 Hello, in this lecture, I will again go back to the correlation function g of r t, whose double Fourier transform over time and momentum gives me the s q omega, the scattering law that I tend to measure in an experiment. I will try to familiarize you with g of r t and its physical significance and also since s of q omega and g of r t are Fourier space of each other, I will also introduce you to the design requirements of spectrometers depending on the kind of dynamics or kind of structure that you want to see because you need to decide it beforehand and that will dictate how the detects are positioned, how energies are measured everything. So, and then I will take you through a tour in the various neutron sources, major and our own source and various components that are present in a neutron scattering setup starting right from the source to the detector. I will take up all these description of all these of course, a little briefly to introduce you to the proper experimental facilities. Now, I go over to the lecture. So, this lecture so, as I said I will discuss the physical significance of the correlation function. I will also discuss the design of experiments versus resolution and then I will introduce you to elements of scattering setups, the source, the beam transportation, the beam monochromatization, two neutron detectors. I will touch upon all of them briefly, if not in this lecture then in the next lecture. So, I had indicated that s of q omega is a double Fourier transform of g of r t. So, d 3 r e to the power i q dot r is one integration was space that takes me to the q space and d t e to the power minus i omega t over time that takes me to the omega space. So, I measure instruments in which specifically I measure neutrons at a certain angle and if it is a attempt to understand dynamics then I also try to find out what is the energy of the neutron because the dynamical process in a system like phonons, diffusion, molecular vibrations they can exchange energy with the neutrons. Why it is not possible with x-rays is the fact that if I take a 1 angstrom x-ray the energy is about 10 kilo electron volts. So, the to measure the change of milli electron volt on a kilo electron volt x-ray is difficult if not impossible. Actually, presently in synchrotron sources using very high resolution backscattering instruments one can measure energy transfers even of that order, but in general I can broadly tell you that the energy transfer of milli electron to measure using a photon of kilo electron volt is like measure trying to measure a milli volt change in voltage using a volt a voltmeter which is calibrated for mega electron volt mega volt measurements. So, now let us come to the correlation function which describes the time space correlation between the scattering units. I mention this consider a wave phonon is a wave the position of one particle at time t allows me to predict the position of another particle at time t prime. Let me just tell you we know that if there is a chain of atoms let us just take a linear chain of atoms in a system. I know if there is a phonon then the phonon possibly goes like this the phonon that means the displacement of this is correlated to the displacement of this particle it is also correlated to the displacement of this particle and this particle. So, at any instant t if I know the position of this particle r or I call r comma t that is more conventional then position of all other particles at any other position any other position r prime at a time t prime is known for a phonon. So, that means this phonon dispersion relation gives me g of r t for this system. Now coming back to static system if I consider g of classical r t. Now at t equal to 0 if I consider this has got two parts because this is the sum of delta functions r 0 minus r j t. Now at I sum over all the partners and do a this is a ensemble averaging. The ensemble average is inherent because as I told you earlier when I do a measurement I do it over finite time that means I take frame after frame on to the detector and that does the ensemble averaging automatically. So, this ensemble average is shown by the brackets as I shown you. Now this has two parts a particle of course it is in its own location it will have a delta r, but then for all other partners of it there is that of a delta function peaks whenever r j minus r 0 hits another atom. So, let me again the same chain of atoms in the static thing. I have got this chain of atoms I have this chain of atoms this means one it has got a delta function at the origin and any particle can be the origin. So, it is a delta r for this for all the particles, but also whenever the distance this distance this is the let us call it the this is the jth particle jth particle jth. So, whenever the distance is such that there is a particle there I have got a delta function peak and this delta function peak gives me not only the delta r I have got plus I have got a pair correlation function g of r what do you mean by pair correlation function and these are ensemble average. A simplest example suppose I take a case of molten sodium chloride I am taking a simple example. So, there are Na plus and Cl minus ions at any instant of time if I look at this molten salt if I take a sodium I can assure you statistically they are surrounded by a circle of chlorine minus because their ions and they attract each other and then surrounded by the chlorine ions who will have maybe possibly another shell of sodium and statistically. So, this shell compared to this shell is more clearly defined as you go farther and farther the this shells that I am showing you they get diffuse and more diffuse and finally, I have constant density. So, starting with one sodium as origin I have got a pair correlation function which will possibly look something like this. So, I have got a peak the delta function peak here which I am not plotting now the pair correlation function should have a peak then smaller peak and then it will continue. So, I have got a first in the first shell I have got a peak if call it r 0 then whenever delta r r minus r r equal to r 0 I have this shell and I will have a peak. And this is actually later when I discuss that if the structure of liquid in our system it will come naturally I will introduce you to it more in more details at that time. But what I mean is that in a static system in a crystalline system you have got sharp delta function peaks. But does not mean that when we have a liquid or amorphous system there are no peaks there are diffuse, but there are very clear short range order in such systems and that can be found out from the GFR that we can measure in an experiment. But GFR I have taken out time. So, I am not doing now in for this experiment this is not a dynamics targeted experiment we are trying to look at the structure. So, this is if I may say a brief introduction to GFR or pair correlation function for a static case and for a dynamic case as I told you that given a phonon an acoustic phonon a transverse acoustic phonon which I was drawing for you transverse acoustic phonon then the displacement of all the particles are correlated and their GFR is also well known. So, you can find out s of q omega for this also this is a dynamics case the dynamic case. So, with this I have introduced you to GFR T and for a static case and for a dynamic case. Now, I will come to resolution of an experiment and what can I do about it. So, because q omega and R T. So, q and R q is basically momentum transfer h cross q is the momentum transfer. So, it is a momentum space and R is real space and they are related we know by the uncertainty relation delta P delta R should be of the order of h cross and or higher the greater than equal to h cross. Similarly, the energy transfer is h cross omega and energy and time they are also an uncertain relation because of Fourier inverse of each other in this expression. So, delta A delta T should be greater than equal to h cross or of the order at the best. Now, in an experiment the delta K is basically the range of momentum transfer that we measure in the experiment because that gives me the limit of momentum transfer or momentum and that is the gives me the limit of uncertainty for a given system and that tells me twice pi by h should tell me that how much is the quantum mechanical space resolution that I can achieve let me be little more specific. So, we need to design or choose the experiment depending on what you want to see. So, let me just write that for a diffraction experiment let us say the quantum mechanical quantum mechanics says the resolution in an experiment is 2 pi by q max 0 to q max measurement range and q max again let me go back to the board we know that q is equal to 4 I am sorry 4 pi by lambda sin theta when the actually the beam has deviated by 2 theta. So, half of that is sin theta and this is q is equal to 4 pi by lambda sin theta this comes from a very simple geometry this is K 1 K 1. That is a thread you are seeing this is K 2 outgoing this is your q vector if this is equal to this is 2 theta and the related q is 4 pi by lambda sin theta. So, for then I know when I say q max q max it is proportional to theta of how much theta I am covering in an experiment theta max if I am doing the if lambda equal to constant I make this comment over here because from this expression you can see they can I can also change the q by changing lambda I can go to smaller and smaller lambda and q will become larger and larger and that is what is done in pulse spallation neutron sources, but that will come later what I mean is that here in an angle dispersive experiment your q max depends on maximum angle of the experiment. Then I can evaluate let us say I want to go for a wave vector transfer of 10 angstrom inverse with 1.2 angstrom incident neutrons. One needs to go to scattering angles of around 140 degree 2 theta equal to 140 degree. In a typical powder diffractometer we have 10 to 12 angstrom inverse range. So, you can see that we had to have detector moving from 0 to 140 degree and this gives me a quantum resolution of 0.6 angstrom for measurement of the lattice parameters and that is reasonable because often lattice parameters are in the range of this. Now, suppose I want to broaden the resolution that means, I do not want to see the atomic structure I want to see some larger conglomeration. So, I make my resolution 30 angstrom this is not improving the resolution it is making the resolution poorer then and I am using let us say a 4 angstrom neutron. All these have meaning later why I am using 4 angstrom neutron because with a 4 angstrom neutron for having a resolution of 30 angstrom we need to go to about 4 degrees. If I use shorter wavelength neutrons shorter wavelength neutrons then to go to this kind of smaller q values a theta also has to be much smaller you can see that in that expression I wrote theta equal to 4 degree. Now, you have a direct beam and you have to measure up to 4 degree which is sorry. So, if I need to go to 4 degree is a reasonably small angle 4 degree. If we have to go to 1 degree let us say 1 degree because I want to go to 0.2 angstrom inverse using a shorter wavelength neutron then such small angles are difficult because you also have a direct beam which has got a beam width of certain angular spread and the direct beam might start getting into this. So, that is why I said that I will be using 4 angstrom neutrons and that is called a cold neutron. I will tell you how to how we get cold neutrons and so, we have to go to 4 degree. So, this one the top one the yellow rectangle this is a typical test for a powder diffractometer for crystallographic structure where we are going up to wave vector transfer 10 angstrom inverse. Similarly, for a resolution of 30 angstrom other way round we need to go to 0.2 angstrom inverse look at the difference 10 angstrom inverse 0.2 angstrom inverse with the 4 angstrom neutron we need to go to an angle of around 4 degrees. So, this is an example that I have taken which is a typical small angle neutron scattering of sands case. Here we do not see the medium as consisting of atoms and crystal crystallographic lattice, but we are looking at conglomerations which may be 30 angstrom sometimes it can be even large 100 angstrom 200 angstroms and these are in homogeneity center system that we can be studying. So, the need of what do we want to see dictates a spectrometer. So, and we have to design the spectrometer various spectrometers depending on this need. So, I will come back to this in the next part of the talk. So, what we can see with neutrons in conglomerates matter.