 So, we have already told two experiment, first one is turn experiment, second one is Jutman and Ko, Jutman and Ko, third experiment just I want to tell, which is the turn improved experiment. Actually, in this experiment, turn has improved his earlier experiment, so that is the reason it is known as turn improved experiment. So, in 1947, so turn Esterman and Simpson, they have arranged a more precise experiment for verifying the law of distribution of velocities. This experiment is based on the method of molecular beam, where the free fall of the molecules is observed in the gravitational field. Actually, there they have exploited the free fall of molecular beam in the presence of gravitational field. The experimental arrangement is schematically shown in figure, I will also explain it. The apparatus consists of an oven just like Jutman and Ko with a narrow horizontal opening A. Here, the working substance they have taken cesium, cesium is taken as the source of atoms, which is heated in the oven V. A slit S is placed at a distance of 1 meter from A, T is a thin tung strand wire placed at a distance 1 meter from the slit S. Basically, it serves as a target itself. The entire arrangement A, S and T are all along strictly horizontal line. The entire arrangement is enclosed in a highly evacuated chamber. So, obviously, you understand it. When Stern has done his first experiment, that time the whatever the vacuum he has maintained, this time he has maintained higher degree of vacuum. So, that is the reason it is known as improved experiment. Main thing is the vacuum of this chamber, that is the main idea actually. So, the cesium atoms flow out of the oven through A. A is the narrow opening. In the absence of gravitational field, they will strike the target at T directly. This is obvious. If there is no gravitational field, so it will hit directly to the target T. However, due to gravitational field, the atoms travel along a parabola. That we know the trajectory of the particle in the presence of a gravitational field is a parabola. So, obviously, so in the presence of gravitational field, the atom travel along a parabola. The atom emerging from A with a velocity horizontal along the x axis will not pass through the slit S and will not reach the target. The atom emerging from A at small angle theta will pass through S and strike the target. The tungsten wire target is heated by an electric current passing through it. When cesium atoms strike the wire target, they get ionized. These positively charged ions leaving the target get into the negatively charged cylinder surrounding the target. Thus an electric current of ion passes through the wire and the cylinder which can be measured with accuracy. The ionic current gives the number of atoms hitting the target. This is the definition of current. Moving the target in a vertical direction such as position T prime and T double prime. The ionic current and hence the number of atoms hitting the target is measured at different heights. We find the number of atoms having different velocities. The atom hitting at T prime have velocity higher than those at T double prime. This gives us the distribution of atoms with velocity. This observation experimental observation which is in complete agreement with Maxwell's law of distribution of velocity. This is the experiment which proves very accurately Maxwell's law of distribution of velocity. So now next couple of minutes we will try to we will try to deduce some equations of some law from Maxwell distribution law. First calculation of mean velocity, mean square velocity and most probable velocity. The mean value of C n is defined as in general average values of c to the power n is 0 to infinity c to the power n f of c d c. f of c d c is the probability of finding c n variable c n c plus d c where f of c is given by the Maxwell distribution. Then we get c to the power n is 4 pi a q 0 to infinity. If you will substitute f of c e to the power minus b c square c to the power n plus half to d c. So, if you will substitute b c square equal to x, then average values of c to the power n is 2 pi a q times b to the power n plus 3 by 2 0 to infinity e to the power minus x x to the power n plus 1 by 2 d x. This is just you substitute the variable b c square equal to x and if you will do a little bit mathematics you will get this expression. This is a standard integration and this standard integration can be written in terms of this gamma function. So, it is given in any mathematical physics book. So, this value is 2 by pi to the power half b to the power minus n by 2 gamma of n plus 3 by 2. This gamma is nothing but the gamma function. Its value is tabulated in any mathematical physics book. From this one can calculate the mean velocity, mean square velocity etcetera. So, let me first calculate mean velocity, then mean square velocity and then we will calculate some other velocity and try to see a relation among them. So, first we want to calculate the mean velocity. That means the mean velocity c. So, in that case in the in our general expression we just put n equal to 1. That means c to the power 1 which is c in the equation what we will get it. Average values of c we will get 2 pi b to the power minus of gamma of 2. Gamma of 2 means gamma function having argument 2. If you will put the value of gamma 2 which is nothing but the 1 as we know. So, which is nothing but the 2 into pi b to the power minus half. So, we know b is m by 2 k t. So, if you will put again the values of b in the above expression what we will get. The average values of the mean velocity is 8 pi 8 k t by pi m to the power half. This is the mean value of a velocity c which is obviously proportional to the temperature and the more the temperature the average values of velocity will be more and it is inversely proportional to the mass of this atoms or molecules. So, if the mass will be large then the average value mean velocity will be small. If the mass is small then the mean velocity will be large and that happens in the in our atmosphere also. If I want to calculate the mean of the square velocity then I have to take the mean of c square then in that case I have to put n equal to 2. So, if I put n equal to 2 then I will get the mean of c square is just I will substitute values of n equal to 2 I will get 2 by pi to the power half b gamma of 5 by 2. Just we will know what is gamma of 5 by 2. So, which is nothing but the 3 by 4 square root of pi that we know gamma of n equal to n gamma of n minus 1. So, if these values are tabulated in mathematical physics book if I will see it then I will get the value which is nothing but the 3 by 2 b. If I will put the values of b which is m by 2 k t which is 3 k t by m. So, this is the mean of the square which is c square which is square root of 3 k t by m sorry 3 k t by m. If I want to calculate the r m s velocity which is the square root of the mean of the square velocity in that case it will be square root of 3 k t by m. So, in both cases the mean velocity as well as the r m s velocity in both cases velocity is proportional to the temperature which is physically which is which is physically which physically looks which looks feasible and which is always inversely proportional to the mass of this molecule or atom which is also physically it looks physically feasible. So, then I will see the ratio of mean of the velocity and the r m s velocity which I will get it. So, 8 by 3 pi to the power 1 by half which is 0.921. So, that means the mean velocity is smaller than the r m s velocity by the amount 0.921 percent. So, now let us calculate some other velocity which is very useful quantity in this area which is known as the most probable velocity which is the values of c at which f of c is maximum. I have already shown you when we have discussed the figure f of x versus x where I have shown you when f of c will be when the probability is 1 at what values of the variable it is 1. So, that this is known as the most probable velocity that time we have already calculated. So, the other way to calculate it if the f of c is maximum then d f by d c should be the 0 you know what is f of c is 4 pi a q e to the power minus b c square into c square this is the probability. So, if you will this if you will demand that probability is maximum then d d c of this d of f of c should be the 0. If I will differentiate it I will get the equation 8 pi a q c e to the power minus b c square into 1 minus b c square equal to the 0. Since a is a constant since a q c e to the power minus b c square not equal to 0 then we have 1 minus b c square equal to 0 or c equal to b to the power minus half. So, which is nothing but the if you will inverse it then which is 2 k t by m. So, this is the most probable velocity. So, till now we have calculated the three velocities first one is the mean velocity second one is the square root of the mean of the square velocity which is nothing but the root mean square velocity third one is the most probable velocity. Let us see what is the relation among these three velocities. So, if you will write it most probable velocity in terms of the root mean square velocity then it will come out c m p equal to root over 2 k t by m then just multiply 2 third inside and 3. So, we will get 3 k t by m into 2 third. So, which is nothing but the square root 2 by 3 times root mean square velocity since square root 2 by 3 is less than 1. So, obviously most probable velocity is less than the root mean square velocity. So, now if you will compare the above 3 velocities we get a beautiful relations which is nothing but the root mean square velocity is just greater than the mean mean velocity which is greater than the most probable velocity. So, this is the beautiful relation among the three velocities. These three velocities sometimes depending on the situation these three velocities plays an important role. So, one must know the form of these three velocities. So, now I will calculate different quantities using the Maxwell law of distribution of velocity. First I want to calculate what is the mean kinetic energy. We have already calculated the mean kinetic energy in my earlier lectures on kinetic theory of gases, but there I have done in some empirical way because there I have taken there is no concept of temperature as such. The concept of temperature was a foreign element in the kinetic theory of gases where we have we had brought the temperature through the experimental equation of state. If you remember my earlier lectures there temperature was in foreign element, but here just let me see how the temperature has come here temperature will come automatically. So, let us see let a particle of mass m move with a velocity c. Then its kinetic energy is e equal to half m c square, but anyway let me say some few words. Here we are talking the Maxwell distribution of velocities for the monoatomic gases. So, obviously here the kinetic energy although it is not I have told you, but it is implicit here kinetic energy is nothing but the translation kinetic energy. So, hence the its mean will be e average is average of half m c square. Then we know the definition of average definition of average of any quantity is integration that quantity we have already told you average values of c n is integral c n f of c d c 0 to infinity. So, same thing is here also 0 to infinity half of m c square f of c d c half of m m by 2 I will come out of this integration. So, what I will get it once I will substitute f of c f of c is the it will come from the Maxwell law of distribution of velocity. So, if you will substitute values of f of c. So, what we will get 4 pi a q 0 to infinity half m c square e to the power minus b c square c square d c. So, half of m will come out. So, finally, we will get 0 to infinity e to the power minus b c square c to the power 4 d c. Then if you will integrate that equation that integration this is a very standard integration if you will integrate that equation what you will get you will come 3 by 4 b which is 3 by 2 k t here it will come automatically. Now, you understand the difference of my earlier lecture to these lectures in my earlier two lectures if you remember correctly their temperature was a purely foreign concept. The concept of temperature has come from the experimental equation of state from there we have made a statement that the average kinetic energy is a interpretation of the average kinetic energy as a temperature. But here the average values of kinetic energy will come automatically the temperature has come that is the beauty of it. Third thing we want to calculate what is the momentum distribution of particles or molecules. Suppose, somebody could ask what is the probability of finding the molecule having momentum p and p plus d p. So, answer is very straight forward if you know from the distribution of velocities that means Maxwell's law of distribution of velocities tells that velocity of molecules having velocity v and v plus d v. So, just you know velocity and momentum at least for non-relativistic particle you know the relation between momentum and velocity p mass times velocity just you rewrite the Maxwell distribution of velocity in terms of the coordinate in terms of the variable momentum. So, you will get the Maxwell distribution of the moment of the particle so that is what we are going to do it. So, according to kinetic theory of gases the number of molecules having the velocity lying between c and c plus d c is given as d n c equal to n c time d c which is 4 pi n m by 2 pi k t to the power 3 by 2 e to the power minus m c square by 2 square to d c c square d c. The momentum of a particle is mass times velocity here mass times velocity c please do not confuse this c with the velocity of light in free space since we have no other options left. So, we are forced to use the velocity c this velocity is completely the velocity which is nothing to do with the velocity of light in free space. So, be careful. So, otherwise you cannot write p equal to m times c as you know this is the purely non-relativistic expression using this expression the number of particles having the momentum lying between p and p plus d p is expressed as n p d n p equal to n p d p is 4 pi n 2 pi m k t to the power minus 3 by 2 e to the power minus p square by 2 m k t into p square d p. This is the distribution law of momentum of molecules. Fourth energy distribution of a molecules now somebody could ask no I am not interested the velocity distribution of molecules I am not interested the momentum distribution of molecules I know the energies you tell me what are the number of particles having energy e and e plus d a. So, let me so energy of a molecule obviously in that case we assume that they are non interacting in the sense that there are no interactions among them. So, that means e energy is nothing but the kinetic energy however we can incorporate the potential energy also means we can incorporate the interaction among them also. So, we will do it in the next couple of minutes the energy of a molecule is half e equal to half m c square the interaction among them I have not been incorporated. So, if you will substitute this e equal to half m c square in the velocity in the Maxwell distribution of velocity we obtain an expression. So, if you will substitute in the equation 54 we will get. So, number of molecules having the energies between e and e plus d e is d n e equal to n e d e equal to 2 n pi to the power minus half k t to the power minus 3 by 2 e to the power minus e by k t e to the power half d e. This is the distribution of energies of molecules lying between e and e plus d e. So, finally we have got Maxwell distribution of velocities Maxwell distribution of momentum Maxwell distribution of energy. So, depending on the situation we can use the Maxwell distribution law if I needed the Maxwell distribution of velocity then I will use that if I need Maxwell distribution of energy then I will use that expression. So, depending on the situation I will use the Maxwell distribution law for velocities for momentum for energy. So, now I will ask a very fundamental I would like to address a very fundamental question. Suppose if the available energy of the system is e. So, then may I address a question how this energy is distributed among its different degrees of freedom. This is a very fundamental question which can be asked to somebody. So, if the available energy of the system if the system has a thermal equilibrium or at equilibrium then how this available energy of the system which is maintained at equilibrium is distributed among its different degrees of freedom. So, let me start let me first discuss what are the degrees of freedom for the different system. So, let me understand what are the different degrees of freedom for the different systems. When you consider a particle in space we use three Cartesian coordinates x, y, z to locate the position of the particle. Hence it has three degrees of freedom. So, its kinetic energy is given as half m x dot square plus y dot square plus z dot square which is nothing but the summed over the kinetic energies in the x direction, y direction and z direction respectively. Half m x dot square is the kinetic energy in the x direction, half m y dot square is the kinetic energy in the y direction etcetera. Suppose if the particle is constant to move in a plane the kinetic energy is given by half m x dot square plus y dot square. So, it has only two degrees of freedom as it is not free to move in all direction in this case not in the z direction. So, it has only two degrees of freedom. Similarly, if it is moving in a straight line it has only one degree of freedom and its energy is given by the kinetic energy in the x direction itself. So, degrees of freedom of a dynamical system will be given by the motion of the particle or system in which in dimension. Suppose if its motion is a three dimensional then its degrees of freedom is three. If the motion is constant to move in a plane then its degrees of freedom is two. If the particle is moving in a straight line then its degrees of freedom is one. If the system contains n particles then it is necessary to have three n coordinates. Hence the system has three n degrees of freedom. These degrees of freedom are translational when the particles are assumed to be point passes. So, suppose for a monatomic gas so if it consists of n number of monatomic n number of molecules or atoms are there. So, all these are have these are only translational degrees of freedom. Since each atoms of molecules has three degrees of freedom so there are n molecule. So, there will be three n degrees of freedom will be there. However, a rigid body cannot can not only move, but rotate about an axis passing through itself. In general a rigid body can rotate about any three mutually perpendicular axis through any point fixed in itself known as the principal axis of inertia. So, the kinetic energy of rotation is given as e root equal to half i 1 omega 1 square plus i 2 omega 2 square plus i 3 omega 3 square where i 1 i 2 i 3 are the principal moment of inertia and omega 1 omega 2 omega 3 are the corresponding angular velocities. Obviously, it means that omega 1 is the angular velocity in the direction x axis keeping y and z direction fixed. Similarly, omega 2 omega 3 represents the angular velocities 1 and 2 axis fixed respectively. Hence, the position of the rigid body can be completely fixed by three position coordinates of the center of gravity, three angles of orientation of the body. Thus the rigid body has six degrees of freedom in which there are three translational degrees of freedom, three rotational degrees of freedom. So, there will be two energy energy, one energy will be associated with the translational degrees of freedom, other energy will be associated with the rotational degrees of freedom. Thus we can define the degrees of freedom of the system as the total number of independent quantities necessary to specify completely the state of motion of the system and not on the motion itself. Thus a molecule of a mono atomic gas has three degrees of freedom only and which are translational in character. A diatomic molecule has five degrees of freedom where there will be three translational and two rotational provided the relative separation between the two atoms which forms a diatomic gas are kept constant. A triatomic molecule has six degrees of freedom where three translational and three rotational. So, let me summarize once again a mono atomic gas has three degrees of freedom which are only translational, a diatomic gas has five degrees of freedom where three translational and two rotational and triatomic gas has six degrees of freedom where three are translational and another three are rotational. Extension of Maxwell distribution law as we have already told you that whatever we have derived Maxwell distribution law of velocities momentum and energy it is only for the translational motion, but now for a rigid body suppose for a diatomic gas triatomic gas or in general for any polyatomic gas it has all energy is translational as well as kinetic energy. So, let me derive Maxwell distribution law for any diatomic or any polyatomic gases. The Maxwell distribution law was originally derived for molecules of mono atomic gas having translational motion only. Let us now extend this law to a polyatomic gas in thermal equilibrium. If the distance between atoms in a molecule is fixed the total energy of a molecule is given by E equal to E t plus E r. So, in principle for any polyatomic gas the total energy consist of two parts one is due to the translational motion another is due to the rotational motion where the energy of the translational and rotational are given as follows E t equal to half m x dot square plus y dot square plus z dot square and E r is half i 1 omega 1 square plus i 2 omega 2 square plus i 3 omega 3 square. So, these are the two energies rotational and translational motion. The molecules of the gas will collide with one another and the energy corresponding to any degree of freedom for a molecule will vary from 0 to infinity. So, therefore, we can extend the distribution law to include the other modes of motion. So, in case of polyatomic gas the probability distribution for velocity can be modified as follows. D w equal to a t into a r exponents e to the power minus e t plus e r by k t d x d y d z d omega 1 d omega 2 d omega 3. It is actually product of two terms one is due to the translational motion another is due to the rotational motion where a t as we know already which is m by 2 pi k t to the power half a r is nothing but i 1 by 2 pi k t to the power half i 2 by 2 pi k t to the power half i 3 by 2 pi k t to the power half as we know that moment of inertia i is the equivalent of mass in the translational motion. So, that is the reason a t consist of product of three terms i 1 by 2 pi k t to the power half i 2 by 2 pi k t to the power half i 3 by 2 pi k t to the power half. Boltzmann expressed this distribution law in more general form using generalized coordinates q 1 q 2 q 3 q n and generalized momenta p 1 p 2 p 3 p n where n is the number of degrees of freedom according to Gibbs distribution the probability that a particle in the range q n q plus d q p n p plus d p is given as d w is f of p q d p d q which is nothing but the a e to the power minus e as a function of p q by k t d p d q where a is a constant and e is the energy of a particle and is the sum of kinetic energy and potential energy that is e equal to t plus e of q with t is nothing but t is summed over i j a i j p i into p j. The quadratic expression in the piece can be transform into a sum of squares in the form which can be written as t equal to half summed over alpha x i square if you will split the summation then it is half alpha 1 x 1 square plus alpha 2 x 2 square plus alpha 3 x 3 square where x i are the new variables called the momentoids and alpha i are the coefficient so that is the beauty in terms of in terms of a new coordinate the kinetic energy which is which is written as which is written as summed over a i j product of p i p j which can be written as a square of sum term so that it can be easily handled. So, in terms of new variables f of x q x q d x d q is a exponential minus e of x q by k t d x d q where d n is n a exponential minus e by k t d x d q where d n is a exponential minus d x is d x 1 d x 2 d x 3 d x n this is the generalized form of Maxwell distribution law. Now let us see what will happen in case of some ordinary Cartesian coordinates in the Cartesian coordinates the energy of a particle can be written as e equal to p square by 2 m plus e of x y z where p square by 2 m is the kinetic energy e of x y z is a measure of interaction or potential energy. If p x p y p z are the components of p along x y z direction then p square is p x square plus p y square plus p z square and the probability that a particle is in the volume element d p into d r d p means in the volume element in the momentum space d r is the volume element in the coordinate space is given by d w equal to d w p into d w r d w p in the momentum space d w r in the coordinate space where d w p is a e to the power minus p square by 2 m k t d p where d w r is b e to the power minus u by k t d r. Here d w p and d w r are the probability of momentum position coordinates respectively and a and b are the constant which can be determined by the normalization conditions of d w p and d w r respectively. First normalization condition is integral d w p equal to 1 if you will substitute values of d w p then it will come a e to the power minus p square by 2 m k t d p equal to 1. Second normalization condition is integral d w r equal to 1 if you will put the values of d w r then it will come b e to the power minus u by k t d r equal to 1. So, in Cartesian coordinate d p is d p x d p y d z d r is nothing but the d x d y d z. So, if you will substitute these values so first equation number 76 in integral d w p will come out a integral minus infinity to plus infinity e to the power minus p x square plus p y square plus p z square by 2 m k t d p x d p y d p z integrating over all momentum values equal to 1. Next we will give the values of a a is nothing but the 2 pi m k t to the power minus of which is understandable because in one dimension a is 1 by 2 pi m k t in three dimension obviously it is a product of three quantities. So, a is 1 by 2 pi m k t to the power 3 by 2. Using the values of a the probability distribution for momentum components is finally given as d w p is 2 pi m k t to the power 3 by 2 e to the power minus p x square plus p y square plus p z square by 2 m k t d p x d p y d z. So, this is the probability distribution for momentum component. So, converting from momentum to velocity using the relation momentum equal to mass time velocity we get the probability distribution for velocity as d w c means probability of having the velocity c is d w c is m by 2 pi k t to the power 3 by 2 exponential e to the power minus m u square plus v square plus w square by 2 k t d u d v d w where u v w as we have already told earlier is the velocity component in the along x y z direction. So, p x is nothing but the m times u p y is nothing but the m times v p z is nothing but the m times w. If the n be the number of particles per unit volume then number of particles per unit volume having the velocity components lying between u and u plus d u v and v plus d v w and w plus d w is given as d n is n times m by 2 pi k t to the power 3 by 2 e to the power minus m u square plus v square plus w square by 2 k t d u d v d w this is the Maxwell distribution law of velocities. Now, let me calculate what is the probability in the coordinate space d w r we have to calculate. If n be the number of particles in the system the number of particles for a volume element d x d y d z in the vicinity of the point x y z is given as d n equal to b times n e to the power minus u of x y z by k t d x d y d z this is called the Boltzmann distribution. It describes the distribution of particles in the coordinate space as a function of the potential energy basically whenever we have extended the Maxwell's law. So, we have got two kinds of Maxwell distribution one tells the Maxwell distribution of velocities or momentum in the momentum space or velocity space which are almost same. So, another distribution Maxwell distribution which is in the coordinate space what is the probability of finding particle between x n x plus d x y n y plus d y z and z plus d z which will be driven by its potential energy. Let us assume that the potential energy u is 0 at the point some reference point which is x naught y naught z naught where the concentration of the particle is given by n naught is d n by d x naught d y naught d z naught which is the volume element around that point x naught y naught z naught. Then from this above equation d n equal to b n e to the power minus u by k t d x d y d z then n naught will be given as d n by d x naught d y naught d z naught which is nothing but the b n then or other way around b naught b into n is nothing but the n naught. Then you substitute the values of b n in equation number 85 what you will get is d n is n naught e to the power minus u by k t d x d y d z. Or the particle concentration at the point x y z is nothing but n is d n by d x d y d z equal to n naught e to the power minus u by k t where n naught is the particle concentration at some reference point where potential energy was 0. So, if some point potential energy is 0 if the particle concentration is n then one can ask what is the concentration of particle at some arbitrary point x y z where the potential energy u is as a function of x y z. Then Maxwell distribution tells that n is n naught e to the power minus u by k t. So, that means the concentration of particle at some particular point will be driven by its potential energy at that particular point. So, finally the particle concentration of a system having at having at equilibrium at the point x y z is n equal to n naught e to the power minus u by k t. This is the most commonly used form of Boltzmann equation. It determines the fraction of particles which in the condition of thermal equilibrium have energy u. The fraction n by n naught apart from the magnitude of this energy depends only on the temperature that is the beauty.