Penning Trap Motion

This note is based on a cylindrical Penning Trap, of which TAMUTRAP, at Texas A&M's Cyclotron Institute, is the largest in the world. Read more about the specifics of TAMUTRAP here.

Overview

What does a Penning Trap do?

A Penning Trap is a device meant to hold one single charged particle in empty space, keeping anything from touching it. This particle can be anything from antimatter to, in the case of TAMUTRAP, ions of unstable nuclei. These ions are held for the sake of measuring properties that vanish within an extremely tiny timescale. In this note, let’s try to reckon with the needs and constraints of the physical machine itself to see if we can land on a mathematically complete explanation of these competing forces and resultant motion.

Motivating the Derivation

To carry out its purpose, a Penning Trap needs to provide confinement in all three directions simultaneously, using purely electric and magnetic forces (we cannot touch the trapped particle). As we will see, this proves a bit of a fun challenge!

With TAMUTRAP being part of TAMU’s Cyclotron Institute, let’s first implement a uniform magnetic field along the (arbitrarily chosen) z-axis, as cyclotrons do. In such a field, a charged particle, now under influence of the Lorentz Force, will simply undergo cyclotron motion, circling around in the xy-plane with acceleration perpendicular to velocity. So, the x and y-directions are handled, but the particle is still free to slide along the z direction freely.

insert cyclotron motion graph here?

We can try to remedy this with an electric field that pushes a particle back towards our midplane along the z direction, a restoring force. This, however, is where Gauss’s Law, $\nabla \cdot \mathbf{E}=\frac{\rho }{{ϵ}_0}$ steps in. In free space, there is no enclosed charge, ($\rho =0$) and so divergence of the electric field must similarly be zero: $\nabla \cdot \mathbf{E}=0$. Physically, we cannot just have a purely axial restoring force. Any field that pulls toward the center along z is forced to push outward radially. With this, the magnetic field must pick up the slack, further providing radial confinement to fight this defocusing effect.

Derivation

Assumptions

Lorentz Force: $\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$, or $m\ddot{\mathbf{r}}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$

Gauss’s Law: $\nabla \cdot \mathbf{E}=\frac{\rho }{{ϵ}_0}$

B-field

Since the B-field is just along one axis, let’s say the z-axis, then $\mathbf{B}=B\hat{z}$. Now, we can do that cross product: $\mathbf{v}\times \mathbf{B}=\left|\begin{array}{ccc}\hat{x}& \hat{y}& \hat{z}\\ \dot{x}& \dot{y}& \dot{z}\\ 0& 0& B\end{array}\right|=\boxed{(\dot{y}B,-\dot{x}B,0)}$

E-field

We want to stop the particle from sliding along the z-axis, and we will use the E-field to do so. For a particle to always return to the origin, the easiest scenario is one where $V\sim z^2$, so $E_z\sim -z$, creating as simple restoring force ($F_z\sim -z$). As we are essentially in a free space, following the constraints of Gauss’s law, $\rho$ must be 0 (you cannot have an enclosed charge without enclosure), and so $\nabla \cdot \mathbf{E}=0$, meaning $\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}=0$.

Now, letting $V=c{z}^2$ for some positive c, $\frac{\partial V}{\partial z}=E_z=-2cz$, and so $\frac{\partial E_z}{\partial z}=-2c$. Once again, we are constrained by Gauss’s Law, which says that a purely axial field cannot exist, so there must be x and y-components, which in order to preserve cylindrical geometry around the z-axis, must be equal. With $\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}=0$, $\frac{\partial E_x}{\partial x}=\frac{\partial E_y}{\partial y}$, and $\frac{\partial E_z}{\partial z}=-2c$, then $\frac{\partial E_x}{\partial x}=\frac{\partial E_y}{\partial y}=c$., giving us the full electric field: $\mathbf{E}=(cx,cy,-2cz)$. This is an exact differential equation, so you can simply integrate each component: $V(x,y,z)=-\frac{c}{2}{x}^2-\frac{c}{2}{y}^2+c{z}^2$, or $\frac{c}{2}(2z^2-x^2-y^2)$.

$d=\sqrt{\frac{2z_0^2+r_0^2}{4}}=\frac{1}{2}\sqrt{2z_0^2+r_0^2}$. Why? Where does this come from? We need to match boundary conditions, endcaps and the ring electrodes. We can say the endcap electrode sits at $z=z_0$, with a potential $+U_0/2$. $V(0,0,z_0)=\frac{c}{2}(2z_0^2)=c{z}_0^2$. Then, $c{z}_0^2=\frac{U_0}{2}$, so $c=\frac{U_0}{2z_0^2}$. If $V(r_0,0,0)=\frac{c}{2}(-r_0^2)=-\frac{U_0}{2}$, so $c=\frac{U_0}{r_0^2}$. So, $2z_0^2=r_0^2$, hence $c=\frac{U_0}{2d^2}$. Plugging back in, we get $V(x,y,z)=\frac{U_0}{4d^2}(2z^2-x^2-y^2)$, and $\boxed{\mathbf{E}=\frac{U_0}{2d^2}(x,y,-2z)}$.

Equations of Motion

First, we start with the simplest component of motion: $m\ddot{z}=q{E}_z=q(-2\frac{U_0}{2d^2})z=-\frac{q{U}_0}{d^2}z$, and so $\ddot{z}=-\frac{q{U}_0}{m{d}^2}z$. This is just SHM (with the standard form $\boxed{\ddot{z}=-{\omega }_z^{2}z}$), and so ${\omega }_z^2=\frac{q{U}_0}{m{d}^2}$. We get ${\omega }_c=\frac{qB}{m}$ from Cyclotron Motion.

Now, with $m\ddot{\mathbf{r}}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$, $\mathbf{E}=\frac{U_0}{2d^2}(x,y,-2z)$, and $\mathbf{v}\times \mathbf{B}=(\dot{y}B,-\dot{x}B,0)$, we can start to solve for each component of motion. Starting with the x-component, $m\ddot{x}=q(\frac{U_0}{2d^2}x+\dot{y}B)$, and so $\ddot{x}=\frac{q{U}_0}{2m{d}^2}x+\frac{qB}{m}\dot{y}$, so $\boxed{\ddot{x}=\frac{{\omega }_z^2}{2}x-{\omega }_c\dot{y}}$. Similarly, $\boxed{\ddot{y}=\frac{{\omega }_z^2}{2}y-{\omega }_c\dot{x}}$.

We can package this all together in one linear system: $\dot{\mathbf{u}}=\left(\begin{array}{cccccc}0& 1& 0& 0& 0& 0\\ \frac{{\omega }_z^2}{2}& 0& 0& {\omega }_c& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& -{\omega }_c& \frac{{\omega }_z^2}{2}& 0& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& -{\omega }_z^2& 0\end{array}\right)\mathbf{u}$. For the purpose of visualization, we may as well get position as a function of time as well. The case of the z-component is easy. As $\ddot{z}+{\omega }_z^{2}z=0$, SHM gives immediately gives us $z(t)=A_z\cos ({\omega }_{z}t+\varphi )$.

Big Picture

https://www.desmos.com/3d/dn2q9xaoit

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Related

• Penning Trap
• Lorentz Force
• Gauss’s Law
• Laplace’s Equation
• Cyclotron
• Simple Harmonic Motion