 Hello and welcome to lecture 12 of module 2 of this course on Accelerator Physics. Today we will continue with our studies on longitudinal dynamics of particles. In the previous lecture we learnt that for acceleration using time varying fields, so for phase stability the synchronous phase of the time varying field must be chosen to lie between minus phi by 2 and 0. So even though there is acceleration for the entire positive part which lies between minus phi by 2 and phi by 2, for phase stability the synchronous phase must be chosen to lie between minus phi by 2 and 0. For a particle in the bunch that is not the synchronous particle the stable region for motion extends from some value phi 2 to minus phi s. This phi 2 can be calculated from the value of phi s. The separatrix which is the limiting stable trajectory passes through the unstable fixed point at delta w is equal to 0 and phi is equal to minus phi s and it also passes through the point phi is equal to phi 2. So the width of the separatrix is psi and this is modulus of phi s plus modulus of phi 2 which comes out to be minus phi s minus phi 2. So does the phase width depends upon the choice of the synchronous phase. The phase width is minimum for phi s is equal to 0. So if we choose the synchronous phase to lie at the rest of the wave, so this is phi s is equal to 0 if we choose here then the phase width is minimum and it increases as it goes to as you choose the synchronous phase from 0 to minus phi by 2 it increases. At minus phi by 2 or minus 90 degree the phase acceptance is maximum extending over the full 360 degree phase. So in the previous lecture we derived the equation of motion of the particle which is not the synchronous particle in the longitudinal direction. So we got two equations, two differential equations one in terms of phi and the other in terms of w which is the energy. So here phi s is the phase of the synchronous particle and w s is the energy of the synchronous particle. So this is the electric field with time or phase. So phi is equal to omega t. So this is the time or phase axis and this is the synchronous phase chosen to lie between minus phi by 2 and 0. Now we define small w as equal to w minus w s by m o c square capital A and B two constants. So A is 2 phi by beta s cube gamma s cube by lambda and B is q e 0 t m o c square. So putting these in these equations we can simplify this into this form. So we get A w square by 2 plus B sin phi minus phi cos phi s is equal to h phi where h phi is a constant of motion and A and B are given by these values. Now if we look at this expression we see that the first term is of the form of kinetic energy and the second term is of the form of potential energy and h phi is the constant of motion. So this is this curve will if you plot this in phase space in the longitudinal phase space it will give you the trajectory of the particle depending upon the value of h phi. So this potential energy function can be evaluated. So we see that for phi is equal to 0 this is equal to 0 and there is a maxima for phi is equal to minus phi s and the minima for phi is equal to phi s. So if we plot it here we see that the potential at phi is equal to 0 the potential is 0 there is a minima for phi is equal to phi s and a maxima for phi equal to minus phi s. So if we plot this curve it will come out to be of this form and we see that there is stable motion between minus phi s and some value phi 2 because a potential well is formed here. So for these values of phase the particle will execute stable motion. So all particles other than the synchronous particle that lie within this region between phi 2 and minus phi s. So that means this is phi 2 and this is minus phi s. So all particles these particles will oscillate about they will execute stable oscillations about the synchronous particle. So the separatrix defines the area within which the trajectories are stable and it can be plotted if the constants a and b are null. So we can plot this equation in phase space for phi and delta w this is a longitudinal phase space. So we can plot this equation and the stable motion extends from phi equal to minus phi s and phi equal to phi 2. So this is this curve is called the separatrix. The separatrix is also called the fish and the stable area within is called the bucket. So since it looks like a fish it is called a fish and the all trajectories lying inside the separatrix they are stable they will execute these particles will execute stable motion and all particles lying outside will execute unstable motion. So the area inside is called the bucket. There are two separatrix solutions for w is equal to 0. So for w is equal to 0. So this is the w axis for w is equal to 0. The separatrix cuts the phi axis at two points at phi is equal to minus phi s and phi equal to phi 2. So there are two separatrix solutions for w is equal to 0 which determine the maximum phase width of the separatrix. So this is the maximum phase width of the separatrix. So one solution is phi is equal to minus phi s which is a positive number because phi is itself is a negative number. So minus phi s will be a positive number. So this is this point gives the maximum phase for stable motion. So this gives the maximum point for stable motion. Any particle lying beyond this point will execute unstable motion. The other solution is phi is equal to phi 2. So this point and w is equal to 0. This point gives the minimum phase for stable motion. So any particle with a phase lying below this phase will again execute unstable motion. The equation of separatrix for this case is, so in this expression you can put in the point. So you can put in the first point here w is equal to 0 and phi is equal to minus phi s to calculate the value of h phi and then you can substitute the second value phi is equal to phi 2 and w is equal to 0 to get this expression. So from here you can calculate the value of phi 2 and this can be numerically solved and we can see that phi 2 is a function of phi s. The phase width of the separatrix is then given by phi s modulus of phi s plus modulus of phi 2. So it is this plus this which is equal to minus phi s because phi s is a negative number minus phi 2. So from this expression we can write an expression for phi s in terms of psi. So we get tan phi s is equal to sin psi minus psi divided by 1 minus cos psi. Now if the separatrix width and the synchronous phase are very small as compared to 1 then this can be simplified and written as under this approximation tan phi s is equal to minus psi by 3 and it can be further simplified as if tan phi s is equal to approximately equal to phi s it can be written as psi is equal to 3 phi s. So for small values of psi and phi s the separatrix width is simply 3 times modulus of phi s. So you can calculate phi 2 this will be equal to 2 phi s and the separatrix extends from minus phi s to phi 2, phi 2 is now 2 times phi s. But remember this holds only for very small values of psi and phi s it is not always true. So this is an approximate condition. The phase width of the separatrix ranges from minus phi s to phi 2. So that it depends on the value of phi s. So if you choose the synchronous phase for example as equal to 0 so that it lies at the crest here then the phase width is 0. Now as you go as the phase width as you change the value of phi s from 0 to minus 90 this we can see that the phase width increases. So this is for phi s is equal to minus 30, this is for phi s is equal to minus 60 and this is for phi s is equal to minus 90. Here we see that it is maximum for minus 90 and it extends over the entire range of phase 360 degree of phase. But at 90 degree the field is 0, the accelerating field is 0 so no acceleration. As you go towards the crest the accelerating field increases so the acceleration is small but the phase width increases. So you have to maintain a proper compromise between the choice of synchronous phase for the choice of synchronous phase. Now let us see the longitudinal dynamics with field errors. A linac is designed to provide a specific synchronous velocity profile. So that means let us say the synchronous particle is at the center of this gap here. Now here it sees the right synchronous phase, the right value of the electric phi, it gets the right energy gain and travels from here to here in time t or dy2 depending on whether it is a zero mode structure or a pi mode structure. So this length is designed accordingly. So there is a design velocity with which it travels from the center of one gap to the center of the other gap. So this length, cell length is accordingly designed depending on this velocity. So it is beta lambda by 2 if it is a pi mode structure and it is beta lambda if it is a zero mode structure. Now let us consider an electron. For electron accelerators we know that beta is 0.99 so it becomes constant and it attains a velocity almost equal to the velocity of light. So for electrons now the cell length is constant since beta is constant the cell length all these cell lengths are constant. Now let us see if there is what happens if there is an error in the field. So let us say the accelerator is designed for this blue value of electric field. Now in actual practice due to certain reasons what you get in the accelerator is this orange curve. So this if you inject the electron into the gap to see the right synchronous phase. So it sees the right synchronous phase here but now it sees a field since there is a field error it sees a field that is lower than the design value. So it gets a lesser energy gain than it should have got had it seen the correct value of the field. So it gets a lesser energy gain but being an electron it is already relativistic. It will still move from this gap to this gap in the same time because the velocity is constant velocity is almost equal to C. So it will come to the next gap at the right time and again it will see the same synchronous phase. What is different is now that because it has seen a lower value of electric field the energy gain will be lower. So as it moves from one gap to the other it will arrive at this right time it will see the right synchronous phase but the energy gain in each gap will now be less. So the energy gain changes for a field error in the case of an electron accelerator. So for electrons the field errors cause a shift in the final energy. Now for a linar that accelerates non-relativistic ions let us say protons or even non-relativistic electrons a field error changes the particle velocity gain causes a shift to a new synchronous phase. Now let us see let us say we have a non-relativistic particle that is being accelerated in the linar. So again this blue curve is the designed value and for some reason the actual field profile that we are getting in the linar is this orange curve. So now the particle is injected into the linar at the right time. So it sees instead of seeing this value of electric field it sees this value of electric field which is lower than the design value. So now it gets a lesser energy gain it moves slower and it arrives at the next gap at a later time. So if it arrives at a later time now it sees a different value of the synchronous phase because the velocity has changed for a proton or for a non-relativistic ion with change in kinetic energy the velocity changes. So that is why when it arrives at the next gap it sees a difference different value of the synchronous phase. So we see that this causes the field error causes a shift to a new synchronous phase. Now to see this consider a design energy profile given approximately by this expression. So let us say the energy gain is given by this expression. So you have Q e0 t is the design value and cos phi s is the synchronous phase. Now if the actual accelerating field is e0 t like in this case it is the let us say it is the orange curve and this is different from the design value. A different particle phase can still satisfy the synchronous energy profile and the new synchronous phase is given by cos phi is equal to cos phi s e0 t designed by e0 t. So we can still get if the field value has changed we can still get the same energy gain if you change the synchronous phase to a new value which is given by cos phi s e0 t designed by e0 t. So in this case the final energy remains unchanged. Now at phi is equal to 0 we know that the phase acceptance finishes. So the width of the separatrix is 0 at phi is equal to 0. Now if you increase the phase beyond that if the synchronous phase increase beyond that beyond that there is no phase stability. So the corresponding accelerating field is the threshold field for forming a stable bucket for synchronous acceleration. So that is from here you can calculate the value of the threshold field the field that should not be exceeded in order to have stable motion in the Linux. So e0 t threshold is now given by e0 t designed into cos phi s. So this is obtained by putting phi is equal to 0. Now we have calculated the maximum phase width of the separatrix we can also calculate the energy width of the separatrix. So in this case let us calculate the energy half width of the separatrix. So this is the energy half width of the separatrix and this passes through phi is equal to this is at phi is equal to phi s. So this occurs for phi equal to phi s. So solving the equation for the trajectory we have we can put phi is equal to phi s there. So we get a w square by 2 plus b sin phi minus phi cos phi s is equal to minus b sin phi s minus phi s cos phi s. For w is equal to w max which occurs at phi is equal to phi s we can write this expression. So here we can substitute phi is equal to phi s. So we get this expression. Now simplifying it further we get a w square max by 2 is equal to minus 2b this we can take it on the right hand side minus 2b sin phi s minus phi s cos phi s. So from here we can calculate the values of w max square it comes out to be this. We can substitute the values of a and b which are constants to get the value of w max. So this w max can be calculated. So this w max by definition is delta w by mc square. So this comes out to be under root of 2q e0t beta cube beta s cube gamma s cube lambda divided by pi mc square multiplied by phi s cos phi s minus sin phi s. So this is the energy half width of the separatrix. In this way you can calculate the energy half width of the separatrix. Now let us consider small amplitude oscillations. We know that the particles other than the synchronous particle will oscillate about the synchronous particle if the synchronous phase is chosen between minus pi by 2 and 0. So considering the motion of particles with coordinates close to the synchronous particle. So we consider the particles close to the synchronous particle. For a phase difference that is small relative to the synchronous particle the cos phi can be expanded around phi minus phi s. So we have cos phi we can write it as cos phi s minus phi minus phi s sin phi s minus and so on. Now we can substitute this in this equation d2 phi by ds square is equal to minus ab cos phi minus cos phi s. So we can calculate the value of cos phi minus cos phi s from here. So we substitute this here. So we get the equation of motion for small longitudinal oscillations. So we get d2 phi by ds square is equal to minus ab and we have now substituted this whole value here. Now we can take sin phi s sin of minus phi s outside this bracket. So d2 phi by ds square plus taking it on the right hand side plus ab sin of minus phi s is equal to we are left with phi minus phi s and here we get phi minus phi s square by 2 tan minus phi s is equal to 0. So we have ignored the higher order terms here. So this term here ab sin minus phi s we write it as kl0 square. So kl0 square is equal to ab sin of minus phi s. So kl0 now putting in the values of a and b so we know a and b are constants we put in the values of a and b so we get the values of kl0. So kl0 is the longitudinal phase advance per unit length and corresponding to this we can find out the angular frequency of the small longitudinal oscillation. So this is the frequency with which the particles that are not the synchronous particles oscillate about the synchronous particles. So this frequency omega l0 is equal to kl0 into beta s into c. So if we divide it by omega and take the square we get this value. So just substituting this and rearranging we get this. So here omega is the RF frequency. So omega is omega is equal to 2 pi c by lambda is the RF angular frequency at which the linac is operating. From this expression we see that the longitudinal oscillation frequency is usually small compared to the RF frequency. So this frequency is quite small as compared to the frequency of the applied RF. As the beam becomes relativistic the longitudinal oscillation frequency approaches 0. So as the beam becomes relativistic the value of beta gamma which is in the denominator increases. So this the frequency the longitudinal oscillation frequency approaches 0. So you can see that for electrons that are relativistic we do not have the oscillation of the particles about the synchronous particles. Because all the particles are now travelling with the same velocity which is close to the velocity of light. Now in this expression for longitudinal motion we see that there is a quadratic term. So this is a linear term in phi minus phi s and this is a quadratic term. So this is the lowest order non-linear terms. Other terms we have neglected because they are very small. Now since it is a quadratic term it will produce an asymmetric potential well. And the negative sign means that the non-linear part of the restoring force weakens the focusing. Now this part of the force which is linear so it is with a positive sign. So it is a restoring force. It focuses the beam in the longitudinal direction or in other words under the influence of this term the particles execute stable motion around the synchronous particle. Whereas this term comes with a negative sign. So this is a defocusing term. So it will it will make the beam unstable. But its magnitude is small as compared to the linear term. Now let us calculate the trajectory in longitudinal phase space. Now in the approximation that phi minus phi s is much much less than 1. Sin phi can be expanded around phi s. So we have sin phi as sin phi s plus phi minus phi s cos phi s minus phi minus phi s square by 2 sin phi s and so on. So from here we can take sin phi minus phi cos phi s. So this is equal to sin phi s minus phi s cos phi s and so on. Now we can substitute this in the equation for separate tricks. So this is the expression for separate tricks. So we get so sin phi minus phi cos phi s instead of that we can write this value. So we get aw square plus b instead of this now we have substituted this value. This is equal to h phi. So again some rearrangement we have aw square plus b sin of minus phi s minus phi s multiplied by phi minus phi s square by 2 is equal to h phi. So minus sin phi s plus b phi s cos phi s. So these are all constants. So we have taken this on the right hand side. So the right hand side is a constant here and the left hand side is in terms of small w and phi minus phi s square. So phi s is negative for stable motion. So synchronous phase has to lie between minus phi by 2 and 0. So phi s has a negative value. So this is the equation of an ellipse in longitudinal phase phase. So you can see that this is the equation of an ellipse in small w and phi. So this is the equation of an ellipse which is centered and w is equal to 0 and phi is equal to phi s. So in the longitudinal phase phase phi and delta w you can write this you can draw this equation and this will come out to be an ellipse. So this is centered at w is equal to 0 and phi is equal to phi s. Now we can define a delta phi 0 which is equal to phi 0 minus phi s where phi 0 is the maximum phase here. So we see that the point of maximum phase corresponds to a point on the ellipse with w is equal to 0. So this point is the maximum extent in the phase and it is at this point w is equal to 0. So now we substitute in this expression which we have derived here. So we get so we put here w is equal to 0 and phi is equal to phi 0. So we get b sin minus phi s phi 0 minus phi s whole square by 2 is equal to this constant term. Now putting phi 0 minus phi s is equal to delta phi 0 as we have defined in the previous slide. So from this we can calculate the value of h phi. So h phi comes out to be this value which is now a constant value. Now substituting the value of h phi in this expression we get this aw square plus b sin minus phi s phi minus phi s square by 2 is equal to. So now we are putting the value of h phi here. So this is the value of h phi and these two terms remain. So we see that these terms now cancel out and we are left with only these three terms here. So again we can rearrange this equation and we get 4 aw square divided by b sin of minus phi s del phi 0 square phi minus phi s whole square divided by del phi 0 square is equal to 1. Now this can be written in the form of the equation of an ellipse such that w square by w 0 square. So we write here the other terms we call w 0 square here plus phi minus phi s whole square divided by del phi 0 square is equal to 1. So where now w 0 can be calculated from this expression here. So it is in terms of a and b which is constant. Again we can put in the values of a and b here and we get w 0 square. So this is the value of w 0 square and from here we can take under root of this and we get the value of w 0. So this is the normalized energy on the ellipse that corresponds to phi is equal to phi s.