Physics Friday 122

Part 2: The Vector Potential and Gauge Transformation.

In electrostatics, one often works not with the electric field, but the electric potential. Since the electric field is irrotational in electrostatics ( for all space), it must be the gradient of some scalar function. We choose the negative of this function as the potential, so that ; note that the potential is unique only up to a constant.

For magnetostatics, we have


,

Here, we use the first equation; since the magnetic field is divergence-free, it must be the curl of some vector field, which we call the vector potential:
. In fact, last week, we saw that , already showing that the magnetic field is the curl of a vector field, specifically .
However, just as the scalar electric potential may be transformed by addition of an arbitrary constant without changing the physics, the vector potential is not unique. Noting that the curl of the gradient of any scalar function is identically zero, adding the gradient of an arbitrary scalar field to the vector potential produces the same physics; this transformation is called a gauge transformation. Thus, our general form for the vector potential is
.

Substituting the definition of A into the equation for the curl of the field, we get
,
and using the identity for curl of a curl, we see
.
Thanks to the freedom of gauge transformation, we can choose a scalar φ so that . This is known as the Coulomb gauge. With this choice, we see the vector potential satisfies the (vector) Poisson equation .
Taking the divergence of
,
we see that the Coulomb gauge reduces to , since the integral term has zero divergence. For an unbounded space and the absence of sources at infinity, this reduces to a constant φ, which in turn gives us
.

Also of note, consider an orientable surface S bounded by a simple closed curve C, oriented via the right-hand rule. Then the magnetic flux through S is ; in terms of the vector potential, this is
;
and by Stokes’ theorem, the latter integral can be turned into an integral over the bounding curve C:
;
and the magnetic flux through a simple loop is equal to the line integral of the vector potential over the loop.

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3 Responses to “Physics Friday 122”

  1. Physics Friday 123 « Twisted One 151's Weblog Says:

    […] is of the form of a Dirac delta ; thus , for a constant C (dependent on n and I). Thus, from last week, we see . Now, by our cylindrical symmetry, we can choose the point at which we measure the […]

  2. Physics Friday 128 « Twisted One 151's Weblog Says:

    […] rather than the usual dA to avoid confusion with the magnetic vector potential A). Now, as we noted here, the definition of the vector potential combined with the Kelvin-Stokes theorem tell us that , […]

  3. Physics Friday 130 « Twisted One 151's Weblog Says:

    […] field is static, , and the above reduces to our electrostatic potential. Recall that when we first discussed the vector potential, it was noted that the vector potential is not unique, and that adding the gradient of any scalar […]

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