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

Science Advisor
Homework Helper
Insights Author
12,928
504
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
394
7
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 [1]
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) [2]
where:
S: Poynting vector
dV: volume

Equations [1] and [2] must be equal, so we conclude:
qA= - Integral( S dV)


Can this be true??
 

JK423

Gold Member
394
7
The apove post has probably a mistake.
So, can anyone help me with the question *what is the physical meaning of the 'qA'* term?

Thanks in advance!
 
59
0
The generalized momentum have no meaning. Can be the angular momentum, the mechanical momentum or just something. The problem is that the generalized coordinate can have units different from length. Depending the generalized coordinate the generalized momentum change. The generalized momentum can be associated with translation in the configuration space, but the momentum is associated with translation in the actual space of the problem.
qA= - Integral( S dV)?
Is not true. In EM the momentum of the wave is associated with the poynting vector, but the vector potential is associated with the generalized momentum of the charged particle. That's two totally different things. never confuse mechanical momentum and generalized momentum, are two totally different things.

Well i want to ask a question here. If the configuration space is isotropic, with have some quantity that is conserved and we can call it the generalized momentum i suppose. The problem is that the generalized coordinates are not orthogonal in general, then i don't know how to rotate the system in configuration space and also i don't know the form that takes the generalized angular momentum. Normally in the discussion of rotation the different text use orthogonal configuration space, but that not always the case. My question is:

Is possible to define the generalized angular momentum for any configuration space in general?
that mean like a qxp or something like that. Well maybe using forms and tangential space. Maybe is impossible because of the arbitrariness of the configuration space.
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top