Good morning to everyone, my name is Shu Prasad, I am from Department of Physics IIT Bombay and I will be giving some lectures initially on the quantum ideas, it is basically the quantum mechanics portion of this course, then I will be giving some lectures on special theory of relativity and finally I will end with some lectures on crystallography. Now when this is our first experience of dealing with so many centers, normally as a teacher we are always used to sort of entertaining questions in between the sessions, I am not sure whether this will work very easily with all so many centers, so we will try to experiment and try to see some you know what is the optimum way, in any case if it is not possible to entertain the questions in between, what we would suggest that I will sort of finish let us say about 10 minutes earlier and then try to take those questions whichever are coming and essentially discuss those questions for the last 10 minutes or 15 minutes and after that you know we can sort of if there are still more questions probably I can you can send it and then we look at and try to answer in the next session or one of the later sessions, so that is the way let us try to do the strategy and as we keep on going on probably we will also get a little bit more experienced and then we will tell you what is the optimum way of doing it, okay so normally when we teach in our first year the quantum ideas this is generally treated as a part of course on what we generally call as a modern physics, so basically the students who are coming just at the first year level they have passed out their 12th standard and they have some vague idea there are certain experiments there are certain things which are taught to them in the name of what we call as a modern physics and one of my attempt whenever I teach in the first year in IIT Bombay is to essentially assemble those ideas put them into a proper line or a proper way, so that people can get and can get somewhat you know clear ideas of how things have evolved, so that is the probably the strategy which I am going to follow as far as this particular course is concerned, so many of these ideas these students in 12th standard would have known but still I tend to repeat them just to make sure that you know they understand at least the way I want them to understand also as my experience and teaching for last 35 years I have always felt that it is always a better idea to launch your course at a level where students are fairly comfortable rather than starting with a jump, so that they start feeling right from day one that you know they are sort of lost, so start at a level where the students are fairly comfortable means they feel that they know pretty well and then slowly expand and go into more difficult problems or you know more advanced areas, so with that strategy I have prepared my lectures, so my first transparency which I would talk about this thing is that when we call it modern physics is it really modern you know how modern is the word modern when we say modern physics, so as it appears that it is not really so modern it is more than 100 years old and there was a time when it was realized that many of the ideas of what we now call as a classical physics are not working anymore, there are many observations experimental observations which cannot be explained on the basis of those classical ideas I list what do I mean by those classical ideas now just now, but the basic problem was that you know remember in science we always follow the experiment if there is an experimental observation which has been made we must develop a theory in order to understand it in order to interpret it okay, but that is what it happened at that time there are series of experiments which it was not possible to understand very very clearly, so and most of these experiments probably can be classified into two clear dark areas which I will define just now, so these are two clear dark areas the experiments lying in these areas which people found that they cannot understand on the basis of the traditional classical physics, now when I say classical physics what do I mean about it, so I would say that as far as this particular course is concerned I will say there are three landmarks of the classical physics the first thing that I would say is electromagnetic theory which as all of you know is a very very powerful theory if a professor Ghosh has introduced just the electromagnetic theory and he is going to cover lectures on electromagnetic theory, but basically what I would like to say as far as it pertains to the concept of modern physics that as far as the nature of light is concerned this electromagnetic theory for the first time provided an explanation that that light can be considered as an electromagnetic wave, in fact what was one of the billion things which Maxwell did was to find out an expression of a speed of light in terms of fundamental constants epsilon naught and mu naught which I am sure professor Ghosh will talk about it and using this particular thing this particular constant could match with the experimental value of speed of light which gave a very very firm footing to the concept of light which says that it is actually an electromagnetic wave. So this is what I will treat as the first landmark of classical physics the second landmark as all of you know is a Newtonian mechanics force is equal to mass into acceleration. In fact Newton has three laws of motion but generally it is believed it is that the second law of motion which is the most fundamental law of motion because first law of motion was actually originally interpreted in a different fashion and the third law is a part of more fundamental law of laws conservation of momentum. So it is a second law which is the most important law which is the many times we just say Newton's law which means the Newton's second law of motion. So the traditional mechanics of I mean not only the particles but also of rigid body could be explained on the basis of ideas which were derived on the basis of Newtonian mechanics. The third thing is about the classical statistics a lot of things at that time were interpreted on the basis of what I will call later as equipartition law of energy which was originally probably discussed from the point of view of kinetic theory of gases because that was one theory which was quite a bit in fashion at that time and how Maxwellian distribution of velocities and using that particular thing we could calculate that what is the probability of a particular particle having a particular energy. So this is what I will call as a classical statistics. So these were the three landmarks of the classical physics electromagnetic theory, Newtonian mechanics and classical mechanics. And what I want to say that using these three landmarks there are certain type of experimental observations which one could not explain. So now let us go to what we call as these clouds or the dark areas. So I am calling these two clouds I have classified these two clouds as cloud one and cloud two cloud one was the failure of Maxwell Boltzmann doctrine regarding equipartition of energy. This is something about which I will speak little more and tell what were the type of experiments which this particular Maxwell Boltzmann doctrine is could not explain doctrine is almost like religious preaching. So at that time this particular Maxwell Boltzmann concept of equipartition law of energy was being used almost as a sort of religious preaching that you know you have an experiment at any experiment you apply equipartition law of energy and you will get the answer. And then we realize that it does not work for every system and there are definitely clear cut experiments on which this particular law would not be able to explain the results. The second thing about which we will talk little later probably in the second week is the motion of earth in the ether which eventually related to the special theory of relativity. And what is very very interesting that this particular cloud the second cloud which was about the motion of earth in ether which required the development of special theory of relativity it was very very quick basically in 1905 Einstein put the theory of special theory of relativity. And for this year happens to be the centenary of the discovery of general theory of relativity but you know in this particular course we will be talking only about the special theory of relativity. And this particular thing this cloud could be dissolved very easily on the other hand cloud 1 took lot of time there are lot of experiments there are lot of discussions lot of theories and only then one could actually use the replace the Maxwell-Boltzmann doctrine by a formal quantum mechanics as we will be discussing these things in subsequent lectures. Now they are typically as far as the first cloud is concerned there are three different type of experiment which I will describe rather quickly as I say generally the students in the high school are well aware of these things but nevertheless I want to repeat them just to make them at a familiar ground. So these were three different type of experiments which we are being used by we try to explain the on the basis of equipartition law and we found that we were not successful. So let us sort of go first let me describe little bit about the equipartition law of energy and then we will talk about how these equipartition law of energy were applied to the concept of specific heat of gases and the specific heat of solids and eventually the most notorious problem of 19th century which is called the concept of blackbody radiation. Now let us first describe what is an equipartition law. So as I said this particular thing was originally devised for the case of gases the kinetic theory of gases and if you look at the originated from the kinetic theory of gases but let us not go into the details of those how it was derived let us just mention what is this particular law. Whatever I mean the way this particular law was being applied that you have to calculate something which is called degrees of freedom of a system. Degree of freedom essentially means the total number of coordinates that you require in order to completely specify the system. So if you have a particular specific system let us say it could be gas molecules or whatever it is we will give some examples just now. Then if you want to specify that particular thing completely in a space then the total number of coordinates that you require was considered as degree of freedom. For example if you take a point particle or a gas molecule you normally would require three coordinates to give the values or for example in Cartesian coordinate system you want x, y and z coordinates to completely specify the position of the gas molecule in this space. So this particular gas molecule will have a 3 degree of freedom. Remember gas molecule was always treated in the classical physics as a point particle so it was never sort of assumed that this particular particle could also rotate about its own center of mass. Now the degrees of freedom can be of two different type which also I will explain one is what we call as a translational rotational or other three different type translational and rotational and third could be of the type of vibrational. So there is a translational degree of freedom there is a rotational degree of freedom and there is a vibrational degree of freedom. Now corresponding to each degree of freedom which corresponds to a translation motion or a rotational motion we associate half kT per degree of freedom that is what is the equipartition law that you first calculate the total number of degrees of freedom find out how many of them are translational or rotational and once you know that particular number multiply that by half kT then find out what are of course the remaining will be vibrational and find out how many of those vibrational degrees of freedom are there and multiply them by kT you add the total energy of the system that will be the total energy of the average energy of the system at a given temperature by the system. Now this is what as we said equipartition law let us try to apply this to some simple thing so that we can understand the way it was being applied and its utility or the problem of its application. So let us first consider what we call as a monatomic molecule. So let us just take a case of a single particle okay let us put it like this that this is a particular particle and this assume to be a point particle. Now q require basically because it is a point particle so I assume that it cannot rotate about its own self so only thing it can do is to translate if it translates then it has 3 degrees of freedom as we have just now mentioned. For example if I am using Cartesian coordinate system then it will require x, y and z coordinates of this particular point particle to give its position in the space if I am using spherical polar coordinates then I require r, theta and phi three different coordinates in order to completely specify its position in this space. So for any particular particle you require three degrees of freedom and these three degrees of freedom I will call them as a translational because if the position of this particular particle changes in this space okay then it can only change by translating it. So this particular particle there is no other type of motion possible remember I am assuming this to be point so if its coordinates change it can only change when this particular particle translates let us go up till this particular position. So this is a displacement and there is a translational so these all these three degrees of freedom are translational because if anything can change for example let us suppose this is my x axis it could change position along the y axis or it could change anywhere here whatever it is this particular particle has to translate to this particular point then only its position will change. So I will call these three degrees of freedom are translational which it means that according to equipartition law there has to be half kT of energy associated with this particular particle an average of half kT of energy at a given temperature T. So if this particular particle forms a part of the system then this particular part in this particular thing this particular particle on the average will have a half kT of energy at any given temperature. So I will say half kT where k is of course Boltzmann constant. Now let us take one mole of the substance let us say we take one mole of a monotomic gas for example let us say inert gas like argon helium or whatever it is or neon then this particular one mole of the gas will actually contain Avigadro number of atoms. Now it means that if I have to take the total energy of all the molecules at a given particular temperature then this particular thing has to be multiplied by the Avigadro number. So this is what I have written in this particular transparency that is monotomic molecule you have three degrees of freedom and all of them will be translational. Now if I have been able to find out little I will come to this particular thing little later about the diatomic rigid molecules and diatomic weakly bound things little later well let us look at just the monotomic molecule itself. Now what do I do once I find the total energy of the system one of the biggest thing that I can do is that if I take its derivative with respect to temperature at constant volume I can get specific heat which we call as a molar specific heat because I am talking of one more of the substance. So a molar specific heat is defined as dE dt when we are looking at the total energy of the gas molecule at a given temperature and when the total number of atoms are equal to the Avigadro number of atoms. So this is what I just now wrote that if I have a monotomic gas like inert gas then only translation of molecules is possible then total degree of freedom becomes 3 NA then if I want to calculate its energy then its energy becomes 3 NA multiplied by half kT and if I want to take this specific heat then the specific heat of this particular gas will turn out to I have to differentiate this with respect to temperature. So this T will go away so I will get 3 by 2 NA k which is actually 3 by 2 R because NA multiplied by k which is NA is Avigadro number k is Boltzmann constant is generally called gas constant and the value of the gas constant is given here. So it tells that if we have a monotomic gas then its specific heat irrespective of temperature would always be 3 by 2 R. Remember energy does depend on temperature but the specific heat would not depend on the temperature it will be constant as a function of temperature. Now let us look at the diatomic case just to give an example. So here now you have two sort of atoms which are connected by a bond let us assume that these are connected you know so now I have shown a sort of over enlarge bond length but you know let us just sort of take it that you know it is a sort of schematic for this particular diatomic gas. Now what can happen in this particular case see this particular point is a point atom this particular thing is also a point atom but once I have connected and formed a bond then the combination of these two atoms is no longer a point particle so it is a sort of elongated particle so there is a possibility that this particular particle can also rotate. Now in principle you require 3 coordinates to give to specify this particular particle you require 3 coordinates to specify this particular particle so you require in principle 6 degrees of freedom but remember there is also a possibility here that this particular particle its center of mass does not change but nevertheless the position of these particular particles can change for example this particular particle could get rotated like this now the center of mass need not have changed and still the particular particle this particular molecule would have rotated. Now so what we can say that 3 degrees of freedom are associated with this translational motion because center of mass can translate there are two different ways in which it can rotate for example it can rotate about an axis like this which is going perpendicular to it and there is an axis which is going in plane about it it cannot rotate about its own axis because then it means it is a rotation of point and I say as point you know we have always assumed that it cannot rotate therefore we assume that there are 2 degrees of freedom which are purely rotational because these 2 degrees of freedom there are 2 different ways in which this particular molecule can rotate without altering the position of the center of mass. Now there are 2 possibilities that this particular molecule could be rigid in this sense that this particular bond length does not change at all in that case there will be only one degree of freedom will be lost because we have a constraint that the distance between these 2 atoms will be constant on the other hand there is a possibility that this particular bond is not so rigid in that case the distance between these 2 particles can also change and therefore there is a possibility that the coordinates of the 2 atoms change here without changing of the center of mass but by changing the distance between them but that would be possible only when these atoms vibrate and therefore in that case 1 degree of freedom will be a vibrational degree of freedom. So now let us come to the rigid molecule if we have a rigid molecule in means the bond is not flexible at all in that case as I have said that 1 degree of freedom is lost so we have only 5 degrees of freedom I have a regular number of molecules because I am always talking of a molar specific heat so therefore total degrees of freedom will be 5 Na and their half kT per degree of freedom has to be associated therefore the total energy of the system will be 5 Na multiplied by half kT which means if I differentiate with respect to temperature that the specific heat of this particular gas would be 5 by 2R. So again it tells that if we are using a diatomic gas and if it so happens that the bonds between the atoms of this particular molecule are rigid in that particular case you will find that the specific heat will be independent of temperature and will always give you a value of 5 by 2R. On the other hand if the gas molecules were flexible the atoms were or the bond was flexible and the atoms could vibrate then there is an additional degree of freedom but that happens to be a vibrational degree of freedom so with that additional degree of freedom you have to associate kT of energy so you have 5 degrees of freedom total number of degrees of freedom was 5 Na for Na atoms or rather Na molecules total energy because of this is 5 Na multiplied by half kT then there is a one degree of freedom per molecule because of vibration there are Na molecules so there are total Na degrees of freedom and because they happen to be vibrational therefore there is a kT energy so this total energy becomes Na times kT so you will find that this becomes 7 by 2 Na kT you differentiate with respect to temperature and you will find out that the specific heat will be 7 by 2R. So if it happens to be a diatomic gas and if the gas molecules are non-flexible then I should get the specific heat to be 5 by 2R and if the molecules happen to be flexible molecules or the bond has happens to be flexible bond then the specific heat will be 7 by 2R in both cases it will be temperature independent. Now if you look at the experimental data see remember at that time when this particular things were conceptically partition law was being floated it was we did not have experimental facilities to measure as a function of temperature specifically at lower temperature whatever was being talked was more or less the measurements which were being done at room temperature and when the experiments were performed a very good agreement was formed for the case of monotomic gases people were happy that equipartition law works. When remember we were talking as I mentioned that equipartition law was being used almost as a doctrine for everything equipartition law is the final solution. If you take diatomic gases one does get a value of 2.5R so you are happy okay that is what you had expected so you can say that probably the diatomic gases the bonds are too strong they are rigid and they are not flexible at all so I am getting 2.5R. But when you started measuring the specific heat at different temperatures you realize that actually for many of them you find the specific heat increases as a function of temperature when you raise the temperature you will find that the specific heat is not constant. So remember if the bonds were rigid I should have got according to equipartition law a fixed value of 2.5R at all temperatures the specific heat would not have depended on the temperature. But now what we find experimentally that if it increases if you increase the temperature the value sort of increases and tend to approach to a value of 3.5R as if at higher temperature the bonds have become flexible but while at low temperatures they were not flexible. Let us look at hydrogen gas which is probably one of the most interesting result hydrogen gas happens to be the lightest diatomic gas I am talking about hydrogen molecule gas the lightest gas as you know. Now when you make a measurement of hydrogen gas remember many of these measurements were made much later because as I said the facility of measuring specific heat at lower temperatures probably were not available in the 19th century many of these measurements were done at later time and it was found that for hydrogen gas you always got a value of 1.5R around 100 degree Kelvin and when you increase the temperature remember room temperature is around 300 degree Kelvin then you find approximately a value of 2.5R around room temperature. Then as you start heating you find out the specific heat again tends to increase and tends to be approximately 3.5R around 1000 Kelvin okay at that temperature the gas dissociates and it is no longer a diatomic gas so you cannot apply your theory anyway. So whatever measurements you can do in hydrogen gas so long this is a diatomic molecule you find that the specific heat shifts from 1.5R to 2.5R and to 3.5R it is very very interesting 1.5R I would have got for a monotomic gas okay hydrogen gas even though it is a diatomic seems to be showing a value 1.5R at low temperature and it shows 2.5R corresponding to a non-flexible band bonds around room temperature and tends to show a value approximately corresponding to a flexible bond when we are going to higher temperatures. Let us now come to the case of solids which is again a very very interesting case which we call as a DuLong and Petit law and this is generally again taught in high school of course at that time when we teach in high school we always tell them that this is the product of molecular weight and the specific heat for all solids is 3R but because I am using a molar specific heat and not the specific heat per gram so therefore I do not have to multiply it by the atomic weight or molecular weight. So the DuLong and Petit law would say that the specific heat for all the solids should be equal to 3R and this particular thing this value of 3R is based on equipartition law of energy and let us it is very very simple to derive it and this is what I have done derived this on the basis of equipartition law in this particular transparency. Now let us take one more of a solid for example you can take copper nickel or sodium or for that matter any solid. Now I am taking one mole of the substance so it will have avocado number of atoms so we have avocado number of atoms then as I said corresponding to each atom you will have 3 degrees of freedom so total degree of freedom has to be 3 and a but remember unlike the case of gas molecules where these atoms are free to move here in the case of solids all these atoms are strongly bond by the bonds with their nearest neighbors therefore this particular atom cannot really move out of this particular solid it can keep on going somewhere as so long it remains in the solid state all it can do is can change its relative position with respect to its nearest neighbors which essentially means that it can only vibrate. So translation is not possible as far as this atom is concerned this rotation is also not possible only thing which is possible that this particular atom can vibrate because in fact if you remember in the solids one of the simplest picture is that the atom is actually you know sort of connected with some sort of springs with its nearest neighbors so it can only vibrate when it vibrates it means all these degrees of freedoms have to be only vibrational degrees of freedom. Therefore with this 3 and a I multiplied by kT and not half kT once I multiplied by kT the total energy at a given temperature becomes 3 and a multiplied by kT and if I differentiate with respect to temperature I will get a total specific heat which is equal to 3R this is what DuLong and Pettit Law said that for all solids the specific heat should be equal to 3R. Now this is as far as the law is concerned but we must see what happens when you take an experiment. Now again with whatever data was available at that time it shows a pretty good agreement with almost all the solids and the value did turn out to when I say RT it means room temperature it should not get confused with gas constant multiplied by temperature it actually means it should be R dot T dot I should have written which essentially means at room temperature. So good agreement at room temperature for a large number of solids there had been some exceptions like diamond which did not show the value of 3R but for most of the solids it did show a value if not 3R fairly close to 3R so it was fairly successful in explaining the specific heat material. Until you started having data at lower temperatures and then you find that specific heat for all the solids including diamond tends to go down as you decrease the temperature and of course you can never reach 0 degree Kelvin but you can go to very close to 0 degree Kelvin and you will find that the specific heat of solids tends to 0 as temperature tends to 0. So irrespective of the fact remember DeLonge-Pathet law said that it should be 3R at all temperatures but here we see experimentally that this does not turn out to be equal to 3R. It does tend to be temperature dependent and at lower temperature of course at room temperature it does show approximately 3R but at low temperature it tends to go to 0 and for high temperature it always approaches the value of 3R. In fact even for the case of diamond if you would have heated it to a higher temperature at it is not at room temperature but at temperatures higher at the room temperature you will find that this particular specific heat does actually turn out to be equal to 3R. So high temperature limit for solids is 3R. So what DeLonge-Pathet law gave was actually a high temperature limit and the third thing which is a very very important law was that at low temperature you will find that specific heat actually is proportional to T cube which is generally called DeWeis T cube law because DeWeis was the first person who on the basis of simplified assumption could explain this particular T cube law why specific heat should be proportional to T cube. So basically there are 3 different type of experimental observations which says that at low temperature the specific heat for all the solids is 10 to 0 at temperatures high enough when we say high enough it depends on typical solid okay but at high enough temperature the specific heat of solids will go to a value of 3R and at low temperature you will find that specific heat is proportional to T cube okay. In principle none of these phenomenon can be explained on the basis of DeLonge-Pathet law or equipartition law of energy which showed that the value of a specific heat must be equal to 3R for all the solids at all temperatures. Let me just put a disclaimer here see here when we are talking about specific heat this is only because of vibration there are many other elementary excitations in solid which can also have its own contribution to the specific heat but at least at room temperature the vibration for contribution is the dominant contribution. So it is just a sort of disclaimer to say that just to be correct. Now if you have seen that all the examples which I have given you so far where the examples in which most of the experimental data was obtained at much later time see remember helium was liquefied by K Merling's own somewhere in 20th century earlier 20th century and then only it was if you could get facility of measuring at temperatures typically of the order 4 to 5 degree Kelvin because helium had a boiling point of 4 degree Kelvin 4.2 degree Kelvin otherwise you know we could not make any measurements at temperatures low enough. So whatever we talked about as the issues with the classical equipartition law those issues we are I mean probably realize much later now today we can talk but at that time probably it was not we could not have talked about it okay but this was the problem about the black body radiation was the main problem which at that time was considered as a very very severe problem. As probably all of you know that black body is a sort of an idealized system in which all the radiation which falls on it is totally absorbed there is no radiation which is transmitted or no radiation which is sort of reflected it absorbs all the radiation that is that that is falling on it it is sort of idealized system but remember in physics we always start with the idealistic system because that turns out to be a simpler system and we will try to experiment with a system which was as close to the ideal system as possible and they tried to find out how the radiation gets emitted from it. So basically the idea is that any body which becomes hot always emits radiation at any temperature and this radiation has certain type of specific qualities first of all the total amount of the radiation that is emitted at a given temperature and the second thing is the spectral distribution of the radiation what are the frequencies it emits as all of you know that if you take any particular solid okay if we heat it and you know just put our hand close to it so you will find that it is radiating but you can only feel the heat it means basically the emission is taking place only in the infrared region but you heat quite a bit let us take an iron rod and heat to a comparatively higher temperature you will find that it becomes red hot it means now it is also emitting light it means the radiation is also has shifted to a visible range. So you do find that the spectral distribution changes as a function of temperature and this black's body radiation problem was an attempt to actually calculate the spectral distribution of the radiation which is being emitted by any black body. Now eventually people applied equipartition law of energy when they try to apply equipartition law of energy by using certain method of standing waves they try to calculate the total degree of freedom. So they actually counted the total modes of standing waves and they named it as a degree of freedom of the system okay and this turns out to be given by this particular factor and they applied equipartition law of energy on this particular thing calculate this is what is called MSA power it requires a little bit of calculation that this particular moment my idea is not to explain this thing just to tell about the problems specifically the problem of black body radiation because at that time in 19th century this was the most serious problem which was being considered. Now if you do this particular thing this is what is generally called the universal black body radiation curve this is an experimentally observed fact that this particular if you plot e by t to the power 5 as a function of lambda t this shows a graph which is like this which is initially increases and shows a maximum and then again decreases. Now if you go to this particular what we call as the Rayleigh Jean's law you plot this particular thing as a function of lambda t you will see that this is just 1 upon lambda t to the power 4 because all others are constant. So as you reduce lambda t to lower values this essentially will diverge and at lambda t is equal to 0 this will become infinity. So obviously this particular law is not able to explain this particular curve which is called universal black body radiation curve and as I have mentioned earlier that this was considered as a notorious problem there are many other attempts on different type of sort of assumptions and this particular experimental curve could not be explained at all on basis of any theories and it was really a notorious problem at that time. Until the year 1900 which was as I will call the birth of modern physics or birth of quantum mechanics let me put like that to be more exact it was the birth of quantum mechanics. When Planck on the basis of two simplified assumptions remember they were assumptions they were not based on any further theory at that time no other theory was known it was just a purely in hypothesis he made two assumptions and using that he rederived what we call as equipartition law of energy and these assumptions were the following assumptions which I am going to just read out for you and this provided the first explanation of the black body radiation curve. The first assumption is that the energy levels of harmonic oscillators which were supposed to be giving rights to these particular radiations in the black body they are quantized it means whenever there is energy transfer which takes place it cannot take place in any arbitrary unit this has to be always a multiple of a particular given value of energy which we now call as a quantum of energy. So, energy of harmonic oscillators are quantized all transfers of energy will take place only in a multiple of a given packet of energy. Second thing which is much more important assumption is that this quantum is proportional to the frequency of the oscillator. So, whatever is the energy transfer which is taking place this particular energy transfer will take place only in multiple of this quantum h nu and this nu depends on frequency and this h is generally considered I mean this is a constant which we now know is called Planck's constant which has been named after Planck's constant this is a parameter which you can use to fit in your experimental data. Now this particular idea of quantization which was originally proposed by Planck which was later used picked up by many other people. Einstein provided the first theory of specificity of solids which is called Einstein's theory of specificity of solids and for the first time it could he could explain I mean he could not explain the complete thing but he could explain at least two features of the experimental features of the specific heat. One is that specific heat goes to 0 as it tends to 0 and the second is that the specific heat tends to a value of 3 R when temperature is raised. These two aspects one could explain, Debye explained later on the third aspect but the basic thing that they picked up this particular idea of the Planck in order to explain many more things. The specific heat of diatomic gases could also be explained on the basis of quantization which also required of course little much later ideas when the people realize that even the rotational motion and the vibrational motions are also quantized and therefore it is not always possible at a given temperature to excite a particular type of either vibration or rotational motion. Therefore the specific heat can really vary as a function of temperature sometimes showing as if it is only translation sometimes showing no vibration third time vibration of a non-rigid atoms or non-rigid bonds. And of course as you know later it was also applied by Bohr atom in the case of Bohr's theory again the idea of quantization was used of course Bohr adopted the quantization to the angular momentum but once he applied this particular idea to the angular momentum because he was using only the classical theory later almost everything got quantized the atomic radii got quantized the energy got quantized though originally he applied the idea with the quantization of angular momentum. Now let us look at the concept of quantization per se what is this quantization idea actually it is not very surprising per se if we talk philosophically many things in nature are quantized for example you know if I have to give you books I can give you only one book or two book or ten books or if I have to give you pens I can give you one pen two pen ten pen okay I cannot give you 1.5 pen I cannot give you 1.3 pens okay similarly let us say our monetary system is quantized okay for example whatever transaction we make we may make a transition of let us say 1 million rupees to somebody but it has to be a multiple of only a paisa you cannot give somebody 1.3 paisa 1.2 paisa 1.9 paisa okay. Now for example our unit is let us say 1 paisa but if you go to United States there the least unit is 1 cent okay 1 cent is not equal to 1 paisa okay their quantization is also different from our quantization but nevertheless their monetary system is also quantized our monetary system is also quantized okay. So similarly what happens in the case of harmonic oscillators that it depends on the frequency if one particular oscillator oscillates with the frequency nu then the quantization is n times h nu if it is the frequency of with which the oscillator is vibrating as I say nu 1 then the quantization will be n times h nu 1. So as far as the quantization concept is concerned is nothing very very surprising because we are used to the system of quantization at least philosophically. Let us look in the case of scientifically scientifically also there are many systems which are quantized though this word quantization we have never used okay even when we derived or even Rayleigh and Gies derived they had put a boundary condition that the displacement of the or the amplitude at the two ends of the container in which these vibrations are taking place must be 0 okay. So in a very large number of systems we apply boundary conditions that is what I have written through application of boundary condition we somehow quantize things okay. Let us take example of vibration of strings which probably much simpler see let us suppose there is a vibration of string which we all know we probably teach in high schools okay. Now it is like a musical instrument where you have just put one particular string and this under tension and the value of tension is let us say t. Now if I pluck this particular string I can set certain type of stationary waves in this particular system and only those waves which have zero displacement at this point and a zero displacement at this particular point can actually be sustained in this particular string okay. So you have either this particular type of wave which is possible another wave which is possible like this okay there could be other waves which are something like this but they all require that the displacement at this particular point and at this particular point is 0. So I must have this condition only those wavelengths can propagate in this particular vibration of strings where l is equal to n lambda by 2 condition is satisfied okay which of course because this particular lambda depends on the frequency okay this is the condition it means it there are only certain frequencies which can travel in this particular string. Now if you look at this particular thing this is also a sort of a quantization because there are only fixed values of lambdas only fixed values of frequency which can travel in this particular string and this particular thing is given by this particular value. So nu has to be n times this particular thing okay whatever is this particular value of this particular expression okay so this is also in a way a quantization because you can have only selected frequencies which are traveling. Though at that time we never talked that we never called this as quantization but the fact is that you do not allow all the frequencies to travel in this particular string they are only fixed values of frequencies which are allowed to be travel in this particular string which satisfy the boundary condition that the displacement here is 0 and the displacement here is 0. Therefore this is sort of a quantization so this is what I have written in a way the frequencies that can be excited in this string are sort of quantized. The boundary condition has limited the wavelengths in the string to discrete values this also limits the frequencies to discrete values as the two are related to each other through what is called a dispersion relation this particular word I will use little later so I have just sort of introduced dispersion relation. See whenever I quantized I quantized wavelengths and because frequency depends on wavelength okay therefore frequency is also get quantized. So this is what I call a dispersion relation is a relationship between the frequency and the wavelength as far as this is concerned we will see later that this could also be treated as a relationship between energy and wavelength okay but generally a relationship between frequency and wavelength which depends on what type of system that we are talking the systems can be could the wave excitations could be of various different types in type in a solid and each could have its own dispersion relation not everything can could have a very simple dispersion relation as we will be just seeing even in the case of quantum particles. Now this was one set of experiments there were some parallel set of developments of different type which were also taking place at that particular time the year the first 10 years in let us say in the 20th century from 1901 to 1910 saw lot of developments in a totally different type of fashion which we call as a parallel development of different type when the special theory of relativity came about which I will be talking later okay and then we realize that we started talking about the nature of light. As I said using the classical mechanics I mean not the classical mechanics necessary but the classical ideas that one of the pillars of the classical concepts were electromagnetic theory where light was sort of agreed upon to be taken as an electromagnetic wave. So we have tried to apply all the things of what we know about a wave to light also light was considered as a wave okay of course there the wave was what was varying in a light is electric field and the magnetic field. Now if a light like any wave would have a frequency it has a nu it was later realized that I mean as in if you take a relativity it gave you concept of some particles which can have a zero rest mass and it was shown that those particles with zero rest mass can also have energy and can also have a moment. At that time there was an experiment of photoelectricity experiment which sort of proved that the light could also be treated as a particle and these particles can be treated as particle with zero rest mass. In fact as we will be seeing in relativity we can show that a particle with zero rest mass can exist and it means they can have an energy they could have a momentum but then they must travel with the speed of light. Now obviously photons travel with the speed of light so I can treat this photon as a zero rest mass particle and photoelectric effect experiment sort of told us very clearly that the energy of this particular particle would be actually equal to h nu. In fact the special theory of relativity predicted that it will have a momentum of h nu upon c as we will be seeing later that photoelectric effect experiment actually cannot give an idea about the momentum. We require a different experiment which is called we call as a Compton effect experiment by which we can really sort of see sort of let me say prove that the photon really has a particle which also has a momentum which is given by h nu by c. So this actually came as a support also of a special theory of relativity and also prove that there is a dual nature of light could there are many experiments like diffraction interference which can be understood only on the basis of wave picture. It is very difficult to understand them on the basis of particle picture which we will try to do in the later part of the course and you will see what type of difficulties we face when we try to do these experiments. So we had in the first 10 years of the 20th century a series of experiment where we found that the light seems to be showing a dual nature. Sometimes there are certain experiments which can be explained only on the basis of wave picture. There are certain other experiments like Compton effect photoelectric effect experiment which can be sort of explained only on the basis of the particle nature of the light. Now let us try to assemble our thoughts. See we just now said that electromagnetic waves showed dual nature their experiments which can be explained only on the basis of wave nature of the light their experiments which can be explained only on the basis of particle picture of light. So let us put the second thought nature loves symmetry if photons can behave like particle or a wave depending upon the experiment is it not logical that we assume that massive particle I have used the word massive because photons are the particles which have a zero rest mass but they are the same equations but they are special in the sense that their rest mass is 0. But even if we have rest mass nonzero it means the particles are massive then is it possible that they can also show some wave property. So if nature loves symmetry then if photons tend to behave both like particle and wave is it not logical to assume that particles could also behave like waves. I mean let me put like that let me sort of qualify this statement earlier what classically we thought as a wave showed a particle nature. So is it not possible that what we classically thought as a particle could also show a wave nature. This is third thought if at all this happens that a particle could have wave properties then like we have talked about these waves stationary waves which are traveling in a string these particular waves can also be put to certain specific boundary conditions like the boundary condition that we had put in the case of vibration of a string where I had put the displacement at the two ends to be equal to 0. So these waves can also be subjected to some boundary conditions depending upon the problem and if it says so happens then this particular boundary condition in principle could limit their wavelengths to discrete values just like a simple example which I have given you of vibration of a string that is what happened there that in principle if this string would have been infinite in length all the vibrations should have been possible but because I have clamped the string therefore only selected wavelengths in principle I will not say it could travel but could be excited because the waves are stationary waves could only be excited in these strings. So I am limiting their wavelengths by applying a boundary condition. So could it happen that a massive particle or massive a particle which we considered classically as a particle and if it has a wave property and this wave property could be subjected to certain type of boundary condition then that particular thing could be limited to certain discrete wavelengths and if it so happens that the energy of the particle depends on this particular wavelength for example like photon we know that the energy depends on ash nu which nu is depends on wavelength. So the energy of the photon does depend on wavelength. So if it so happens that this particular particle which has now a wave property and which is being subject to certain boundary condition would also lead to a quantization of energy because by putting boundary condition I have limited its wavelength and by limiting its wavelength I have been able to limit this energy to certain discrete values. So this was the forethought that if these particles could be shown to have a properties probably we can understand why there could be a quantization of energy. Using this particular concept we probably for the first time proposed in 1924 that the massive particles could also possess a wave property. Now if you remember the year this is 1924 original Planck's idea had come in the year 1900 it took almost 24 years which is quite a bit of time. 24 years is essentially the half of the life probably 3 for the fourth of life of an academic person. It took such a long time for the broadly to give this particular idea that a particle particle which is massive can also possess a wave nature and he gave an expression for its wavelength. Now in order to make it consistency we know that for a case of photon the momentum has to be equal to ash nu upon C this C upon nu can be written as lambda so this becomes H by lambda. So remember this expression P is equal to ash nu by C is valid only for photon but if I write in terms of wavelength P is equal to H by lambda what De Broglie proposed that whether a particle which is either photon or a massive let this particular equation be correct equation the wavelength be related to the momentum by this particular expression lambda is equal to H by P remember this H is the same Planck's constant which Planck had used to explain his black body radiation even in the photoelectric effect we have always used the same value of H same Planck constant. So this particular expression now is valid both for photon and for a massive particle this is what he proposed. In fact he had some clues for the same because he was trying to actually fit a wavelength De Broglie wavelength along the circumference of the Bohr model. Of course as we now know today this particular Bohr model is not correct the only thing which is correct about the Bohr's model is its energy we cannot talk of the electron really does not move in a circle as was proposed by Bohr. So these ideas we are not really are not really correct in today's world but at that time this is the way he arrived at. Now just to give an idea that if I take a 200 gram ball which is moving with the speed of 100 meters per second that is a typical cricket ball then its wavelength will be 3.3 into 10 power minus 35 meters is such a small wavelength that we will never be able to measure it by any experimental method. So it means this particular wave nature for a cricket ball you will never be able to determine you will not be able to perform any experiment at least whatever we know today that where you could measure such a small wavelength of the order of 10 power minus 35 meter in fact that is the reason you do not require really quantum mechanics to explain the motion of a cricket ball but if you take what we call as a thermal electron a thermal electron I mean that a particular electron which has a energy of 3 by 2 kT where T is room temperature. So if I take room temperature at room temperature the value of kT is approximately 1 upon 40 electron volt 0.025 electron volt if I take that particular energy and calculate its wavelength its wavelength will turn out to be of the order of 6.25 into 10 power minus 9 meters which is typically of the order of an angstrom the angstrom type of distances are known because the typical inter atomic distance in a solid is of the order of an angstrom. Therefore it is possible to create a grating which has typical dimension of the order of one angstrom and that grating will be actually the grating of the atoms like x-rays can be diffracted through a crystal therefore it will be possible for an electron to get diffracted through a crystal and therefore we will be able to see we will be able to observe its wave property. That is what Davieson and Gerber did later in 1925 and later by G.P. Thamson they succeeded in diffracting electrons through a solid and they essentially used exactly the same 2D sin theta is equal to n lambda which was derived by Bragg's to explain x-ray diffraction if this is the formula that we use in x-ray diffraction about which we will talk in the later part of the course also okay. Same formula is being used for electron diffraction the only difference is that this particular lambda will be used by using the Broglie wavelength. Therefore the electrons can be diffracted today we know that electrons or for that matter particles can be diffracted. The x-ray diffraction electron diffraction and neutron diffraction are very standard techniques which are being used by modern day experimental phases the only thing it depends on the cost the x-ray diffraction is much more simpler almost every I mean in a variety Bombay we have multiple machines performing x-ray diffraction but if you are looking of electron diffraction you will have only one or two units and if you are talking of neutron diffraction in the entire country we have only one facility in BRC where you can perform the neutron diffraction experiments but the fact is that electron and neutron which were classically being considered as a particle. So remember x-ray was classically being considered as a wave electron and neutrons were being classically considered as particles okay all three of them can be diffracted like x-ray can be diffracted for diffraction of x-ray it is not surprising because classically we thought this is a wave but it is surprising that electrons and neutrons can be diffracted because classically we thought them to be particles and with particles we do not associate things like diffraction interference type of phenomenon. This is some of the electron diffractions just to give you a feeling which have been performed in an IIT Bombay by some of our students. So you can see very very clear rings these are the rings which are because of the diffractions from different planes so like x-ray you can have diffraction from different planes here also you can have diffraction from different planes these are polycrystalline materials so you get different rings so you can see really that electrons can be diffracted we are for sure we know we have been using them in a very very useful sense the electron diffraction phenomena we are very sure that electrons really which are classically treated as particles can be diffracted. Now let us try to go little bit into the depth when we are saying that there is a wave nature associated with the particle let us talk about what we mean by an ideal wave in the classical sense. Now an ideal wave is something which we can write in terms of this I mean instead of displacement I am writing psi because as you will be seeing quantum mechanics we are using this particular thing called psi is a sin kx minus omega t many times in high school instead of k we write this as 2 pi by lambda instead of omega we write as 2 pi f what to make this equation look comparatively cleaner generally we always write this in terms of k k is actually 2 pi by lambda and omega is equal to 2 pi times frequency so this omega is what we call as angular frequency and this k is generally called as a wave vector. So using this particular thing a displacement of this particular type will represent an ideal wave A will represent the maximum displacement of the particle so A will be maximum displacement minus A will be the least displacement or maximum in the negative direction and this particular wave can be for all values of x and time this should be the displacement. So remember as position changes the wave will change the displacement will change for a given time or for a given value of x at different times again the displacement will change. So an ideal wave normally is shown either as a function of x for a given for a fixed time or as a function of t for a given value of x. So here what I have shown that this particular wave this particular displacement psi I have plotted as a function of x for a given time.  Now if I look at a different time you will find that displacement moves like this. So this particular point which was having 0 displacement moves slightly to the right this moves to the slightly right this particular point where which was having maximum displacement as time progresses we will find that this displacement goes to the right direction and then you say that this particular wave is actually travelling wave and because this particular point which is here moves to the right hand direction as the time increases I would say that this particular wave travels in plus x direction. So if I have to look in which direction this particular wave travels basically I have to look at a particular point which is let us say for example maximum displacement and find out that as time increases what is the value of x whether the value of x will increase or whether the value of x decrease depending on if this as time progresses this particular point would have moved backwards then you would have known that this particular wave would have travelled in minus x direction. Now if I have a right equation in this particular form as you know that if I t if I increase t then this x has to be decreased to make this particular value same. So remember I need not have used this particular point I could have used this particular displacement value also and watched it what happens as time increases if x has to increase for this having same value of psi here then this is a wave which is moving in plus x direction. So if I want psi to have same value and I want to increase time then x also has to increase so that psi remains same it means this represents a wave which is travelling in plus x direction. On the other hand had there been a plus sign here then in order that this psi remains same and tie t increases x must should have been decreased in that case this wave would have been travelling in minus x direction. So as you can see that basically depends on the relative sign of the x term and the t term if their signs are opposite then this wave will be travelling in plus x direction if their signs are same then this will be a wave which will be travelling in minus x direction. This discussion I am giving because very very important when we are doing quantum mechanics to define to find out the direction of the wave motion because certain of the boundary conditions will depend on the direction of the motion of the wave therefore it is important to describe about it that is what I am trying to sort of explain at this particular point. So looking at the relative sign of x and t term I will be able to find out in which direction the wave is moving. So this is what I said this is what I call as a phase speed of the wave take a point at t is equal to 0 for which psi is equal to 0 the time increased by 2 delta t what will be the value of delta x to maintain psi is equal to 0. So I have increased time by delta t I am taking psi 1 minus psi 2 what will be the value of delta x if I have to find out the speed then what I have to do I have to actually find out how much this particular point moves in a given time that is what I am trying to calculate. So I calculate psi 1 at a given value of time then I take psi 2 at a given value of time then find out what will be the value of in order to make this delta psi 0 because I want to look at the same particular point let us say at this particular point. So I want this particular psi not to change but I increase my t and I have increased my x so that this particular psi remains same. So if I do this particular case in order to make this difference 0 you will find that delta x upon delta t which is the velocity the speed of the wave which we call as a phase speed will be given by omega by k just solve this particular equation you get omega by k. So omega by k will be the phase speed of this particular wave and I am using this specifically the word phase speed because just now we will define a sort of a different speed which is a very very important thing for us to realize. So phase speed of the wave is given by omega by k which is nothing but you know because omega you write as 2 pi nu and k you write as 2 pi by lambda so just nu multiplied by lambda which we know classically that this is the equation of the speed of a wave. Now I have just now discussed how do we find out the direction of the motion of the wave if a wave has this particular nature then this particular wave is moving in minus x direction if there was a negative sign then this is a particular wave which is moving in plus x direction if I had plus omega t and minus k then again this will be a wave which is moving in plus x direction. So basically the relative sign of the x term and t term will decide in which direction the wave travels. Now what I wanted to say that this particular wave that I have really looked at it strictly speaking is an ideal wave it means it must be present for all values of x and must be present for all values of time. We know that in nature no such wave exists whatever type of wave that you are talking will always decay after certain time it will always decay after certain position. Let us take very standard waves like you know ocean waves they may appear to be very very long very very big waves but they always have a finite strength finite limit as far as position is concerned it does not happen that the entire ocean the only one wave comes okay they always break their various type of waves which are coming. So any particular wave will always have finite dimension there is nothing like infinite in that sense. So any wave group is a realistic wave group is not really a type of wave an ideal wave but something which is similar to this particular thing. If there are some questions I think let us take some question before we start about this wave thing. 1001. Sir in this expression for displacement of the wave yes we have psi equal to a sin x kx minus omega t that is right normally we denote this by y equal to a sin x kx that is right. So yeah what is the difference between y and psi over here. Yes let me try to explain this particular thing see at the moment I have not introduced the wave function I have not talked about anything which is imaginary I will talk about that later. So whatever we are writing in the normal equation as y I have just called this y as psi as of now because when we are talking of quantum particles when we try to give the value I will try to give a physical interpretation to this psi then we will talk about this psi. So at this now I am this particular expression which I am writing is for any wave and this particular wave in principle could be let us say water wave could be a sound wave could be a light wave and instead of writing for example if I would have written light wave I would have written e is equal to something like that or b is equal to something like that okay in place of that I have just written a common factor psi. So let us just take this particular psi as something which represents a displacement it could be just a in electromagnetic wave electric field variation okay. So let us not at this particular moment interpret this anything beyond what we in a normal wave we call a displacement or a y okay when we will give it meaning we will talk about its imaginary part etc later. Okay sir fine sir thank you. Welcome. One two double five Monsignor. I have some questions. Yeah please go ahead. From 15th slide can you go further here you saw that g lambda d lambda is equal to 8 d lambda divided by the lambda 4. Yeah that is right. Black body radiation is there any correction it is a fully dependent upon a temperature versus a wavelength. Yes that is right. There is no temperature term is there. That is right. See this is what we call as actually I mean these days we call this is a density of state but this is what at that time was called the degrees of freedom of the system okay. So these are sort of total number of waves which are allowed to be traveling between lambda and lambda plus d lambda inside the black body cavity okay. So this was essentially like degrees of freedom what I have written in the case of diatomic for example gas as let us say N a 5 N a okay or 3 N a okay. So this is that number. So this is only degrees of freedom this has to be multiplied by k t okay in order to generate in distance to the what we call as the energy distribution okay. Then you do some mathematical manipulation then this E is what is we call as m s a power okay. E is eventually calculated from that particular expression and then find out how much power is likely to be radiated. So there are quite a few steps involved between this and this which of course one can look into the any standard book on you know black body radiation where they will give this particular thing. So this is sort of what I call as a degree of freedom of the system. From this particular degrees of freedom you first multiply by k t then convert this particular energy density to what we call as m s a power. Once you do that then you will get this particular expression. So one more case. Sir in slide 34. 34. This is psi is equal to a sin k x minus omega t I think it is a a sin omega t minus k x. See that is what I said there are various ways of writing it you can write omega t minus k x also that will also represent the wave traveling in plus x direction okay. It depends on what phase you want to talk it okay there are multiple ways of writing the waves you can write a sin k x you can write cos k x okay you all will be representing a wave that does not bother I mean certain textbook may prefer to write in a particular direction some textbook may prefer to write in terms of cosine some textbook may prefer to write in terms of sin some people may give k x minus omega t some people may give omega t minus x but all I am trying to say is that basically it looks at the relative sin. If I you are writing omega t minus k x the only thing which is which is different is the phase okay otherwise it represents exactly the same motion okay and it also represents a wave which is traveling in plus x direction. If you write k x plus omega t then only you have to be careful because that will represent a wave which is traveling in minus x direction. Sir thank you. Thank you. 1, 3, 3, 4. Seliguru Institute. Slide number 34.  And positive x axis. Yes. So please explain why it is a class omega t when it is moving along negative x axis. That is right that is right because if you have k x plus omega t then in order to increase time k has to be decreased in order to have the same value of psi. If I look at the same value of psi if I want it later time the same value of psi then it will go down okay. It means this particular wave will travel in minus x direction. Another question. Yes please. Slide number 14. There might be another equation called Vines Displacement Law or Vines Distribution Law that is for Displacement Law. That is right. So please explain. Yeah see I have not explained Vines Displacement Law. See there are various attempts of explaining the blackboard dilation curve okay two of the famous things which were one was Rayleigh Jean's law another was Wien's law and there is a reason why people also talk of Wien's law. Wien's law as far as I know was not directly based on the equipartition law of energy okay. The advantage of Rayleigh Jean's law was that first of all it does not have any constants to fit in. All these things are fundamental constants. There is no parameter which requires a fitting okay. In spite of that this fitted the later part of the curve very very well. This particular part of the curve very well. Only thing it starts showing it will diverge. While Wien's law could explain this particular part of the curve very well but it did not explain later part of the curve. In fact it went almost about this thing. So there are various attempts but these two attempts were partly successful because at least one of them was able to explain the lower part of the curve. The second part could explain at the higher part of the curve but there was not a single expression which could explain this as well as this okay and that is what Planck basically did that this was in fact he got the idea probably both using the Wien's idea and the idea of from your Rayleigh Jean's law that if he makes energy quantized and make it proportional to the frequency he will be able to get one single expression which will fit the entire curve. Okay thank you. Welcome. Maha Institute 1096. What is the physical significance of propagation vector k? How can we understand the propagation vector k? Okay see in principle when you are talking of a three dimensional wave there are two things which are associated one is magnitude another is direction. So in principle direction of the k will define in the direction in which the wave is actually travelling okay and its magnitude will give its wavelength. So now we are talking of only one dimensional case. So basically k will give only its wavelength okay and looking at its relative sign with respect to the x direction okay of that term I will be able to find out whether the wave travels in plus x direction or a minus x direction. So at this particular moment because we are talking of one dimension the k is just 2 pi by lambda but when we are talking of three dimensional case it becomes little more complicated that is why it is called wave vector because the direction will essentially give you the direction of the wave in which it travels and then it has to be associated with a particular wavelength okay which will again be given by a similar expression. Sir good morning. Good morning. My question is that as we know gases behave like good conductors at low pressure. So how can we relate this concept using kinetic theory of gases? Well that is a slightly question which is slightly in a different type because then you have to talk about the ionization things and things see the things that you know normally if you are taking purely a gas okay I do not think that will conduct you know unless you try to ionize it and you know that becomes a totally different type of physics when you are talking it is not really related to a standard kinetic theory of gases thing. So it is essentially you have to create a sort of ionization in the gases which is completely simpler when you have a lower pressure and therefore it will be easier for you to pass the current. So this becomes a different part of the physics not really the standard kinetic theory of gases in that sense. Thank you so much sir. Thank you. Sir you talked about quantization in nature at different levels. So is there any specific number of levels in which we can find quantization? Always depends on problem probably we will try to do some of those problems depending upon the time. See for example if you take a particle in a three in a one dimensional box they are infinite number of levels but if you take a finite box you will always a finite number of levels and there are methods by which you can find out how many number of levels will be present. So this particular problem when we do a formal quantum mechanics we will be discussing in detail how to find out the different type of levels alright. So in principle in a given problem we can find out I mean depends on the type of problem you can always find the total number of the levels okay this particular number could be infinite for example for a given problem or it could be finite but you can always find the total number of levels that is not an issue. Yeah one two double nine Dronachary Institute. So there was a question from the center first question about the displacement psi k sin kx minus omega t yes and she was asking that why it is not k sin kx. So in this case the displacement varies with position as well as time that's correct when we talk in terms of traveling waves yes and we set the origin and we calculate the displacement after time then this term comes kx minus omega t. See the thing is that you know we always define with respect to an infinite system a particular origin. My question is yes we have talked in terms of degrees of freedom yes in case of potential and vibration we have to take the value half kT yes but in case of vibration yes we have to take the value kT that's right. So what is the what is actually what we know we have different modes of vibration yes and we consider one mode as one degree of freedom yes right. So can you suggest some basic concept why this value is kT in case of vibration for single degree of freedom okay and why it is half kT in case of a rotational motion rotational degree of freedom. Okay I think I have sort of understood your question let me try to explain and you let me know whether I have answered your question correctly or not. See the question is something like that see why it's half kT in case of rotational and translational because if you look at the derivation of the Maxwellian distribution and for example the Maxwell Boltzmann distribution or for that matter let's say even equipartition any system which has only possibility of kinetic energy will give you half kT any system which has potential energy as well as kinetic energy both of the both the energy is involved then that will have kT energy. So like for example when there is a vibration when there is a vibration there has to be a force okay without a force a particle particle will not be able to vibrate and therefore there has to be a potential energy also involved for example if you take a simple harmonic oscillator okay there is a potential energy term there is a kinetic energy term and the total energy sometimes the energy is completely kinetic sometimes the energy is total potential and this keeps on exchanging between them. Now if you calculate the averages you find that half kT comes from the average kinetic energy and half kT comes from the average potential energy alright. Now in a system where there is no possibility of having potential energy you have only average kinetic energy which gives you half kT. So for example if a particle translates it is not vibrating so there is no potential energy involved so it will have only half kT. So basically the average of kinetic energy turns out to be equal to half kT and the average of potential energy turns out to be half kT. So when there is a vibration in which both kinetic energy and potential energies are involved then the total energy becomes kT otherwise it just remains half kT because there is no potential energy contribution. Now as far as its link with your the normal wave equation is concerned see energy there in the classical concept is always dependent upon the the amplitude okay. So if you want to link these two ideas okay then only you can say that if the energy has to be have a discrete value essentially it would mean in a classical sense that the amplitude value of this particular wave will also be quantized. So its a value will take only discrete values because classically the wave energy depended on the amplitude of the wave okay. If I want to link the two ideas I don't know whether I have answered your question. Thank you sir please. Welcome. Nristan Institute. Oh good afternoon sir. Yeah. Take the reference of slide number 16 sir. TBR curve. Here? Yeah. My question sir that you know this curve T power of y and lambda T as x. Yes. This curve must be symmetrical about y axis but it is not right here. Okay let me first put a disclaimer this particular graph I have plotted myself this is not really may not be 100 percent current graph if you want to look at the graph I will look into a standard textbook okay this I have only you know sort of roughly drawn but I still do not understand why it should be symmetric about y axis because lambda T can only have positive values and we see power only has to be positive are you saying that this should be symmetric about this particular point this particular line. Yeah sir. Actually experimental curve is not symmetric about that line in fact it has a longer tail it has a much longer tail. Why sir why. Okay see remember this is an experimental graph this is not the plot of the equation which I have written earlier that equation will give you a diverging curve only. Sir take the reference of slide 15. Yes see that is what I am saying. Look at the expression e t power 4 y spice. Yes the numerator of this right hand side is actually constant. Yes. If you take left hand side lambda T as a single variable x put it negative value you will always get positive value right hand side y so y putting minus x and you are getting positive value curve must be symmetrical about y axis. The graph that I have plotted is not a graph of this equation that is the first thing that you should understand alright. That graph is an experimental graph this is what you get from Rayleigh Jean's law. So that particular graph is not a representation of this equation okay. If you plot this equation remember here also e has to be positive m s a power has positive temperature has to be positive lambdas have to be positive all these things are positive. So this particular graph has to be always plotted only in the first quadrant. Now let us come to this particular equation. If at all I wanted to plot this particular graph this particular thing will be just inversely proportional to lambda T to the power 4. If I am plotting as lambda T at high value of lambda T it will tend to 0. At lower value of lambda 2 it will diverge because at lambda T is equal to 0 it should show you infinite value. So it should go like this the curve should be something like this alright. But what I have plotted here is an experimental graph which you actually find out by using experiment. And of course let me sort of tell this is an approximate graph alright. But here there is no symmetry because this is what you have observed experimentally. But again I insist all these factors are always positive. So this graph can always be plotted only on the first quadrant. There is no meaning of m s a power which is negative. It means it is absorbing rather than emitting okay. Similarly we always talking of wavelength which are positive. We are always talking of temperatures which are positive. I hope it is clear. I think we will stop here if you have questions you can please send us on the on the Moodle forum. Moodle forum and then we will try to answer in the next session.


I'm delighted to now welcome Anne, and Anne is also working for MSF, and I think she qualified fairly recently and straight away went to work for MSF, so that must be a good thing to do. And Anne is going to talk about cervical cancer screening in Zimbabwe. I look forward to your talk. Thank you very much. Good morning everyone. So, the aim of this presentation is to approximate the rates of precancers and cancerous lesions that are found among our HIV positive women, and also to determine the effectiveness of our program. We did a retrospective review of all our routine data collected data, and because of this we didn't have to go for an ethical review. So, MSF started in Epworth, which is a suburb to Harare by the 2006 in response to the HIV crisis. And our current focus is HIV treatment failure, its pediatric HIV, and other certain co-infections such as multi-drug resistance TB. The cervical cancer screening program was started in 2013 and handed over to the Ministry of Health in January 2015. So, why did we decide to start a cervical cancer screening program? It's one of the defining diseases, but also it's one of the most common forms of cancer in Zimbabwe and black women, accounted for more than 5,000 new cases per year, out of which 60% are HIV associated. But as most women can't afford to go to a health facility for diagnosis or treatment, it's a gross underestimation. So, because of this, the Zimbabwe Minister of Health decided to start scaling up cervical cancer screening services using visual inspection with acetic acid and cervicography, but so far the access is quite limited. So, here's one of the nurses who are recruiting HIV positive patients in the waiting area, and it was open to all sexually active HIV positive women. They could go for screening on the same day, and the nurses use the camera to take a picture of the cervix and then a monitor for visualisation. Depending on the size of the lesion, the patients were either given cryotherapy on the spot by the nurse or referred for a loop electro-surgical excision procedure at a central hospital, and MSF was paying for the procedure. In cases where the nurses were suspecting cancer, they just offered advice on where to turn next, but MSF weren't covering either investigations or treatment for that. So, in the one-and-a-half years that we did the screening programme, we screened more than 3,000 women for the first time and more than 300 full-up screenings. And considering that the HIV positive cohort of both male and female patients was more than 6,000 patients, we managed to cover the majority of our patients. 16% of them had positive lesions out of the rich 64 required cryotherapy and 46 were referred for a leap. The nurses also found 26 cases where there were suspecting cancers and these ones were referred to hospital for further treatment. So, as you can see, the nurses were seeing more than 150 patients per month, which is approximately 10 patients per day. There's a small peak here in September and October due to that we started referring patients from a neighbouring clinic, but that stopped in November and December and that's where it goes down. I want you to remember this because I'm coming back to it on the latest slide. As to the positivity rate, it was initially quite low towards 10%. And from the start we had a communication with Newlands Clinic, which is a Swiss aid supported HIV clinic, which was one of the first to introduce cervical cancer screening in Zimbabwe. And their positivity rate was 30%. So, they were advising us to send our nurses for retraining at their centre, which we did in July 2014. And after that, you can see that our positivity rate went up to 30% as well. As for the treatment, the green line here stands for all the patients who were with positive lesions. And as you can see, there's a peak in September and October, both as we start finding more patients and also screening more patients, which goes down in November-December due to that we started screening less, but the positivity rate was the same. The light blue staple are the women that were referred for LEAP, not treated, just referred. And the blue staple are the ones who underwent cryotherapy and that gap in between are the women who did not receive cryotherapy. So, out of the 342 that we referred for cryotherapy, only 60% received treatment. And initially that number was even lower towards 30%. And this was due to that the women were booked for cryotherapy at a later date, because after doing a procedure, you have to abstain from sexual intercourse for four weeks. And it was considered prudent that they were allowed to go home and discuss with their partners before undergoing the procedure. However, in August 2014, we introduced a see-and-treat approach where the women were signing a former consent for cryotherapy before doing the screening. And with that, we had a much better treatment rate and the screening rates were the same. But as you see, there was still 20% who did not receive treatment and that was because the cryotherapy broke down for one month during the autumn. Out of the 265 patients referred for LEAP, only 76 were seen at the hospital for a first visit. And out of these, only 52 received an initial biopsy result. When they saw them, they decided that 30% actually needed LEAP, but out of them, only 25 received LEAP. Now, as I said, the Zimbabwe and Minister of Health have been scaling up their cervical cancer screening services, but the access to LEAP has not been scaling up to the same extent. There was a huge queue initially and then the LEAP machine broke down in September 2014 and it's still broken down. So in addition to this, they also had a system to do a biopsy before doing the treatment and that's to make sure that people don't get unnecessary procedures. But as you see, they also lost patients on the way. At Newlands Clinic, they do LEAP at the clinic and they do LEAP for all patients that the nurses send and that gives them better treatment rates. As to the histology results, 3 fourths had abnormal cytology and 55.8% had high-grade lesions and 16.3% said low-grade lesions and 3.8% showed up with cervical carcinoma and these patients have been referred to hospital for further treatment. Sadly, we did not do a biopsy for all our patients with positive screening results, but if we consider that the 26 patients where the nurses were suspecting that they had cancer, that they actually had cancer and add that to the two patients with confirmed cancers to make the prevalence of cancer among the patients who did biopsy, who were waiting for LEAP and put that on all the patients who never received LEAP, then we could project and further eight cases among our LEAP patients. If we then add these numbers together, we would get a prevalence of 1.1% and this should be compared to the national prevalence which is 0.035%. HIV causes chronic sinusitis and this leads to more infection with the human papillomavirus with multiple forms, with more oncogenic forms, less self-healing and higher progression rates to cancer and lesions that are more difficult to treat. Because of this, it is recommended that they come for yearly screening but only 25% came back after a year. It is also recommended that they come back six months after going through a procedure, but only 13% came. This is the figures after we handed over to the Minister of Health and as you can see, the patients are still coming for screening, although not to the same amount as before. The positivity rate is high still and all the patients who require cryotherapy receive it. But the nurses are flagging that the ones that they are referring for LEAP can't afford to go for LEAP. Our future plan is to open up a VAIC in one of the neighboring clinics in Donbu and that started in April 2015. In conclusion, we can say that we had a high prevalence both of pre-cancers and cancer lesions among our HIV positive women and that the staff training was very important for us to identify those lesions. However, one of our continuing challenges is that such a low percentage of patients that come for re-screening after one year come after procedures and also to show up before procedures. So, in order to actually see an effect on morbidity and mortality we would have to make a follow-up much later on. But it's important to remember that the patients we treated might actually have self-healed and also some of the patients that we treated can still develop cancer as many of the HIV positive patients need several treatments and there was quite a low percentage who came for re-screening after six months. So, what I would find really interesting for MSF to consider for the future is to look more into human papillom and virus vaccination especially for similar settings such as Zimbabwe. Thank you very much. Thank you very much for that very nice presentation about a very actually important problem in developing countries of viral cancer and particularly so among HIV positive women. So, I'm sure there will be a number of questions from the floor. Would you care to raise your hand and also maybe introduce yourself and let us know where you are working? Yes, sir, in the blue shirt. Thank you. I'm Isaac, Deputy Program Manager for MSF in Tokyo and I also happen to be Zimbabwean. So, maybe I can say hi, ma'am. Ma'am, we are actually considering starting a project similar to this in one of our missions and my question is actually one of your questions so maybe you could give us your opinion on whether you think HPV vaccination could have a role because we are actually considering it as part of our project. What is your opinion? You have asked the question already. Well, I completely agree. I think it's very important and I know that I was reading yesterday about Gaby that they want also to give more access to vaccination in third world countries and according to them the price was $4.5 so it's not very expensive either to do it and that would be a short-term intervention that could give a long-term effect and I think especially for HIV positive adolescents which in our project they have about 1,000 and it's very important to give them HPV vaccination for the women. Yes, we have a question from the online forum. Firstly, I have MSF Zimbabwe who say yay on Twitter and then I have a question from the Lancet ID who asks regarding your research if the false positive rate was estimated? As I said, we didn't do biopsy for all our patients who had positive results. We only have the biopsy result for the mountain leap and as you can see there, there were three fourths who were positive but these are the big lesions. The other ones are small lesions. We haven't assessed that. For interest, the other positivity like we found one TB serviceitis, one sister somiasis so there were other pathologies but not necessarily cancer. Thank you. I would like to add maybe that that is something to consider in future work because this is very clearly a very topical area of work and could do with more evidence as suggested so that might be nice to consider in your various programmes. Yes. Could you speak up a little bit? Sorry, Nellie Staderini. I'm the SRH referend for OCG working on cervical cancer as well in Mozambic. I have two questions. The first one is you said that you take some pictures and I would like to know what do you do with those pictures. Do you send to people who also can compare the diagnosis? And the other thing is how many providers did you have and did you have any difference regarding the different providers? For the result. Thank you very much. The pictures were saved in a computer system so we had a computer filing system and that's where they were saved. So whenever you enter a new picture every year then you can follow the development of those pictures. At the new clinic we don't have a computer system so then we just file them on a computer and then you can search for that picture the next time the patient comes. As for the providers there were mainly three nurses who were providing the screening and only two of them went for the retraining so the other one has still not gone for the retraining and as you saw there was a significant difference between the patient and the patient so we had to give it away before and after the retraining. We also did quality control every month. Over weeks I went through the pictures with the nurses and they were both, you know, we went through them together and when there were pictures that were not good or pictures that were a bit doubtful then we called back the patients for re-screening. Right. Given the time, although you did stick to your time I'm going to ask you to keep the rest of your questions still the panel discussion. Thank you very much again for the next presentation.


Here's a famous question. If the Sun were to suddenly disappear, how long would it take us to know? Obviously, it would take us 8 minutes for our eyes to notice the difference since that's how long it takes light from the Sun to reach Earth. But what about gravity? After all, the Sun's gravitational exertion does play a huge role in anchoring our solar system together, so the absence of a Sun would be felt by all of us when the orbit of our planet went out of whack. But the question is how long would it take us to feel it? Would we feel the effects instantly or would it take some time? Stick around to find out. In 1687, Isaac Newton published his Principia Mathematica. In it, he laid out his views on space, time, and gravity, stating that space and time are two separate entities and that gravity was something innate in an object that acted over any distance instantaneously. A couple of hundred years later, Albert Einstein then came along, claiming that space and time were actually one entity and that they were interwoven together in this sort of fabric that permeated the universe called spacetime. Moreover, Einstein believed that the presence of matter and energy could curve the fabric of spacetime and that it was this warping of spacetime that we perceive as gravity. So, do we have proof of Einstein's claim? Well, in 2004, we launched Gravity Probe B, a spacecraft which had four gyroscopes. Should Einstein's theory be true, Gravity Probe B's four gyroscopes would have minute changes in the direction of their spin as Earth's gravity tugged at the spacecraft as it spun. If Einstein was wrong, however, then the gyroscopes would continue to be pointed in the same direction. Fortunately for Einstein, the gyroscopes did have a detectable direction change, establishing a link between gravity and the warping of spacetime. But it doesn't end there. Einstein's theory also states that gravity isn't an instantaneous force, but rather gravity actually travels at the speed of light in a vacuum. This can be a little confusing to understand, so let's use an example. Let's say we had a ball about half the size of Earth. With the size and mass that large, it would warp the fabric of spacetime by a considerable amount. Now, let's say this ball suddenly disappeared. What would happen next? Well, obviously the warp space would go back to its normal shape, but not at every point at same time. The space returns to being normal at a central point, which then sends out disturbances like ripples in a pond. These disturbances in spacetime called gravitational waves travel out at the speed of light. So now that we have a better understanding of what gravity is and how fast it travels, we have some clue to what our answer is to the disappearing sun question. It would take eight minutes for us to feel the effects of the sun's disappearance as the sudden removal of our sun would send out gravitational waves rippling through the fabric of spacetime at the speed of light, which would eventually reach our Earth. But what would happen when those gravitational waves did reach our Earth? Well, gravitational waves have hit the Earth, and we have even been able to detect and measure them. But if space really is wobbling around like waves in the ocean, which you can generally feel, why don't we feel gravitational waves whenever they hit Earth? Well, spoiler alert, it's because this phenomenon is incredibly quick and tiny. But what I think really makes this interesting is how we were able to detect them, and the links at the team at the laser interferometer Gravitational Wave Observatory, or LIGO, had to go in order to actually get these results. If that also interests you, head over to my channel, Canubis, to learn more about this. It will absolutely blow your mind as to how much care and precision went into this. And of course, be sure to subscribe to both our channels.


We're going to do an experiment to determine the molar heat of vaporization of water. We're going to mass an empty calorimeter. We're going to add about 70 milliliters of water to the calorimeter, and we're going to remass the calorimeter and the water. We're going to use our CBL and temperature probe to measure the initial temperature of the water. We have a steam generator containing water. I'm going to cap off the steam generator and we're going to bubble steam through our calorimeter. The water will condense to a liquid and transfer the energy to the water in the calorimeter. We'll record the final temperature. We'll let the calorimeter cool to room temperature and remass it to determine the final mass of the water in the calorimeter.