 Hello and welcome to lecture 9 of module 2 of this course on Accelerator Physics. In today's lecture, we will continue with the transverse dynamics of beams. In the last lecture, we learned about the transverse dynamics. So the beam in general tends to diverge in the direction, transfers to the direction of motion due to various reasons. So it could be due to the inherent divergence of the beam, it could be due to the defocusing RF forces, the RF fields or it could also be due to space charge effects which is the Coulombic repulsion due to the same charge particles in the beam. So due to these reasons, the beam tends to diverge in the transverse direction. So the beam has to be focused and brought back to the axis. Beam can be focused in the transverse direction using a magnetic or electric quadrupole. So we learned about the focusing of the magnetic as well as electric quadrupole in the last lecture and we saw that a quadrupole focuses in one direction but defocuses in the other direction. So you need to use a combination of two quadruples in order to focus in both the direction. So a combination of two quadruples of opposite polarities can be used to focus the beam in both the transverse directions. We also saw how a solenoid focuses and we saw that these solenoids focus the beam in both the transverse directions. We also derived the transfer matrices from the equation of motion for the quadrupole both the focusing quadrupole and the defocusing quadrupole and for the drift space. And in this way, we can derive the transfer matrices for all the elements in a accelerator system whether it is an RF gap or a dipole magnet or any other system. So transfer matrices for all elements can be derived using the equation of motion. By using the matrix multiplication method using transfer matrices of the various elements can be used to find the final coordinates at the end of the elements if we know the initial coordinates of the particles and the elements of the transfer matrices. So we saw that if we know the transfer matrix and if we know the initial coordinates of the particles, we can find out the final coordinates of the particles. So various focusing lattices are used for focusing in a linear accelerator. So for example in a drift tube line, you can see the picture of a drift tube line here. So these are the drift tubes. In between the drift tubes, there is this gap where the RF field is applied. So the beam gets accelerated in the gap and when it enters inside the drift tube, since fields cannot penetrate inside the drift tube, it does not experience any accelerating field. Inside these drift tubes, quadruples are used for focusing the beam. So there are quadruples inside this drift tubes, they are used for focusing the beam in the transverse direction. So you have to use a combination of two quadruples so that you can focus in both the directions. So let's say there is a focusing quadruple here, this will be a defocusing, there will be another focusing and then a defocusing quadruple. So different types of focusing lattices are used for focusing, this is known as a photolattice. So in a photolattice, there is a focusing quadruple followed by, so this is a drift space followed by a defocusing quadruple, then a focusing, defocusing and so on. And remember if this is focusing, this quadruple is focusing in the x direction, for the y direction it will be defocusing. But the overall combination of such quadruples will always produce focusing in both the directions. We can also use a 4-4-doto type of lattice for focusing where in one drift tube we have a focusing quadruple, in the next one again a focusing quadruple and then followed by two defocusing quadruples and so on. A 4-4-4-doto type of lattice can also be used for focusing. So there are, you can arrange the quadruples in various combinations as long as you have both a focusing quadruple and a defocusing quadruple and the net result will then be focusing in both the directions. FD type of quadruple focusing can be used or is generally used between cryo modules containing 5-cell or multi-cell elliptical cavities. So here for example these shown here is this is a 5-cell elliptical cavity. So there are three groups of 5-cell elliptical cavities, they are inside a cryostat. So we have already learnt about this type of cavities. They are used for acceleration. So this is inside a cryostat, this is a superconducting cavity. In between two cryostats we have warm quadruples or normal conducting quadruples that are used for focusing. So again they are arranged in an FD type of manner. So this type of lattice can be used for focusing in between cryo modules containing multi-cell cavities. And solenoid based focusing lattice in a cryo module with single spoke resonator cavities. Again single spoke resonator cavities are superconducting cavities and so these are the cavities, the SSR cavities and this is the cryo module. So they are placed inside the cryo module and in between the SSR cavities solenoid magnets are used for focusing the beam. So everywhere we have periodic focusing, the focusing has to be throughout and the focusing is periodic. So periodic focusing lattices are used for focusing the beam in the linear. So in the last lecture we derived the equation of motion for the quadrupole. So this is for the focusing quadrupole and this is for the de-focusing quadrupole. Here k stands for k is a constant and it is qg by mvz where q is the charge, m is the mass, vz is the velocity in the z direction and g is the gradient of the quadrupole used. So we can solve this to find out the trajectory of the charge particle under the influence of this type of quadrupole. Similarly for a drift space we can write the equation of motion as x double prime is 0 and y double prime is 0 because here now there is no focusing or de-focusing force k is equal to 0. So this is the equation of motion for a particle inside a drift space, a field free region. Similarly for a dipole magnet we can write the equation. So a dipole magnet is a magnet with two poles, it bends the beam in one direction. So here rho is the radius of curvature of the bend. So in the x direction you can write the equation as x double prime plus 1 by rho square x square is 0 and in the y direction there is no force so y double prime is equal to 0. So in general we can write the equation of motion of the particles inside any system as x double prime plus kxx is 0 and y double prime plus kyy is 0 in the x and y directions. Now here depending on what is the element whether it is a quadrupole magnet, a focusing quadrupole or a de-focusing quadrupole or it is a drift space or an RF gap the kx will take values. Let us consider for example a photo lattice. So here we have a focusing quadrupole, a drift space, a de-focusing quadrupole and a drift space again followed by the same thing. So this is the length of the period denoted by L. So here we notice that the period repeats itself after L. So kx is periodic in L. So we can write kxs is equal to kxs plus L and similarly for y. s is the longitudinal coordinates which is also z. So in general we can write the equation of motion for a periodic system as x double prime plus kx is equal to 0 with ks is equal to ks plus L. So here k is now a function of s. So depending on where the charged particle is whether it is inside a quadrupole or inside a drift space or a de-focusing quadrupole k will take values. So this equation is known as the Hill's equation. The equation of motion of a single particle also known as the Hill's equation is similar to that of the equation of a harmonic oscillator. So the harmonic oscillator is x double prime plus kx is equal to 0. The only difference here is that here for the simple harmonic oscillator k is constant. So we can write the solution for the simple harmonic oscillator x is a cos root ks. So here the amplitude a is constant. So you can see here the amplitude is a constant. Also the phase advance which is under root ks is a constant. The motion of the particle under the influence of a constant force like this in phase space is an ellipse. So you can find out x prime and from here you can eliminate cos and sin. So you get an equation of an ellipse. So the equation in phase space is that of an ellipse. So from this we can try to understand the solution of the Hill's equation. So let us solve the Hill's equation. So this is the Hill's equation and for a periodic system we can write it as ks is equal to ks plus l where l is the length of the period. Now from Floke's theorem which states that the amplitude and phase functions satisfy a periodic boundary condition similar to that of ks. The solution can be written as let us say under root epsilon beta s cos phi s plus phi 0. So here we notice that the amplitude is now a function of s. For the simple harmonic oscillator the amplitude was constant. Here the amplitude is a function of s because k is a function of s. So here the epsilon is a constant and beta s is a periodic function given by the focusing properties of the lattice that is quadruple. So beta s has the same periodicity as that of the lattice. So we can write beta s is equal to beta s plus l. Now we can differentiate this with respect to s. So we have two terms here which depend on s. One is the beta s, the other one is cos phi s. So in first case we can keep the cos phi s term constant and differentiate with respect to beta and in the second case we can keep the beta term constant and differentiate with respect to phi. So we get this differentiating this again. So now we get six terms differentiating this with respect to s and now we can substitute the value of x double prime in equation 1 that is the Hilse equation. So Hilse equation is x double prime plus ks x is equal to 0. So if we substitute this here this is what we get. So we see that there are some terms with cos and some terms with sin. Now we can equate the coefficients of sin and cos separately to 0. So doing that we are left with two equations. So we have two equations now in terms of beta and phi. So let us see the first equation. So here we have an expression in terms of beta and phi. Let us take under root epsilon beta constant. So we are left with this expression and from this expression we can write beta prime phi prime plus beta phi double prime is equal to 0. This can be rearranged as beta prime by beta is equal to minus phi double prime by phi prime. So this can be solved and we get beta phi prime is equal to constant and this constant can be taken to be 1. So from here we can write phi is equal to integrating over the whole period from 0 to L ds beta s plus phi 0 which is some constant. So here phi s is called the phase advance between 0 and L. So 0 and L L here is the period it is one period of the lattice. Now here we define two functions alpha and gamma in terms of beta. So alpha is simply minus beta prime by 2. So we take the derivative of beta and with a minus sign divided by 2. So this is the definition of alpha and gamma is 1 plus alpha square by beta. Now since beta is periodic in s, alpha and gamma will also be periodic in s. So alpha, beta, gamma are all periodic in s. Now let us take the second term, the second equation here and equating it to 0. So we get this again taking under root epsilon beta common here. So we are left with this expression which goes to 0. So this is an expression in terms of beta, k and phi. So here beta is the amplitude function, k is the focusing parameter or the force constant and phi is the phase advance. So this expression can be written in this form here. It can be rearranged to write in terms of beta and gamma in this form here and in terms of alpha, beta, gamma in this form here. So here alpha, beta and gamma, these are called the twist parameters and they depend on s. Now let us see the phase space for the particle in the case of the particle satisfying the Hill's equation. So the Hill's equation again is x double prime plus ks x is equal to 0. The solution is of the form under root epsilon beta cos phi. So from here we can write cos phi as equal to epsilon by under root x upon under root epsilon beta. Now taking the derivative with respect to s, so we have x prime is equal to, we get a cos term and a sin term. Now here beta prime, we can write it as minus 2 alpha and phi prime can be written as from here, phi prime can be written as 1 by beta. So substituting the values of beta prime and phi prime here, we get this expression for x prime and from here we can find out the value of sin phi. So now we have an expression for cos phi and an expression for sin phi and we know that sin square phi plus cos square phi is equal to 1. So substituting these values, we get this expression. So we get this equation here and then simplifying it even further, we get 1 plus alpha square x square plus 2 alpha beta xx prime plus beta square x prime square is equal to epsilon beta. Now we have defined gamma as 1 plus alpha square by beta. So we can put 1 plus alpha square is equal to beta gamma in this expression here. So we put this here and we simplify this and we get this equation. So in terms of x and x prime, so we see that the motion in phase space is again an ellipse. So here epsilon is the constant of motion, it does not depend on s but alpha, beta and gamma depend on s. So here the ellipse varies with s. So the motion in phase space is still an ellipse for the particle satisfying the Hils equation and but here the alpha, beta, gamma they depend on s or the position. So here we have used alpha is equal to minus beta prime by 2 and gamma is equal to 1 plus alpha square by beta. So this is the general equation of an ellipse. All the particles in the beam will satisfy this equation because all the particles in the beam will experience the same force. So they will all satisfy this equation. For alpha is equal to 0, so in this if we put alpha as equal to 0, this term goes to 0 and we get an upright ellipse. So upright ellipse means we get an ellipse which is like this or an ellipse which is like this. So for the case of Hils equation, we can see that here since k is now a function of s and it is periodic in s, so the motion will be, it will execute, the particle will execute pseudo harmonic oscillations which you can see here. So the solution is now of the form of A under root beta s cos phi s, so the amplitude also depends on s in this case here. The amplitude is modulated with s. So beta is the, beta is a periodic envelope function. So this is denoted by beta, so here this is A under root beta and if you calculate the phase advance which we have already done, so we found the phase advance to be equal to integration over 0 to s ds by beta s and this is now non-uniform. So and the phase space, motion of the particles if you calculate, you see that you get an ellipse but now the coefficient of the terms the x square xx prime x prime square terms, this is now, these are now functions of s, so it is an ellipse but now the ellipse is changing with s. So we can now compare the simple harmonic oscillator versus the Hils equation, in a simple harmonic oscillator k is a constant, here k is a function of s and it is periodic in s, here the solution is A cos root ks, here the solution is A under root beta is cos phi s. So here we see that the amplitude is a function of s, so in this case for the simple harmonic oscillator the amplitude is constant whereas in this case the amplitude is modulated with s, so amplitude is A under root beta s and beta s is a periodic envelope function. So it has the same periodicity as that of the lattice. The phase advance for the simple harmonic oscillator is under root ks which is a constant whereas the phase advance here is integral 0 to s ds by beta s which is nonuniform. The phase space is an ellipse for both the cases for the simple harmonic oscillator and for the Hils equation but here the ellipse is changing with s because V twist parameters alpha, beta and gamma are functions of s. Now in real life the beam consists of many charge particles having coordinates xi and xi prime where i stands for the ith particle, so general equation of the ellipse in x, x prime phase space is this, so all particles will satisfy this equation, all particles in the beam, all the i particles in the beam they will satisfy this equation. So here beta gamma minus alpha square is equal to 1, so each particle in the beam will satisfy this equation. At a given location s in trace space x, x prime each particle will lie on an ellipse defined by the twist parameters alpha, beta and gamma. So each particle you can see here will lie on the ellipse, so for all these ellipses the twist parameters alpha, beta, gamma are the same. The alpha, beta, gamma tell you about the shape and orientation of the ellipse, so for all these particles since all the particles satisfy this equation, so for all these particles the twist parameters are the same. So ellipses of all the particles are concentrated, the area of each ellipse for each of this case the area of each ellipse depends on the value of the constant of motion which is epsilon i for that particle. So what is different for each particle, it is the constant of motion epsilon. So the outermost ellipse if you see this is the outermost ellipse, it defines the maximum beam size and the beam divergence. So what is the maximum beam size? This is the outermost ellipse, this green ellipse is the outermost ellipse and this is the maximum x, if you see this location here, this is the maximum x. So this defines the maximum beam size and this is the maximum x prime, so this defines the maximum beam divergence. So let us now consider the outermost ellipse. So the twist parameters are the same for all the ellipses, only the epsilon, the constant of motion is different for different particles. So beam phase space contours in a linear, they have approximate shape of an ellipse which we have just derived and seen. So general equation of the beam ellipse is given by gamma x square plus 2 alpha xx prime plus beta x prime square is equal to epsilon. So here alpha beta gamma are the twist parameters and they are related by gamma is equal to 1 plus alpha square by beta. The motion of the particles is along curves of constant Hamiltonian epsilon and so here you can see that this is the outermost ellipse and this defines the maximum beam half width. So xm is equal to under root beta epsilon, so this is equal to under root beta epsilon and this is the beam maximum beam half width. Similarly this defines the maximum beam divergence and this is equal to under root gamma epsilon, this is the beam half divergence. Now emittance of the beam is defined as the area of the outermost ellipse, so this is the outermost ellipse, you take the area of this ellipse and divide it by this is known as the beam emittance. So beam emittance is a figure of merit of the beam and its units are in meter and so meter and radiance. So here epsilon corresponds to the constant of motion of the outermost particle. Now Lieuble's theorem says that if x is the transverse position and px is the transverse momentum for any particle, then for a group of particles integral px dx is equal to a constant. So by Lieuble's theorem the integral of px dx is equal to constant. Now what is px? px is the momentum in the x direction, so we can write it as mass into vx. Now mass is in relativistic mechanics defined as gamma into the rest mass, where gamma this is not the risk parameter, it is the relativistic gamma and vx can be written as dx by dt. Now here again dx by dt can be written as dx by dz into dz by dt. So we can write this as gamma MO dx by dz is simply x prime and this is vz. So we have beta gamma MOC x prime, vz can be written as beta c. So now substituting the value of px in this expression integral px dx is equal to 0, so we can take MOC beta gamma outside the integral and inside the integral we are left with x prime dx. So this is a constant, this whole thing is a constant by Lieuble's theorem. Now integral x prime dx is simply the emittance because it is the area of the ellipse. So an integral x prime dx is the emittance of the beam. So from here now this says that this expression says Lieuble's theorem says that this is a constant which can be absorbed in the constant here. So beta gamma into emittance, so beta gamma into the emittance is equal to a constant. Now if there is acceleration beta and gamma are the relativistic parameters, so if there is acceleration beta and gamma will change. So beta will increase, gamma will increase. So in other words this parameter will increase. So this parameter multiplied by the emittance is a constant. So in the presence of acceleration emittance is not a constant. What is constant is emittance multiplied by beta gamma. So emittance multiplied by beta gamma is known as normalized emittance and this normalized emittance remains constant during acceleration. If there is no acceleration then the emittance is also a constant but during acceleration the emittance is no longer constant. The normalized emittance which is beta gamma multiplied by the emittance that is a constant. So beta gamma epsilon is the normalized emittance and it is conserved even during longitudinal acceleration. The emittance epsilon is conserved if there is no longitudinal acceleration.