# Lagranian:Generalized momentum

Hi,
I am wondering how to physically interpret the generalized momentum quantity derived from the Euler-Lagrange equations. For some Lagrangians is it equal to the actual momentum for the particle, however i have noticed that for a relativistic particle moving in an electromagnetic field the generalized momentum is not equal to the actual relatvistic momentum.
Could someone explain why this is so, or maybe explain the physical significance of the extra term in the generalized momentum for this EM field case?
Thanks

Ray

dextercioby
Homework Helper
The relativistic particle "couples" or interacts with the free relativistic field and this coupling generates a new term to the momentum. The idea is that the Lagrangian's derivative wrt to the generalized coordinate doesn't have a physical significance, but it has a geometric one.

Daniel.

JK423
Gold Member
I have a question on this.
As the OP says, the generalized momentum (P_g) of a charge in an electromagnetic field is different from its *mechanical* momentum (P_m). Specifically we find that:
P_g=P_m - qA 
where A is the vector potential. This iis the momentum conserved.
However, if we work differently by starting from the Lorentz force, after some calculations (for example following Griffiths), we find that the conserved momentum is:
P_g=P_m + Integral( S dV) 
where:
S: Poynting vector
dV: volume

Equations  and  must be equal, so we conclude:
qA= - Integral( S dV)

Can this be true??

JK423
Gold Member
The apove post has probably a mistake.
So, can anyone help me with the question *what is the physical meaning of the 'qA'* term?