Generalized momentum - Physical meaning

In summary, the particle's momentum is conserved due to the qA term. The physical meaning of qA is unknown.
  • #1
JK423
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We know the the generalized momentum is
P=mu + qA
Can someone explain to me, what's the physical meaning of the quantity 'qA'?
The particle's momentum that we measure is just 'mu', right?
 
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  • #2
It has to do with the Hamiltonian formulation of classical mechanics with velocity dependent potential energies.
 
  • #3
Yes i know, but what is the physical meaning of the 'qA' term?
 
  • #4
I will give you an answer as soon as you tell me what the physical meaning of [itex]\mathbf{P}[/itex] is.
 
  • #5
Ok, we know that for a free particle the generalized momentum coincides with the *mechanical* momentum 'mu'. That's not true for motion of charge in a EM field due to the 'qA' term. We know that the generalized momentum is the one that is being conserved and not the 'mu'.
'mu' is the particle's momentum that we measure.
What is the 'qA' term?
 
  • #6
How is the generalized momentum conserved? I didn't know that.
 
  • #7
Well yes..
The reason i ask is to confirm that the 'qA' term represent the momentum of fields
 
  • #8
It does not. q is the charge of the particle. the momentum density of the fields is given by [itex]\mathbf{g} = \mathbf{D} \times \mathbf{B}[/itex]. Also, what is 'well yes..' supposed to mean?
 
  • #9
You mean that if we have 2 charges interacting with each other (closed system) then the generalized momentum is not conserved?
 
  • #10
I did not say that. How do you go on finding [itex]\mathbf{A}[/itex] for a system of 2 interacting charges?
 
  • #11
I don't know and i don't care actually.
If you start from the Lorentz force, and you do some calculations you end up to a conservation of momentum law, which shows that it's not the mechanical momentum of the charges that is conserved but the mechanical+field momentum. The field's monentum density is the poynting vector as you mentioned.
In the example of the two charges for example, if you do this procedure and integrate in all space (So that the maxwell tensor term disappears) you will find that the charges' + field's momentum is conserved.
If you do the Lagrangian way, i suppose that you'll find that the generalized momentum is conserved. These two 'total momentums' derived with two different ways should be equal, right? If this is so, then the 'qA' term should be equal to the integral of the poynting vector.
Is this correct?
Thats why i want to know what the physical meaning of the 'qA' term is. It seems to me that it has to do with the field's momentum.
 
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  • #12
" I did not say that. How do you go on finding LaTeX Code: \\mathbf{A} for a system of 2 interacting charges? "
Maybe using the Liénard–Wiechert potentials ?
 
  • #13
JK423 said:
We know the the generalized momentum is
P=mu - qA
Can someone explain to me, what's the physical meaning of the quantity 'qA'?
The particle's momentum that we measure is just 'mu', right?

When it comes to theoretical mechanics it is not always possible to find physical meaning of involved quantities. The Lagrangian for instance: up to now no physical interpretation has been found. It has dimensions of energy, but it is not any meaningful energy of the system.
Let alone generalized moments, which are derivatives of Lagrangian w/r generalized coordinates. What does it mean to divide something with dimensions of energy by something with no dimensions (like radians)?

Some cases will happen to have a definite physical meaning, for example when the generalized momentum coincides with the Newtonian momentum. In general, however, it is not the rule.

Specifically referring to the term qA you are asking about I didn't find any physical meaning in it at first glance because even the vector potential A hasn't physical meaning (unless you are able to find a physical meaning to a quantity whose rotational is the magnetic flux B). Nevertheless if you look at P=mu-qA you could certainly say the qA is the momentum that a moving particle with the same mass (and with no charge in order not to be influenced by A, let's call it M) would have to carry in order to totally stop your charged particle after a collision with M if all the speed is transferred from the charged one to M.
 
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  • #14
Fernsanz said:
Nevertheless if you look at P=mu-qA you could certainly say the qA is the momentum that a moving particle with the same mass (and with no charge in order not to be influenced by A, let's call it M) would have to carry in order to totally stop your charged particle after a collision with M if all the speed is transferred from the charged one to M.

I correct myself. This is wrong, let me think it better, lol.
 
  • #15
Maybe quantum mechanics provides the answer:

The wave equation for a point particle in an external electromagnetic field given by the potentials [itex](\Phi, \mathbf{A})[/itex] is:

[tex]
i \, \hbar \, \frac{\partial \, \Psi}{\partial t} = \frac{1}{2 m} \, \left( \frac{\hbar}{i} \, \nabla - q \, \mathbf{A} \right)^{2} \, \Psi + q \, \Phi \, \Psi
[/tex]

Taking the complex conjugate of this equation, we get:

[tex]
-i \, \hbar \, \frac{\partial \, \Psi^{\ast}}{\partial t} = \frac{1}{2 m} \, \left( -\frac{\hbar}{i} \, \nabla - q \, \mathbf{A} \right)^{2} \, \Psi^{\ast} + q \, \Phi \, \Psi^{\ast}
[/tex]

Multiplying the first one by [itex]\Psi^{\ast}[/itex] and the second one by [itex]\Psi[/itex] and adding them up, we get:

[tex]
i \, \hbar \, \left( \Psi^{\ast} \, \frac{\partial \Psi}{\partial t} + \frac{\partial \Psi^{\ast}}{\partial t} \, \Psi \right) = \frac{1}{2 m} \, \left\{ \Psi^{\ast} \, \left(\frac{\hbar}{i} \, \nabla - q \, \mathbf{A} \right)^{2} \, \Psi - \left[\left(-\frac{\hbar}{i} \, \nabla - q \, \mathbf{A} \right)^{2} \, \Psi^{\ast}\right] \, \Psi \right\}
[/tex]

It is obvious that the term in the large parenthesis on the lhs is:

[tex]
\frac{\partial}{\partial t} \left(\Psi^{\ast} \, \Psi \right)
[/tex]

What is not so obvious, but can be checked by straightforward calculation using nabla calculus rules is that the term in the braces on the rhs is:

[tex]
-\nabla \cdot \left[ \hbar^{2} \, \left(\Psi^{\ast} \, \nabla \Psi - \nabla \Psi^{\ast} \, \Psi \right) + \frac{2 \, q \, \hbar}{i} \, \mathbf{A} \, \Psi^{\ast} \, \Psi \right]
[/tex]

Recalling that [itex]\rho(\mathbf{r}, t) = \Psi^{\ast}(\mathbf{r}, t) \, \Psi(\mathbf{r}, t)[/itex] is the probability density for finding the particle at position [itex]\mathbf{r}[/itex] at time [itex]t[/itex], we see that the probability current density for a particle in an external electromagnetic field is given by:

[tex]
\mathbf{J} = \frac{\hbar}{2 m i} \, \left(\Psi^{\ast} \, \nabla \Psi - \nabla \Psi^{\ast} \, \Psi \right) - \frac{q}{m} \, \mathbf{A} \, \Psi^{\ast} \, \Psi
[/tex]

The first term is present even in the absence of an electromagnetic field. The second term is of the form [itex]\mathbf{J} = \rho \, \mathbf{v}[/itex], where [itex]\mathbf{v} = -q \, \mathbf{A}/m[/itex].

If I can dare to say, it is as if the presence of a vector potential [itex]\mathbf{A}(\mathbf{r})[/itex] leads to dragging of charged particles with the background velocity that depends both on their charge and on their mass. The presence of a magnetic field [itex]\mathbf{B}[/itex] produces a vorticity of this background velocity field:

[tex]
\mathbf{\omega} \equiv \nabla \times \mathbf{v} = -\frac{q}{m} \, \left(\nabla \times \mathbf{A}\right) = -\frac{q \, \mathbf{B}}{m}
[/tex]

If we look at the formula for the vorticity of the velocity field of a rigid body rotating with angular velocity [itex]\mathbf{\Omega}[/itex] ([itex]\mathbf{v} = \mathbf{\Omega} \times \mathbf{r} \Rightarrow \mathbf{\omega} = \nabla \times \mathbf{v} = 2 \, \mathbf{\Omega}[/itex]), we see that the vorticity of the "dragging" velocity field for the particle in a magnetic field is the same as if there is a rotation with half the cyclotron frequency of the particle ([itex]\omega_{c} \equiv |q| |B|/m [/itex]) and with the direction of the angular velocity in the opposite (same) direction as the direction of the magnetic field if the particle is positive (negative).
 
  • #16
What, now we are counting on Quantum physics to interpret Classical physics?? Hahah ;)
Thx Dickfore, i'll take a look at it later (im in my exams-period :S)
 
  • #17
JK423 said:
We know the the generalized momentum is
P=mu - qA
Can someone explain to me, what's the physical meaning of the quantity 'qA'?
The particle's momentum that we measure is just 'mu', right?

Where did you find that?
It's the very thing I'm working on at the moment!
A is the vector potential. qA is electrodynamic momentum.
 
  • #18
I'm not sure if the following statement will be satisfactory as an "interpretation," but let me give it a shot. I see the 'qA' term as being the contribution to the momentum due to the interaction between a charged particle and the electromagnetic field. This is in contrast to the 'mu' term, which is the contribution to the momentum due to the movement of the massive particle. The vector sum of the two quantities thus being the actual (canonical) momentum.

As an example, consider a stationary, charged particle in an purely magnetic field. I do this so that there is nothing (e.g. an electric field) to perturb the particle into motion. We now have a closed system. So, even though the particle is not in motion (mu=0), we still have a non-zero (canonical) momentum P=qA. Only in the limit of zero charge do we retrieve the more intuitive result from kinematics, P=0 for a stationary particle.

The following is how I came up with my interpretation. It's not necessarily a complete argument because I don't really conclude or wrap anything up in the end. Hopefully it makes sense though:
If you look at the Lagrangian for a charged particle in an electromagnetic field, then you will see that two terms appear that can be, together, interpreted as the potential energy of the system: 'qV' and 'qu.A'; where 'V' is the electric potential. It is the latter term that gives rise to the 'qA' term in the canonical momentum.
 
  • #19
For electrodynamic momentum alone, neglecting any other form:-

[tex]
Phase \ \psi = \int(\omega_{1} - \omega_{2})dt = \oint k.dl
[/tex]
Potential difference is energy(=frequency) difference so.
[tex]
Flux \ \phi = \int Vdt = \oint A.dl
[/tex]

When the mass contribution is included you get the form you are discussing.
 
  • #20
AJ Bentley said:
Where did you find that?
It's the very thing I'm working on at the moment!
A is the vector potential. qA is electrodynamic momentum.
See Griffiths for example.
What do you mean by 'electrodynamic' momentum?

cmos said:
I'm not sure if the following statement will be satisfactory as an "interpretation," but let me give it a shot. I see the 'qA' term as being the contribution to the momentum due to the interaction between a charged particle and the electromagnetic field. This is in contrast to the 'mu' term, which is the contribution to the momentum due to the movement of the massive particle. The vector sum of the two quantities thus being the actual (canonical) momentum.

As an example, consider a stationary, charged particle in an purely magnetic field. I do this so that there is nothing (e.g. an electric field) to perturb the particle into motion. We now have a closed system. So, even though the particle is not in motion (mu=0), we still have a non-zero (canonical) momentum P=qA. Only in the limit of zero charge do we retrieve the more intuitive result from kinematics, P=0 for a stationary particle.
Something seems to be wrong with this interpretation.
Take the Lorentz force. It's a differential equation actually. Solve it, and find the velocity of the particle. Now you see that in order to find the velocity, you need to insert the electric and magnetic field in the equation. Which means all the interaction of the charge with the EM field is accounted in the function of the velocity (u). So, the term 'mu' contains all the interaction with the EM field, meaning 'how the particle is going to move in space'.
So i think that you cannot interpret it that way.

Actually i think that i found some solution to this interpretation problem but i cannot work on it right now due to the exams-period I'm in.
There are two ways of 'treating' the equation of Lorentz force.
The first is to substitute all the fields E, B with their potentials V,A. Doing this you will find:
d/dt(mu+qA) = -grad(qV-qu.A) [1]
There is another way, where we substitute the charge 'q' with the charge density ρ, placing a volume integral, and we're working with the fields only substituting charges and currents, ρ and J, with their fields E,B. Doing so we will find:
d(mu)/dt = -d/dt (volume integral of poynting vector) + (surface in integral of Maxwell tensor). [2]
The quantity '(volume integral of poynting vector)' represents the momentum stored in the fields E,B.
In order to give a physical meaning to the 'qA' term, it seems to me that we must do the following. We start from 'qA', and replace q-->ρ+volume integral. Then, we substitute ρ with E (1st Maxwell equation), and try to bring this equation to a form that we can compare it to equation [2]. We'll see what is the connection between this term and the momentum stored in the fields.
 
  • #21
I presume we cross-posted.
Following on from my last.

Total momentum is :-

[tex]\hbar k = p = p_{el} + p_{mv} = qA + mv[/tex]

So v is a direct measure of the imbalance between total momentum hk and electrodynamic qA

Current density is just the movement of N charges per volume : J = qnV

So we can write this in terms of wave vector k, vector potential A and the current density J

[tex]J = \frac{qN}{m} ( \hbar k - qA)[/tex]

This is all from Prof. Mead 'Collective electrodynamics'

My problem here is the dimensions qA has the units of angular momentum - the mass momentum does not. I'm struggling to reconcile that.
 
  • #22
Interesting discussion.

AJ Bentley said:
Total momentum is :-

[tex]\hbar k = p = p_{el} + p_{mv} = qA + mv[/tex]

That equation applies only to a single particle, doesn't it? The [tex]\hbar k[/tex] doesn't and can't represent a many-body system because the individual waves for each particle will in general not have a simple or a fixed relationship between each other in terms of frequency and phase, right?
AJ Bentley said:
So v is a direct measure of the imbalance between total momentum hk and electrodynamic qA

Current density is just the movement of N charges per volume : J = qnV

So we can write this in terms of wave vector k, vector potential A and the current density J

[tex]J = \frac{qN}{m} ( \hbar k - qA)[/tex]

This is all from Prof. Mead 'Collective electrodynamics'

That's a nice derivation but I have to believe it doesn't quite work out because of the misuse of the [tex]\hbar k[/tex] factor except in situations where a stream of a single type of particle flows extremely coherently.
 
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  • #23
PhilDSP said:
That equation applies only to a single particle, doesn't it?

Repeat of original post deleted.

EDIT

I should have read my own post more carefully.

Yes, this is a very coherent system. It's electron flow in a semiconductor.
 
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  • #24
AJ Bentley said:
I presume we cross-posted.
Following on from my last.

Total momentum is :-

[tex]\hbar k = p = p_{el} + p_{mv} = qA + mv[/tex]

So v is a direct measure of the imbalance between total momentum hk and electrodynamic qA

Current density is just the movement of N charges per volume : J = qnV

So we can write this in terms of wave vector k, vector potential A and the current density J

[tex]J = \frac{qN}{m} ( \hbar k - qA)[/tex]

This is all from Prof. Mead 'Collective electrodynamics'

My problem here is the dimensions qA has the units of angular momentum - the mass momentum does not. I'm struggling to reconcile that.

I came up with the correct units of momentum. This is one thing I also wished to resolve in Mead's development. The second issue is that the momentum is a scalar. For two axially aligned superconducting rings, P, or at least dP/dt, should be the vector component of electrodynamic momentum aligned with the mutual axis.
 
  • #25
The electromagnetic momentum is interesting but not Lorentz covariant, and not manifestly covariant (this means valid in curved spacetime) as is the rest of electromagnetism expressed in differential p-forms.

The magnetic potential, A is missing the electric potential, phi and the charge, q is missing its other half, electric current.

So qA begs to be generalized. qA should only show up by itself as a result of experimental constraints.

Now, Carver Meade's use of qA seems to develop by implicity integrating over the volume of a superconducting ring, so it may be reasonable to look for covariant forms containing rho times A, where rho is the charge density.

Amu = (phi, Ai)
Amu = (-phi, Ai)

Jmu = (rho, Ai)
Jmu = (-rho, Ai)

Latin is used to subscript the spatial dimensions (1,2,3). Greek subscripts all of spacetime, where zero indexes time (0,1,2,3).

So the idea is to find diffenent ways to combine the one-forms Amu and Jmu to obtain another differential p-forms.

An interesting one is

*(*Jmu wedge_product Anu) = Ji A_i - rho phi​

The Hodge star operator (*) in front of Jmu turns the one-form J into a 3-form. The leading star operator turns the four-form (*Jmu wedge_product Anu) into a scalar density (a zero-form). But the units are not momentum nor momentum density. It is a scalar density with units of energy per unit spatial volume.

A better candidate with units of momentum per unit spatial volume is

Jmu wedge_product Anu = - rho A + J phi + J x A

Integrating over a spatial volume element, rho A becomes qA. The bold type indicates a spatial vector. I.e.: Ai = A.The result is a type (0,2) tensor density. Without normalizing units, three of the independent terms are momentum density and the other three are energy density.
 
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  • #26
I've been obtuse.

q (phi, A) is a perfectly well behaved vector on Minkowski space time.
 

FAQ: Generalized momentum - Physical meaning

What is generalized momentum and how is it different from regular momentum?

Generalized momentum is a concept in physics that extends the idea of regular momentum to include systems with more complex dynamics, such as those with constraints and non-conservative forces. It is defined as the derivative of the Lagrangian with respect to the generalized coordinates, while regular momentum is defined as the product of an object's mass and its velocity.

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