Angular momentum from the Lagrangian

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The discussion centers on deriving the angular momentum component along an axis from the Lagrangian, specifically referencing Landau and Lifgarbagez. It explains that the angular momentum, M_z, can be determined by differentiating the Lagrangian with respect to the angular velocity. The proof involves applying a finite rotation to an isolated system, demonstrating that the change in the Lagrangian remains zero due to the isotropy of space, leading to the conclusion that the time derivative of angular momentum is also zero. The clarification emphasizes that the derivation relies on the invariance of the Lagrangian under infinitesimal rotations, linking conservation laws to symmetries in physics. This understanding resolves the initial confusion about the relationship between the Lagrangian's properties and angular momentum conservation.
Geofleur
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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!
 
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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.
 

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