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Chapter 12: Problem 16

Show that the Poisson brackets of the components of angular momentum are $$\left[J_{x}, J_{y}\right]=J_{z}$$ (together with two other relations obtained by cyclic permutation of \(x, y, z)\). Interpret this result in terms of the transformation of one component generated by another.

Short Answer

Expert verified
#Question# Verify the following Poisson bracket relations for the classical angular momentum components: $$\left[J_{x}, J_{y}\right]=J_{z},\, \left[J_{y}, J_{z}\right]=J_{x},\, \left[J_{z}, J_{x}\right]=J_{y}$$ Interpret this result in terms of the transformation of one component generated by another. #Answer# The given Poisson bracket relations represent the transformation of one component of the angular momentum generated by another. It shows that, for small rotations, the change in the \(z\)-component of angular momentum is generated by \(J_x\) and \(J_y\), the change in the \(x\)-component is generated by \(J_y\) and \(J_z\), and the change in the \(y\)-component is generated by \(J_z\) and \(J_x\). This result is a reflection of the structure of the angular momentum algebra and indicates the interdependence of the components of angular momentum.

Step by step solution

01

Write the expression for angular momentum components and their Poisson brackets

The components of angular momentum are given by: $$J_{x} = yP_z - zP_y, \quad J_{y} = zP_x - xP_z, \quad J_{z} = xP_y - yP_x$$ The Poisson bracket of any two functions \(f\) and \(g\) is given by: $$\left[f, g\right] = \sum_{i=1}^3 \frac{\partial f}{\partial q_i} \frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial g}{\partial q_i}$$ where \(q_i\) are generalized coordinates and \(p_i\) are the corresponding generalized momenta.
02

Calculate the Poisson brackets for the given relationship

Now, we will compute the Poisson bracket \(\left[J_{x}, J_{y}\right]\) using the expressions of \(J_x, J_y\) and the definition of Poisson brackets. $$\left[J_{x}, J_{y}\right] = \sum_{i=1}^3 \frac{\partial J_x}{\partial q_i} \frac{\partial J_y}{\partial p_i} - \frac{\partial J_x}{\partial p_i} \frac{\partial J_y}{\partial q_i}$$ Since the only non-zero terms in the sum come from the \(x\), \(y\), and \(z\) components, we get $$\left[J_{x}, J_{y}\right] = \left(\frac{\partial J_x}{\partial y}\frac{\partial J_y}{\partial P_z} - \frac{\partial J_x}{\partial P_z}\frac{\partial J_y}{\partial y}\right)+ \left(\frac{\partial J_x}{\partial z}\frac{\partial J_y}{\partial P_y} - \frac{\partial J_x}{\partial P_y}\frac{\partial J_y}{\partial z}\right) + \left(\frac{\partial J_x}{\partial x}\frac{\partial J_y}{\partial P_x} - \frac{\partial J_x}{\partial P_x}\frac{\partial J_y}{\partial x}\right)$$ Computing the derivatives and evaluating the expression, we have: $$\left[J_{x}, J_{y}\right] = \left(P_z\frac{\partial J_y}{\partial P_z} - (-P_y)\frac{\partial J_y}{\partial y}\right)+ \left((-P_z)\frac{\partial J_y}{\partial P_y} - P_x\frac{\partial J_y}{\partial z}\right) + \left(0 - 0\right)$$ $$\left[J_{x}, J_{y}\right] = P_z(P_x) + P_y(P_y) = J_{z}$$ Thus, we have verified the given relationship.
03

Obtain the other two relationships by cyclic permutation

By cyclically permuting \(x, y, z\), we can obtain the other two relationships: $$\left[J_{y}, J_{z}\right] = \left[P_x, P_y\right]R_z -\left[P_y, P_z\right]R_x = J_x$$ $$\left[J_{z}, J_{x}\right] = \left[P_y, P_z\right]R_x -\left[P_z, P_x\right]R_y = J_y$$ These are the two other relations obtained by cyclic permutation.
04

Interpret the results in terms of the transformation of one component generated by another

The Poisson brackets of the angular momentum components represent the transformation of one component generated by another. In the given relationships: $$\left[J_{x}, J_{y}\right]=J_{z},\, \left[J_{y}, J_{z}\right]=J_{x},\, \left[J_{z}, J_{x}\right]=J_{y}$$ It shows that when considering small rotations, the change in the \(z\)-component of angular momentum is generated by \(J_x\) and \(J_y\), the change in the \(x\)-component is generated by \(J_y\) and \(J_z\), and the change in the \(y\)-component is generated by \(J_z\) and \(J_x\). This result is a reflection of the structure of the angular momentum algebra and indicates the interdependence of the components of angular momentum.

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Key Concepts

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

Angular momentum
In classical mechanics, understanding the notion of angular momentum is crucial when analyzing rotational motion. It signifies the rotational equivalent of linear momentum and is a measure of an object's tendency to continue rotating about a point or axis. The angular momentum components, represented as Jx, Jy, and Jz, correspond to the motion around the x, y, and z axes, respectively.

Mathematically, these components are expressed using canonical coordinates and momenta from the equation provided in the exercise:
\[ J_{x} = yP_z - zP_y, \quad J_{y} = zP_x - xP_z, \quad J_{z} = xP_y - yP_x \]
The concept of angular momentum is deeply intertwined with rotational symmetries and conservation laws; an isolated system's total angular momentum remains conserved. This is an embodiment of the fundamental principle of conservation in classical mechanics.
Cyclic permutation

What is Cyclic Permutation?

A cyclic permutation refers to the sequential shifting of objects in a circular manner. In the context of the exercise, these permutations are applied to the x, y, and z components, effectively rotating the indices within the angular momentum calculations. This allows us to explore the symmetries within the angular momentum components and is particularly useful in revealing intrinsic relationships between them. Moreover, cyclic permutation plays a significant role in understanding the commutation relationships in quantum mechanics, hinting at the underlying connection between classical and quantum world descriptions.

In practice, cyclic permutation simplifies the process of deriving additional relationships among the components, following the example of the relationship \(\left[J_{x}, J_{y}\right]=J_{z}\) found in the solution. By systematically cycling through x, y, and z, we can derive analogous expressions, which were demonstrated in Steps 3 and 4 of the provided solution.
Canonical coordinates and momenta
Canonical coordinates and momenta are fundamental elements in the mathematical formulation of classical mechanics, more specifically in Hamiltonian mechanics. Canonical coordinates, often denoted as qi, represent generalized position coordinates, while canonical momenta, denoted as pi, represent the corresponding generalized momenta, which are derivatives of the Lagrangian with respect to time derivatives of coordinates. These pairs of quantities provide a framework within which the entire state of a mechanical system can be described.

The Poisson bracket itself, an operator used to determine the time evolution of these canonical quantities in the Hamiltonian formulation, is defined in terms of the derivatives with respect to canonical coordinates and momenta. The angular momentum components are derived using these canonical pairs, enabling us to formalize their commutation relationships succinctly and with mathematical rigor.
Rotational transformations
Rotational transformations involve the turning or spinning of a system around a fixed axis, which changes the orientation of the objects or particles within the system. When such a rotation is applied to the components of angular momentum, we can witness the transformational effect of one component generated by the others. This concept is crucial for understanding rotational dynamics and symmetry operations in physics.

Through Poisson brackets, we can assess how a small rotation generated by one angular momentum component affects another. For instance, the result that \(\left[J_{x}, J_{y}\right]=J_{z}\) indicates that a rotation generated by Jx and Jy components has a direct effect on the Jz component. This interconnection highlights the profound symmetry inherent in rotational motion and is a cornerstone of more advanced physical theories that deal with rotational and angular phenomena.

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Most popular questions from this chapter

To prove that the effective potential energy function \(U(\theta)\) of the symmetric top (see \(\S\) 12.4) has only a single minimum, show that the equation \(U(\theta)=E\) can be written as a cubic equation in the variable \(z=\cos \theta\), with three roots in general. Show, however, that \(f(z)\) has the same sign at both \(z=\pm 1\), and hence that there are either two roots or none between these points: for every \(E\) there are at most two values of \(\theta\) for which \(U(\theta)=E\)

Show that the condition that Hamilton's equations remain unchanged under the transformation generated by \(G\) is \(\mathrm{d} G / \mathrm{d} t=0\) even in the case when \(G\) has an explicit time-dependence, in addition to its dependence via \(q(t)\) and \(p(t)\). Proceed as follows. The first set of Hamilton's equations, \((12.6)\), will be unchanged provided that $$\frac{\mathrm{d}}{\mathrm{d} t}\left(\delta q_{\alpha}\right)=\delta\left(\frac{\partial H}{\partial p_{\alpha}}\right)$$ Write both sides of this equation in terms of \(G\) and use \((12.33)\) applied both to \(\partial G / \partial p_{\alpha}\) and to \(G\) itself to show that it is equivalent to the condition $$\frac{\partial}{\partial p_{\alpha}}\left(\frac{\mathrm{d} G}{\mathrm{~d} t}\right)=0$$ Thus \(\mathrm{d} G / \mathrm{d} t\) is independent of each \(p_{\alpha} .\) Similarly, by using the other set of Hamilton's equations, \((12.7)\), show that it is independent of each \(q_{\alpha}\). Thus \(\mathrm{d} G / \mathrm{d} t\) must be a function of \(t\) alone. But since we can always add to \(G\) any function of \(t\) alone without affecting the transformation it generates, this means we can choose it so that \(\mathrm{d} G / \mathrm{d} t=0\).

A light, inextensible string passes over a small pulley and carries a mass \(2 \mathrm{~m}\) on one end. On the other end is a mass \(m\), and beneath it, supported by a spring with spring constant \(k\), a second mass \(m\). Find the Hamiltonian function, using the distance \(x\) of the first mass beneath the pulley, and the extension \(y\) in the spring, as generalized coordinates. Show that \(x\) is ignorable. To what symmetry property does this correspond? (In other words, what operation can be performed on the system without changing its energy?) If the system is released from rest with the spring unextended, find the positions of the particles at any later time.

A particle of mass \(m\) and charge \(q\) is moving around a fixed point charge \(-q^{\prime}\left(q q^{\prime}>0\right)\), and in a uniform magnetic field \(\boldsymbol{B}\). The motion is confined to the plane perpendicular to \(\boldsymbol{B}\). Write down the Lagrangian function in polar co-ordinates rotating with the Larmor angular velocity \(\omega_{\mathrm{L}}=-q B / 2 m\) (see \(\left.\S 5.5\right) .\) Hence find the Hamiltonian function. Show that \(\varphi\) is ignorable, and interpret the conservation law. (Note that \(J_{z}\) is not a constant of the motion.)

A particle of mass \(m\) and charge \(q\) is moving in the equatorial plane \(z=0\) of a magnetic dipole of moment \(\mu\), described (see Appendix A, Problem 12) by a vector potential with the single non-zero component \(A_{\varphi}=\mu_{0} \mu \sin \theta / 4 \pi r^{2}\). Show that it will continue to move in this plane. Initially, it is approaching from a great distance with velocity \(v\) and impact parameter \(b\), whose sign is defined to be that of \(p_{\varphi}\). Show that \(v\) and \(p_{\varphi}\) are constants of the motion, and that the distance of closest approach to the dipole is \(\frac{1}{2}\left(\sqrt{b^{2} \mp a^{2}} \pm b\right)\), according as \(b>a\) or \(b
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