the simple reason that if you compute del v dot v cross b this quantity is what appears ultimately it is 0 okay it is identically 0 if you differentiate with respect to the components of v this quantity here it is immediately obvious that this quantity is 0 because write this as del i with respect to v so del delta v i epsilon ijk vj vj vk okay and of course the derivative of this fellow with respect to delta v i is a Kronecker delta ij and that contracts with this and gives you zero. So this formalism trivially extends to the case of a magnetic field even though this is a velocity dependent force it does not matter you can still write it you see now what we would like to do is to derive yeah exactly so what we are going to do is to write there is no energy that is gained by there is no there is no work done by magnetic field on a charge particles okay now what we are going to do is to see where conservation laws come from okay so let us look at the following let us write I need a symbol for this whole mess what should I call it let me call capital phi of r and t equal to an integral over d3 because I have always got v1 sitting here this is v1 v1 etc this is v1 so let me call this v1 some function little phi of v1 times f of r let me define it is like a moment of this distribution with respect to the v variable if I put phi equal to 1 or v or v squared v cubed etc I get the moments of this distribution but it could be an arbitrary function of if I plug that in here multiply both sides by little phi of v1 and integrate over all values of v1 okay what happens so let us do d3 v1 phi of v1 f of v1 t right here and the first term well we can write down the first term directly it is delta over delta t that is this term okay plus what should I write think of what is the most notationally the simplest way to do this without messing around is write it out and see where it gets us I can certainly pull this fellow out that is true and I want to be a little cautious I want to be a little cautious well bear with me for a minute I want to be a little cautious I do not want to so phi v1 here and then f over m assuming this to be a function of r alone dot gradient v1 equal to well this part is okay but equal to on this side we have d3 v1 d3 v2 d sigma let me just forget about this stop our moment of this distribution d sigma and then phi of v1 u and all the rest of it on this side now let us come to terms with what is going on here in the absence of an external force this is 0 no external force implies this is 0 there is no f so you have this plus this term here now what is this term look like it says phi of v1 sitting here and then there is a v1 I want to simplify this term so I would not be able to write well this term here is gradient r dot well messing around with the notation I want to write this as a current integral d3 v1 phi of v1 times f this quantity here is a vector quantity and the v1 goes away so let me call this equal to j some current density related to this function so let me put it as j phi and it is a function of r and t and am I right in calling this del dot j I mean I can write this as gradient r grad f dot v1 and bring the gradient all the way out of the integral I put this as v1 and then take this grad outside because there is nothing to do the integral and this quantity I called j whatever it is so I am getting an equation which says this last grad r dot j which is related to the function phi t equal to something on the right hand side but look at what happens on the right hand side there is a phi of v1 here but I know that if I interchange v1 and v2 the rest of it does not change I can interchange variables of integration this becomes phi of v2 but u does not change because it is mod v2 minus v1 so let us keep that in our mind this quantity on the right hand side is the same if you put a phi of v2 here so I can add the 2 and take the average then I will interchange the initial and final conditions this fellow is going to become v1 prime and this is d3 v1 phi v1 prime v2 prime but in the process I can now change variables of integration because d3 v1 d3 v2 is equal to d3 v1 prime d3 v2 prime exactly as we did for the Boltzmann for the H theorem but this becomes phi of v1 prime and this term here becomes f2 f1 minus f2 prime f1 prime so there is a minus sign and then I interchange v2 prime and v1 prime so once again I get phi of v2 prime out here with a minus sign so on the right hand side this quantity now we are in good shape this quantity is equal to an integral one fourth because I took 4 of these fellows here one fourth d3 v1 integral d3 v2 times integral d sigma d sigma which is a function of this wherever you are and u times u that is also sitting there time times f2 prime f1 prime minus f2 f1 prime. Times the following phi of v1 plus phi of v2 minus phi of v1 prime minus phi of v2 prime because all this remained invariant including this and this fellow change sign when I interchanged initial and final states I have taken that into account here and I have added up all 4 ways of writing the right hand side and divided by 4 so this is an exact equation where this current was defined as d3 v or v1 we do not care phi of v times f of r so this is identically this looks like a conservation law if the right hand side was 0 so now the argument is if this function phi of v is a function such that this quantity is identically 0 in a scattering process then you have a conservation law automatically so now you turn it around and say if phi of v is such that phi of v1 plus phi of v2 is equal to phi of v1 prime plus phi of v2 prime plus then delta capital phi of rt over delta t plus the divergence with respect to r the usual divergence of j phi equal to 0 that is an equation of continuity. Now we list all the quantities that are invariant in this scattering what is the simplest quantity that you can think of that is unchanged in scattering well phi equal to constant if I put phi equal to 1 that is a constant that is a good 1 plus 1 is equal to 1 plus 1 so that certainly a constant of the motion a trivial constant of the motion what is it going to conserve well if phi equal to 1 phi of if phi of v if phi of v is equal to 1 then the corresponding phi of rt equal to integral d3 v1 times f of r v1 t and that is equal to the number density at the point r at time t. So our first conservation principle is the conservation of number of mass if you like of matter right so this is immediately going to tell us that delta n over delta t the function of r and t if there is a local density fluctuation then there has got to be a flux of particles across right this is equal to plus delta j of rt equal to 0 where this j is equal to and now all we have to do is to plug in the formula for what this j was if you recall this quantity was equal to integral d3 v right times what what do we have we had a v definitely we had a v and then what else and then an f we had phi is 1 in this case phi is 1 so we had a phi of v and then an f right and there was just a v and that is it what is this representative of we have been careless a little bit about putting in dimensional quantities like mass and so on we should be a little careful but it is okay as it stands what is it what is it represent this is the momentum current of m this is the momentum current so this is the usual continuity equation for matter in the usual hydrodynamic language the current is rho times v the density times the velocity that is exactly what you have density so that is the equation of continuity for number density what happens if you put phi equal to the energy incident if there is one more thing that is conserved I never said that phi has to be a scalar it did not have to be a scalar I got to be a little more careful about how I manipulate these vector symbols but phi could have been a set of three quantities a vector components of a vector v itself because I know that in a collision process I know that in this process if since it is equal masses v1 plus v2 equal to v1 prime plus v2 prime component for component right so certainly that is going to be again a conservation principle here right and what are we going to so we put phi of v equal to v itself and in difference to notation let us put a vector there then this fellow is a vector out here and it is equal to v so this is the momentum if you like and each component of it satisfies a continuity equation on the right hand side this j would now become a two index object because for each component each of the components of phi you are going to get since you are going to integrate with the phi here there is going to be a v i v j on the right hand side so you are roughly going to get something like delta over delta t now fix these factors I have not looked at it carefully enough so d3 v f of r v t v i any index i is going to plus this is going to look like some del i some current j i j equal to 0 where this thing is going to involve an integral d3 v v i v j f of r v t apart from some factors this is the kinetic part of the so called stress tensor what happens if I put phi of v equal to half m v square that is the energy right then there is going to be an energy flux density on this side so I leave you to write down that equation so there are basically 5 conservation principle rules that we can write down immediately phi of v equal to 1 for the number then phi of v equal to v that gives you 3 more for the momentum current and then you have for the kinetic energy one more so the choices are phi of v equal to 1 phi of v vector equal to v itself and phi of v equal to half m v square and they will give you the corresponding current continuity equations from those continuity equations you can extract the transport coefficients such as the diffusion coefficient viscosity thermal conductivity electrical conductivity if it is charged then electrical conductivity and so on and then you put a magnetic field you can construct the Hall conductivity. So the business of kinetic theory is to compute all these transport coefficients and they would follow from several conservation principles in this fashion. Now let us turn to what happens to the Boltzmann equation itself and I would like to make I would like to complete this so I would like to make some comments on this on the on the solution of the Boltzmann equation in the simplest case. So let me schematically explain how this is done or maybe defer this till tomorrow since I have already run out of time basically the point is let me just say it in words the point is that you have an equation for this quantity f of r v t which is non-linear and integral differential equation and you assume a small deviation from equilibrium. So you write this as basically equal to f equilibrium of v which satisfies detail balance plus a correction a small correction which tells you how far away from equilibrium you are in the distribution sense. So you write plus a g of r now put this into the Boltzmann equation this portion of course the delta over delta t does not act on it at all the left hand side is 0 for this portion. This portion will be acted on on the left hand side and on the right hand side you are going to have a product of 4 of these guys 2 at a time and a difference. So you write this for v 1 v 2 v 1 prime v 2 prime and then take f 2 prime f 1 prime minus f 2 f 1 the term which is 0th order in g will cancel out by detail balance because that is the definition of the equilibrium part and what will be left will be combinations of f 1 f equilibrium with a g with a different index then you put in delta functions in v to take out a g common with some given variable and then delta functions to fix what the actual velocity argument is for each of these and then you try to solve that linearized Boltzmann equation you throw away terms like d square now if you want a systematic approximation scheme then you would have to keep systematically the second order higher order etc. But the linear term itself is hard enough and what I will do tomorrow is take the time to actually write this expression out and then we will see how to extract transport coefficient from it. In particular we would like to see if we can touch base with what we did earlier with the Langevin equation for diffusion for instance we had a formula for the diffusion constant in terms of the friction constant temperature etc. We would like to see how that comes out or if at all it does from the Boltzmann equation. We also had a Fokker-Planck equation for the velocity conditional velocity. Now there we made assumptions about the mass of that particle Brownian particle being much bigger than the rest of it the molecular mass we have made no such assumption here. In fact we are talking about elastic collisions here. So we would like to see how this differs from that in what happens and what happens in the simplest approximation. We linearize the Boltzmann equation and make a lot of simplifying assumptions essentially saying there is no positional dependence everything is uniform distribution in space and only the velocity is distributed and let us see what it does what it gives