Problem 15

Question

To investigate the stability of the motion described in the preceding question, evaluate the second derivatives of $U$ at $\rho=a, z=0$, and show that they may be written $\begin{gathered} \frac{\partial^{2} U}{\partial \rho^{2}}=\frac{q^{2}}{m}\left[B_{z}\left(B_{z}+\rho \frac{\partial B_{z}}{\partial \rho}\right)\right]_{\rho=a, z=0} \\ \frac{\partial^{2} U}{\partial \rho \partial z}=0, \quad \frac{\partial^{2} U}{\partial z^{2}}=-\frac{q^{2}}{m}\left[B_{z} \rho \frac{\partial B_{z}}{\partial \rho}\right]_{\rho=a, z=0} \end{gathered}$ (Hint: You will need to use the $\varphi$ component of the equation $\boldsymbol{\nabla} \wedge \boldsymbol{B}=\mathbf{0}$, and the fact that, since $B_{\rho}=0$ for all $\rho, \partial B_{\rho} / \partial \rho=0$ also.) Given that the dependence of $B_{z}$ on $\rho$ near the equilibrium orbit is described by $B_{z} \propto(a / \rho)^{n}$, show that the orbit is stable if \(0

Step-by-Step Solution

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Question: Determine the condition for the stability of the circular orbit of a charged particle in the z = 0 plane, given that the magnetic field depends on the coordinate ρ near the equilibrium orbit as $B_{z} \propto(a / \rho)^{n}$, where a is a constant, and \(0

1Step 1: Use the curl equation

To evaluate the second derivatives, we need to use the given hint, which involves the equation for the curl of the magnetic field $\nabla \wedge \boldsymbol{B} = \mathbf{0}$. We are given that $B_{\rho} = 0$ and $\frac{\partial B_{\rho}}{\partial \rho} = 0$. In cylindrical coordinates, the curl equation can be written as: $$ \nabla \wedge \boldsymbol{B} = \left(\frac{1}{\rho} \frac{\partial B_z}{\partial \rho} - \frac{\partial B_{\phi}}{\partial z}\right) \boldsymbol{\rho} + \left(\frac{\partial B_{\rho}}{\partial z} - \frac{\partial B_z}{\partial z}\right) \boldsymbol{\phi} + \frac{1}{\rho}\left(\frac{\partial \left(\rho B_{\phi}\right)}{\partial \rho} - \frac{\partial B_{\rho}}{\partial \phi}\right)\boldsymbol{z} $$ The $\phi$ component must be zero according to the hint: $$ \frac{\partial B_{\rho}}{\partial z} - \frac{\partial B_z}{\partial z} = 0 $$ Since $B_{\rho} = 0$, $$ \frac{\partial B_z}{\partial z} = 0 $$

2Step 2: Evaluate the expressions

Now, we will evaluate the expressions: $$ \frac{\partial^{2} U}{\partial \rho^{2}}=\frac{q^{2}}{m}\left[B_{z}\left(B_{z}+\rho \frac{\partial B_{z}}{\partial \rho}\right)\right]_{\rho=a, z=0} $$

3Step 3: Determine the dependence of $B_z$ on $\rho$

We are given that the dependence of $B_z$ on $\rho$ near the equilibrium orbit is described by $B_{z} \propto(a / \rho)^{n}$. Therefore, we can write: $$ B_z \sim \left(\frac{a}{\rho}\right)^n $$ Differentiating with respect to $\rho$, we get: $$ \frac{\partial B_z}{\partial \rho} \sim -n \left(\frac{a}{\rho}\right)^{n+1} $$ Now, we substitute these expressions back into the second derivatives: $$ {\frac{\partial^{2} U}{\partial \rho^{2}}}_{\rho=a, z=0}=\frac{q^{2}}{m}\left[B_{z}\left(B_{z}+\rho \frac{\partial B_{z}}{\partial \rho}\right)\right]_{\rho=a, z=0} = \frac{q^{2}}{m} B_z^2 \left(1 - n\right) $$ and $$ \frac{\partial^{2} U}{\partial z^{2}}=-\frac{q^{2}}{m}\left[B_{z} \rho \frac{\partial B_{z}}{\partial \rho}\right]_{\rho=a, z=0} = -n \frac{q^2}{m} B_z^2 $$

4Step 4: Check the stability

For the motion to be stable, the second derivative of the potential energy with respect to $\rho$ and $z$ should be positive and negative, respectively. So we need to satisfy: $$ {\frac{\partial^{2} U}{\partial \rho^{2}}}_{\rho=a, z=0} > 0\quad \text{and} \quad \frac{\partial^{2} U}{\partial z^{2}}_{\rho=a, z=0} < 0 $$ From the expressions above, we have: $$ \frac{q^{2}}{m} B_z^2 \left(1 - n\right) > 0\quad \text{and} \quad -n \frac{q^2}{m} B_z^2 < 0 $$ Since $\frac{q^2}{m} B_z^2 > 0$, we can deduce that $1 > n > 0$. Therefore, for the circular orbit to be stable, the condition \(0

Key Concepts

Potential Energy DerivativesCylindrical CoordinatesMagnetic Field CurlStability Analysis

Potential Energy Derivatives

When we analyze a system's stability, understanding how its potential energy changes is crucial. A common method is by examining the second derivatives of the potential energy, which can indicate points of equilibrium or instability. In this context, we're dealing with a function called potential energy, denoted as $U$. The derivatives we're interested in are:

$\frac{\partial^{2} U}{\partial \rho^{2}}$
$\frac{\partial^{2} U}{\partial \rho \partial z}$
$\frac{\partial^{2} U}{\partial z^{2}}$

The second derivative of potential energy with respect to $\rho$ gives insights into radial stability, while the mixed partial derivative $\frac{\partial^{2} U}{\partial \rho \partial z}$ helps determine if cross-dependencies affect stability. Lastly, $\frac{\partial^{2} U}{\partial z^{2}}$ informs about the stability in the axial direction. If these derivatives satisfy certain conditions, namely positivity or negativity, we can declare the system as stable or unstable at certain points.

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system that is particularly useful in physics for dealing with problems involving symmetry around an axis. They are denoted as $(\rho, \phi, z)$.

$\rho$ is the radial distance from the origin to the projection of the point onto the $xy$-plane.
$\phi$ is the angle between the positive $x$-axis and the line connecting the origin to the projection of the point onto the $xy$-plane.
$z$ is the height above the $xy$-plane.

This coordinate system is excellent for problems with an axis of rotation, as it simplifies many equations that would otherwise be quite complex in Cartesian coordinates. For our exercise, these coordinates help describe the stability of an orbit influenced by a magnetic field, specifically enabling the calculation of derivatives and curl.

Magnetic Field Curl

The curl of a magnetic field is a vector operator that describes the rotation or the 'twist' of the field in space. The operation involves a mathematical construct, denoted as $abla \wedge \boldsymbol{B}$, that helps in understanding how the magnetic field changes around any given point.In the context of this exercise, we focus on the simplified case where the magnetic vector doesn't depend on certain coordinates, allowing us to exclude terms with $B_{\rho}$. By setting $abla \wedge \boldsymbol{B} = \mathbf{0}$, we assume that the magnetic field is static, meaning there's no real 'rotation' of the field along that part of the system. This simplification makes it easier to evaluate potential energy derivatives, as it constrains certain variables, narrowing down the expression to more manageable forms.

Stability Analysis

Stability analysis involves evaluating whether small perturbations or changes can grow over time, leading to instability, or if they diminish, leading to stability. In physical systems, particularly those involving circular or magnetic motion, this analysis focuses on the derivatives of potential energy.To determine stability, consider:

If $\frac{\partial^{2} U}{\partial \rho^{2}} > 0$, small changes in radial position $\rho$ will result in restoring forces, indicating stability.
If $\frac{\partial^{2} U}{\partial z^{2}} < 0$, this indicates that axial changes will be opposed, again suggesting stability.

The condition \(0

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