Undergrad Time derivative of the angular momentum as a cross product

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SUMMARY

The discussion centers on the time derivative of angular momentum, specifically for a particle with mass m and magnetic moment defined as ##\boldsymbol{\mu} \equiv \gamma \boldsymbol{L}## in a magnetic field ##\boldsymbol{B}##. The Hamiltonian is given by ##H=\frac{p^{2}}{2 m}-\gamma \boldsymbol{L} \cdot \boldsymbol{B}##, leading to the time evolution equation ##\frac{d \boldsymbol{L}}{d t}=-\gamma \boldsymbol{B} \times \boldsymbol{L}##. The identification of the term ##-\gamma \boldsymbol{B}## with angular velocity ##\boldsymbol{\Omega}## is crucial, resulting in the equation ##\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\Omega} \times \boldsymbol{L}##, which is linked to Larmor precession.

PREREQUISITES
  • Understanding of classical mechanics and angular momentum
  • Familiarity with Hamiltonian mechanics
  • Knowledge of magnetic moments and their role in physics
  • Basic concepts of vector calculus, particularly cross products
NEXT STEPS
  • Study the derivation of Larmor precession and its implications in quantum mechanics
  • Explore Hamiltonian dynamics in greater detail, focusing on angular momentum
  • Investigate the relationship between magnetic fields and angular momentum in particle systems
  • Learn about the physical significance of angular velocity in rotational dynamics
USEFUL FOR

Physicists, students of classical mechanics, and anyone interested in the dynamics of angular momentum in magnetic fields will benefit from this discussion.

AndersF
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I am trying to find the equations of motion of the angular momentum ##\boldsymbol L## for a system consisting of a particle of mass ##m## and magnetic moment ##\boldsymbol{\mu} \equiv \gamma \boldsymbol{L}## in a magnetic field ##\boldsymbol B##. The Hamiltonian of the system is therefore

##H=\frac{p^{2}}{2 m}-\gamma \boldsymbol{L} \cdot \boldsymbol{B}##

I have found that the time evolution equation for the angular momentum is

##\frac{d \boldsymbol{L}}{d t}=-\gamma \boldsymbol{B} \times \boldsymbol{L}##

However, the solution identifies the term ##-\gamma \boldsymbol{B}## with the angular velocity ##\boldsymbol{\Omega}##:

##\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\Omega} \times \boldsymbol{L}##

I do not understand what the justification is for making this identification. This last equation looks familiar to me, but I'm not sure where I've seen it... Could someone give me some guidance on this?
 
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I think he introduced a new parameter defined as
\Omega:=-\gamma B.
 
Okay, but does the formula ##\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\Omega} \times \boldsymbol{L}## have any special meaning, being ##\boldsymbol{\Omega}## the angular velocity?
 

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