Is the lagrangian always K.E-PE?

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SUMMARY

The Lagrangian is not always represented as the difference between kinetic energy (K.E) and potential energy (PE). In the case of a charged particle in a vector potential, the Lagrangian takes the form L = 1/2MV·V + qV·A - qW, where V is velocity, A is the vector potential, and W is the scalar potential. The Hamiltonian derived from this Lagrangian simplifies to H = 1/2MV·V + qW, indicating that the term qV·A, which represents interaction energy, does not appear in the Hamiltonian due to its nature as a conservative force, resulting in zero work done.

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Is the lagrangian always K.E-PE?
I don't think so?
so WHAT REALLY is the physical significance of the lagrangian?
Superficially it appears to be KE-PE,but the lagrangian for a charged particle in a vector potential is 1/2MV.V +qV.A-qW.
where V=velocity
A=vector potential
W=Scalar potential.
If the hamiltonion for the above lagrangian is calculated it turns out to be 1/2MV.V+qW.
So what does the term eV.A represent?
It appears to be the interaction energy.
But why does it dissapaer in the hamiltonion.
 
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The term qV.A represents the interaction energy between the particle and the vector potential, which is also known as the Lorentz force. It disappears in the Hamiltonian because it is a conservative force and its work done is zero. Therefore, it does not contribute to the energy of the system.
 

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