 In this lecture we will see the relations between transfer matrices and the fixed parameter we have obtained. And also we will analyze one of the basic optical cells known as FODO cells. Also we will introduce a new parameter namely dispersion which takes care of the dynamics of the particles which have momentum deviated than the design. So we will start this. Now solving the Hilse equation we have seen two methods. One is the piece by solution in which we have breakdown the optics in the smaller pieces and then in each piece keeping the key constant we have opportunity solution. This is similar to what we solve differential equation in simple numerical method. This means we probe the equation in different smaller segments and each segment we solve the equation. And second and more elegant way was to introduce the solution using some optical parameters, these parameters. And when we were obtaining the solution by this method we have got invariant of motion namely as atoms. In this case if k is periodic then we can obtain solution in which twist parameters are also periodic. Means in the case of periodic k there may be possible range of solutions among which we obtain those choose those parameters which are periodic nature. We say this is the best solution. If we know values of twist parameter two locations can we connect the particles trajectories or particles coordinate between these two parameters or between these two locations. If yes we can have a relationship between the twist parameters and matrices because matrix connect the coordinates between these two points. And if we can connect these two coordinates at these two locations in the twist parameters we can have one-to-one relation between the matrices and twist parameters. So, we are going to do that. Now as Hill's equation we have a general solution in this fashion x is equal to a root beta cos phi plus b root beta sin phi. Here beta and phi are the function of the s and because this is a second order differential equation. So, we have two constants of integration a and b with these constants are determined using the usual conditions. And if we differentiate this x with respect to s we get x prime and which is a alpha root beta cos phi because we have when we differentiate this beta root beta we will get one upon two root beta d beta by ds and one upon two beta prime is minus alpha which is written here minus alpha upon root beta cos phi. Then we will differentiate this phi. So, we will get minus 3 upon root beta sin phi d phi by ds and d phi by ds is one by beta. So, this root beta and one by beta will give you the one upon root beta which is zero one by root beta. Similarly, we will differentiate this term. So, we will get alpha upon root beta sin phi and b upon root beta cos phi. So, this gives you the x prime and as I have put earlier that this a and b can be determined using the initial conditions. So, let the initial condition like this. Let the initial phase phi is equal to zero and the amplitude function of this parameter beta has the value beta zero at that location. And the initial coordinates of the particle be x0 and x0 prime. So, if here we put x is equal to x0 and a then root beta is root beta zero because it is the initial. Then cos phi phi because we are taking initial as zero. So, cos phi will be one and here it is the sin phi. So, initial phi means zero. So, this term will be zero. So, we will have x0 is equal to a root beta which is written. Similarly, we will obtain the initial x prime. x prime initial is x prime initial zero because sin phi itself is zero but phi is equal to zero. So, these are the two equations which relates initial coordinates to initial parameters. And using these two equations we can obtain the a and b. So, now here constant values of a and b are as a function of initial parameters initial coordinate at zero and initial this parameters beta zero and alpha zero. Now, we can put the values of this a and b in the solution in this solution. So, we do that after a little bit of algebra means when we will put the value of this a and b in this equation and arranging the terms together of the x0 and x0 prime. You will get x is equal to root beta upon beta zero cos phi plus alpha zero sin phi and coefficient of x is a prime zero will be beta root beta sin phi square with square root. Here beta you can say suppose this is the orbit of the sin proton and this is our initial point where we are defining beta zero alpha zero and phi is equal to zero. This is our initial point initial and second point is say here where we are talking about beta alpha is that this parameter and between these the phase advance is phi. So, this beta and this beta are at this location and these are at this location and this beta zero and this beta zero are the parameters at initial point. So, now here you can see that x has been written down in terms of x0 and x prime zero means in terms of the initial coordinate. So, now the coordinates of here x and x prime here the coordinates were x0 and x0 prime. So, now x has been written down in terms of x0 and x prime zero. We did this using the transform matrices. Similarly x prime can also be written down using these values of a and b and when we do a little bit of algebra we got this equation. So, now these two relations give us the coordinates at another point here in terms of the coordinate of the initial point. So, this is similar to the transform matrices. So, if we compare these terms using the transform matrices from here to here means suppose we have m1 to transform matrix from point 1 to 2 which we can obtain using the multiplying transform matrices of each element which occurs between these two points then that matrix and this matrix should be the same. So, now this you can see is exactly in the shape of transform matrices but this transform matrix contains the twist parameters as the elements means we have now transform matrices in terms of twist parameters. We can also obtain the transform matrices in terms of the magnetic strength the magnetic effective length L etcetera which we did earlier using the piecewise solution then these two matrices must be equal because these both are the solution of the same equations. So, if we do that we equate these elements we get the connection between transform matrices and twist parameters. Now, here you can see that if suppose this is a orbit of the synchrotron and this is our initial point as I said initial point and this is our final point or point 2. If we move the point 2 and reach here after one term in this case beta should be equal to beta naught and alpha should be equal to alpha naught. When you will do this you will get beta upon beta naught here beta upon beta and this will be cancelled out. So, you will have cos phi plus alpha sin phi. Here beta beta naught will become beta because beta naught equal to beta and here 1 plus alpha alpha naught upon beta beta naught this term will become minus 1 plus alpha square because alpha naught is equal to alpha and beta naught is equal to beta. So, root beta beta naught will be beta. So, this is gamma which is written here and similarly these beta naught and beta will be cancelled out. So, we will have cos phi plus alpha sin phi. So, this matrix is for the one term where we have periodic solution of twist parameters. Since if we want to know the transfer matrix of one complete term in terms of this parameter where we are choosing the periodic solution then transfer matrix is this. Now, this matrix can be written down in a very interesting form. Our matrix was cos phi plus alpha sin phi beta sin phi minus gamma sin phi and cos phi minus alpha sin phi. This was our one term matrix which we obtained earlier. Now, this can be written down in this fashion in the form of summation of two matrices. The first matrix is cos phi 0 0 cos phi and second matrix is alpha sin phi beta sin phi minus gamma sin phi minus alpha sin phi. So, you can take common cos phi here and here sin phi. So, we will do that. So, 1 0 0 1 and alpha beta minus gamma minus alpha sin phi. Now, you can see that this is the unit matrix 1 0 and 0. And if we multiply this matrix with itself then what will we get? We will see this alpha. We will get alpha square minus gamma beta here and alpha square, you know that alpha square plus 1 is equal to gamma beta. So, alpha square minus gamma beta will be minus 1. So, this will be minus 1. This element will be 0, this element will be 0 and this element again will be alpha square minus gamma beta which is equal to minus 1. So, here you get the identity matrix and here this matrix is that kind of matrix whose square gives you the identity matrix with minus sign. Means we can write down this in this version. The square of this matrix is minus unit matrix. Means one term matrix can be written down as unit matrix cos phi plus j sin phi. This is similar which we write down as unit is to eta phi is equal to cos phi plus eta sin phi. Means one term matrix can be written down in the Euler form. In the Euler form what is the use of this? Suppose you want to know m one term power m means after n terms what will happen? Using this relation you can obtain m power of m one term very easy. And another important property of this matrix is the trace. This matrix again I am writing for it as cos phi plus alpha sin phi here beta sin phi here minus gamma sin phi and here cos phi minus alpha sin phi. Now you can see the trace will be cos phi plus alpha sin phi plus cos phi minus alpha sin phi and this will give you 2 cos phi. Now for bounded motion phi should be the real quantity and phi should be the real quantity means 2 cos phi must be lesser than 2 and greater than minus 2 because cos phi should be cos phi should be less than 1 and greater than minus 1 so 2 cos phi should be greater than minus 2 and less than plus 2. So modulus of trace of our matrix must be less than 2 for stable motion. What does that mean? Suppose we are connecting various magnets for making the objects like this and now by putting matrices of each element so first element is drift then focusing lens then drift then de-focusing lens then drift then dipole magnet then drift then focusing lens then drift then de-focusing lens then drift and we multiply these matrices to get an overall transfer matrix and if you found that the trace is greater than 2 means no bounded oscillatory beta term motion is possible in that kind of objects. So we have to change the strength of our magnets so that trace becomes less than 2 and in case of synchroton no motion will be possible or synchroton cannot be run if trace is greater than 2 for that matrix. So one term matrix in the synchroton should have trace less than 2 for stable motion. So we have to check whether our matrices overall matrix produces a trace less than 1 if not then we have to change our optics accordingly. Now in one complete term means we started from the initial point and in synchroton we again reached to this point this is one complete term if the phase advance of this one complete term is phi this is the phi so phi by 2 pi this will give you the number of betatron oscillations in one complete term the number of betatron oscillations in one complete term is known as betatron tune and this betatron tune is very very important parameter for analyzing the beam dynamics and synchrotons we will see a little bit about of betatron tune when we will study about the chromaticity in later chapters there will be two betatron tune one for the horizontal plane and one for the vertical plane now we will analyze the tools which we have learned so far using those tools we will analyze some optics so let the very simple optics here some focusing quadrupole, then drift space then defocusing quadrupole, then drift space then focusing quadrupole, then drift space then defocusing quadrupole then drift space like that this is repeated means this is a periodic optics in which this basic cell you can say this basic cell been repeated many times. Here it is the focusing quadruple so f this is drift space means open space you can say o this is again a defocusing quadruple so you can say this is d and again open space o. So this kind of optics is f, o, d, o photo optics so this cell is photo cell and now you can see that in this figure there is a repetition of four such cells and it may continue like that. Here we are assuming a very simple optics means focusing quadruple and defocusing quadruple have the same strength in magnitude. Sine wise will be different focusing quadruple has minus k strength and defocusing quadruple has the k strength. So the strength in magnitude are same for both the quadruple magnets and length of each drift space is same. This drift space has the same length as of this disk space and all other drift space also have the same length as of this. So now if we draw once cell only which is repeated then this will be look like so from point a to point b we have one photo cell and here you can see that when we obtain the cell in this fashion in this cell this quadruple magnet is half only and this quadruple magnet is also half and this is the full quadruple magnet in this cell. The remaining half will be in the next cell and this remaining half will be in the previous cell. So now designate this drift space by L. This is the length of the disk space. Consider these quadruple magnets as the thin lines means we are ignoring the length of these quadruple and taking it as a point focusing lens. So from a star to symmetry point I am talking about this point symmetry point you can see that this is the symmetry point means here you have focusing quadruple length and de-focusing then you put a mirror at this location P you will get the complete optics. So P is the symmetry point. So up to symmetry point if we calculate it the remaining will be the same. So M of half de-focusing quadruple M of drift space length L and M of half focusing quadruple M means matrices. So these will be the transfer matrices and in this order if we will multiply we will get the matrices to the symmetry point. So let us do that. The first half cell will be first of all the particle will see the focusing lens. So 1 0 minus 1 by F1 then a drift space of length L and then de-focusing quadruple of length L. Up to the half cell de-focusing quadruple is also half. The remaining half will be after symmetry point. So here what we are writing down F this is the focal length of the half quadruple. For full quadruple the focal length will be half of this. So after multiplying these three matrices we get this matrix. And you can see that this element is with minus sign means this shows overall focusing in this particular plane. In later half after the mirror symmetry the order of the focusing and de-focusing magnet versus and then and when we multiply it we get matrix similar to the above here plus sign and here minus sign. Now if we multiply these two matrices together we get a complete matrix of the photocell. And so complete matrix of the photocell will look like this. These two elements are equal and this is 2 L1 plus L5 and this again shows the focusing characteristic of this photocell. Now if solution is periodic if you want to see that solution which is periodic in nature then this matrix must be equal to one term matrix. Means what we can do this element will be equivalent to beta sign for and this beta will be the beta at the location of the starting point of the cell where beta in the focusing plane will be the maximum. Why maximum we will see that. So we have made this kind of matrix this is our cell this is focusing quadrupole this is de-focusing quadrupole this is again a focusing quadrupole. So after focusing quadrupole before focusing quadrupole we will have like this and this magnet will focus the wing so size will again reduce so this will give a location of maximum beta means at this location d beta by ds will be 0 means at this location alpha will be 0. Similarly alpha will be 0 here and after this lens after de-focusing quadrupole again wing will become divergent so this will have a beta minimum point this point is beta minimum point and here again alpha will be 0. So now if we will compare element by element so beta sign phi will be 2l1 plus f and because here alpha is 0 and we know that gamma is equal to 1 plus alpha square by beta alpha is 0 so gamma will be 1 by beta. So at this location we can write down that this quantity which is equal to minus gamma sign phi because alpha is 0 at this location so this will be minus 1 by beta sign. So this element should be equal to beta sign phi and this element should be equal to minus 1 by beta sign phi and we did this these are the two relations. Now if we divide this equation by this equation then sign phi will be cancelled out and only beta max will be eliminated with square sign and here you will get 1 plus alpha f upon 1 minus alpha. So this gives you the value of maximum beta at this point this value of maximum beta in terms of magnetic or optics geometry means what is the length of the digital base what you what is the focal length here chosen for the quadrupole using these parameters you can obtain what is the value of beta max here this beta max is for a particular solution which is periodic in nature. So this beta max will give you the periodic solution of this prototype so using these two transfer matrices comparing these two matrices one obtained by piecewise solution and other one obtained using the quiz parameter we can get this solution or values of these quiz parameters at the different locations of the optics. If we multiply instead of dividing these two equation beta will cancel out and we get the value of the phase sign phi. So sign phi will be this so now we get beta as well as phase advance in this cell. Similarly m11 element with one turn matrix in one turn matrix we write down cos phi plus alpha sign phi and beta sign phi then minus gamma sign phi and then cos phi minus alpha sign phi because at that location alpha is 0 so this will only cos phi and cos phi is written here it is 1 minus 2 L square and because cos phi must be smaller than 1 and greater than minus 1 when you put these limits you will get the focal length of half quadrupole should be greater than the drift space length means if you want to make a stable photocell this condition has to match means the chosen quadrupole should have the focal length greater than the length of the drift space between the two quadruples. Now as cos phi is 1 minus 2 sin square phi by 2 so sin phi by 2 is L by f why are we why we are doing this kind of calculation here from cos phi to sin phi by 2 because sin phi by 2 will give the phase advance of the half cell and at the mirror symmetry we have seen that there will be beta minimum so by knowing these elements we can obtain what is the value of beta minimum in this photocell we have obtained the values of beta max now we will obtain the values of beta because up to the half point the starting point will have beta max and at the symmetry point we will have beta mean so e square root of beta mean beta max and multiplied with sin phi by 2 because we have either here at this location now one term matrix will not be used here here we will use complete matrix that is root beta 2 by beta 1 cos phi plus alpha 1 sin phi and root beta 1 beta 2 sin phi this matrix will be used here because up to half cell no periodic solution will be there periodic solution will be for full set so this is written here basically so using this now we know what is the value of beta max put that value of beta max here put the sin phi by 2 from here and you will get beta minimum so now you have beta max now you have the values of beta mean and the phase advance means you have obtained values of these parameters completely at different locations of this cell