Constant of motion in velocity dependent motion

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SUMMARY

The discussion centers on the Lagrangian mechanics of a point particle in an electromagnetic field, specifically addressing the constant of motion related to conjugate momenta. The correct Lagrangian for nonrelativistic motion is expressed as L = (m/2) v² - qΦ(x) + (q/c) v · A(x), where Φ represents the scalar potential and A denotes the vector potential. The confusion arises from the combination of scalar and vector potentials in the Lagrangian formulation. The participant successfully clarifies their initial confusion regarding the implications of velocity-dependent potentials.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with electromagnetic theory
  • Knowledge of conjugate momenta in classical mechanics
  • Proficiency in Gaussian (Heaviside-Lorentz) units
NEXT STEPS
  • Study the derivation of the Lagrangian for charged particles in electromagnetic fields
  • Learn about the role of scalar and vector potentials in classical mechanics
  • Explore the concept of conjugate momenta and its applications
  • Investigate the implications of velocity-dependent potentials in Lagrangian dynamics
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This discussion is beneficial for physicists, students of classical mechanics, and anyone interested in the dynamics of charged particles in electromagnetic fields.

vaibhavtewari
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I have little confusion, we know that if Lagrangian is from from variable [itex]\theta[/itex] then conjugate momenta [itex]P_{\theta}[/itex] is a constant of motion. When it comes to velocity dependent potential like [itex]L=1/2mv^2+qv\times B[/itex] how will this differ ?
 
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This cannot be right since you ad a scalar and a vector. The correct Lagrangian for the nonrelativistic motion of a point particle in an external electromagnetic field is given by

[tex]L=\frac{m}{2} \vec{v}^2-q \Phi(\vec{x})+q \frac{\vec{v}}{c} \cdot \vec{A}(\vec{x}).[/tex]

where [itex]\Phi[/itex] is the scalar potential, and [itex]\vec{A}[/itex] is the vector potential of the electromagnetic field. I've used Gaussian (or Heaviside-Lorentz) units.
 
You are right, thanks. I actually resolved what I was asking.
 

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