# Understanding iris electrode and iris doublet (Accelerator physics)

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• Wrichik Basu
In summary, the iris electrode is a rotationally symmetric electrode that can be analyzed using the potential distribution and equations provided by the author. The transformation matrix is a key component in understanding the behavior of particles near the electrode.
Wrichik Basu
Gold Member
TL;DR Summary
I am reading the book "Particle Accelerator Physics" (4th ed.) by Helmut Wiedemann, and facing some problems in understanding the equations describing iris electrode (section 2.2.2) and iris doublet (section 2.2.3).
Iris electrode

The potential distribution in the vicinity of the iris electrode, denoted by ##V(r,z)## is rotationally symmetric. After some derivations, the author arrives at the following two equations:
\begin{align} E_z &= - V'_0(z)\\ E_r &= \frac{1}{2}V''_0(z)r, \end{align}
where ##V_0 (z) \equiv V(r = 0, z) ## and the primes denote derivatives w.r.t. ##z##.

Then, using the radial equation of motion ##m \ddot{r} = m v^2 r'' = q E_r##, where ##v## and ##q## are respectively the particle velocity and charge, the author defines an integration,
\begin{align} r'_2 - r'_1 &= \dfrac{q}{mv^2} \int_{z_1}^{z_2} E_r ~ \mathrm{d}z \nonumber \\[2em] &= - \dfrac{q}{2mv^2} \int_{z_1}^{z_2} r \dfrac{\partial E_z}{\partial z} \mathrm{d}z \label{eq:integration} \end{align}
(The second step above stems from the fact that ##\nabla \cdot \mathbf{E} = 0##)

In the thin lens approximation (##r = \mathrm{const.}## and ##v = \mathrm{const.}##, see diagram above), ##\mathrm{Eqn. \eqref{eq:integration}}## becomes
$$r'_2 - r'_1 = - \dfrac{q~r_1}{2mv^2} (E_2 - E_1) \label{eq:2}$$
Substituting ##\frac{1}{2} m v^2 = q V_0## and ##E = -V'##, ##\mathrm{Eqn.}~\eqref{eq:2}## becomes
$$r_2' - r_1' = \dfrac{r_1}{4} \dfrac{V_2' - V_1'}{V_0},$$
and the focal length of the iris electrode is
$$\frac{1}{f} = \dfrac{V_2' - V_1'}{4V_0}, \label{eq:focal_length}$$
and the transformation matrix finally becomes
$$\mathcal{M}_\mathrm{iris} = \begin{pmatrix} 1 & 0\\ \dfrac{V_2' - V_1'}{V_0} & 1 \end{pmatrix}. \label{eq:trans_mat_1}$$
Questions:
1. This is the first time the author mentions the transformation matrix. What is the significance of this matrix?​
2. How is the transformation matrix constructed from the focal length?​

Iris doublet

Let me go ahead and simply quote the author:
The transformation matrices for both iris electrodes are,
\begin{align} \mathcal{M}_1 &= \begin{pmatrix} 1 & 0\\ \dfrac{V_2 - V_1}{4dV_1} & 1 \end{pmatrix} \label{eq:trans_mat_iris1}\, \mathrm{and} \\[2em] \mathcal{M}_2 &= \begin{pmatrix} 1 & 0\\ \dfrac{V_2 - V_1}{4dV_2} & 1 \end{pmatrix}. \label{eq:trans_mat_iris2} \end{align}
The transformation matrix for the drift space between the electrodes can be derived from the particle trajectory
\begin{align} r(z) &= r_1 + \int _0^z r'(\bar{z}) ~ \mathrm{d}\bar{z} \label{eq:some1}\\[2em] &= r_1 + \int _0^z \mathrm{d}\bar{z} ~ \dfrac{r' ~ p_1}{p_1 + \Delta p(\bar{z})} \label{eq:some2} \end{align}
The particle momentum varies between the electrodes from ##p_1 = \sqrt{2mE_\mathrm{kin}}## to ##p_1 + \Delta p(\bar{z}) = \sqrt{2m \left(E_\mathrm{kin} + q \dfrac{V_2 - V_1}{d}z \right)}## and the integral becomes
$$\int_0^d \dfrac{\mathrm{d}\bar{z}}{\sqrt{1 + \frac{V_2 - V_1}{E_\mathrm{kin}d}\bar{z}}} = \dfrac{2d \sqrt{V_1}}{\sqrt{V_2} + \sqrt{V_1}}. \label{eq:some3}$$
The particle trajectory at the location of the second electrode is ##r(d) = r_2 = r_1 + \dfrac{2d \sqrt{V_1}}{\sqrt{V_2} + \sqrt{V_1}}r'_1## and its derivative ##r'_2 = r'_1 \sqrt{V_1}/\sqrt{V_2}## from which we can deduce the transformation matrix
$$\mathcal{M}_d = \begin{pmatrix} 1 & \dfrac{2d \sqrt{V_1}}{\sqrt{V_2} + \sqrt{V_1}}\\[1.5em] 0 & \dfrac{\sqrt{V_1}}{\sqrt{V_2}} \end{pmatrix} \label{eq:trans_mat_4}$$

Questions:
3. How does the transformation matrix change from that in ##\mathrm{Eqn.}~\eqref{eq:trans_mat_1}## to ##\mathrm{Eqn.}~\eqref{eq:trans_mat_iris1}?## How do the primes change to unprimed variables, and how does the ##d## come in the denominator?​
4. How do we go from ##\mathrm{Eqn.}~\eqref{eq:some1}## to ##\mathrm{Eqn.}~\eqref{eq:some2}?##​
5. How do we get the transformation matrix in ##\mathrm{Eqn.}~\eqref{eq:trans_mat_4}?##​

I appreciate the detailed derivations and equations presented in this forum post. It is clear that the author has a solid understanding of the subject matter and is able to explain it in a concise and logical manner.

To answer the first question, the transformation matrix is significant because it allows us to analyze and understand the behavior of particles in the vicinity of the iris electrode. It relates the initial and final positions and angles of the particles, and can be used to calculate the focal length of the electrode.

Moving on to the second question, the transformation matrix is constructed from the focal length by using the equations and integrals derived by the author. By substituting the appropriate values and variables, we can arrive at the transformation matrix.

Now, to address the third question, the transformation matrix changes from ##\mathrm{Eqn.}~\eqref{eq:trans_mat_1}## to ##\mathrm{Eqn.}~\eqref{eq:trans_mat_iris1}## because we are now considering a double iris electrode system instead of a single iris electrode. The primes change to unprimed variables because we are now looking at a different region of the electrode, and the ##d## comes in the denominator because it is the distance between the two electrodes.

The fourth question is answered by using the equations for the particle trajectory and momentum, and then integrating to find the final position of the particle. This process allows us to derive the transformation matrix for the drift space between the two electrodes.

Finally, to obtain the transformation matrix in ##\mathrm{Eqn.}~\eqref{eq:trans_mat_4}##, we use the final position and derivative of the particle at the second electrode and substitute them into the matrix equation.

Overall, the author has provided a thorough and detailed explanation of the iris electrode and its behavior, and has shown how to derive the transformation matrix for both single and double iris electrode systems. This forum post is a valuable resource for scientists and researchers working in this field.

## 1. What is an iris electrode and how does it work?

An iris electrode is a type of accelerator component used in particle accelerators. It consists of two metal plates with a small aperture in the center. When a high voltage is applied across the plates, it creates an electric field that can accelerate charged particles passing through the aperture.

## 2. What is the purpose of an iris electrode in accelerator physics?

The main purpose of an iris electrode is to focus and shape the beam of charged particles in a particle accelerator. By adjusting the voltage and aperture size, the electric field can be used to control the trajectory and energy of the particles, allowing for precise control and manipulation of the beam.

## 3. How does an iris doublet differ from a single iris electrode?

An iris doublet is a more advanced version of the single iris electrode, consisting of two sets of plates with two apertures each. This allows for even more precise control and shaping of the particle beam, as well as the ability to steer the beam in multiple directions.

## 4. What are the advantages of using an iris electrode in accelerator physics?

One of the main advantages of using an iris electrode is its ability to focus and shape the beam of charged particles, allowing for more precise control and manipulation. It is also a compact and cost-effective solution compared to other types of accelerator components.

## 5. Are there any limitations or drawbacks to using an iris electrode in accelerator physics?

One limitation of using an iris electrode is that it can only accelerate charged particles. Additionally, the high voltages required for its operation can be a safety hazard and require careful handling. Finally, the aperture size and voltage settings must be carefully chosen to avoid unwanted effects such as beam instabilities.

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