 So, let us analyze the potential energy term. So, potential energy term let us call it V phi is given as V sin phi minus phi cos phi s. So, V is a constant and phi s is the synchronous phase and phi is the phase of the arbitrary particle. So, at phi is equal to 0, we can calculate V phi is equal to 0. So, phi is equal to 0, this goes to 0 and this goes to 0, V phi is equal to 0. So, here this is the E z field. So, this is the E z field, this is the synchronous phase, the synchronous phase has to lie between 0 and minus phi by 2. So, this is the synchronous phase chosen between minus phi by 2 and 0. So, if we take phi is equal to 0, V phi comes out to be equal to 0. Again, let us differentiate V phi with respect to phi. So, we get V cos phi minus cos phi s. So, putting dV by d phi is equal to 0, we get phi is equal to plus minus phi s. So, plus minus phi is equal to plus minus phi s, these will be points of maxima or minima. So, let us calculate which is a maxima and which is a minima. So, we get, differentiating V phi again, we have d2 V phi by d phi square is equal to minus b sin phi. Now, let us apply the first, let us take the first value phi is equal to phi s. So, we get d2 V phi by d phi square, this is equal to minus b sin phi s. Now, phi s is a negative value because it lies between minus phi by 2 and 0. So, this quantity will be greater than 0. So, phi is equal to phi s is a point of minima for the potential. So, phi is equal to phi s at phi is equal to phi s, we should have a minima for the potential. So, we saw that the potential is 0 at phi is equal to 0, at phi is equal to phi s, we should have a minima. Now, next we calculate at phi is equal to minus phi s. So, here d2 V phi by d phi square is equal to minus b sin minus phi s and phi s is a negative number. So, this turns out to be less than 0. So, this is a maxima at phi is equal to minus phi s. Now, phi s lies here minus phi s will lie at this point. So, there is a maxima at phi equal to minus phi s. So, we see the potential will have a form like this, there will be a maxima at phi equal to minus phi s. And then, since there is a minima here, the potential keeps on increasing like this in the other direction. So, we see that there is a potential well for phi line between minus phi s and some value which we can write as phi 2 here. So, there is a potential well for phi equal to minus phi s and phi equal to phi 2 and we expect that the motion of the particles will be stable if they lie within this region. So, for us given phi s, the stable region for phase motion extends from phi 2 to minus phi s. So, coming back into this expression again. So, here the first term represents the kinetic energy, the second term is like the potential energy and this is the constant of motion. So, this expression represents the trajectories of particles in the longitudinal phase space. So, the longitudinal phase space is represented by phi and delta w. So, delta w is the change in energy of the particle with respect to the synchronous particle. So, this depending on the values of the constant of motion, different particles will have different trajectories in the longitudinal phase space. So, you can see these are the trajectories of different particles. For phi equal to phi s that is the synchronous particle, we see that there is no motion. So, because this is the ideal particle. So, it arrives at the gap at the right time and gets the right energy gain. So, there is no change in energy with respect to the synchronous particle. So, delta w is 0 and phi is always equal to phi s. The other particles will execute oscillations about the synchronous particle. So, we have seen the form of the potential. So, this is the electric field, this is the synchronous phase and we see that at synchronous phase, the potential has a minima at phi equal to 0, the potential is 0. It has a maxima for phi equal to minus phi s and the extent of the stable motion extends from minus phi s to phi 2. So, this is the extent of the stable motion. So, all particles that lie within this phase they will have stable motion. So, the stable region for phase motion extends from phi 2 lying between sorry phi lying between phi 2 and minus phi s. So, the separatrix which is a limiting stable trajectory passes through the unstable fixed point delta w is equal to 0, phi is equal to minus phi s. So, the separatrix is the limiting curve which is a limiting stable trajectory. So, all trajectories within the separatrix are stable and all trajectories that lie outside the separatrix are unstable. So, in other words it means that all the particles that have phases within the separatrix. So, lying within the limit phi equal to minus phi s and phi 2 they are stable they will execute stable oscillations about the synchronous particles and lying outside the separatrix these are unstable. So, the separatrix passes through the unstable fixed point delta w is equal to 0, phi is equal to minus phi s. So, this point is the unstable fixed point because the potential has a maxima here. It also passes through the point phi 2. So, this phi 2 can be obtained numerically by solving the equation h phi 2 is equal to h of minus phi s. So, since the this equation is satisfied for the separatrix. So, for the separatrix we can find out the value of h phi by putting in the value of minus phi s here. So, at minus phi s delta w is equal to 0. So, we can find out the value of h phi. The separatrix also passes through the point phi is equal to phi 2 and here also delta w is equal to 0. So, having found out the value of h phi we can put in the values of delta w equal to 0 and phi is equal to phi 2 and we can find out the value of phi 2 in terms of phi s. So, this way h phi 2 can be solved numerically to find out the value. The phase width of the separatrix is defined as psi which is modulus of phi s plus modulus of phi 2. So, the phase width is this. So, it is defined as the modulus of phi s. So, this is the modulus of phi s plus the modulus of phi 2. So, this gives you the total phase width of the separatrix. The particles that lie within this will execute stable motion. So, for a given phase synchronous phase minus phi s the particles that have phases lying between phi s minus phi s and phi 2 they will execute stable oscillations around 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 known. It is also called the fish and the stable area within is called the bucket. So, this stable area inside the separatrix this is known as the bucket. Now, there are two separatrix solutions for w is equal to 0 which determine the maximum phase width of the separatrix. So, the two solutions are phi is equal to minus phi s and phi is equal to phi 2. So, one solution which is for phi is equal to minus phi s which is a positive number for stable motion because phi s is a negative number. So, phi minus phi s will be a positive number. So, this point gives the maximum phase for stable motion. So, this gives maximum phase for stable motion. The other solution is phi is equal to phi 2 and w is equal to 0. So, this gives the minimum point for stable motion. So, the phase for stable motion is if you see in the graph for electric field lying from between minus phi s and phi 2. So, the equation of separatrix for this case is given by sin phi 2 minus phi 2 cos phi s is equal to minus sin phi s minus phi s cos phi s. Okay. So, from here this can be solved to find out the value of phi 2 as a function of phi s. If you know phi s you know the synchronous phase you can find out the value of phi 2 and then from there you can find out the phase width of the separatrix. So, phase width is simply phi s modulus of phi s plus phi 2. So, the phase width is phi s plus this phi 2. So, the phase width of the separatrix is then given as so, phi s plus phi 2 remove the modulus you will get minus phi s minus phi 2. So, from here phi 2 is equal to minus of phi s plus sin. So, you can take sin on both sides. So, sin of phi 2 is equal to minus sin phi s plus sin then expanding the right side right hand side. So, we have sin a plus b is equal to sin a cos b plus cos a sin b. So, we can write sin phi 2 as equal to minus sin phi s cos psi minus cos phi s sin psi. So, putting this in the expression here for phi 2. So, we can get the value of tan phi s is equal to sin psi minus psi 1 minus cos psi. So, we have an expression for phi s in terms of the phase width of the separatrix. So, this is the phase width of the separatrix and this is the synchronous phase phi s. Now, when the psi is much much smaller than 1 and phi s is also much much smaller than 1 we can expand sin psi and cos psi in this way. So, sin psi is psi minus psi cube by 6 plus so on and cos psi is 1 minus psi square by 2 and so on. And we can substitute in this expression for tan phi s by keeping only the initial 2 terms. So, we can write tan phi s is equal to sin psi minus psi. So, sin psi minus psi will be equal to minus psi cube by 6 and 1 minus cos phi this will be equal to psi square by 2. So, this we are left with psi minus psi by 3. So, we get that tan phi s is equal to minus psi by 3. So, we see that this is a good approximation even up to taking modulus of phi s approximately equal to 1. So, then with tan phi s as phi s. So, if phi s is also very small. So, we have psi is equal to 3 times phi s. So, it is if the values of the phi s and psi are very small. So, we are left with a very simple expression that the separatrix phase with this 3 times the modulus of the synchronous phase. Okay. So, it extends up to so if this is phi s. So, here and here. So, if phi 2 is 2 times phi s and there will be phi s here. So, this extends up to 3 times of phi s. So, but this formula holds only when the values of psi s and phi are small much, much smaller than 1. Now, if psi s is equal to minus 19. So, that means if you take the synchronous phase to lie in this is pi by 2 this is 0. If you choose the synchronous phase to lie here and this region the electric field will be 0. So, there is no acceleration. However, we can calculate the phase width. So, the phase width acceptance is maximum here extending up to full 360 degree. So, it will lie from since phi s is equal to minus 90. So, minus phi s will be equal to 90 and lie up to minus 270 here. So, full 360 degrees of phase acceptance phase acceptance is maximum. So, in principle DC beam can be full DC beam can be accepted here. So, we can calculate the phase width for different values of the synchronous phase. So, phase width of the separatrix ranges from minus phi s to phi 2 and so thus we can say that the phase width choice of it depends upon phi s. So, for 0 if we have the phase phi s is equal to 0. So, that means lying here the phase width is minimum it is equal to 0 as phi s decreases. So, as let us say it is minus 30 here this red line is minus 30. So, this is the separatrix corresponding to minus 30 as it goes to minus 60 this is the phase width for the separatrix for minus 60. So, the phase width keeps on increasing the size of the separatrix goes on increasing and at phi s is equal to minus 90 it is maximum ranging to full 360 degrees acceptance. So, we see that the phase width depends upon the choice of the synchronous phase. So, finally summarizing for acceleration with 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 phase motion it extends from phi 2 to minus phi s and this phi 2 is a function of phi s. So, if you know phi s you can calculate this value of phi s. The phase width of the separatrix is given as modulus of phi s plus modulus of phi 2 which is minus phi s minus phi 2. So, the phase width depends upon the choice of phi s. The separatrix which is a limiting stable trajectory it passes through the unstable fixed point delta w is equal to 0 and phi is equal to minus phi s. It also passes through the point phi is equal to minus phi 2. The phase width is minimum for phi s is equal to 0. So, if you choose the synchronous phase to be 0 so, where the acceleration is maximum. So, here the phase width is minimum and it increases as phi s changes from 0 to minus 90 degree. Now normally electrons where since we know that for electrons the velocity becomes constant almost equal to the velocity of light because the electrons become relativistic. The synchronous phase is chosen to be 0 because the particles do not execute oscillations about the synchronous particle. At phi s is equal to minus 90 the phase acceptance is maximum extending over the full 360 degree. So, in the next lecture we will study more about longitudinal.