 Hello, welcome to this lecture of bio mathematics. We have been discussing vectors for the last few lectures and also we discussed calculus, how do we do differentiation, integration and all that. Today, we will discuss an application of calculus and vector, the idea of that we learned in calculus and the ideas we learned in on related to vectors. We can combine together and learn a very important equation in biology or a very important concept in biology that is related to something related to membrane potentials. So, this is our aim today. Our aim is to basically understand a new interesting idea related to membrane potentials and to understand that, we will be using the ideas that we learned from calculus and the ideas that we learned from vectors. So, today's topic is basically. So, today we are going to discuss applications of calculus and vector algebra in biology and the important concept I think we are discussing is called the Nernst equation. So, this is an equation that relates the potential across a membrane to the concentration of ions. So, if you have some ions, you know that there is NaCl, KCl, salt solutions in the cell. So, they are in water typically K appear as K plus, they in solution in cells they will be K plus and C L minus. So, there will be ions of K plus and C L minus and you know that there are they can go K plus for example, can go across the membrane cell membrane and this can lead to some kind of electrostatic potential. So, as you learn in school when you have charges there is a potential can develop and this can lead to some electrostatic potential this movement of ions, the flow of ions across the membrane channel and to really understand. So, the Nernst equation is an equation. Many of you might have heard of this equation called Nernst equation. This governs, this tells you if we have some amount of ions on one side of the membrane and some amount of. So, you have a membrane and you have some amount of ions here and some amount of ions here. What is the potential across this membrane? What is the equilibrium potential? You can have you can put some amount of ion here, some amount of ion here and then they will reach an equilibrium because they can flow this way and that way and this flow will lead to an equilibrium and what is the potential at equilibrium across this membrane? That is what Nernst equation will tell you. So, this is what we will understand. This is an interesting and important idea in biology which is used in many places. So, let us start thinking about what we just said that is ions across a membrane like both sides you have a membrane separating the space into two parts. You can imagine one part was inside the cell and the other part is outside the cell if you wish. Essentially, what you have is a membrane that is separating physically, separating two regions in space that is outside the cell or inside the inside the cell outside the cell or either way you like. So, there is a physical separation that is important concept that is membrane and then membrane is a semi permeable membrane that that. So, what does that mean? So, let us have a look at it. So, let us have a look at this picture here in the slide. So, you can see this is k C L k plus and C L minus ions and here also k plus and C L minus ion. Now, if you look at here you can see that there are more k plus ions in this side. So, if you count here this is just a pictorial representation. So, you can if you count here you can see four of them here, but it does not mean four. So, you can see three here. So, you can what does it only mean what I want to represent in by this picture is that in this part of this partition you have more k plus than this part. So, the concentration of the k plus or C L. So, equal number of k plus and C L. So, 4 k plus and 4 C L minus in this side. So, on total this part of the box is neutral equal number of k plus and equal number of C L minus here also equal number of k plus and equal number of C L minus. So, this is also neutral. So, the concentration of k C L here is C 1 and the concentration of k C L here is C 2 and as you see in this picture C 1 is greater than C 2 that is the concentration here or the number of ions in a given volume is more here and less here. So, the volume is equal in both sides. So, the number is more here. So, more concentration here and less concentration on this side and this green line this is the line with dashed lines or there is holes here. So, this through this hole this smaller ion k plus can pass through. So, what does this mean the k plus can go either way. So, this is a semi permeable membrane it does not allow both k plus and C L minus to go through it only allows k plus to go through either way k plus can go from here to here and here to here whichever is permitted by the laws of physics it will go that way. If it is feeling some force if it is pulled in this way it will go this way if it is pulled this way it will go this way. So, now we have to see the physics and find out whether which way it is going, but remember C 1 is greater than C 2 that is more concentration here and less concentration here. Whenever there is something having more concentration in one side and less concentration in other side you know the you know that something like called diffusion will happen diffusion is the concentration will change things will flow from higher concentration to a lower concentration something like there is there is a diffusive flow. So, this will happen now because you know that one side it is more concentration the other side is less concentration. So, as you can see from your intuition things will go from higher concentration to lower concentration. Why this all this we will understand later why is this happening we will learn about diffusion little more carefully in the coming lectures, but at the moment you just need to understand that it will go from higher concentration to it will flow from higher concentration to lower concentration which is intuitively probably you know. So, that is what shown here in this arrow. So, the only k plus and the k plus again will go from here to here there will be a flow of k plus ions. So, flow as we said is a vector. So, to say something about the flow as you can guess we will have to use some ideas about vectors because flow of k plus ion in this direction is a vector. So, what what does this mean? So, there is a flow and how do we say this mathematically how do we say this in mathematically how do we say that there is a flow from here to here before that let us just understand a bit more about the concentration. So, you can see that concentration decreases as we go from here to here. So, let us plot for a minute here. So, have a look at this here. So, let us plot a minute here the concentration and the x axis. So, you have this here. So, more concentration here less concentration here. So, the more concentration here less concentration here. So, what does it mean? There is like lot of particles here and less and less particles here. So, this is many, many, many, many ions here and as it goes it will be like less and less. So, there is a higher constant c here is greater than c. So, if you take this as the x axis. So, if this is you take this as the x axis as you go along the x axis the concentration is decreasing. So, the concentration. So, let me plot concentration versus the distance from the left end. So, at the left end there is big large concentration. So, let me mark here at the left end you have large concentration and as we go little here the concentration decreases and further you go concentration decreases and further you concentration decreases. So, the concentration decreases it need not decrease in this particular way it can decrease this way also. So, we do not know which way is decreasing, but it is clear that the concentration is decreasing as we go along. That much is clear. Now, if let us imagine the first case. The first case is, you know, you, we learned in calculus that if we have a straight line like this, this is an equation y is equal to minus m x plus c, where this m, the slope is negative. What does it mean? As x increases, the c decreases. So, that is why this is negative. d y by d x, the change in x and change in y, this is minus m. It is a negative number. So, d y by d x is negative. So, that means here d c by d x is minus m. This is a negative number. So, this is less than 0. It is negative. So, this much. So, as we said, we also, we can also represent this by partial derivative del c by del x. This also means if c, which is a concentration, can be a function of the distance, the space and it can be also a function of time. So, if we are only finding the derivative with respect to x, it is a partial derivative. So, we will, we can represent del c by del x. So, but in this lecture, most of the time, we are not considering time. We are, we want to discuss the equilibrium properties of the system, where the time is independent of time. So, even though we will write d c by d, del c by del x, the partial derivative, essentially it is same as the full derivative d c by d x here, because we are not considering our time. So, we will interchangeably use del c by del x and d c by d x. We will interchange and use del c by del x and d c by d x. But anyway, the point is that d c by d x is negative. If you take this or even if you take this, the d c by d x, the c decreases. So, d c, as we go along the x, the change in c is less, it is negative. That is, the del c, d c is negative. The c change, the c decreases. So, this is the point to remember. The point to remember is that d c by d x is negative. So, keep this point in mind. Keeping this in mind, let us look at. Now, what do you want to look at? We want to look at, so let us look at, we want to look at the flow. So, we have this, we have this membrane and we have ions here and ions here and c 1 is greater than c 2 and we want to know about this flow from here to here with a larger concentration to a. So, this I, we said is this length of a diffusion. So, you have a diffusive flow, if you wish. So, let me call this J d. It is the diffusive current and it is a vector. It has a direction. So, we know that it has, the direction of this has to be along the x direction. We know from this. So, there is some magnitude and some direction and we do not know what the magnitude is and we know the direction. It has to be along this because here it is more concentration. Here it is less concentration. So, it has to be along this. We can also guess that more the concentration difference, more the flow is. If it is a large concentration here and small concentration here, the more it will flow. So, this is proportional to del c by del x. This much is also clear. The more the concentration, the more the flow. This is clear because the more the concentration difference, if the del c by del x is the large, like that means a huge difference in concentration, there will be, the flow will be more. If the del c remains, if del c approaches 0, that means if delta c goes to 0, the flow will go to 0. That is clear. So, there are two things. The direction is this, we know. It has to be proportional to this. But in this case, we just discuss that del c by del x is negative. So, if we say that J d is some constant times del c by del x, x, this is negative. So, this will be del c by del x is negative. So, what is, what would this mean? This would mean, so this is some proportionality constant which we have to find. So, this would mean that this is a negative quantity. So, negative x, this would mean that this goes along the, this direction because del c by del x is already negative. We found that. So, if this is negative, this would say that it goes in the minus x direction because negative times x cap would mean that it goes in the negative x direction. But that is not true. It is going in this direction. So, there has to be a negative sign here, extra. So, that del c by del x is negative. There is a negative sign. So, totally it is positive. So, J d is along the x cap. So, just from mere common sense, we deduce this much. What did we, how did we deduce? First we said, first we said that c 1 is greater than c 2. So, it will flow from here to here. So, if since this is the x axis direction. So, we call this, this direction as x cap direction. The direction of the flow, J d is the current or the flow. So, this is the current of ions in this direction. So, this current due to this concentration difference has to be in the direction of x. That much is clear. It is also clear that it has to be proportional to del c by del x. Del c by del x is the change in the difference in concentration. The change in concentration as we go along x axis. So, if you look at, if you look at, at this point at this interface and look, this is the concentration difference between here and here. The more the concentration difference, the more the flows. We are interested to calculate the flow at this interface. At this particular interface, how much is the flow? That is our interest. Whether it is flowing this way or this way and how much is the flow? That is the thing that we are interested in. So, if you look at this point, the more the concentration difference, the more the flow. Now, you can write this proportionality constant as some constant into del c by del x. Now, del c by del x is negative, which we just said, because since the concentration decreases as we go along the x, del c by del x is negative. So, if this is positive, it will mean some negative x cap. That means it would mean, the direction of the g d is along this, which is not true. So, you have to add a negative sign here. So, this leads to our next slide, where we said the current j d is minus d times del c, where del is something, which we learned in the last, in the previous lecture. So, one of the lecture, we said del is, for one dimension, it is del c by del x x cap. Del is a vector, it is a gradient vector. So, we said that del is a gradient vector. This means del by del x x cap in 1 d. In 2 d, the del would mean del by del x into x cap plus del by del y y cap. In 3 d, the del would mean del by del x into x cap plus del by del y into y cap plus del by del y del z del by del z into z cap. So, there is, this is in 1 d, 2 d and 3 d. So, this is 2 d and this is 3 d. So, 2 d and 1 d. One dimension, 2 dimension, 3 dimension. So, if you are talking about flow in a plane, if things are flowing in all the direction in a plane, we have to apply this derivative, because there is, things concentration can change along x and y. And if it is flowing in 3 d, all over the 3 d, then you have to use the third derivative that we discussed. But here, for simplicity, we will only consider flow along a line, because there is easy to understand and whatever we learn is very easy to extend to 3 dimension, if you wish. It is easy. So, here we will use 1 d to make everything simple, so that it is easy for you to understand. And once you understand the simple idea, you can always go ahead and do the more little more complicated generalized version of 3 d. So, let us look at here. So, what we said so far is that the current, the flow due to the diffusion or the concentration gradient, the concentration gradient flow is essentially minus d del c by del x, x cap. So, this is the current. Now, what will, what will this lead to? So, what did we say? So, we had, we had this situation, where k plus will go from here to here, because it is more concentration here and less concentration here. And what would that lead to? That would lead to this. So, more k plus here and more k plus here and less. So, it will lead to the k plus going here, but only k plus can go. So, here this will be more and more positively charged. This side of the semi permeable membrane will be more and more positively charged. And here there are more and more C l minus, because C l minus cannot go. So, what will we end up with? We will end with more positive charges here and more negative charges here. What will happen if we have more positive charges here and more negative charges here? This k plus will be attracted back by this negative charges. This negative charges will attract this positive charges. So, we have more C l more negative charge here. So, that will attract this positive charges. So, there will be an electrostatic attraction on k plus, just because many k plus flew, they were flowing in this direction. So, the concentration or the number of k plus increased here. So, the number of positive charges increased here. So, number of, therefore, number of negative charges increased here, as you can see from the picture. Do look at this picture carefully. So, now this positive, for example, this k plus has 4 c l minus here and 3 c l minus here. So, this k plus will feel more attraction towards this side on an average. So, this k plus would want to go this way, because there is an electrostatic attraction. So, what will happen? So, the k plus will start now flowing in this way. So, now how much is this electrostatic attraction? How much will be the flow of k plus in this direction due to the electrostatic attraction? So, let us see how much that will be. So, what we have now, what we have now is more positive ions here and a few negative ions only and more negative ions here and only a few positive ions. So, this positive ion will have to, will be feeling an electrostatic force and there will be a flow. They will flow, they will flow with a, so if something, so there is a force. So, if you push something in water, so this is water here of course, this is water everywhere. So, if you something, you are pushing this electrostatic forces or pulling this electrostatic forces by a force and this force is an electrostatic force. So, you know that if you push anything in water by a force F, especially in bar, in bar, very small ions in highly viscous. So, compared to, so this is something called a low Reynolds number regime, but does it mean is that friction is very high. So, high friction regime, because there is, compared to the size of the ions, the friction is very large. So, that means, this small ions will be feeling lot of friction. So, the force, whatever the force you feel, the electrostatic force. So, you will, each of this plus ions will be feeling some electrostatic force. This electrostatic force will be proportional, will be, you know anything moving in this spherical particles, the force, the drag force is 6 pi eta A v, where v is the velocity of this. So, this will move with a velocity v given by this formula. If you have this much force, electrostatic force, the velocity will be given by this formula F is equal to 6 pi eta A v. So, now, F e is the electrostatic force and this is the drag force. So, when they are equal, there will be like a, there will be a uniform flow, there will be a steady state flow or is a steady flow and this will, the velocity of this flow will be given by this v, which is related to this electrostatic force by this formula. So, we know that there is a electrostatic force. Now, what is this electrostatic force? How much is this electrostatic force? How much is this electrostatic force? There is, there will be an electric field. So, if you have an electric field E, the force is charge times electric field. So, force F is q times e, q e charge times electric field is the force. So, this is the, this is the force. So, the force is charge times electric field and this will, this force will be equal to this flow, this 6 pi eta A v, because there will be So, the velocity of the flow will be given by this. So, if we know the velocity of the flow, so what do we know? We know the velocity of the flow. We know that F is, the electrostatic force will be equal to 6 pi eta A v. That means v is F e divided by 6 pi eta A. So, we know that velocity is a vector, force is a vector. This way we have vector signs. Now, we also know the direction of the velocity. It will go from more positive charge to less positive charge or here more negative, I should say, more negative, more positive. So, in this way it will flow. It, this will be the velocity direction will be in this direction and the force direction also will be in this direction. Now, the current due to the electric electrostatic interaction, what will be the current? We found out in the previous case, the current, we found out the current due to the concentration difference. We called it J D. Now, we will find out the current due to the electrostatic interaction and we will call it J E. So, this J E has to be proportional to this velocity, because more the velocity or more the force, the more the current. You know that velocity is proportional to the force. So, the more the force, more the current. So, by just looking at the dimension, by dimensional analysis, we will know that this proportionality constant has to be concentration. So, the proportionality constant here, so J has to be written as C v, where C is the concentration at this, at this particular point. At this, any point, the flow will be the concentration at that point, times the velocity at that point. If you do not understand how the C is coming, just look at the dimension of velocity and current, the J and you will see that they have, they are related to this C v. Now, known this, let us look at this slide. So, this is the essential message, which I am giving here, that the current J is C v and the force is 6 pi eta A v, which is, the force is also Q times E, where E is the electric field. And from this, we know that Q v is Q E by 6 pi eta A. I just take Q E this way. So, you get Q E is equal to 6 pi eta A. So, essentially what you have is J E is C times Q E by 6 pi eta A. So, just I substitute this v in this equation. So, v is Q E by 6 eta A. So, I substitute for this v from here for this velocity. I just simply substitute this. So, I get J E is C times Q E by 6 C j y. So, now, we found out two currents J E and J D. This is simple algebra. I just rearrange all this and then I can get this. So, I have now two currents. The diffusion or the concentration gradient will lead to a current in this direction and illiterative static attraction will lead to a current in this direction. So, we have two currents. One current in this way, which is the concentration gradient, which we said and the other current as this flows in this way. The more, as the concentration gradient leads to some flow in this way, the more positive charge will build up here and more negative charge here, which will generate an opposite flow. So, there will be two flows this way and this way. So, when this both this flow balance, that means the flow this way is equal to the flow this way or the total flow here, if the total flow is 0. So, when the flow is equal, the total flow is 0 because it is now it is flowing this way and flowing in this way. So, the vector sum of the flow at this particular point, when that becomes 0, when the vector the flow the sum of the total flows is 0 in this along this membrane, you have equilibrium. We call it an equilibrium. We can call it an equilibrium. That means the flow in this way balances the flow in this way. So, when this, if 5 of them go this way and 5 of them go this way, 5 of them go this way and 5 of them go this way, 5 ions go in, let us say in 1 minute, 5 ions go this way and 5 ions go this way, then on an average the number of ions on both sides will remain the same because same number goes this way and same number of goes this way. So, we can call it equilibrium because the number on an average will remain the same and the flow will be equivalent opposite. So, when the flows are equivalent opposite, what we get is equilibrium. So, we want to know at equilibrium what happens, what is the force and what is the concentration difference etcetera at equilibrium, we want to know that. So, let us go ahead. So, we understand what is the, so at equilibrium the flows should balance, the flow in this direction should be equal to the flow in this direction. The flows should be equal and opposite. So, let us go here. So, when equilibrium, when both the currents are balanced, we reach equilibrium. So, and let us assume that C 1 and C 2 are the equilibrium concentration of k plus ions at either side of the membrane. At one side at equilibrium that is when the flows are balancing, when the flows are balancing on an average there is C 1 be the concentration on one side of the partition, the other side of the partition we have C 2 in the concentration at equilibrium. So, these are equilibrium concentration, this is not the initial concentration initially we had some concentration. So, now at let me call this initial concentration C 1, 0 and C 2, 0. So, now this is the C 1 and C 2 the concentrations at equilibrium. So, if C 1 and C 2 are the concentrations at equilibrium, the question we want to ask is that will there be any potential across the membrane or not. So, C 1 and C 2 may not be equal, we do not know. If they are not equal what will be the potential? That is what we are going to calculate. So, let us go and calculate this, but before that let us understand the idea of equilibrium a bit more. So, what did we say just now? We said that the flow there is two flow there is J D and in this way there is J E. So, they have be equal and opposite or in other words at this point the total flow the J D plus J E has to be 0, the total flow has to be 0. Then only we will get this equilibrium. So, at equilibrium what we expect is that this flow will balance this flow, the total flow at this point will be 0. In other words J D will be equal to minus J E the flows will be equal and opposite. So, this minus sign gives you some direction. So, there will be equal and opposite. So, that is what we, we want that equilibrium. So, now let us have a look at a few more details. So, what did we say a minute ago? We said that when there is charges there will be an electric field. So, there is one more idea that you want to understand about electric field and charges etcetera. So, from the Coulomb's law we know that the force, the Coulomb force is if you have charges q 1 q square if you have two charges you have q square by k r square. This is the, if you have two charges q of, then this will be the electrostatic force. If it is plus and minus it is, it will be minus q into minus, it will be minus sign. So, but this will be the force. Now, we also just a minute ago we said that q E is equal to f. So, what does this mean? E is equal to f by q. So, which is q by k r square. This is our E and we said in previous lectures that the force is minus del E by del r, del energy by del r, d energy by del r. So, let me call phi as the energy here. So, let me, we said that f is generally the force is derivative of energy. So, we said that force is derivative of energy. So, if you use that idea that the force is derivative of energy in this, what you will end up is that electric field is derivative of electrostatic potential with a minus sign. But there has to be some direction associated with this. So, there has to be some unit vector. So, let me have some unit vector here in some particular direction. So, in other words E can be written as minus del phi, where del is, as we said this means minus del by del x of phi along the x. So, this is the electric field. So, phi, where phi is the electrostatic potential. So, the potential energy. So, if you know the potential energy, the electric field and potential energy are related in this particular way. So, knowing this, so this is some simple idea that we remind ourselves from, that we have to remember from electrostatics. So, if we know that, if we know that electric field is derivative of potential, we can go back and do the algebra of all the equations that we learnt. So, we have to just remember that electric field is minus del phi, where phi is electrostatic potential. This is another vector relation. So, the E is a vector and del is a vector and phi is a scalar. Potential is like, but energy is a scalar. So, this is just like force is derivative of energy, something equivalent of that here. So, now we can substitute this E in the J E we had. So, J E, we said that, we had said that J E is q c E by 6 phi eta a and E is substituted with del phi. So, for electric field, wherever we see electric field, we substituted as minus del phi and del phi is del d phi by del x, d phi by del phi by del x with x cap and there is a minus sign here, which I put here. This only means that, this current is along the minus x direction. It is along this direction. That is what we have been telling all the while, that the current is along this particular direction. So, now what does this mean? So, we have a current, we have an expression for the current, we have an expression for the J E and we just said that, at equilibrium 0 net flow, there is no net flow at equilibrium. This means, I we just said J D plus J E equal to 0. That is the net flow at the partition is 0. That is J D equal to minus J E. Now, we can substitute for J D and J E. So, we go back and, if you go back and look the expression for J D, we had written minus d d c by d x x cap and J E was 6 into, sorry c into c by 6 phi eta a into d phi by d x x cap. So, we might have changed the partial derivative to full derivative, because the phi and c at this moment only depends on x. It does not depends on time. It does not depend on time or any other quantity at this moment. They all at equilibrium and we say equilibrium is independent of time. There is no idea of times like forever. This is true. That is for irrespective of the time, this is true. So, d c by d x is d phi by d x. So, now, we have such an equation. This is a differential equation. So, there is derivative here and the derivative here. Now, what we can do? We can take this d x this side and this c this side. So, if I take this c to the left hand side, I will get d c by c. So, there is a way here d c by c and I take this d x this way. So, this will d x into d x. So, and essentially what you have is d c by c is equal to minus q by 6 by eta a d into integral phi 1 to phi 2 d phi and here c 1 to c 2 d c by c. So, you can integrate like this and you essentially end up with this integral. So, what would this lead to? If you do this algebra, what you get is that long c 2 by c 1 is equal to minus q by 6 by eta a d into phi 2 minus phi 1, if you do this integral. So, there is a minor typo here. So, the right way to write this equations, let me write. So, have a look at this equation. So, let us look at this equation. So, this equation let me write here. So, if you write here you have minus d d c by d x x cap is c by 6 phi eta a d phi by d x. So, this is also x cap. So, this quantity in the brackets they will be equal. So, this would now I can take this c here and this d x there. So, I take this c here. So, you have minus d d c by c is equal to I say 1. There is a q here. So, q by 6 phi eta a. So, there is a typo. So, q here. So, this is the correct one. q by 6 phi eta a into d phi by d x times d x. So, this is the equation. Now, we have to integrate. Now, I can if you wish I can take this d also this side. So, I can write this d here if you wish. But, I can integrate both sides this c 1 to c 2 and this phi 1 to phi 2. So, if you do this integral properly you know that integral of 1 by c is log c. So, let us let us do this carefully this integral. So, I take this d this side and rewrite this equation. So, if we rewrite this equation what you get we have integral c 1 to c 2 d c by c. So, that is the and I take minus sign this side. So, if I take the minus sign this side what you get is minus q times 6 phi eta a d 6 phi eta a d and there is this integral. Integral phi 1 to phi 2 d phi by d x into d x. So, you know that this d c by d s is log c. So, log c in the limits c 1 to c 2. So, integral of this d phi by d x d x is equal to minus q 6 phi eta a d into phi in the limits phi 1 to phi 2. So, this is just d phi by d x d x this integral is essentially phi because if you integrate d phi by d x you will get phi. So, you have phi in the limits phi 1 to phi 2 and log in the limits c 1 to c 2. So, if you have this integral you can say that apply this limits. . So, you have log log c 2 minus log c 1 that is the if you apply limits on the left hand side you get this minus q 6 phi eta a d phi 2 minus phi 1. If I apply the limit here it is phi 2 minus phi 1. So, log a minus log b is log of a by b. So, I can write this thing log of a minus b as log of c 2 by c 1 is minus q by 6 phi eta a d into phi 2 minus phi 1. So, as some of you are familiar with this equation can already see the Nernst equation appearing. So, what is this? Log c 2 by c 1 is equal to minus q by 6 phi eta a d into I can write delta phi. Delta phi is nothing but phi 2 minus phi 1 if you want I can I can write it as delta phi if you wish. So, this is the change let me write it phi 2 minus phi 1 itself. So, the change in potential is related to the change in concentration. So, this is potential difference. So, difference in potential between the two sides is related to the change in concentration in this particular way. So, this is essentially the Nernst equation. Now, if you want I can beautify this constant a bit by substituting something for d. So, let us do that. So, this is essentially the Nernst equation we have already obtained the Nernst equation. Now, I am just rewriting in a different more familiar fashion. So, let us have a look at here. So, what we have is this and now there is something called famous Einstein's relation. Einstein in his 1905 very famous paper derived the relation between diffusion coefficient d and temperature and viscosity. So, K b t is K b is Boltzmann constant t is temperature eta is the viscosity a is the radius a is the radius of the particle. So, if you have this many variable. So, this is the relation between the diffusion coefficient d and temperature and eta which is the viscosity. So, if you just borrow this relation and substitute here for this particular d what you would get is that you will get that ln c 2 by c 1 is q by K b t because as you can see here d is K b t by 6 pi eta a. So, if I substitute here 6 pi eta a and the 6 pi eta a they will cancel each other and you will end up with this equation q by K b t into phi 1 minus phi 2. So, if I call phi 1 the with the phi 2 minus phi 1 as gel phi I can write this particular relation. There can be a phi 1 minus phi 2 as gel phi I can write this particular relation. So, if del phi is K b t by q into ln c 2 by c 1. So, this is the Nernst equation basically here what is happening is that the potential difference is related to the concentration across the membrane. Repeat this slide. So, what we essentially will end up getting is this equation as you see here log of c 2 by c 1 which is the concentration on both sides c 2 and c 1 are the equilibrium concentration at either side is related to is equal to q by K b t into phi 1 minus phi 2 phi 1 and phi 2 are the potential this is the potential difference across the membrane. So, phi 1 minus phi 2 is the potential difference. So, if I call this del phi delta phi. So, the delta phi is equal to K b t q by K b t by q into ln c 2 by c 1. So, the potential difference across the membrane is related to the concentration difference in other words. So, this potential difference across the membrane is related to the concentrations of irons equilibrium concentrations of irons across the membrane. So, this is at equilibrium if there is a concentration difference c 1 and c 2 they will lead to some if the equilibrium concentrations c 1 and c 2 that will have a potential difference that will develop a potential difference delta phi given by this particular equation. So, this equation is called the Nernst equation. So, we learned how to derive the Nernst equation. So, to summarize what we said so far in summary we first that the flow of irons due to concentration difference we discussed. We said that they go in a particular direction and in the opposite direction there will be a flow due to the. So, this flow of irons due to concentration difference will lead to some kind of charge differences. It will lead to positive charge end up in ending up in one side and the negative charge ending up in other side and that will lead to a counter flow of irons due to electrostatic interactions because this will lead to charge separation and charge separation will lead to electrostatic interactions and there will be a flow in the opposite direction. So, when these two flows balance we get equilibrium and at equilibrium there is a potential difference across the membrane and the potential difference is given by this particular equation delta phi is equal to k B T by q in the long c 2 by c 1. So, this is the potential difference and this is the famous Nernst equation. So, by just knowing by using the ideas from vector as well as calculus. So, vector algebra or direction vector and ideas from calculus we could derive this equation the Nernst equation which is a very important equation in biology because the potential across this membrane the membrane potential is the thing that drives lot of movement of irons across the membranes and this can this has lot of role as we go along we might come and discuss some part of this role of this in biology. So, at this with this lecture we are stopping here we are concluding with this Nernst equations. So, we understand the Nernst equation and we will go ahead and learn more new things in the next coming classes. So, bye.