Homework 5 Discussion

Some General Comments:

• grade: 22 possible
  + pts per problem as indicated
  + ave score (as-is): 15/22 (will be re-scaled)

Cavity Problem 1 (6 points)

Consider a pillbox cavity designed for frequency \(\omega\) and particle velocity \(\beta\) with peak surface electric field of \(E_0\).

1. Calculate the radius \(R\).
   \[ \omega_0 = \frac{2.405c}{R} \longrightarrow R = \frac{2.405c}{\omega_0} \]
2. What are the Electric fields on the beam axis?
   \[ E_z(\rho,\phi,z,t) = E_0\;J_0\left(\frac{2.405c}{R}\right)e^{\pm i\omega t} \\ E_z(\rho=0) = E_0e^{\pm i\omega t} \]
3. What’s the maximum voltage this cavity can achieve?
   \[ at~\rho = 0, ~~~ V_0 = \int_0^L |E_z(z)| dz = E_0L \]
4. What’s the maximum voltage a beam could get?
   • Assume constant \(\beta\); also, symmetric integrals are your friend

\[ V_{acc} = \int_0^L E_0 e^{i\omega t}dz = \int_0^L E_0 e^{i\omega z/\beta c} dz \\ = \frac{\beta c}{i\omega}\int_{-L/2}^{L/2} E_0 e^{i\omega z/\beta c} (\frac{i\omega}{\beta c}dz) = E_0\frac{\beta c}{i\omega}\left( e^{i\omega L/2\beta c} - e^{-i\omega z/2\beta c} \right) \\ = 2E_0 \frac{\beta c}{\omega}\sin\frac{\omega L}{2\beta c} \]

5. What \(L\) optimizes this beam voltage?

\[ \frac{dV_{acc}}{dL} = 2E_0 \frac{\beta c}{\omega}\cos\frac{\omega L}{2\beta c} \cdot \frac{\omega }{2\beta c} = 0 \longrightarrow \cos(~)=0 \longrightarrow \frac{\omega L}{2\beta c} = \pi/2\\ L = \frac{\pi\beta c}{\omega} = \frac{\beta}{2}\frac{2\pi c}{\omega} = \frac{\beta\lambda}{2}. ~~~~~~~~~ (\lambda f = c,~~ \lambda\omega = 2\pi c) \]

6. What’s the ratio of Maximum Beam Voltage/Maximum Cavity Voltage?

\[ \frac{V_{acc}}{V_0} \equiv T = 2E_0 \frac{\beta c}{\omega}\sin\frac{\omega L}{2\beta c} \cdot \frac{1}{E_0 L} = \frac{2\beta c}{\omega L}\sin \frac{\omega L}{2\beta c} \\ @~ L=\beta\lambda/2 \longrightarrow T = 2/\pi = 0.637 \]

Cavity Problem 2 (4 points)

Now consider an equivalent of a pillbox cavity made with rectangular waveguide with sides \(a\), \(b\), and \(c\).

1. Draw the fields of the accelerating mode; it should look pretty similar to a pillbox cavity.
   
2. What mathematical form do these fields take?
   Guess: \[ E_z = E_0\cos(k_x x)\cos(k_y y) e^{i\omega t} \] Boundary Conditions: \[ E_z = 0 ~ @ ~ x = \pm a/2, ~ y = \pm b/2 \\ \cos(k_x a/2) = 0 \longrightarrow k_x = \pi/a \\ {\rm and~~} k_y = \pi/b \] Hence, \(\vec{k} = (\pi/a)\hat{x} + (\pi/b)\hat{y}\) and \[ \vec{E} = \hat{z}E_0\cos(\pi x/a)\cos(\pi y/b)e^{i\omega t}\\ \vec{B} = \frac{i}{\omega}\vec{k}\times\vec{E} \]
3. What’s the frequency of this cavity? Hint: Don’t overcomplicate it.

\[ c^2\left[(\pi/a)^2 + (\pi/b)^2 \right] = \omega^2 \\ \rightarrow ~~~\omega = \pi c\sqrt{1/a^2 + 1/b^2} \]

Cavity Problem 3 (12 points)

For a half-wave resonator:

1. Calculate the Geometry Factor, given \(f\), \(a\), and \(b\).
   First: \[ G = \frac{\omega U}{P_d/R_s} \\ U = \int_V\frac12\frac1\mu_0|B|^2dV \\ \vec{B} = \frac{E_0a}{c\rho}\sin(\pi z/L)e^{i\omega t}\hat{\phi} \\ U = \frac{1}{2\mu_0}\frac{E_0a^2}{c^2}\int_z^b\rho d\rho\int_0^{2\pi} d\phi \int_{-L/2}^{L/2} dz \;\frac{\sin^2(\pi z/L)}{\rho^2} \\ \rightarrow U = \frac{\epsilon_0\pi E_0^2 a^2\lambda}{4}\ln(b/a) \\ \] Secondly,

\[ P_D/R_s = \frac12\int_S|\vec{H}|^2 dA \\ = 2\cdot\frac12\mu_0\int_0^{2\pi}d\phi\int_a^b\rho d\rho B^2|_{z=L/2} ~~({\rm 2~end~walls}) \\ + \frac{1}{2\mu_0}\int_0^{2\pi}d\phi\int_{-L/2}^{L/2}dz \rho B^2 |_{\rho=b} ~~({\rm Outer~Conductor}) \\ + \frac{1}{2\mu_0}\int_0^{2\pi}d\phi\int_{-L/2}^{L/2}dz \rho B^2 |_{\rho=1} ~~({\rm Inner~Conductor}) \\ \rightarrow P_D/R_s= \frac{\pi E_0^2a^2\epsilon_0}{\mu_0}\left[ \frac\lambda4(1/a+1/b)+2\ln(b/a) \right] \]

So, finally,

\[ G = \frac{\omega U}{P_d/R_s} = \frac{\pi c\mu_0\ln(b/a)}{\left[ \frac\lambda2(1/a+1/b)+4\ln(b/a) \right]} \]

2. Calculate \(R/Q\) in terms of the \(TTF\) (transit time factor).

\[ R/Q = \frac{(TTF)^2 8\ln(b/a)}{\pi^2\epsilon_0 c} \approx TTF\cdot\ln(b/a)\cdot 300[\Omega] \]

3. (BONUS) Show that \(\frac{\beta\lambda}{2} = (b+a)\).

1. Why does this make sense?

Gap-to-gap = \((a+b)/2~\cdot~2\) should equal \(\beta\lambda/2\) to get synchronization between gaps.

2. Not trivial! Integral of \(sin(x)/x\) is not closed; either be more clever or demonstrate numerically?

4. Assuming c above is proven, how would you go about optimizing \(b-a\)?

\(G\) can be optimized for \((b-a)/(b+a)\), but \(R/Q\) always pushses toward larger gap.

\[ \frac{\partial}{\partial[(b-a)(b+a)]}\left( \frac R Q \right) = 0 \] gives “an” answer, but peak surface fields are also important.

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1. Northern Illinois University and Fermi National Accelerator Laboratory↩