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Noether's theorem asserts a connection between invariance principles and conservation laws. In Section 7.8 we saw that translational invariance of the Lagrangian implies conservation of total linear momentum. Here you will prove that rotational invariance of \(\mathcal{L}\) implies conservation of total angular momentum. Suppose that the Lagrangian of an \(N\) -particle system is unchanged by rotations about a certain symmetry axis. (a) Without loss of generality, take this axis to be the \(z\) axis, and show that the Lagrangian is unchanged when all of the particles are simultaneously moved from \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}\right)\) to \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}+\epsilon\right)\) (same \(\epsilon\) for all particles). Hence show that $$\sum_{\alpha=1}^{N} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}}=0.$$ (b) Use Lagrange's equations to show that this implies that the total angular momentum \(L_{z}\) about the symmetry axis is constant. In particular, if the Lagrangian is invariant under rotations about all axes, then all components of \(\mathbf{L}\) are conserved.

Short Answer

Expert verified
Rotational invariance implies \( \sum_{\alpha=1}^{N} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}}=0 \) and conservation of angular momentum \( L_{z} \).

Step by step solution

01

Establish the Condition of Rotational Invariance

We start by using the given condition that the Lagrangian \( \mathcal{L} \) is invariant under the transformation of the azimuthal angle \( \phi_{\alpha} \) for each particle about the \( z \) axis by some small angle \( \epsilon \). This means that if we rotate all the coordinates \( \phi_{\alpha} \) to \( \phi_{\alpha} + \epsilon \), the Lagrangian remains unchanged: \( \mathcal{L}(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}) = \mathcal{L}(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha} + \epsilon) \).
02

Apply Taylor's Expansion to the Lagrangian

Given that \( \mathcal{L} \) is unchanged, we can apply a first order Taylor expansion on \( \mathcal{L} \) around \( \phi_{\alpha} \). The expansion gives: \( \mathcal{L}(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha} + \epsilon) = \mathcal{L}(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}) + \epsilon \sum_{\alpha} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}}(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}) + \mathcal{O}(\epsilon^2) \). Since \( \mathcal{L} \) does not change, \( \epsilon \sum_{\alpha} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}} = 0 \).
03

Conclude the Result from Small Angle Rotation

As this result holds for arbitrary small \( \epsilon \), it implies \( \sum_{\alpha=1}^{N} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}} = 0 \). This confirms that the partial derivatives with respect to the azimuthal angles sum to zero due to the rotational invariance.
04

Analyze Using Lagrange's Equations

Lagrange's equations tell us \( \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_{\alpha}} \right) - \frac{\partial \mathcal{L}}{\partial q_{\alpha}} = 0 \). For \( q_{\alpha} = \phi_{\alpha} \), this becomes \( \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{\phi}_{\alpha}} \right) = \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}} \). Using the result from Step 3, we have: \( \frac{d}{dt} \left( \sum_{\alpha} \frac{\partial \mathcal{L}}{\partial \dot{\phi}_{\alpha}} \right) = 0 \).
05

Relate to Conservation of Angular Momentum

The term \( \frac{\partial \mathcal{L}}{\partial \dot{\phi}_{\alpha}} \) is the definition of the component of angular momentum \( L_{z,\alpha} \) for each particle. Therefore, \( \sum_{\alpha=1}^{N} L_{z,\alpha} = L_{z} \) is constant. Thus, if \( \mathcal{L} \) is invariant under such rotations, then the total angular momentum \( L_{z} \) is conserved. Similarly, if \( \mathcal{L} \) is invariant under rotations about all axes, all components of \( \mathbf{L} \) are conserved.

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Key Concepts

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

Conservation Laws
Conservation laws are fundamental principles in physics stating that certain quantities remain constant throughout the evolution of a system. These laws arise from the symmetries found in the physical world. For instance, one of the most familiar conservation laws is the conservation of energy. It tells us that the total energy in an isolated system remains consistent across time.

Noether's theorem beautifully ties together symmetries and conservation laws. It highlights how each symmetry in the laws of physics corresponds to a conserved quantity. When a system is unchanged under a specific transformation, a conservation law emerges.
  • Translational Symmetry: Conservation of linear momentum.
  • Rotational Symmetry: Conservation of angular momentum.
  • Time Symmetry: Conservation of energy.
These connections are profound, offering a unified framework to understand the behavior of systems under different conditions.
Lagrangian Mechanics
Lagrangian mechanics is a reformulation of classical mechanics focusing on a function called the Lagrangian, denoted as \( \mathcal{L} \). This function summarizes the dynamics of the system using the difference between kinetic energy \( T \) and potential energy \( V \), defined as \( \mathcal{L} = T - V \).

In contrast to Newtonian mechanics, which uses forces, the Lagrangian framework uses energies to describe motion. This approach often simplifies the analysis of complex systems, particularly those involving constraints and symmetries.
  • Generalized Coordinates: Lagrangian mechanics operates using generalized coordinates \( q_i \), which can represent any parameter describing the system.
  • Formulating Equations: Lagrange's equations, derived from the principle of stationary action, provide the equations of motion:
  • \[ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \right) - \frac{\partial \mathcal{L}}{\partial q_i} = 0 \]
These equations allow us to solve for the path a system follows, considering both forces and constraints seamlessly.
Rotational Invariance
Rotational invariance refers to a scenario in which a system's physical properties don't change under rotations about a particular axis. This concept is crucial in understanding why conservation laws, like conservation of angular momentum, exist.

When a system's Lagrangian remains unchanged under a small angular displacement, it suggests that there is a symmetry. For this exercise, considering the system's invariance around the \( z \) axis tells us crucial information about the dynamics of the system.
  • Azimuthal Angles: In a rotationally invariant system, all particles' azimuthal coordinates \( \phi_\alpha \) shifted uniformly do not impact the Lagrangian.
  • Implication of Symmetry: This invariance indicates that a certain quantity related to this symmetry remains consistent over time.
Such symmetries form the foundation for understanding sophisticated systems in physics, from atomic structures to galaxies.
Angular Momentum Conservation
Angular momentum conservation is an essential principle in physics. It implies that the angular momentum of a system remains constant if no external torque acts on it. Thanks to Noether's theorem, angular momentum conservation stems from the system's rotational symmetry.

Consider an \( N \)-particle system. If the Lagrangian is unchanged under a rotation, the angular momentum about that axis does not fluctuate, mathematically confirming:
\[ \sum_{\alpha=1}^{N} L_{z,\alpha} = L_z \text{ is constant.} \]

  • Components of Angular Momentum: When analyzing angular momentum along different axes, each contributes uniquely to the system's state.
  • Conservation Significance: Angular momentum's constancy provides insights into the behavior of rotating objects, critical for studies ranging from particle dynamics to celestial mechanics.
Overall, this conservation law explains why objects maintain their spin or rotational motion in various environments.

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Most popular questions from this chapter

The center of a long frictionless rod is pivoted at the origin, and the rod is forced to rotate in a horizontal plane with constant angular velocity \(\omega\). Write down the Lagrangian for a bead threaded on the rod, using \(r\) as your generalized coordinate, where \(r, \phi\) are the polar coordinates of the bead. (Notice that \(\phi\) is not an independent variable since it is fixed by the rotation of the rod to be \(\phi=\omega t\).) Solve Lagrange's equation for \(r(t) .\) What happens if the bead is initially at rest at the origin? If it is released from any point \(r_{\mathrm{o}}>0,\) show that \(r(t)\) eventually grows exponentially. Explain your results in terms of the centrifugal force \(m \omega^{2} r\).

A mass \(m_{1}\) rests on a frictionless horizontal table. Attached to it is a string which runs horizontally to the edge of the table, where it passes over a frictionless, small pulley and down to where it supports a mass \(m_{2} .\) Use as coordinates \(x\) and \(y\) the distances of \(m_{1}\) and \(m_{2}\) from the pulley. These satisfy the constraint equation \(f(x, y)=x+y=\) const. Write down the two modified Lagrange equations and solve them (together with the constraint equation) for \(\ddot{x}, \ddot{y},\) and the Lagrange multiplier \(\lambda\). Use (7.122) (and the corresponding equation in \(y\) ) to find the tension forces on the two masses. Verify your answers by solving the problem by the elementary Newtonian approach.

Using the usual angle \(\phi\) as generalized coordinate, write down the Lagrangian for a simple pendulum of length \(l\) suspended from the ceiling of an elevator that is accelerating upward with constant acceleration \(a\). (Be careful when writing \(T\); it is probably safest to write the bob's velocity in component form.) Find the Lagrange equation of motion and show that it is the same as that for a normal, nonaccelerating pendulum, except that \(g\) has been replaced by \(g+a\). In particular, the angular frequency of small oscillations is \(\sqrt{(g+a) / l}\).

Consider the well-known problem of a cart of mass \(m\) moving along the \(x\) axis attached to a spring (force constant \(k\) ), whose other end is held fixed (Figure 5.2 ). If we ignore the mass of the spring (as we almost always do) then we know that the cart executes simple harmonic motion with angular frequency \(\omega=\sqrt{k / m} .\) Using the Lagrangian approach, you can find the effect of the spring's mass \(M,\) as follows: (a) Assuming that the spring is uniform and stretches uniformly, show that its kinetic energy is \(\frac{1}{6} M \dot{x}^{2} .\) (As usual \(x\) is the extension of the spring from its equilibrium length.) Write down the Lagrangian for the system of cart plus spring. (Note: The potential energy is still \(\frac{1}{2} k x^{2}\).) (b) Write down the Lagrange equation and show that the cart still executes SHM but with angular frequency \(\omega=\sqrt{k /(m+M / 3)} ;\) that is, the effect of the spring's mass \(M\) is just to add \(M / 3\) to the mass of the cart.

Write down the Lagrangian for a one-dimensional particle moving along the \(x\) axis and subject to a force \(F=-k x\) (with \(k\) positive). Find the Lagrange equation of motion and solve it.

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