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A beam of particles is moving along an accelerator pipe in the \(z\) direction. The particles are uniformly distributed in a cylindrical volume of length \(L_{\mathrm{o}}\) (in the \(z\) direction) and radius \(R_{\mathrm{o}} .\) The particles have momenta uniformly distributed with \(p_{z}\) in an interval \(p_{\mathrm{o}} \pm \Delta p_{z}\) and the transverse momentum \(p_{\perp}\) inside a circle of radius \(\Delta p_{\perp} .\) To increase the particles' spatial density, the beam is focused by electric and magnetic fields, so that the radius shrinks to a smaller value \(R\). What does Liouville's theorem tell you about the spread in the transverse momentum \(p_{\perp}\) and the subsequent behavior of the radius \(R ?\) (Assume that the focusing does not affect either \(L_{\mathrm{o}}\) or \(\Delta p_{z}\).)

Short Answer

Expert verified
The transverse momentum spread increases as the radius decreases.

Step by step solution

01

Understanding Liouville's Theorem

Liouville's theorem states that the phase space volume occupied by a set of particles remains constant over time in a Hamiltonian system. This means even though the distribution of particles in coordinate space (like position) evolves, the distribution in momentum space evolves accordingly to keep the overall phase space volume constant.
02

Initial Phase Space Volume

Initially, the phase space volume for the transverse direction can be described by the cylindrical area \(\pi R_{\mathrm{o}}^2\) for position and the momentum circle \(\pi (\Delta p_{\perp})^2\), giving a volume of \((\pi R_{\mathrm{o}}^2) \cdot (\pi (\Delta p_{\perp})^2)\)."
03

Final Phase Space Volume

After focusing, the spatial distribution becomes \(\pi R^2\) for position. According to Liouville's theorem, the phase space volume must remain constant, so \(\pi R^2 \cdot (\Delta p_{\perp, \text{final}})^2 = (\pi R_{\mathrm{o}}^2) \cdot (\Delta p_{\perp})^2\).
04

Solving for Final Transverse Momentum

Solving \(\pi R^2 (\Delta p_{\perp, \text{final}})^2 = \pi R_{\mathrm{o}}^2 (\Delta p_{\perp})^2\) yields \(\Delta p_{\perp, \text{final}} = \Delta p_{\perp} \frac{R_{\mathrm{o}}}{R}\). This shows that as the radius \(R\) decreases, the transverse momentum spread \(\Delta p_{\perp, \text{final}}\) increases to conserve the phase space volume.
05

Conclusion on the Behavior of R and Momentum

Liouville's theorem indicates that shrinking the radius \(R\) due to focusing must result in an increased spread in transverse momentum \(\Delta p_{\perp, \text{final}}\). As the physical size of the beam decreases, energy (spread in momentum) must increase proportionally.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Space Volume
In the realm of classical mechanics, Liouville's theorem plays a pivotal role in explaining how particles behave within a given space. This is particularly focused on the concept of phase space volume. Phase space is a mathematical construct that combines position and momentum coordinates into a unified space. For a particle beam, each particle has its own position and momentum, and plotting these creates a point in phase space. A collection of particles forms a "volume" in this space.

According to Liouville's theorem, in a Hamiltonian system, this phase space volume remains constant over time. Even as particles are subject to various forces and constraints, the overall volume of their positions and momenta does not change. This principle applies to the cylindrical volume of particles moving through an accelerator, where both position (defined by the radius) and momentum (defined by momentum intervals) create a distinct phase space profile. When particles are focused, changes to the radius alter the configuration, but not the volume, requiring adjustments to transverse momentum.
  • Phase space combines position and momentum.
  • Liouville's theorem maintains volume constancy.
  • Changes in one (position or momentum) necessitate adjustments in the other.
Transverse Momentum
Transverse momentum refers to the component of momentum perpendicular to the primary direction of motion. In the case of particles moving through an accelerator, the transverse momentum is the momentum perpendicular to the primary motion along the z-direction. Changes in transverse momentum are crucial when considering how focusing in particle beams works.

When the radius of the beam is reduced using electric and magnetic fields, the focus alters the configuration, shrinking the beam's physical size. However, due to Liouville's theorem, even as the radius decreases, the transverse momentum must spread out more to keep the phase space volume constant. This spread in transverse momentum ensures that the system is conserving its overall energy distribution, a crucial aspect for maintaining stability in accelerators.
  • Transverse momentum acts perpendicular to motion direction.
  • Focusing affects beam radius and momentum distribution.
  • Momentum spread adjusts to conserve phase space volume.
Hamiltonian System
Hamiltonian systems are a cornerstone of classical mechanics, describing systems that are conservative, meaning they conserve energy over time. In the context of particle physics, a Hamiltonian system allows for the precise modeling of particles moving through an accelerator.

The beauty of Hamiltonian systems lies in their ability to elegantly track how energy and momentum interact under physical constraints. Liouville's theorem is directly applicable, as it assumes the system is Hamiltonian. In such systems, focusing a beam using external fields does not add or subtract energy from the overall system; rather, it redistributes energy across different components, like altering the transverse momentum when the beam's radius changes.
  • Hamiltonian systems are energy-conservative.
  • Essential for modeling particle physics scenarios.
  • Energy redistribution occurs without changing total energy.

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Most popular questions from this chapter

In the Lagrangian formalism, a coordinate \(q_{i}\) is ignorable if \(\partial \mathcal{L} / \partial q_{i}=0 ;\) that is, if \(\mathcal{L}\) is independent of \(q_{i}\). This guarantees that the momentum \(p_{i}\) is constant. In the Hamiltonian approach, we say that \(q_{i}\) is ignorable if \(\mathcal{H}\) is independent of \(q_{i}\), and this too guarantees \(p_{i}\) is constant. These two conditions must be the same, since the result " \(p_{i}=\) const" is the same either way. Prove directly that this is so, as follows: (a) For a system with one degree of freedom, prove that \(\partial \mathcal{H} / \partial q=-\partial \mathcal{L} / \partial q\) starting from the expression (13.14) for the Hamiltonian. This establishes that \(\partial \mathcal{H} / \partial q=0\) if and only if \(\partial \mathcal{L} / \partial q=0 .\) (b) For a system with \(n\) degrees of freedom, prove that \(\partial \mathcal{H} / \partial q_{i}=-\partial \mathcal{L} / \partial q_{i}\) starting from the expression (13.24).

13.13 \star\star Consider a particle of mass \(m\) constrained to move on a frictionless cylinder of radius \(R\), given by the equation \(\rho=R\) in cylindrical polar coordinates \((\rho, \phi, z) .\) The mass is subject to just one external force, \(\mathbf{F}=-k r \hat{\mathbf{r}},\) where \(k\) is a positive constant, \(r\) is its distance from the origin, and \(\hat{\mathbf{r}}\) is the unit vector pointing away from the origin, as usual. Using \(z\) and \(\phi\) as generalized coordinates, find the Hamiltonian \(\mathcal{H}\). Write down and solve Hamilton's equations and describe the motion.

Evaluate the three-dimensional divergence \(\nabla \cdot\) v for each of the following vectors: (a) \(\mathbf{v}=k \mathbf{r}\) (b) \(\mathbf{v}=k(z, x, y),(\mathbf{c}) \mathbf{v}=k(z, y, x),(\mathbf{d}) \mathbf{v}=k(x, y,-2 z),\) where \(\mathbf{r}=(x, y, z)\) is the usual position vector and \(k\) is a constant.

The general proof of the divergence theorem $$\int_{S} \mathbf{n} \cdot \mathbf{v} d A=\int_{V} \nabla \cdot \mathbf{v} d V$$ is fairly complicated and not especially illuminating. However, there are a few special cases where it is reasonably simple and quite instructive. Here is one: Consider a rectangular region bounded by the six planes \(x=X\) and \(X+A, y=Y\) and \(Y+B,\) and \(z=Z\) and \(Z+C,\) with total volume \(V=A B C .\) The surface \(S\) of this region is made up of six rectangles that we can call \(S_{1}\) (in the plane \(x=X\) ), \(S_{2}\) (in the plane \(x=X+A\) ), and so on. The surface integral on the left of (13.63) is then the sum of six integrals, one over each of the rectangles \(S_{1}, S_{2},\) and so forth. (a) Consider the first two of these integrals and show that $$ \int_{S_{1}} \mathbf{n} \cdot \mathbf{v} d A+\int_{S_{2}} \mathbf{n} \cdot \mathbf{v} d A=\int_{Y}^{Y+B} d y \int_{Z}^{Z+C} d z\left[v_{x}(X+A, y, z)-v_{x}(X, y, z)\right] $$ (b) Show that the integrand on the right can be rewritten as an integral of \(\partial v_{x} / \partial x\) over \(x\) running from \(x=X\) to \(x=X+A .(\text { c) Substitute the result of part }(b)\) into part (a), and write down the corresponding results for the two remaining pairs of faces. Add these results to prove the divergence theorem (13.63).

Consider a particle of mass \(m\) moving in two dimensions, subject to a force \(\mathbf{F}=-k x \hat{\mathbf{x}}+K \hat{\mathbf{y}}\) where \(k\) and \(K\) are positive constants. Write down the Hamiltonian and Hamilton's equations, using \(x\) and \(y\) as generalized coordinates. Solve the latter and describe the motion.

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