# About the magnetic vector potential

1. May 1, 2009

### snoopies622

While I understand the mathematical definition of the magnetic vector potential field A ( $$\bf {B} = \nabla \times \bf {A}$$ ), I don't have an intuitive grasp of its physical meaning.

For the (scalar) electric potential the matter seems rather simple. The dimensions of φ are energy per unit charge, and one can use it to easily find the amount of energy it takes to move a charged particle from one electric potential to another by subtracting the difference in the two potentials and multplying that by the charge.

This kind of approach does not work with magnetic potential, however. The dimensions of A are momentum per unit charge, but the change in a charged particle's momentum as it moves from one magnetic potential to another is not equal to its charge multiplied by the difference in the two vector potentials.

To take a simple case, let

$$\bf {A}=(0, 0, y)$$

then

$$\bf {B} = \nabla \times \bf {A} =(1,0,0).$$

A charged particle could move in a circular path around the x axis with constant speed v as long as r = mv/qB, which in this case is mv/q. The difference in its momentum as it moves between (0,0,r) and (0,0,-r) would be (0,-2mv,0), but the A vector is (0,0,0) at both locations - it doesn't change at all.

So my question is, what is the physical meaning of the A field? What does it represent?

Last edited: May 1, 2009
2. May 1, 2009

### jackiefrost

I always thought there was no "real" physical significance, being more a mathematical or computational convenience. But - in Feynman's Lectures, Vol 2 - Lecture 15, R.F. does try to show that under certain circumstances, from the viewpoint of QED, A and the scalar electrical potential phi "becomes more fundamental (than E and B) in a set of equations that replace Maxwell's Equations...". [The quotes were around "real" since even B has kind of ghostly reality to it since it is entirely a motion-related phenomena. I don't know enough QED to go any further about his inference].

3. May 1, 2009

### cabraham

The A vector. known as the magnetic vector potential, represents the direction and magnitude of the magnetic force acting on a dipole. If a loop carries a current I, then A will be a family of closed loops surrounding the wire, inside and outside the loop. A dipole, such as a compass needle, placed in the vicinity of the current carrying wire, would point in a direction consistent with A.

Since there is no known "magnetic monopole charge", an H field does not accelerate a magnetic monopole, as opposed to an E field, which accelerates an electric monopole. In the magnetic domain, we describe the influence an H field exerts on a dipole. Hence we must specify magnitude as well as direction.

As far as electric charges and fields are concerned. The negative of the time derivative of A is equal to the component of E due to induction. If charged particles are present, the other component of E is given by -grad V. Hence, E = -dA/dt - grad V.

A looks like a facsimile of I, the current. A is parallel to I at corresponding tangent points on the A curve and I curve.

Does this clear it up?

Claude

4. May 1, 2009

### dslowik

When you calculate the change in the particles momentum at those two points, and say that: yet A does not change, You imply that momentum of the entire system, particle + fields, is not conserved. But this example is artificial since you do not calculate the fields due to the accelerating particle. You have the same problem if you use E and B vs. the potentials.

The Aharanov-Bohm effect shows that A affects the action/phase even if B is zero, and this can be detected in systems quantum systems like the double slit Feynman shows in fig 15.7. The action/phase is what's more real; E and B come out in describing the extremal phase paths, i.e., the classical paths, i.e., Maxwells eqs. and Lorentz force law.

5. May 1, 2009

### snoopies622

I like these responses. Unfortunately right now I only have time to partially respond to one of them:
In the example I gave, the torque that acts on a magnetic dipole would be the same everywhere (because the magnetic field is uniform) but the A field points in opposite directions depending on which side of the x-z plane the dipole is located, and it is zero on the plane itself.

6. May 1, 2009

### Ben Niehoff

The problem is that $\vec B$ is not a conservative field (its field lines form closed loops), so you can't simply subtract the endpoints of the path. Instead try integrating

$$\int_{t_0}^t \vec r(t') \cdot \vec A(\vec r(t')) \; dt'$$

along the path of the particle. You should find that this function behaves like a sort of generalized potential energy, except that it is path-dependent!

7. May 2, 2009

### snoopies622

The LaTeX in the Preview Post isn't working, but I'll try doing this anyway..

In this case I get

$$\theta = \frac {v}{r} t$$

$$\vec {r} = (0, y, z) = ( 0, sin \theta, cos \theta)$$

$$\vec {A} = (0, 0, y) = (0, 0, sin \theta)$$

$$\vec {r} \cdot \vec {A} = cos \theta sin \theta$$

$$\int _{0} ^ {\pi} cos \theta sin \theta d \theta = | _{0}^{\pi} \frac {1}{2} sin ^2 \theta = 0$$

again. I guess it makes sense that the potential energy of a charged particle moving in a magnetic field is the same at one place as it is at another. Is this what you had in mind?

8. May 2, 2009

### snoopies622

That is true - I didn't think of taking into account the fields created by the particle. But conservation of momentum was not my concern, really. It was only that I didn't see a connection between a change in the A field at the particle's location and the change in the particle's momentum, and I thought perhaps their should be one since the dimensionality of A is momentum per charge.

Last edited: May 2, 2009