1 Lectures 10 and 11: resonance cavities

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1 1 1 Lectures 10 and 11: resonance cavities We now analyze cavities that consist of a waveguide of length h, terminated by perfectly conducting plates at both ends. The coordinate system is oriented such that the waveguide extends along the z direction from z = 0 to z = h. The waveguide modes that can propagate in the waveguide are reflected at both ends and that creates a standing wave in the z direction. Only waves with certain frequencies can exist in the cavity. We can treat cylindric cavities and rectangular parallelpiped cavities analytic. We use Comsol to treat cavities with other shapes. Since the cavity is a finite waveguide we can decompose the resonances in TEmodes and TM-modes. For the TE-modes we solve Helmholtz equation for the H z component and for the TM-modes we solve Helmholtz equation for E z. In cavities used in accelerators the TM-modes are the most common ones. For this reason we only consider TM-modes here. The TE-modes are treated in the same way. In order to determine the resonances and the corresponding electromagnetic fields we need boundary conditions. On the walls of the waveguide we use the same boundary conditions as in the analysis of waveguide modes, E z = 0 and ˆn H z = 0. We also need boundary conditions for E z and H z at z = 0 and z = h. One can derive the following boundary conditions E z (x, y, 0) = E z(x, y, h) = 0 z z H z (x, y, 0) = H z (x, y, h) = 0 (1.1) 1.1 TM-modes in a finite rectangular waveguide Let us first derive the resonance frequencies and the corresponding fields for a rectangular waveguide a b that is terminated at z = 0 and z = h. We do this with the method of separation of variables. The differential equation is 2 E z + k 2 E z = 0 (1.2) This can be seen as an eigenvalue problem where we look for the eigenfunctions E z and the eigenvalues k 2. First we assume that all solutions can be written as a product of three functions as H z (x, y, z) = F(x)G(y)P(z). We plug this into Helmholtz equation and get The next step is to divide by FGP GP 2 F x 2 + FP 2 G y 2 + FG 2 P z 2 + k2 FGP = 0 (1.3) 1 2 G G y P 2 P z + 2 k2 = 1 2 F (1.4) F x 2 We now make the interesting observation that the left hand side is independent of x and the right hand side is independent of y and z. The only way that this can

2 2 happen is that both lhs and rhs are constant. We call this constant kx 2 and get the equation for F as 2 F x 2 + k2 x F = 0 F(0) = F(a) = 0 (1.5) The general solution to the equation is F = A 1 cos k x x + B 1 sin k x x. The condition F(0) = 0 says that A = 0 and the condition F(a) = 0 says that sin k x a = 0. This can only be satisfied if k x = mπ where m = 1, 2, a The remaining equation for G and P is which can be rewritten as 1 2 G G y P 2 P z + 2 k2 = kx 2 (1.6) 1 2 P P z + 2 k2 kx 2 = 1 2 G (1.7) G y 2 We see that the lhs is independent of y and the rhs is independent of z and hence both lhs and rhs are constant. We let this constant be ky 2. This gives 2 G y 2 + k2 yg = 0 G(0) = G(b) = 0 The solution is G = B 2 sin k y y with k y = nπ b equation for P is 2 P z 2 + (k2 k 2 x k2 y )P = 0 P (0) = P (h) = 0 (1.8) where n = 1, 2, The remaining (1.9) We introduce kz 2 = k2 kx 2 k2 y. The general solution to the equation is P(z) = A 3 cos k z z + B 3 sin k z z. The condition P (0) = 0 says that B 3 = 0 and the condition P (h) = 0 tells us that k z = lπ where l = 0, 1, 2, h We have now found out that the eigenfunctions are ( mπx ) ( nπy ) ( ) lπz E z = A sin sin cos a b h (1.10) and that the wavenumber k is given by (mπ ) 2 ( nπ ) ( ) 2 2 lπ k = kx 2 + k2 y + k2 z = + + (1.11) a b h We call these solutions cavity modes. The resonance frequencies are given by f mnl = c 2π k = c (mπ ) 2 ( nπ ) ( ) 2 2 lπ + + (1.12) 2π a b h where m = 1, 2, 3..., n = 1, 2, 3..., and l = 0, 1, 2, 3....

3 3 1.2 TM-modes in a circular cylindric cavity Consider a circular cylinder with radius a and length h. Also in this cavity only waves with certain frequencies can exist. For each resonance frequency there is an electromagnetic field and we call this field a cavity mode. In order to find these cavity modes we use cylindrical coordinates r c, φ, z. There are certain cavity modes that are of special interest for accelerators. These are the ones that have an electric field that along the axis of the cavity is directed in the z direction. The reason why these are of interest is that the electric field can be used for accelerating particles that travel along the axis. One can prove that these modes are TM-modes and that they are axially symmetric, i.e., there is no dependence on the azimuthal angle φ. To find these TM-modes we solve Helmholtz equation for E z (r c, z). In cylindrical coordinates the equation reads 1 E z r c + 2 E z r c r c r c z + 2 k2 E z = 0 (1.13) The boundary conditions are E z (a, z) = 0 E z (r c, 0) z = E z(r c, h) z = 0 (1.14) We use the method of separation of variables to solve this equation. We assume that all solutions can be written as a product E z (r c, z) = R(r c )P(z). This gives or, equivalently P 1 r c r c r c R r c + R 2 P z 2 + k2 RP = 0 (1.15) 1 2 P P z + 2 k2 = 1 1 R r c (1.16) R r c r c r c The lhs is independent of r c and the rhs of z and thus both of them are constant. We call this constant kr 2 and get 1 R r c + kr 2 r c r c r R(r c) = 0 c R(a) = 0 (1.17) This is Bessel s differential equation. It has the solutions (bounded at r c = 0) R(r c ) = AJ 0 (k r r c ) (1.18) where J 0 (x) is the bessel function of order zero. The boundary condition says that we need to have k r such that J 0 (k r a) = 0. The Bessel function has infinitely many zeros. You may take a look at the Bessel function in Matlab. The commands x=linspace(0,30,10000); plot(x,besselj(0,x))

4 J 0 (x) x plot J 0 (x) from x = 0 to x = 30, see figure. As you can see it looks a little bit like a damped cosine function. The first five zeros are given by J 0 (ξ n ) = 0, n = 1, 2, 3... where ξ 1 = 2.405, ξ 2 = 5.520, ξ 3 = and ξ 4 = This means that k r can take the values k r = ξ n /a. We then determine the function P(z). It has to satisfy P (z) + (k 2 k 2 r )P(z) = 0 P (0) = P (h) = 0 (1.19) We let k z = k 2 k 2 r. Then the solution is given by P(z) = B cosk zz where k z = lπ. We have now found both the resonance frequencies and the corresponding h electric field E z. They are given by f nl = c 2π k = c 2π (ξn ) 2 + a ( ) 2 lπ h E znl (r c, z) = A nl J 0 (ξ n r c /a) cos(lπ/h) (1.20) We have not considered the modes that vary with φ. If we do that we see that the modes have φ dependence cos mφ, where m = 0, 1, The modes are then denoted TM mnl. The axially symmetric modes have m = 0 and that is why the modes we have treated here are denoted TM 0nl modes The TM 0n0 modes One can show that when l = 0 there are no other components of the electric field than E z. These modes are of special interest. So for l = 0 the modes in the cavity are given by E(r c, z) = A n,0 J 0 (ξ n r c /a)ẑ (1.21)

5 5 The mode with the lowest resonance frequency is the one with n = 1. This is the one that is used in accelerators. The reason why this mode is so important is that when l 1 there is a standing wave in the z direction. We have seen that in a waveguide the phase speed is larger than the speed of light, and this is also the case for the standing wave in the circular cylinder. If we like to use such a mode for accelerating particles we run into problems. We like the particle to have an accelerating force along the entire cavity. Since the phase speed is larger than the speed of the particle this is not possible since we cannot let the particle run with the same speed as the field. The cavity will not be an efficient accelerator. If we instead pick one of the l = 0 modes then we can adjust the length of the cavity such that during the time the particle is inside the cavity it experiences an accelerating force in the positive z direction. This is also why we need to send bunches of particles through the accelerator. The time between two bunches is equal to one (or several) period of the accelerating mode. The cavities in accelerators are not exactly circular cylinders but rather deformed circular cylinders. When they are deformed the modes adjust to the deformation and one can still recognize the modes from the circular cylinder. That is why one talks about the TM 010 mode also for deformed cavities. The deformation is such that they are still axially symmetric since one like to have an axially symmetric field in the region where the particles travel. 1.3 Q-value for a cavity The walls in a resonance cavity are not perfectly conducting and this leads to power losses which also means that cavities have a bandwidth that is not zero. The Q-value (Quality factor) is a measure of the losses in the cavity. The bandwidth of an cavity and the Q-value are defined and related as B = f + f = f 0 Q = bandwidth time average of the stored energy in the circuit at resonance Q = 2π the dissipated energy during one period at resonance = Q-value (1.22) where f 0 is the resonance frequency and f + and f are the frequencies where the amplitude of the resonance has dropped to -3 db compared to the value at f 0, i.e., the amplitude is A/ 2 where A is the amplitude at f 0. We don t give a derivation of these relations. You find similar derivations in the analysis of resonance circuits in books on circuit theory. 1.4 Analyzing resonance cavities with Comsol The resonance cavities can be analyzed by Comsol. There are two different cases that are of interest:

6 6 1. If the cavity is axisymmetric and if we are only interested in axisymmetric resonances, then we use 2D axisymmetric in COMSOL. That is a very efficient and accurate solver. 2. If we are interested in non-axially symmetric modes in a axially symmetric cavity, or if the cavity is non-axially symmetric, then we have to use the threedimensional solver. We now give an example of the first case. Example Consider the axisymmetric TE- and TM-modes of a sphere with radius a = 10 cm. We use the fact that the sphere, as well as the modes, are axisymmetric. 1. First choose 2D axisymmetric>radio frequency>electromagnetic waves >Eigenfrequency. 2. Draw a circle with radius a = 0.1 m and put its center at (0, 0). 3. Draw a rectangle that covers the right half of the circle and do Geometry>Boolean Operations> Intersection. The computational domain is now a half circle. 4. Choose Air as material. 5. Go to Electromagnetic waves and choose perfect conductor as boundary condition for the circular line. The symmetry axis has the condition Axial Symmetry by default. 6. In Study>Eigenfrequency we set the frequency to e.g., 1 GHz. This is the frequency where COMSOL starts to look for an eigenfrequency. We can also choose the number of resonances that it will determine. 7. The mesh size is Normal by default. If we need a better accuracy then we choose a finer mesh. 8. We now let Comsol solve the problem. 9. COMSOL calculates the lowest resonant frequencies and their electric fields. There might be spurious solutions that are unphysical. The resonance frequency for these solutions are either very far from 1 GHz, or even complex, and the corresponding field plots are fuzzy. If we are interested in the Q-value for a mode we can determine by using the impedance boundary condition. We then add the material of the metal in Material. We let the boundary of the circle have this material. We then choose Impedance boundary condition under Electromagnetic fields. The problem is now non-linear in the sense that the boundary condition depends on the eigenfrequency. Comsol can handle this if we add a linearization point. We right click

7 on Eigenvalue solver and write 1e9 in the box below transform point. With that value Comsol can linearize the problem and get an eigenvalue. For this application the value of the linearization point is not crucial. We let Comsol solve the problem. The Q-value is a pre-defined quantity that we find if we go to Results>Derived values>global evaluation, choose the frequency that we are interested in, and and choose quality factor. 7