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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

A small cart (mass \(m\) ) is mounted on rails inside a large cart. The two are attached by a spring (force constant \(k\) ) in such a way that the small cart is in equilibrium at the midpoint of the large. The distance of the small cart from its equilibrium is denoted \(x\) and that of the large one from a fixed point on the ground is \(X,\) as shown in Figure \(7.13 .\) The large cart is now forced to oscillate such that \(X=A \cos \omega t,\) with both \(A\) and \(\omega\) fixed. Set up the Lagrangian for the motion of the small cart and show that the Lagrange equation has the form $$\ddot{x}+\omega_{0}^{2} x=B \cos \omega t$$ where \(\omega_{\mathrm{o}}\) is the natural frequency \(\omega_{\mathrm{o}}=\sqrt{k / m}\) and \(B\) is a constant. This is the form assumed in Section 5.5, Equation (5.57), for driven oscillations (except that we are here ignoring damping). Thus the system described here would be one way to realize the motion discussed there. (We could fill the large cart with molasses to provide some damping.)

Consider a double Atwood machine constructed as follows: A mass 4 \(m\) is suspended from a string that passes over a massless pulley on frictionless bearings. The other end of this string supports a second similar pulley, over which passes a second string supporting a mass of \(3 m\) at one end and \(m\) at the other. Using two suitable generalized coordinates, set up the Lagrangian and use the Lagrange equations to find the acceleration of the mass \(4 m\) when the system is released. Explain why the top pulley rotates even though it carries equal weights on each side.

A pendulum is made from a massless spring (force constant \(k\) and unstretched length \(l_{\mathrm{o}}\)) that is suspended at one end from a fixed pivot \(O\) and has a mass \(m\) attached to its other end. The spring can stretch and compress but cannot bend, and the whole system is confined to a single vertical plane. (a) Write down the Lagrangian for the pendulum, using as generalized coordinates the usual angle \(\phi\) and the length \(r\) of the spring. (b) Find the two Lagrange equations of the system and interpret them in terms of Newton's second law, as given in Equation (1.48). (c) The equations of part (b) cannot be solved analytically in general. However, they can be solved for small oscillations. Do this and describe the motion. [Hint: Let \(l\) denote the equilibrium length of the spring with the mass hanging from it and write \(r=l+\epsilon .\) "Small oscillations" involve only small values of \(\epsilon\) and \(\phi,\) so you can use the small-angle approximations and drop from your equations all terms that involve powers of \(\epsilon\) or \(\phi\) (or their derivatives) higher than the first power (also products of \(\epsilon\) and \(\phi\) or their derivatives). This dramatically simplifies and uncouples the equations.]

(a) Write down the Lagrangian for a particle moving in three dimensions under the influence of a conservative central force with potential energy \(U(r),\) using spherical polar coordinates \((r, \theta, \phi)\). (b) Write down the three Lagrange equations and explain their significance in terms of radial acceleration, angular momentum, and so forth. (The \(\theta\) equation is the tricky one, since you will find it implies that the \(\phi\) component of \(\ell\) varies with time, which seems to contradict conservation of angular momentum. Remember, however, that \(\ell_{\phi}\) is the component of \(\ell\) in a variable direction.) (c) Suppose that initially the motion is in the equatorial plane (that is, \(\theta_{0}=\pi / 2\) and \(\dot{\theta}_{0}=0\) ). Describe the subsequent motion. (d) Suppose instead that the initial motion is along a line of longitude (that is, \(\dot{\phi}_{0}=0\) ). Describe the subsequent motion.

Consider two particles moving unconstrained in three dimensions, with potential energy \(U\left(\mathbf{r}_{1}, \mathbf{r}_{2}\right) .\) (a) Write down the six equations of motion obtained by applying Newton's second law to each particle. (b) Write down the Lagrangian \(\mathcal{L}\left(\mathbf{r}_{1}, \mathbf{r}_{2}, \dot{\mathbf{r}}_{1}, \dot{\mathbf{r}}_{2}\right)=T-U\) and show that the six Lagrange equations are the same as the six Newtonian equations of part (a). This establishes the validity of Lagrange's equations in rectangular coordinates, which in turn establishes Hamilton's principle. since the latter is independent of coordinates, this proves Lagrange's equations in any coordinate system.

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