Angular momentum from the Lagrangian

Click For Summary
SUMMARY

The discussion centers on deriving the angular momentum component along the z-axis using the Lagrangian formalism as described in Landau and Lifgarbagez, Vol. 1. The formula for angular momentum is given by Mz = Σa ∂L/∂˙φa, where φ represents the angle of rotation. The proof on page 19 illustrates that applying a finite rotation to an isolated system results in no change in the Lagrangian, confirming the conservation of angular momentum. The discussion emphasizes the connection between infinitesimal rotations and the partial derivatives of the Lagrangian.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with angular momentum concepts
  • Knowledge of the Euler-Lagrange equation
  • Basic grasp of rotational dynamics
NEXT STEPS
  • Study the derivation of angular momentum conservation in Lagrangian mechanics
  • Explore the implications of Lagrangian invariance under rotations
  • Learn about the Euler-Lagrange equation in detail
  • Investigate the relationship between infinitesimal transformations and conservation laws
USEFUL FOR

Students and professionals in physics, particularly those focusing on classical mechanics, Lagrangian dynamics, and conservation laws in rotational systems.

Geofleur
Science Advisor
Gold Member
Messages
426
Reaction score
177
In Landau and Lifgarbagez, Vol. 1, it says "the component of angular momentum along any axis (say the z-axis) can be found by differentiation of the Lagrangian:

M_{z} = \Sigma_{a} \partial L/\partial \dot{\varphi_{a}}

where \varphi is the angle of rotation about the z axis. This is evident from the proof of the law of conservation of momentum..."

They are referring to the proof on pg. 19, where a finite rotation is applied to an arbitrary isolated system, the resulting change in the Lagrangian is required to be zero (from isotropy of space), and the time derivative of the angular momentum is thereby shown to be zero.

My question: How is the former evident from the latter? I've spent quite a while thinking about it and it seems anything but evident right now!
 
Physics news on Phys.org
Geofleur said:
In Landau and Lifgarbagez, Vol. 1, it says "the component of angular momentum along any axis (say the z-axis) can be found by differentiation of the Lagrangian:

M_{z} = \Sigma_{a} \partial L/\partial \dot{\varphi_{a}}

where \varphi is the angle of rotation about the z axis. This is evident from the proof of the law of conservation of momentum..."

They are referring to the proof on pg. 19, where a finite rotation is applied to an arbitrary system, the resulting change in the Lagrangian is required to be zero (from isotropy of space), and the time derivative of the angular momentum is thereby shown to be zero.

My question: How is the former evident from the latter? I've spent quite a while thinking about it and it seems anything but evident right now!

Applying a finite rotation is the same as adding up many many many small rotations. (Infinitesimally) Small rotation is related to the partial derivation:
\frac{\partial L}{\partial \phi_j}
If the Lagrangian is invariant under rotation, then the above partial derivative is zero. Using Euler-Lagrange equation, this means the total time derivative:
\frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}_j} = \frac{\partial L}{\partial \phi_j} = 0
So angular momentum in the j-th coordinate is a conserved quantity.
 
I think I see it now, but I made a mistake earlier - L&L apply an infinitesimal rotation in their derivation of the constancy of angular momentum.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
5
Views
1K
Replies
17
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
67
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K