 Now, let us see the RFQ, let us calculate the fields in the RFQ in the region where the beam is there. So, the beam is in this very small region of the RFQ. So, this is the beam aperture, the beam passes through this region. So, the wave equation in free space as we have already seen before is del square E plus k square E is equal to 0. So, we can write this as del 2 E by del x square plus del 2 E by del y square plus del 2 E by del z square plus 2 pi by lambda square E is equal to 0. So, in the RFQ, the displacement of the electrodes from the axis is very small as compared to the RF wavelength. So, we can see that the region in which the beam is there. So, if you see the displacement of the veins from the axis, this is very small as compared to the RF wavelength. So, A, this aperture is very very small as compared to lambda. So, in this equation since lambda is very large as compared to the dimensions in x, y and z, this term can be ignored. So, this is known as the quasi-static approximation. So, the quasi-static approximation is valid in the RFQ and the wave equation reduces to the Laplace's equation. So, we just need to solve for the RFQ fields, del square E is equal to 0, just the Laplace's equation needs to be solved. So, del square E is equal to 0. Now, we know that del cross del cross E is equal to del of divergence E minus del square E. Now, since there enough charge free region, divergence of E is equal to 0 and del square E is already equal to 0. So, this term goes to 0. So, we are left with curl of E is equal to 0. So, this implies that E can be written as the gradient of a scalar. So, we can write E is equal to minus del U where U is a scalar potential. So, therefore, in the quasi-static approximation, the time dependent electric field components are derivable from a scalar potential. We can write divergence of E is equal to minus del square U and this is equal to 0. So, using cylindrical polar coordinates, the time dependent scalar potential is written as, so we have this potential which is space dependent and this is the time dependence. The fields are time dependent. So, we have sin omega t plus phi. Now, V r theta z is a solution of the Laplace's equation given in cylindrical coordinates by this expression. The RFQ potential of this structure is a solution of the Laplace's equation in the 3D domain. Now, for the RFQ, the following boundary conditions are valid. So, the potential at some location r theta z is the same as the potential at r minus theta z. So, whatever is the potential here, the potential at this location is also minus theta is also the same. Similarly, V r theta z is equal to V r theta plus phi z. Again, the potential at this location and this location are the same. So, with this boundary conditions, we can write a general solution of this equation. So, this is the general KT potential function from which the electric field in the vicinity of the beam can be calculated. So, this is a general equation. Now, for simplicity, let us just consider the first two terms of this potential. So, we select only the lowest ordered terms from the general solution for the potential. So, we take s is equal to 0 here and s is equal to 0 and n is equal to 1 here. So, just the lowest terms. So, we get what we get is V r theta z is a0 r square cos 2 theta plus a10 i0 kr cos kz. So, this is known as the two term potential. Here a0 and a10 are constants that can be determined by the electro geometry and i0 is the modified Bessel function. So, this is a simplified form of the potential for the RFQ. So, this is known as the two term potential function. Now, let us say at this location z is equal to 0. So, here we have at theta is equal to 0. So, let us take this location as theta is equal to 0. So, the horizontal wing is at a distance a from the axis. So, at theta is equal to 0 that is at the horizontal wing tip, we have V is equal to V0 by 2 this at this instant of time this is at a potential V0 by 2 and r is equal to a. So, we substitute in this expression V is equal to V0 by 2 and r is equal to a and theta is equal to 0. So, we get V0 by 2 is equal to minus a0 a square plus a10 i0 kr and z is equal to 0 here. Now, at again z is equal to 0, theta is equal to pi by 2. So, at this location, so we get for the vertical wing tip V is equal to now minus V0 by 2 and here r is equal to mA. So, when the horizontal wing is close to the axis the vertical wing is away from the axis. So, we get minus V0 by 2 is equal to minus a0 mA square plus a10 i0 krmA. So, from here we can solve these two equations and find out the values of the two constants a0 and a10. So, simplifying we get the value of a0 which is V0 by 2 a square i0 kr plus i0 krmA divided by m square i0 kr plus i0 krmA. Similarly, a10 is V0 by 2 m square minus 1 m square i0 kr plus i0 krmA. Now, this a10 is the acceleration parameter. So, if the it depends upon the value of m, if modulation is equal to 1, so that means both the horizontal wing and the vertical wing are equidistant from the axis. So, in that case this term goes to 0. So, there is no acceleration as m is increased greater than 1. So, the acceleration parameter increases and there is acceleration. Now, it is convenient to define dimensionless constants x and a as. So, this term we define as x and this term we define as a. So, this is a is the acceleration efficiency and x is the focusing parameter. So, now we can write a0 as x V0 by 2 a square and a10 can be written as a V0 by 2. So, then the complete time dependent potential is we can substitute the values of a0 and a10 in the 2 term potential and then for the time dependent potential we can include sin omega t plus phi and we get this as the complete time dependent potential. So, here the time dependent voltages on the horizontal and vertical electrodes are V0 sin omega t plus phi by 2 and minus V0 sin omega t phi by 2 respectively. So, this is the complete time dependent potential where x is given by this. This is the focusing parameter and a is given by this. This is the acceleration parameter. So, in Cartesian coordinates now x can be written as r cos theta and y is r sin theta. So, if you substitute x and y here in place of r we get this expression in terms of Cartesian coordinates. So, now electric field can be found from the expression E is equal to minus del U. So, differentiating this we can calculate E x, E y and E z. Now E z is responsible for acceleration and in E x and E y we see there are two terms. So, the first term is responsible for focusing or de-focusing. So, we know that since it is a quadrupole, so it will focus in one direction, de-focus in the other direction. So, you see that the first term in E x and E y they have opposite sign. So, if it is focusing in the x direction it is de-focusing in the other direction. And then the second term is due to the modulation. So, here the z term it is in the z direction and this it includes the z variable and this causes de-focusing in the transverse direction. So, as we have already seen earlier that the RF field causes some de-focusing in the transverse direction. So, E z component provides the accelerating force on the beam. The first term of E x and E y is associated with quadrupole focusing and the second term leads to transverse RF de-focusing force that acts on the beam when the phase phi is chosen for longitudinal focusing. So, when phase phi is chosen for longitudinal focusing it should lie between 0 and minus pi by 2. Now, vane geometry. So, how is the vane modulated in the z direction? So, geometry of the electrodes is specified from the equipotential surfaces that are at a potential. So, transverse cross sections are approximately hyperbola. So, you can see from here the transverse cross sections are approximately hyperbolas. You can always approximate it in order to reduce the peak surface fields and for ease of fabrication. Now, at z is equal to beta s lambda by 4 that is half way through the unit cell. The RFQ has exact quadrupole symmetry and the dips of the x and y electrodes have radius equal to r0 is equal to a under root a divided by under root of x. So, in practice the electrode contours must deviate from the ideal shape to control the peak surface field and to facilitate machining. However, at the dips that is at theta is equal to 0 and pi for horizontal electrodes and at pi by 2 and 3 pi by 2 for the vertical electrodes. So, these can still be chosen to correspond to the correct equipotential of the two term potential function. So, at the dip the x tip is at y is equal to 0. So, this is given by you can put this as equal to 1 is equal to x by a square x square plus a i0 kx cos kz. So, this will give you the profile in x and z direction. Similarly, for the y dip x is equal to 0 and we get minus 1 is equal to minus x by a square y square plus a i0 k y cos kz. So, this will give you the profile of the vein in the y z direction. So, this RFQ these veins are in a cavity that it is we utilize the electric fields of the electromagnetic waves in a cavity. So, the veins are placed inside a cavity and the operating mode in the RFQ is a TE210 like mode. So, we will try to understand the mode in the RFQ. So, when I say TE210 like mode, remember the TEMN0 mode cannot exist in a cavity. So, we will see how this type of mode is generated and why it is called a TE210 like mode. It is not a TE210 mode, it is a TE210 like mode. So, let us first analyze the transverse direction. So, let us consider a TE21P. So, this is the variation in theta direction. So, two full period variations of the fields in the theta directions and this is one corresponds to the variation in the radial direction. So, this is the TE21P mode. So, if we plot ER with theta from 0 to 2 pi. So, let us start from this point this is 0.1. So, we see here that the radial field is 0. So, here there is no field in the radial direction. The field is all in the theta direction. Now, as we move from here to here at 0.2, the field radial field is maximum somewhere here and then again goes to 0. So, the radial field is maximum somewhere here and at 0.2 it goes to 0. Again, as we move ahead the radial field is maximum at this location in the opposite direction and it is going to 0 at 0.3. So, again at 0.3 it is 0 in between it was maximum, but in the opposite direction. Again, as we move from 0.3 to 0.4 we can see that the electric field is the radial component of electric field is 0 at this location, but in between it was a maximum. So, again like this and then as we go from 0.4 to 0.1 again the field is 0 again at 0.1 and then maximum somewhere in between. So, this is if we see the variation in E r in theta from 0 to 2 pi. So, we see that there are two full period variations in the field, in the field. So, this corresponds to m is equal to 2. Now, if we see the variation of B z field, so we will have a B z field here, a B z field here, a B z field here and B z field here. So, if we see, so it is coming out of the screen. So, the B z field if you see the variation with r from 0 to RC we see that the variation is like this. So, there is 1 0 in the field at r is equal to 0, but this is not counted. So, since there are no more 0s, n takes the minimum value that is n is equal to 1. So, in a TE21 type of mode in a cavity the fields are as if in a electric quadruple. So, you can see that the fields are as if in a electric quadruple, it is as if this is at a positive, this point is at a positive potential, this point is at a negative potential. Again this point is at a positive potential, this is at a negative potential. So, the field lines are originating from the positive potential and moving towards the negative potential. So, this is like the fields are as if in a electric quadruple. Now, if we introduce vanes in this region, so this positive potential here shifts to because the electrodes will be equipotential. So, this positive potential shifts to the surface here. So, we have introduced now vanes here. So, we have this at a positive potential, this is at a negative potential, this is at a positive potential and this is at a negative potential. Now, by introducing the 4 poles or vanes in the cavity, the quadrupole electric field becomes more concentrated near the axis. So, in this region the field becomes more concentrated. So, you can see here in this region near the axis near the where the beam will go through. So, the field has become this quadrupolar field has become more concentrated in this region. Also now, because you have put in a vein, they provide a surface that can be modulated to produce the longitudinal electric field. Now, this is a TE mode. In a TE mode, there is no field along the z direction and for acceleration you need a field along the z direction. So, the field cannot be produced unless you have a surface that you can modulate to produce the field in the z direction. So, by introducing these veins, they provide a surface that can be now modulated in the z direction to produce longitudinal electric fields. So, we have seen that but TE 210 mode cannot exist in the cavity because if P is equal to 0 then BZ, this implies that BZ is constant along z from 0 to L. So, let us say we have a cavity here of length L. So, a TE mode means we have a BZ field like this. So, and BZ is P is equal to 0. So, BZ is constant. So, since it is a cavity, it has to be closed at both ends. Now, when you close it with the end plate at this region and at this end, that is at z is equal to 0 and at z is equal to L, this magnetic field becomes the normal component of magnetic field to this surface, to both these surfaces. So, since it is a normal component, it has to go to 0. So, since BZ is 0 at z is equal to 0 and z is equal to L, if BZ is to be constant. Now, it is 0 here and here and P is equal to 0 implies that BZ is constant. So, if BZ is to be constant along the length, so BZ has to be 0. So, for a TE mode already EZ is equal to 0 and if P is equal to 0, then BZ also goes to 0. So, and we know that all the field components, all the transverse field components that is ERE theta, BR and B theta, they depend on EZ and BZ only. So, they will also be all 0. So, all fields will be 0 and there will be no more. So, for P is equal to 0, there is no mode in the cavity. So, P starts for from 1 for a TE mode and a TE M and 0 mode cannot exist in the cavity. But for the RFQ, we require a TE 210 mode. We need a field. So, these are the four ways of the cavity. We need a field that is constant along z. But what we actually get is a TE 211 mode in which the fields are, there is one half period variation in the fields along the z direction. So, that the fields go to 0 at both the ends to satisfy the boundary condition. So, let us say we have these vanes here, these are the two vertical vanes and these are the two horizontal vanes. Now, if we close it with an end plate here, so this is the BZ, this is the BZ field in the two quadrants. So, if we close it with an end plate here, this field has to go to 0 here at the end plate because it is the normal component of field. And then what we will get is a TE 211 mode. So, this is shown here, these are the two vertical vanes. So, modulations are not shown here, they will be there, but they are not shown here. So, these are the two vertical vanes, this is the bottom vanes, this is the top vanes, this is the beam axis. Now, if you close it with a end wall at this region here, this is the normal component and hence the magnetic field has to go to 0. Now, we want a field which is, which should not vary like this, it should be flat. Because if this field is flat, a BZ field is flat, the fields generated, the EZ component generated here will also be flat here. So, it will in all the quadric, in all the cells it will be, it will have the same magnetic. So, what can be done is that the, so the vein is not directly terminated at the end wall. Instead, some space is left between the vein and the end wall. So, there we leave a space between the vein and the end wall. And in addition, provide some undercuts in the vein. So, undercuts are provided in the veins in the end region to allow magnetic field lines to bend and pass over to the next quadrant. So, the magnetic field lines are coming like this. Now, since they do not see the end wall immediately, they do not go to 0, instead there is some space between the vein and the end wall. So, the magnetic field lines can bend over and pass over to the next quadrant. So, this magnetic field line it comes here. Now, there is a space between the vein and the end wall. So, there is some space here. So, they can bend over and move on to the next quadrant like this. So, here when they bend over, this magnetic field is the, so it is not the normal component of magnetic field. It is, so it is allowed, this component of magnetic field is allowed. So, additionally, veins do not terminate at the end walls, there is some space between the veins and the end walls. So, if you see the cross section at the end, so these are the four veins and there is some space between the veins and the end wall. So, magnetic field lines are coming like this and then they turn over the veins and move into the next quadrant. Similarly, they turn over this vein and move on to the next quadrant. So, again the same thing is shown here from the side view and this is the top view. So, this is the vein, this is the end wall. There is a gap between the vein and the end wall. So, the magnetic field lines have space and they turn over into the next quadrant. So, by providing undercut in the vein at the end region of the RFQ, the TE211 mode is modified into a TE210 like this. So, in a TE211 mode, the field was like this. Now here the, because of providing this undercut, the field, the Z component of the field still goes to 0 here, but in the center region it is more or less flat like a TE210 like mode. So, that is why it is known as a TE210 like mode. So, in general in the RFQ, the TE211 mode is modified into a TE210 like mode. So, this is the picture of the RFQ. So, you can see here these are the four veins and so there is an undercut provided here which you can see here. So, the magnetic field line in this quadrant will be like this. So, it comes here and there will be an end plate here, but there is a gap between the vein and the end plate. So, the magnetic field lines will bend over and pass over to the next quadrant, bend over and pass over to the next quadrant. So, this shows the top view here. So, this is the magnetic field lines they bend over and pass on to the next quadrant. Similarly here they bend over and pass over to the next quadrant. So, veins do not extend to the end plates and in addition they are cut to facilitate turning of the magnetic field lines. So, thus the operating mode in the RFQ is a TE210 like mode. So, thus we utilize the fields of the electromagnetic wave in a cavity which is excited in a TE210 like mode which veins added to accelerate the charge particles in a RFQ. So, we can summarize now the DC beam can be injected into an RFQ. The RFQ first punches the beam with more than 90 percent efficiency and then accelerates it. The beam is focused using the electric quadrupole in the RFQ. The operating mode in the RFQ is a TE210 like mode. So, since it is a TE mode there is no EZ component of the field which is required for acceleration. So, the longitudinal component or the EZ component of electric field is generated in the RFQ by modulating the surface of the veins along the longitudinal direction. The RFQ is a RF liner that can efficiently bunch, focus and accelerate low-energy beam. So, it does all the three functions. It bunches the beam, it focuses the beam and it can accelerate the beam at low energies. So, it is particularly useful for high current beams because the quadrupole focusing in the RFQ is there throughout the length of the RFQ. So, it is particularly useful for focusing high current beams where space charge effects are high. So, let me just speak about the LEIPA accelerator at BRC which is the front end Linux for the Indian ADS system. So, as we saw in the first lecture the accelerator driven system needs a high intensity high current accelerator. So, typically of energy is 1 GB. So, this proton beam goes and hits a spellation target and lots of neutrons are produced. So, there and then there is a reactor, a nuclear reactor which is subcritical. So, subcritical means it is deficient in neutrons. The additional neutrons required to sustain the chain reaction in this subcritical reactor is then provided by this accelerator. So, for such an accelerator, so it is planned to build this accelerator in three phases and the first phase of it LEIPA is being built at BRC. So, this is a 20 MEV proton line. So, it consists of an ion source and then there is an RFQ. So, the ion source accelerates. So, ion source gives beam of 50 KV then the RFQ accelerates it to 3 MEV and then followed by a detail which accelerates it to 20 MEV. So, in between there are solenoid magnets here for focusing the beam before the beam enters into the RFQ and then there are quadrupole magnets to focus the beam in between the RFQ and the drift tube. So, here is the picture of the LEIPA at BRC. So, there is an ion source here which is not seen then there is an RFQ, this is the RFQ. There are solenoid magnets. So, this is a solenoid magnet which is used for focusing the beam. There is a drift tube in here. In between there are quadrupole magnets for focusing the beam and these structures that you can see here these are waveguides that are used for feeding the RF power or for transporting the electromagnetic waves high power electromagnetic waves into the cavity into the RFQ cavity. So, there is a klystron sitting in the next room which produces high power RF. So, this is transported through the waveguide into the RFQ where fields are set up in the TE210 light pattern and the beam is then accelerated. So, we have an ion source at 50 keV the RFQ which accelerates from 50 keV to 3 MeV and then there is a drift tube linac which accelerates 3 MeV to 20 MeV and with this we come to an end to the end of module 2.