Right, so we started looking at the way in which a non-uniform distribution equalizes becomes uniform as a function of time essentially the diffusion problem in the context of the Boltzmann equation. And we have a whole lot of symbols on the board but let me go over it again and then we will complete where we were getting at. So if you recall, we want to look at the distribution, we want to look at F of R V T on time scales where the velocity is essentially thermalized. That is where the diffusion regime is, okay, although we are not explicitly saying so. So we would like to ask what happens if the velocity distribution is Maxwellian essentially but the positional distribution is a function of time starts with some initial given prescribed distribution and then it obviously is a function of time due to diffusion, it equalizes. We are trying to understand that process and understand the time scale which appears in the problem when you try to find out how it goes to this equilibrium situation. So the equation we wrote for F was Delta over Delta T plus there is a V dot Del and that is definitely present, this term Del R is definitely present because we are trying to find out how this function of R changes, okay. So this times F plus an external force but that is 0, there is no external force. So F over M dot grad V, this term acting on F, this term is absent and this is equal to on the right hand side minus 1 over tau times F of R V and T minus whatever it relaxes to, whatever the system relaxes to. Remember that when we looked at the Ranjima problem in phase space, one particle phase space, I said that beyond the diffusion time the joint probability density in phase space factorizes into a portion which was essentially Maxwellian times a portion which satisfied the diffusion equation. I did not set it equal to constant, it satisfies the diffusion equation and we are trying to see if that emerges here, okay. So that is the point that you raised and that is what we are looking for. We are trying to find out what happens, just to recall to you, we had a rho there, we had a rho of R V and T and we said that when T was much much greater than gamma inverse in the Ranjima model in this case, this went to some probability distribution R T times the Maxwellian so in the velocity. That is the one that, that is what we discovered in the diffusion limit and then we found out what is the equation satisfied by this quantity. So it is in that spirit that this is being done. So what we need here is an N of R, T times W of V which is the Maxwellian distribution. So just to set your normalizations, remember that F of R V, T, if you integrate this over R and N, V, you end up with the number density finally. So this quantity here, F equilibrium, this goes to F equilibrium of V, that is the absolute uniform distribution in space. So this is equal to N, a constant, it is just the number density at constant times W of V which is the Maxwellian, the Gaussian distribution in V with the variance which is proportional to K T. So that is what is sitting here except that it is not the equilibrium distribution we are going to. We are trying to find out what is this distribution here. So our target is actually to find out what this quantity is. Well if T tends to infinity in the solution of the diffusion equation, everything becomes 0 in an infinite volume. Here everything will become a uniform distribution, 1 over the volume. That is not very interesting, right. We are trying to find out what is the mode by which, what is the timescale, that is really what we are trying to do. So we put in a timescale here and we are trying to see how this timescale is going to play a role in controlling the timescale on which the thing equalizes, okay. But of course we must also look at another fact which is that when you have a non-uniform distribution and you do a Fourier transform in space, you have all wave numbers. So you can resolve it into sinusoidal fluctuations on all length scales, on all wave numbers, right. Now the question is how does that play a role? So clearly the relaxation time is going to depend on the wave number itself. So they cannot be one relaxation time, right. There has got to be a dependence on the wavelength. So a disturbance which is very long wavelength disturbance is going to take a different time to relax than one which is a very sharp wiggle. That is what should come out, right. So although we put in one constant here, you are going to get a whole family of constants, a continuous family of constants. So it is going to depend on the wave number of the fluctuation. And the question is how, okay. So yeah, that is right, exactly, exactly. Yes, absolutely, yeah. We already saw what is the timescale on which in this one, in this single mode thing, we already saw that this tau is the velocity relaxation time. So now we are looking at timescales much bigger than that tau. But it will depend on what k is, what the wave number is. So this is what we are going to discover, right. Essentially you are going to find a whole family of relaxation times, one for each k. But how does it depend on that k is the question, okay. So this is the target and now we are in business. Let me go through this quickly since we did this already last time. I do a Fourier transform with respect to space and Laplace transform with respect to time. So essentially we are going to write f of r v t equal to well and 1 over 2 pi whole cubed integral 0 to infinity dt integral d3r e to the minus st, that is the Laplace transform and then the Fourier transform was e to the minus ik dot r f tilde of k v ns. Oh, what have I done? d3k and our Fourier transform convention was to put a plus sign there for this k. I should write the inverse transform, let me not do that.  And the inverse transform will have a 1 over 2 pi cube e to the plus ik dot r and this is going to become e to the plus st over the melenne contour, whatever. So what is going to happen here? d over dt of this guy but this fellow if I write it down is going to pull an ik dot r so let us without further ado write this down.  Now we made the assumption that the initial distribution f of r v t, this quantity is equal to some initial distribution in r at 0 multiplied by w of v that was the assumption we had made, right.  We got all the factors right now. So I bring this, this and this to the left hand side and I have s plus ik dot v plus 1 over tau acting on f tilde of k v and s is equal to so this term has been taken into account so has this and so has this I move this to the right hand side this is equal to n initial tilde of k plus 1 over tau n tilde of k and s times w of v. So that takes care of this term and this term. Shouldn't there be a what? Well we are finding the Laplace transform of the derivative time derivative so it is the transform of this function minus the value of the function in time at t equal to 0 which is all I have written but then I did a Fourier transform in space and so let us move this fellow down here this thing divided by s plus ik dot v plus 1 over tau and let us multiply both sides by tau so that it is tau times this plus just that is equal to 1 plus s tau plus this guy, i tau k dot v and this goes to that side. Now of course we need to close this set of equation, this equation so what we do is to integrate over v and integrate over v and I get precisely n tilde of k and s once I finish integration over v because f of r v t if I integrate over v I get n of r t by definition so what we have is integral d 3 v remember that so I integrate and Fourier transform so I am going to get f tilde of k is n tilde of k comma s if I take Laplace transform etc so this tells you that n tilde of k and s is equal to tau n initial tilde of k whatever that be plus n tilde of k and s here in this bracket times an integral and that integral is integral d 3 v w of v divided by 1 plus s tau plus i tau k dot v I have to integrate this over v along that it comes out so let us call this integral something let us call this i some integral and it is a function of k and s the v is gone so I have 1 minus i k and s times n tilde of k s equal to tau n tilde initial of k let me write it properly is equal to tau n initial tilde of k times i of k and s and I divide by this factor 1 minus i that is the solution now everything is known we know this it is a Maxwellian in principle it is something in the denominator it is some kernel so once you do this integral you know this function explicitly it sits in both places and you know this from the initial distribution so therefore I know the Fourier Laplace transform of n of r comma t of this one and that is the one I am trying to find out of course it cannot be inverted analytically this is a terrible mess this itself is bad and then on top of it you have to put that here and then invert both the Fourier and the Laplace transform it is an impossible task okay but what are we trying to get we are trying to find out what happens at long times and what happens on long length scales the scales on which diffusion occurs the modes the longest wavelength the long and the shortest wavelength the largest sorry smallest k which means longest wavelength k is the wave number okay which is where the diffusive modes are so one way to do this is to expand this whole thing in powers of k this is 1 plus something and you say k is going to be near 0 so expand this in systematically in powers of k using the binomial theorem and do the first order second order etc. Now when you do that you can see immediately what is going to happen here so let us try to expand i of k and s equal to let us pull this out of the denominator and then do a binomial expansion in powers of k so that is independent of v so this is 1 over 1 plus s tau and then an integral d3 v W of v times 1 I am going to pull this out and take this out in the denominator here so it is 1 plus something inverse therefore it is 1 minus i k dot v i tau yeah yeah absolutely yeah yeah yeah precisely before what do you say is the first order term and k will vanish all odd things will vanish yeah I just want to show you explicitly how that happens you can see so this divided by 1 plus s tau we must keep the next term because that is the one that is not going to be 0 so it is 1 plus x inverse which is 1 minus x plus x square but then there is a minus sign here because of this guy so minus tau squared k dot v the whole squared divided by 1 plus s tau squared plus dot dot dot now integral d3 v W of v is 1 because it is a normalized max value already so the first term is 1 plus now this term here has a k dot v obviously you should use polar coordinates such that the polar axis is along v then this becomes k v cos theta and you will do an integral on cos theta from minus 1 to 1 d cos theta so that is an odd function and it vanishes it is obvious from isotropic here so the first order term is 0 then this one is the first nonzero term and that is easily evaluated so you have 1 minus 1 over 1 plus sorry so this is equal to 1 over 1 plus s tau minus tau squared k squared divided by 1 plus s tau the whole cubed is 1 factor here and 2 more there times and integral d3 v W of v and then v squared out here times cos squared theta where theta is a polar angle I have assumed that we are going to polar coordinates spherical polar coordinates in v space and chosen the polar angle polar direction to be along that of k because that is the vector that is sticking out right so you are going to get v squared here and then a cos squared plus higher orders yeah yeah I just want yeah I just want to do this in spherical polar coordinates because I do not want to get into Cartesian and count different kinds of integral there is only one integral to be done here right in fact I am not even going to do the integral I am going to leave it to you as homework problem it is a trivial problem but what I want to do is to extract the temperature dependence of this term that is crucial this temperature sitting there in W of v on the Maxwellian I want to know what it looks like that is that is the main point I want to get at right so we should keep this but you see this term here what is it going to be this first of all in the in the phi direction there is a 2 pi so first of all this fellow is m over 2 pi k Boltzmann t to the 3 halves that is the normalization of this W of v and then there is a v square d v and then there is a 2 pi from the 2 pi from the phi integration this is v square and cos square theta and cos square theta is integrated from minus 1 to 1 so it is twice integral 0 to 1 and then it is twice integral from 0 to 1 divided by 3 so it is 2 over 3 times 2 pi is 4 pi over 3 that is why I mean it is obvious you are going to get 4 pi over 3 4 pi over 3 and all that is finished and then you are left with an integral 0 to infinity d v there is already a v square and a v square so it is v 4 e to the minus m v square over 2 k Boltzmann t and there is a Gaussian integral with the v 4 here so it is a trivial thing to do and change variables to m v square over 2 k t then you are going to get if you call that u or something like that there is going to be a 1 over square root of u and then a u square so u to the 3 halves so there is going to be a gamma of 5 halves but you can scale this fellow out completely v square scales like k t so this will go like k t whole square and then there is a 1 over square root of k t here and there is this fellow sitting here so this sum and substance is there is going to be a k t finally whatever be that constant some k t over m or something like that okay I leave that to you as an exercise whatever it is finally let me write the final answer down there is a very well known answer let me write this down this fellow after you put that in here and here and then keep to order k square because that is the order to which you cut it off everywhere except this so the coefficient or whatever this is correcting this initial distribution has a term which is of the order 1 and then order k square and so on and so forth so if you do that this term here becomes equal to it looks like the following an initial tilde of k divided by s plus a function of s let me call it d of s times k square where I have used this d of s deliberately because it has dimensions of the diffusion coefficient this fellow has got dimensions of time inverse and this fellow is length to the minus 2 so this whole thing this therefore must have dimensions l squared over t which is the dimensions of the diffusion constant so I called it d of but it is a function of s where d of s equal to and now it is not surprising you are going to get the temperature dependence from here so it turns out to be k B t tau over m 1 over 1 plus s tau you could have sort of guessed some of these things 1 over 1 plus s tau because that is appearing here goes with the k square so that is the solution now you have to tell me what this is and then I have to invert it with this weight but now let us see we are trying to ask how does the initial distribution whatever it is how does it diffuse out it depends on the poles of this because when I do an inverse the plus transform 1 over s plus lambda as the inverse transform e to the minus lambda t right so it is going to depend on k square out here therefore the leading relaxation is going to be of the form for s tau much much less than 1 because we want long time or small s going to 0 you can neglect this term so this is going to look like n initial of k divided by s plus d of 0 k square and now you know exactly how it is going to relax because I can invert the Laplace transform therefore trivial and it says finally so it says n of r t and n tilde of k and t let me leave the k still is equal to n tilde initial of k e to the minus d k square t where d is k Boltzmann t tau over n so you see it is typical of what it is exactly what we would expect remember in the large one model we got k t over m gamma and the velocity correlation time was tau gamma inverse well here the velocity correlation time is tau in this model and m is the mass of a molecule so the approximation has been sort of gross in that sense that has become essentially the mass of a molecule but this tau is completely arbitrary completely arbitrary we just put that in my hand so that is the diffusive mode as you can see and you can see it is a Gaussian solution because what is the inverse Laplace inverse Fourier transform of just this guy if you make this if you give me a start with the delta function for instance then this fellow here is a constant if you give me a delta function in r this n tilde of k is a constant and then the inverse Laplace transform of k e to the minus k square Gaussian is of course e to the minus r square again Gaussian now this Gaussian has a variance 1 over d t and in r it will have a variance proportional to d t so you have exactly the diffusive behaviour with this diffusion constant d but it is giving you a lot more information this thing is telling you much more it is telling you how other modes in lax in principle so even though it is a single relaxation time approximation because of this complicated way in which k appears and s appears it is actually telling you much more it is going beyond this diffusion approximation so this is how a space dependent relaxation appears so first we said the velocity itself how does it relax given a uniform distribution we found that was rather trivial it was like a kubo Anderson process then we said what happens if you have a non-uniform spatial distribution well it is diffusive this is called diffusive mode if you like now let us ask the other question suppose the distribution in space is uniform and I start with the equilibrium distribution but now I switch on an external force then what is going to happen and let us in the simplest case let us switch on a force which is time dependent but uniform so that the uniform distribution in space does not change however you put on a force external force and these particles are going to get dragged so I expect the Maxwellian distribution will be disturbed and if the force finally goes to a constant force as a function of time then the Maxwellian certainly will be distorted sort of intuitively in the direction of that force right so that direction should be singled out and let us see how that happens okay and we play the same game and practice an even easier thing so let us do that so till now we were looking at problems where you are a little away from equilibrium you relax to equilibrium but now I start with equilibrium and I push it out of it and ask what is the new distribution to which it goes so constant uniform so I have f of r v 0 the initial thing is the equilibrium distribution is equal to n times w and it is a uniform force so uniform force is not going to destroy in initially uniform spatially uniform distribution is going to remain spatially uniform so I do not have to worry about the gradient term with respect to r acting on this thing at all then what happens to the equation that we have so some force f no r dependence here so we have delta f over delta t and this is a function of r v and t plus f of t over m dot the gradient with respect to the velocity that is important of f of r v t and this guy is equal to minus 1 over tau f of r v t minus f equilibrium of v so I want to know how the f is moved away from the equilibrium distribution because that is what I started with at t equal to 0 so this is the difference in the single relaxation time approximation so this is the equation I have to deal with now clearly it is sensible to call this something else equal to some g of r v and t in particular I want to compare the answer we are going to get here with whatever we know from linear response theory remember in linear response theory we found the average velocity divided by the applied force the constant applied force per unit applied force was the mobility and in the case in which the force was a constant time independent we got the static mobility so we want to see what happens now in this case but this is completely kinetic we do not need to carry the r dependence everywhere because there is no term with respect to this guy so we forget about it completely it is uniform everywhere so let us make this f of v comma t f of v comma t and that is it no r dependence the statement is you have a force which is uniform time dependent but uniform you start with an initial uniform distribution there is nothing which is going to make the distribution non-uniform in space okay so this is what we got to work out this fellow here and now let us put that in here so we get delta g over delta t plus f of t over m dotted with gradient with respect to v of f equilibrium that is equal to f equilibrium v plus g of v and t because f is f equilibrium plus g by definition this is equal to minus 1 by tau g of v we want to compare now that is an exact equation right however we want to compare with what happens when the external force is weak that is the whole idea that you want to compare with linear response theory now it is clear that this g is going to depend on the external force there is no force this g is 0 identically you remain in equilibrium so g is going to have a first order term in f second order third order etc etc but if you are going to work to linear response level then this term f times g is already of second order in g right this is the correction which is proportional to f and higher powers and there is already an f here so I am going to drop this in preference to this okay and then we get delta g to first order t the equation is delta over delta t g that is this term plus f of t over m dot the gradient with respect to v of f equilibrium of v equal to minus 1 over tau now of course one brings this to this side and that is the same as saying that g of v comma t equal to e to the minus t over tau times something else let us call it h of v comma t just removing this integrating factor so the first term is minus 1 over tau times g itself and that cancels and then you have e to the minus t over tau delta h over delta t equal to minus f of t over m dot gradient v f equilibrium so move e to the t over tau to the right hand side now what is the initial condition well the initial condition on this f was that the start with it was in the equilibrium distribution so g of v comma 0 is 0 therefore h of v comma 0 is 0 this is the initial condition and we have a formal solution and what is the formal solution now it says h of v t is equal to minus this is the initial condition this we got 1 over m integral from 0 to t dt prime f of t prime dot gradient v times e to the we should not forget that e to the t prime over tau times f of t prime dotted with gradient with respect to v f equilibrium of v that is h so it says g therefore is e to the minus t over time tau times that so this is e to the minus t over tau times and that is the formal solution for g f however is f equilibrium plus g so it says f equal to f equilibrium of v minus this guy and that is the solution we would like to see what happens at long times if I put a constant force so what I need to do is to switch on the force by whatever means I like and the typical way to do this just to get an idea of what the asymptotic behavior is is to do the following it is called adiabatic switching and you do this in quantum mechanics for example when you do perturbation theory you switch on the force slowly and not suddenly so that all the energy levels do not get jiggled up but the same spectrum remains but gets slowly perturbed okay in the same way here is a way to switch on the force so let us suppose that on the time axis you have here f of t magnitude or whatever you want to make it go to a constant value you want to make it go to a constant value this value is f and let us say you start switching it on at t equal to 0 because we took at t equal to 0 the system to be in equilibrium and typically what would happen is that you would switch it on it will do this you can choose any force history you like but I would like to see what happens at long times here so a simple model would be to say that f of t equal to it should start at 0 and should end up with a capital F which I will put at the end inside so 1 minus e to the minus t over some time scale whatever you call it is tau 1 this is typically tau 1 the time scale on which this fellow essentially reaches 1 over e its saturation value or something like that so that is one way to do this you put that in here then you integrate what is that get us so this fellow here becomes this term this term alone let us look at what happens to this so let us look at the whole thing so this implies that f there is no r f of v t equal to f equilibrium of v and then what minus e to the t e to the minus t over tau that was the relaxation time over m and then the time integral is essentially integral 0 to t dt prime e to the minus t prime over tau 1 minus e to the minus t prime over tau 1 and then there is outside there is an f dot v f dot grad v of f equilibrium of v this fellow is independent of time it just comes out as I said trivial integral to do this whole business first term is e to the whatever it is a plus I think this is a plus plus t plus out there plus t plus out there so this first term is going to give you e to the t over tau minus 1 and when you multiply by this it becomes 1 minus e to the minus t over tau the second term has got a smaller term which is a positive term so it is 1 over tau minus 1 over tau 1 if tau 1 is much bigger than tau for instance then this term is going to be positive 1 over tau minus 1 over tau 1 but smaller than 1 over tau therefore when you hit it with this this integral that part will vanish exponentially so the sum and substance is the only thing that remains for t much much greater than tau and tau 1 is f equilibrium of v minus what remains this is going to give you an e to the t over tau leading term and it is going to kill this so you are just going to get 1 over m f constant dot grad v of f equilibrium of v what is that going to be because remember that f equilibrium of v equal to n times this m over 2 pi k t stuff multiplied by e to the minus m v square over 2 k Boltzmann t now what is the gradient of e to the minus r square in physical space in 3 dimensional space just d over dr is going to appear and in the direction of r so the gradient of r square is essentially r vector twice whatever it is now when you differentiate this is going to give you a twice and that twice will cancel here and you get to m over k t in the denominator looking starting to look like the mobility as you can see and then there is going to be a piece which is essentially v so this whole thing is going to look like f dot v apart from some constants there is some constants I leave you to figure out and then f equilibrium so it is going to become proportional to this pardon me constant is m over k b t yeah so there is a 1 over k Boltzmann t and a plus and a plus sign because you are differentiating a minus here so and a plus so we can even write this down plus so constant is m over k b t so this m goes away but remember that when I integrate e to the t prime over tau I get a tau on top so yes so there is going to be a tau over k Boltzmann t let us write it properly so that is how it gets dragged in the direction of the force leading correction to 1st order in f we know the answer should be proportional to f it has got to be a scalar and the only other velocity you have is v itself so it has got to be proportional to f dot v times this this gives you the constant of proportionality okay exactly precisely f dot v is the rate at which power is supplied to the system is the rate at which energy is supplied to the system a work is done on the system by the force so that is a sort of very straightforward interpretation now what is done in transport theory is apart from the complications of solving the Boltzmann equation in various situations is you remember the collision invariance we talked about 1 and then the velocity itself which gives a momentum current and then the energy current itself so what one does is to take each of these currents and from that one can extract transport coefficients such as the shear modulus the thermal conductivity the electrical conductivity etc diffusion coefficient we have already extracted so you have to construct you can even construct the heat current the thermal conductivity because there is a mechanical portion which is the moment of all these things depend on taking this fellow here an integral of v times f of r v t times some something here some function of v as I put it these are the currents and if this fellow is 1 you got the momentum current if this was kinetic energy half m e squared you got the energy current and so on so you take half m e squared minus the average value 3 half k t or something like that and that gives you the heat current you can go one step further and say I will find out what thermal conductivity is what is the heat conduction equation you can derive Fourier's law for the heat conduction equation here within linear response theory by using this local equilibrium approximation you say that the temperature is different at different places but in a very mild way so you write t as a function of r itself then the derivatives are going to spatial derivatives will act on that t and in that manner you will end up with Fourier's law of heat conduction which essentially says delta capital t over little t partial little t is minus the thermal conductivity times del square of t you get that you get a actually we get a formula for the heat current which says it is proportional to the temperature gradient that is what will emerge automatically from this okay and so on. So I am going to call a halt to this portion of it which is a kinetic theory really and we go on now to the very current topic well it has been current for a while but it is still a very crucial one namely we will do dynamic critical phenomena next for that we will start with some notions of equilibrium critical phenomena second continuous phase transitions talk about mean field theory and then we will see how time dependence comes in so that is the next thing we take up.


What's going on everybody got a Kyle here relating sports cards, and we're ripping three boxes for drew drew art One moment. It is a one box. We've got mosaic choice football We've got one box of bowman chrome hobby and plates and patches football Ready. Good luck drew. Let's see what we got my friend See what we got. We'll start with the start with the plates Start with the plates Save the mosaic choice for last. I haven't opened one of these yet. Hey, yes, sir Sam got you right after this right after this Good luck, sir. Looks like the Jags did it Jags got it done this way first up. We got a Micah Parsons base like a Parsons Beast for the Cowboys Behind that we have this one is double coverage to 50. That's going to be a Zay flowers April showers bring Zay flowers the same would say for the Ravens 30 of 50 dual jersey Got a highly revered auto. That's going to be a mod Rashad Highly revered throwback auto to 75 for the old Vikings receiver a mod Rashad 275 This one is Josh Downs RPA for the Colts to 149 and looking pretty solid for the Colts man They've been using them a good bit Josh Downs on the two-color jersey to 149 RPA and on the back we have to 25. It's juju Juju Smith Schuster righty, mr. Drew next up. We'll do a bowman chrome hobby Good luck, my friend. Let's see a super in guys. Do have Sam ripping over on fanatics life He's got some group rakes today if you guys are interested They're doing bowman chrome. They're giving away the red sock spot again today So if you guys just hang out in the chat can be eligible for that giveaway once he starts it They'll do it at the beginning of each bowman chrome once it fills I believe they got one pretty close to filling right now. It's the next thing they're working on You guys um have it on the iphone app or anything. I would suggest just lean it open uh hanging out with Sam a little bit can Potentially get a free red sock spot in a full case. There's a Jose Salas to 150 Atomic fault fall league, excuse me Pick up iconic of carol. We've got a Alex from Erez and Vic to insert for the Mets My yeah first Cock time. Let's go Sam I'm down Michael Harris It came to the league Guanipa Speckle is gonna be first bowman auto jack Panada 299 that'll be for the Royals Nice big on card out of there to 299 Neda Hendry Chavilly Reebus There's a Luis Torres refractor do 499 Josh Allen Josh Allen three turnovers for the eagle or for the Falcons Not very good today not very good first of Luis Torres Good luck. Good luck drew on super fuchsia fuchsia is joey vato to 299 Joey vato and speckle first bowman. It's going to be kinder delgado to 299 That'll be Yankees on the first bowman prospect kinder delgado Guerrero and dunno Come on drew the auto here. I believe it is going to be hit one almost one almost first for the Royals Royals on the base first bowman auto There's another one who had um Base in 23 bowman, but no autos. There's a well been Here you go, then we'll do some mosaic choice for you uh drew then we'll get sam going with some more mosaic There's a shimmer houseway to Paula for the adidas Elian soto andie pahays and edwin rodriguez Roddy mr. Drew got one box of mosaic choice football now Good luck. Good luck. So we've got a max crossby silver This is a refractor or a um mosaic of dorian williams rookie for the bills And auto that is pretty sick. It is champ bailey on these showtime signatures for the redskins Champ bailey for the redskins there Showtime signatures to 25 I like that the champ bailey auto big sleep One is seven out of 25 on the uh commanders commanders champ This one is um yellow to 80. It's going to be sauce Got a rookie jake moody and a rookie's insert will levis um nfl debut number to 80 And our green and red's not numbered. We've got derrius davis and tucker crafts I don't righty there that'll do it for you drew nice hit on that um champ bailey to 25. I like that Nice showtime signatures mosaic. We'll get that right out to you, sir Thank you. Thank you, drew


I also want to start by thanking the organizers for giving this great opportunity to be here. And of course, my talk wouldn't be here if I did not have these fantastic collaborators. We started this work with my former postdoc, Vlad Belly. And today, I'm going to focus on most recent work with Bo Peng and Hamid Shosie. OK, so I don't have to show the next slide for this audience, but nevertheless, maybe there are a couple of people in here for whom I want to emphasize the fact that when you take a capsid, inside we have the genome, which is negatively charged due to the phosphate groups. And the n-capsulating capsid is multiple copies of the same similar kind of protein. And they usually carry positive charges. And that's the reason why we have this title of electrostatics. And of course, broadly speaking, there are two kinds of shapes. One is the icosahedral shape. Like this is a tobacco, mosaic virus, satellite tobacco, mosaic virus. And we had this character already today, earlier today. This is TMB. This is more helical. And then we can have a composite with the phage. And the genome could be single-sanded RNA, single-sanded DNA, double-sanded RNA, double-sanded DNA. And today, I'm going to focus my attention on these icosahedral viruses with single-sanded RNA. And that's the place where electrostatics plays a role. And this beast is an entirely different universality class where electrostatics is not that important to my mind. So where do I start? I get into trouble right away. So this is a typical protein, which is constituting the capsid. And somebody who plays with paper and pencils, this is a total, total intimidating mess. And I see the alpha helices, the beta sheets. Of course, I took a lot of courses in biochemistry and biophysics. But nevertheless, this is very intimidating. There is a hairpin in here. And I look at another character. It's extremely scary to me. Maybe I want to do something. Maybe not to deal with this at all. The second thing is the partner. If you take RNA and you take any of these genomes and this RNA structure, we'll see some discussions about it more in the rest of the conference. Each one is a highly, highly branched structure. And it's a negatively charged. It's a highly branched structure. And of course, all these attributes are contributing the size and shape of this RNA and how that's going to interact with the capsid. This is my problem. How am I going to deal with this? And then in one afternoon, we had a lucky break. Just putting our feet up on the chair, just in front of a blackboard, we were looking at it. We looked at several viruses, RNA viruses. And the amount of RNA packaged was different. The number of vertices could be the same. You can have a variety of them. And what we observed was, generically speaking, the typical motive of the building block, this every protein molecule had this building block. There is a hydrophobic domain. And there is a tail, which carried dominantly a positive charge, which is protruding into the lumen of the virus. And of course, this virus is floating around water. You know, you have a solution with salty conditions. And the outside is also charged. And there is also another domain in here, which is also charged, which is going to be exposed to the exterior world. See, this was a typical motive. Then we thought we'll just do a paper and pencil calculation in about 30 minutes. That was the goal. And you are not going to believe me when I say it's only 30 minutes calculation, but people, experts in here, know how this is done. Let's imagine, just for fun, let's imagine the problem is like this. I have a spherical vessel. And then I have a charge of bristles protruding into the room. And then I have my RNA, which is a flexible polyelectrolyte, which is going to be RNA, single cell RNA is flexible. That's going to be inside this lumen. Let me just do this. And how would I do this? I want to start with a Hamiltonian. So yes, I don't have an internet here. That's OK. But I don't see it on my screen. How do I get rid of that? Let me see. I don't see it in here. Sorry? Yeah, I'm not that well-coordinated with my mouse. But I can tell you, right? Please. All right, so basically, right? So that is the cationic protein brush. See, it is not there. No, go back. Please go back. Yeah, so this is very unique. It's OK. No problem. You can still read it, right? It's not important. You can't read it anyway. So here, the blue is RNA. So it has all the confirmations of this RNA. Single cell RNA is in the blue color. And the bristles are the protein tails. And then the hydrophobic domain is somewhere there. And the red color corresponds to all the confirmations of these protein tails. And then everything else, the electrostatics. There is interaction between blue and blue. There is interaction between red and red. Interaction between blue and red. And there is lots of salt in here. There is also interaction from the salt. So the electrostatics coming from RNA, capsule proteins, and the salt, right? And then what do you do? That's the partition sum. You calculate the free energy, minimize the free energy, and get to the optimal confirmation. This is something you can do analytically. In spite of all the complicated equations I'm showing you, it isn't doable analytically. And the answer is, as some of you already know, is the following. It makes very simple predictions. Prediction number one is that if I look at the, this is a very key equation that I must wait for this gentleman. Yeah, I hope you'll give me extra two minutes. Oh, yes, sure. Not at all. No problem. Because the first equation is a very key equation, which comes out in a very simple-minded way. But you have to do the work. You have to do the analytical work. If you do it, the answer is the amount of DNA, single cell and RNA, which is a packaged number of repeat units that is packaged into this virus is simply proportional to total positive charge carried by all these bruises. That's a simple prediction. This is not a novel idea. If you take any atom in the periodic table, we believe in electroneutrality. When you put one extra electron, that's not as stable as the neutral case. No problem, don't worry. See, this is my life. Don't look at it. So that's a prediction. So this total length of RNA that is packaged is proportional to total positive charge carried by all these tails. The second prediction is the density profile of this polynucleotide away from the wall will have this depletion zone. It'll kick up like this, where reaches the maximum come off. I did not say anything about the sequence. I did not say anything about packaging signal in here. Because in my Hamiltonian, it's very difficult to put in the sequence to do it in the way, in the style with which we do analytical calculations. Then I have to resort to computer and then I have to do all the things which we have done separately later on. So these are the predictions. And these predictions have turned out to be good. Surprisingly, given these approximations, simple calculations, these are the data. These are not simulations. These are mother nature's data over billions of years for all kinds of viruses. They fall on this very nice muscle curve, exactly the way we predicted. Namely, lambda is proportional to total charge. And similarly, if you look at the density profile, we heard references to Jack Johnson, writing from his laboratory. They changed the sequences and the virus are similarly the same. And the whole process takes place independent of the sequence. And they said that genetic content is different. And the diversity of RNA transcripts, but they all package following electron neutrality. Now, so what? This has been a good suggestion, because now you can package a DNA with anything that is possibly charged. It doesn't have to be a capsid protein. You can, for gene therapy, you would take DNA, package it, or put cargo inside the capsid proteins. It doesn't have to be an RNA. It can be anything that is negatively charged. As long as it follows the electron neutrality rule, you can package as much as you want. And philosophically, the message is that the viruses assembled in a way that the total charges have to be balanced and meant. It probably in the early stages of evolution, it was a co-evolution. It's not only central dogma. There is only RNA dictating everything, but also they were going back and forth. And that's what the implication is. Now let's look at the kinetics of assembly. Now the situation is slightly different. So let's just go to the basic silicon-free argument. Suppose I have a placket of some size linear states, or two-dimensional placket. And then you can compose the free energy of formation of this nucleus. And that will have a nucleation barrier. And that is a line tension. That is a bulk gain in free energy. And then based upon this, you can think about nucleation time going like into the one over super saturation or super cooling, depending upon which variable you use. And then once that is bound, it will grow by various entropic and energetic encounters of the various individual components at the perimeter. And then it will grow. And then eventually it will close. And then based upon conventional arguments of the classical nucleation theory, one could make a prediction that the growth rate would also be e to the minus one over t minus t. Similar formulas, but the pre-factors would be different. This is, again, a very simple argument one could make. And then you do simulations. What we did was we did Langevin Dynamics simulations. I must make it very clear here. We use the fluctuation dissipation theorem. And the friction coefficient of an entity is appropriate to that fluctuation dissipation theorem. So we do Langevin Dynamics properly, in view of the common that we heard before. When you do this, what we showed was there is a very narrow range of a temperature and a protein concentration when this assembly will take place. And we were absolutely very grateful to this Spanish team. We do not know who they are. But they did a fantastic site, the mutagenesis work. They did a lot of experiments. And then they came up with the minimum structure that you need in order for this virus to take place. So we piggybacked on their work. And the reference is given in our work. And we could show how this happens. We could follow the kinetics, how the kinetics is taking place. To make a long story short, in here, it also involves a nucleation barrier. So in the absence of the RNA, capsidic proteins alone could assemble if the conditions are right. There is a barrier in here. That barrier is substantially reduced in the presence of RNA. And also, the amount that the barrier is reduced depends upon the RNA sequence. The RNA sequence contributes the kinetics of the assembly, but not to the final structure. That's basically the summary. And sometimes when the virus particle without the genome may not be able to assemble, but the RNA would allow it to assemble. And therefore, in fact, what happens is this analysis of these data tells us that we could use the classical nucleation growth theory for understanding this assembly of these viruses. And in fact, in order to check this, I'm going to be very quick in here, the prediction. This was a classical argument. The time for nucleation will go like this. And that's the law that we saw. And similarly, the growth rate is predicted to be that way. And that's exactly what we saw. Whether we look at the nucleation rate or the growth rate, that the melting temperature or dissolution temperature is the same, and it has to be. So this verifies theoretically that nucleation growth mechanism is operating for the growth kinetics of a virus. Yes, Robin? Right. Not in our case. We can discuss this. Not in our case. We calculated the nucleation rate. Not in our case. It's a matter of definition of how you calculate your line tension and also bulk energy. It depends upon that, right? Right. We can pick. Robin, we can talk about this later. Can we talk about this later? Because I'm just showing simulation data and the analysis. They fit perfectly fine. We can cobble about the magnitude. But it's not at all. I'll show you. It's not at all 100 kT. So we are talking about 6 kT. And I'll come back to this energy later on. Incidentally, the ATP hydrolysis, for some of the people who are not familiar, ATP hydrolysis is 20 kT. And I'm talking about 6 kT, 6 k to the 10 kT. So in these simulations, here, oh, sorry. Here, we took MEM, minus-minut virus, I think it's called. Now, that's the one they had. External data, we wanted to piggyback on them. We took that virus. We take that as a typical example where there was data. So that's what we analyzed. But the point that I'm trying to make is a global point. It is not a particular example that I want to talk about. I'm talking about a global point. That is, nucleation growth mechanism is a good platform to think about the kinetics of viruses. So let me show you this. This is from our friends. I think we saw this graph already. Yes, of course, experimentally, the very beautiful fluorescence correlation spectroscopic experiments clearly show that it was a really two-stage mechanism. And you can go back and look at all the arguments, and look at the data. They match extremely well with the simulation data. And also, this is the most recent work. So what do I want to conclude? This is the conclusion for Robin. The conclusion is, I'm not done yet. This is only the first part. In general, electrostatics dominates single-stranded RNA virus assembly, not double-stranded DNA. It controls single-stranded RNA virus assembly. And the spirit of my argument is captured in the model of the very first book on virology. This is by Luria, a 1953 book. I must read it loud, particularly for younger people here. There is an intrinsic simplicity of nature, and the ultimate contribution of science resides in the discovery of unifying, simplifying generalizations rather than in the description of isolated situations, Robin, in the visualization of simple overall patterns. What do I want to share with you next? I want to know what the fate of this virus is. I made this virus. This is what happens. Now, how does it get demolished? We had a beautiful lecture before. I want to see. I want to understand how to break up this virus. So there are lots of versions. Let me just simply go in here. There are two ways it can happen. One is, it completely dissolves, uncomplexation between the polyanion and polycateon. The other one is, it slithers through.  It slithers, it undergoes a translocation and goes out. And of course, they'll recombine and then do proliferation of the viruses, et cetera, et cetera. But that's the question that I want to understand. And how do I want to understand? This is something that is like a child's game.  And I went about three ways. The first way is, let me take this virus using this electrostatic principle. I have some amount of RNA in here. Now let me arbitrarily consider. Suppose I increase the amount of RNA, length of the RNA, arbitrarily. Let me say that it's going to be an excess. That excess is given by n minus q. q is a total positive charge on the protein. And n is a total negative charge on the RNA. And this v is the volume of my capsid of the virus. So theta is my order parameter. So what I want to know is, what is the free energy of this virus as a function of theta? How does that happen? When is it going to be non-stable? How much price I have to pay to package more and more of my RNA into the virus? The second thing that I want to do is, OK, so let me take a complex of a partially charged polymer and negatively charged polymer. And they are complex. And I want to break it up. How do I break it up? I bring an invader. I bring a competitor. And I see whether that is going to pry this blue out of this red. And I make a partner with the red. And what is the force involved? How do I think about it? And what are the energetics involved? What are the appropriate conditions? The other strategy is to look at a translocation to the exterior. Because I already told you, the virus outside of the virus is also polar, because it has to be water friendly. So inside it is going to be capsid here. And then I want to put it out. And how does that happen? What are the energetics involved? In order to decompose. So now the question is, I want you to ponder over the following question. Suppose I have a virus like this. And I have my brissel. That's my starting point. And then I put my DNA, my RNA, sorry. It's not ready enough. My RNA. And I'll make it a little longer, make it a little charged. But then there are also, as we talked about, there are lots of proteins in here. Lots of enzymes in here. Then put some enzymes here, some enzymes here. And of course, they're soft. Plus, minus, plus, minus. And outside, plus, minus, plus, minus. Let's take general. Because I can always put the concentration of these added proteins to be 0 in my general problem. That's the easy limit. So let's think about this. How would we do this? So there are conformational fluctuations of the tails, the RNA, and also these proteins that I'm adding. But most importantly, it turns out, is the small ions, which are mobile, they create osmotic pressure. The translational entropy of these small ions is dominating much more than the conformational contributions to the free energy. It turns out, if you do the calculations. But now I have to make sure that I keep the, down in equilibrium. Namely, the chemical potential of the salt inside is same as the chemical potential outside. I have to make that condition. Otherwise, I'm not doing it right. Let's do this calculation. Something remarkable happens. If you do the calculation, what I find is a following. This is the free energy density as a function of theta that I access the genome that I'm putting in. The global free energy minimum corresponds to that number. That's about 0.2. That value of theta is being 0.2, 0.21, something like this. But then, this is for a particular salt concentration. Now I add the salt concentration. I increase the salt concentration. When I increase the salt concentration, that minimum turns out to be a little bit lower. But still, that's a global free energy minimum. Then when I go to about 0.25 molar monovillant salt, then that minimum goes away. It undergoes a phase transition. It is not a stable global free energy minimum anymore. This line is no genome at all. So it becomes unstable. The system becomes unstable. So when I add more and more salt, the virus doesn't like to have the genome inside. A very simple calculation, but it is kind of indicative of the stability of the system. Then you can calculate the susceptibility, et cetera, et cetera. So that's the conclusion number one. Now let me go to the second case, namely, uncomplexation. I think it's unstable. It becomes unstable. Unstable means that the encapsulated RNA inside the virus is unstable state. I'd love to see that. I would love to see those data, please. Already I was worried that 0.25 molar is very high. But my biologist friend told me that that's OK. There is lots of changes in the inside of the cell. But 0.6 molar, that's very interesting. Please wait for a couple of more slides, which may be related to a comment that you are making. But this is inside the virus. I'm talking about this is inside the virus. Let me go to the second question. Suppose I have a complex. And I want to uncomplex this. So let's think about this for a second. We all know that when you have two charges, we know the force between them. That is coulomb. That is 1 over r squared. And we know that in an electrolyte solution, the force between two charges is given by Debye-Hockel approximately, or Passan Boltzmann. It's going to look like this. It turns out what would be the force between two topologically coordinated charged entities? One is a poly cation, poly anion. None of those, obviously, cannot be, because the topological correlation contributes. And then we compute them. We calculate them. Analytically, we can calculate from the free energy. And also compute them to verify our approximations of reason number one. Then I make this complex. Then what I do is, as I'm repeating myself, bring an invader. And that invader is going to kick this blue out and then make the new pair. And I want to know what the substitution time is. Because that substitution time is critical for me to understand how I'm going to break up my virus inside itself, if this mechanism were to be operated. So let me show you one movie here. Here is a, first of all, this is a complexation. So in this complexation, the counterions are absorbing on the backbone of the chain here, green counterions. And then here is white counterions there. And when they are mating, the counterions are being released. And I used to think the electrostatic attraction is a dominant one in dictating the process. It turns out the release of counterions is actually dominating in the complexation mechanism. Because they are getting a lot of entropy. The release counterions have experienced a lot of translational entropy. And that is creating the complex. Now I have made this complex. Then what I do is, I break it. Break it means I bring an invader. And the blue and red is a complex. And now green is an invader. Please look at this. Lots of counterions are on the green. And that's a pair. And the pair is struggling to remain as a pair.  And here also the counterions are condensing there. It turns out, eventually the breaker will take place. It turns out the dominant force responsible for the substitution is counterion release only in that domain where the attack is taking place. However much charge you bring, it doesn't matter at all. Only at a place where the new chain is attacking, how much counterions that are released, that's the one that is contributing. So based upon the simulations, we looked at the time needed for substitution. And of course, when the ratio of the new chain, length of the new chain versus the previous chain is 1, it's infinity. The substitution will not take place at all. It turns out. And then eventually it goes down. This is 30, 30 length, this is 60, 60 length. And when the ratio is about 50% larger, 1.5, of the length of the new chain to the earlier chain, then beyond that, there is no gain in bringing a longer chain. There's no gain at all. So all that you have to do, if you want to break up this complex, you need to bring a chain which is about 50% longer. Doing more than that is not necessary. Now what is the driving force for this? Again, as I was repeating myself, this is entropy, release of counter ions. That's why electrostatic is not playing that important role in this substitution process, release of counter ions. Then we thought, instead of this, why don't you just challenge this complex with a salt? Because we thought it will loosen up. That's related to your earlier question. Suppose, then what we did was, here is the simulation. There are lots of salt ions from moving around. And if you're challenging it with a 0.5 molar sodium chloride type salt, then it breaks apart. Not 0.25 molar that came with the analytical calculation. With this simulation, 0.5 was good enough for short chains. We are talking about only 30 monomer chains. For that, 0.4, if you did 0.4 molar, it would not break apart. We have to go to very high salt here for this mechanism. So now let me go to the third one. How do I bring the polymer from the interior to the exterior? So here is the problem. Here I go back to theory. So what we do is, let me start with the inside. There's a charge density of the polymer, which is q. Charge density on the surface with the sigma. Salt concentration inside and outside is the same. And r, because I want to make sure the dynamic remains all OK. And r is the radius of this vesicular sphere. And then I want to take this to the outside. I want to calculate the free energy difference. And the way we do, as the experts in the audience know, is write it as a time-dependent short-engine equation. Vp is all the potentials coming from the polymer, polymer, polymer, polymer in here. And ves is between the polymer and the surface. And then for this particular problem, ves and the potential from the surface is written as a diva-huckel. And depending upon whether the exterior problem or interior problem, it's all standard textbook material. And v0 is an important quantity here. Various parameters come into one combined quantity. Sigma is a charge density on the surface. Q is a charge density on the polymer. This is a persistent length of the polymer. Kappa is the inverse diva length, depends upon concentration. And lb is the beer of length. For water, it's about 0.7 nanometers. So that is the setup of the problem. Then what we do is you convert this after massaging these various variables, convert it into this time-independent short-engine equation, and rx, the potential, the effective potential is like this. And then here, when you solve for the eigenvalue problem, then you can calculate the entropic contribution to the free energy. And then this is the entropic contribution to free energy. But the key variable here is this combined parameter. This is my interaction parameter, sigma, q diva length. And then there's a chain swelling, which is, you know, because when you change the salt, the polymer chain will change its confirmation. And that's the beer of length. So let me show the results. There are lots of puzzling results I just want to share as a challenge. We are not to fade it out in terms of analytical work, but the numerical simulations are very good. So the potential inside and outside, of course, has to be continuous. That's the way it is. And then the effective potential has a turning point. People who work with the quantum mechanics will know this. This is really a painful exercise. There's an int inside, outside, inside. There is region 1, region 2. We use a WKB. We're technically challenging. That's what we had to do. We did that. So inside what we found was this parameter 24 times pi times sigma critical charge density q, beer of length, has a function of inverse divide length in units of the persistent length, monomer length. There is a universal law that we get. If this quantity is higher than this curve, if it's above this curve, we will have complexation. And remarkably, that condition depends upon kappa as a 5 half power law. In the outside, asymptotically, the exponent turns to 11 over 5. Again, above these curves for different sizes of these viruses, we will have adsorption. And below that, there will not be any adsorption. Then there is one more remarkable thing that comes out of these calculations, which is if you consider the RNA inside, flexible polyelectric inside, there is a spontaneous selection of a virus size. This came out as a big surprise to us. And that spontaneous size, kappa times that radius of this virus depends upon the interaction parameter would be as a power law, 2 third power law, roughly 2 third power law. There must be an elegant analytical derivation, which I don't have yet. And we can trace the origin of this behavior, why there is a spontaneous selection and solicitor, but there is a spontaneous selection. And now let's look at taking the polymer from inside to outside. Again, we were very surprised. There could be a kubble there. The change in free energy to take the RNA from inside to outside for identical conditions is only about few kT. And of course, that depends upon solid concentration, but it is really modest driving force for expulsion. This is fantastic news. When you think about it philosophically, this is really fantastic. That's why it happens so easily. But this is what this calculation tells me. And it's really very small. That doesn't mean that there is no barrier between the initial and final state. In order for it to go through, it has to go through an entropic barrier. This is another life of me where I have been thinking about a polymer translocation. Christian knows that. I have been doing this. So the way the RNA is going to go through this capsid wall from interior to exterior, there are lots of free energy landscapes. You can write focal point description, da, da, da. And that's pretty much under control in my domain. And there is a barrier. That barrier is about 6 kT. It's not humongous. That's about 6 kT for these kind of polymers. So with these arguments, then let me conclude by saying that. So regarding the fluctuations in the genome size, Don and Iquibrim tells me that there is a phase transition in genome size. There is beyond certain amount of genome. According to this argument, it is not stable. An uncomplexation by competitor, substitution driven by counter and release. And the threshold length for effective substitution is really finite, very small. And I believe this uncomplexation is probably taking place in the cellular environment by salt gradient, not by bringing a competitor. I thought there was going to be some competitor which is going to break it, but I do not know. But I think that seems to be possible. And finally, in terms of translocation in exterior, the criteria for adsorption on the interior and exterior can be very easily derived. Only a few kT as a driving force for expulsion. I think I'm pretty much done. Once more, I love this quote from Luria. I must say that again. And I'm very grateful to my students. Thank you very much. OK. I'm yours. Now kill me.