# Generalized momentum - Physical meaning

1. Jul 6, 2010

### JK423

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?

Last edited: Jul 6, 2010
2. Jul 6, 2010

### Dickfore

It has to do with the Hamiltonian formulation of classical mechanics with velocity dependent potential energies.

3. Jul 6, 2010

### JK423

Yes i know, but what is the physical meaning of the 'qA' term?

4. Jul 6, 2010

### Dickfore

I will give you an answer as soon as you tell me what the physical meaning of $\mathbf{P}$ is.

5. Jul 6, 2010

### JK423

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. Jul 6, 2010

### Dickfore

How is the generalized momentum conserved? I didn't know that.

7. Jul 6, 2010

### JK423

Well yes..
The reason i ask is to confirm that the 'qA' term represent the momentum of fields

8. Jul 6, 2010

### Dickfore

It does not. q is the charge of the particle. the momentum density of the fields is given by $\mathbf{g} = \mathbf{D} \times \mathbf{B}$. Also, what is 'well yes..' supposed to mean?

9. Jul 6, 2010

### JK423

You mean that if we have 2 charges interacting with each other (closed system) then the generalized momentum is not conserved?

10. Jul 6, 2010

### Dickfore

I did not say that. How do you go on finding $\mathbf{A}$ for a system of 2 interacting charges?

11. Jul 6, 2010

### JK423

I don't know and i dont 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.

Last edited: Jul 6, 2010
12. Jul 6, 2010

### |squeezed>

" 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. Jul 6, 2010

### Fernsanz

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 transfered from the charged one to M.

Last edited: Jul 6, 2010
14. Jul 6, 2010

### Fernsanz

I correct myself. This is wrong, let me think it better, lol.

15. Jul 6, 2010

### Dickfore

Maybe quantum mechanics provides the answer:

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

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

Taking the complex conjugate of this equation, we get:

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

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

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

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

$$\frac{\partial}{\partial t} \left(\Psi^{\ast} \, \Psi \right)$$

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:

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

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

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

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

If I can dare to say, it is as if the presence of a vector potential $\mathbf{A}(\mathbf{r})$ 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 $\mathbf{B}$ produces a vorticity of this background velocity field:

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

If we look at the formula for the vorticity of the velocity field of a rigid body rotating with angular velocity $\mathbf{\Omega}$ ($\mathbf{v} = \mathbf{\Omega} \times \mathbf{r} \Rightarrow \mathbf{\omega} = \nabla \times \mathbf{v} = 2 \, \mathbf{\Omega}$), 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 ($\omega_{c} \equiv |q| |B|/m$) 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. Jul 6, 2010

### JK423

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. Jul 6, 2010

### AJ Bentley

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. Jul 6, 2010

### cmos

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. Jul 6, 2010

### AJ Bentley

For electrodynamic momentum alone, neglecting any other form:-

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

When the mass contribution is included you get the form you are discussing.

20. Jul 6, 2010

### JK423

See Griffiths for example.
What do you mean by 'electrodynamic' momentum?

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: