Week 7 Lex 2: Magnetostatics -II ( Magnetic Vector Potential)

Just as in Electrostatics where the formulation of a scalar potential V ( scalar field)  made the problem of determination of the field simpler because E=-∇ V in Magnetostatics also a Potential formulation is possible. The electrostatic potential formulation for V was a consequence of the fact that ∇ X E =0, and this is ensured as Curl ( Grad V)=0 identically for any scalar field.  For magnetostatics ∇. B=0 implies that the magnetic field B=∇ X A where A is a vector field and called the Magnetic Vector Potential since Div( Curl A)=0 identically for any vector field ( proved in the vector calculus part of the course).


So ∇ X B=∇ X ( ∇ X A)=∇(∇. A) -∇² A=μ₀ J. Notice however that there is an inherent non-uniqueness of the Vector Potential A as A + ∇χ where χ is a Scalar field gives the same magnetic field B=∇ X A + ∇ X (∇ χ) and the second term is identically ZERO because Curl ( Grad χ)=0 for any scalar field χ. This transformation of the vector potential by the gradient of scalar field is called a Gauge Transformation. ( gauge transformations
are at the heart of modern fundamental physics and electromagnetism is an example of the simplest type of gauge theories which describe the 3 forces weak, strong and the electromagnetic


So there are an infinite number of vector potential A differing from each other by a gauge transformation, all of which gives the same magnetic field upon taking the Curl of A due to the reason mentioned above. We can use this non-uniqueness of A to fix the divergence of the vector potential as ∇.A=0 9 once this is fixed the vector potential is unique). This is called a Gauge condition or a Gauge choice. There can be many such gauge choices but we will only consider the condition Div A=0. For this gauge choice then the first term of the RHS of ∇ X B vanishes and we have the equation ∇²A=-μ₀ J.


Notice that this equation ( IN CARTESIAN CO-ORDINATES ONLY ) are actually 3 Poissons equations for the three cartesian components of the vector potential A current density J. These equations are exactly like the Poisson equation in electrostatics for the scalar potential V except that the source term on the RHS is a cuurent density J instead of a charge density ρ. So it is obvious that these equations will have the same solution as that for the electrostatic scalar potential provided we identify ( replace) ρ/ε₀=μ₀J_x,y,z whichever component is appropriate for the RHS. 

Note that this holds only for a Caretesian coordinate system because in a curvilnear co-ordinate system it is easy to see that the original equation do not break up into 3 Poissons equations. So even if you use curvilinear co-ordinates for solving teh integrals for the vector potential A, the current density J must first be expressed in Cartesian coordinates ( Check the DIPA problems and Grifiths examples.)


So if we know the solution to a corresponding electrostatic problem we can simply read off the solution of the magnetostatic problem with the replacement as mentioned above. The trick is this is to identify the correct electrostatic problem. Check out exaples in Griffiths and the DIPA problems.


Magnetostatic Boundary Conditions : The magnetic field B follows certain boundary conditions just like electrostatic fields. These follow just as in the electrostatic case from the integral form of the equations ∇.B=0 and the Amperes Law ∇ X B=mu₀ J . Just as in electrostatics the integral form of these equations evaluated over a infinitesimal Gaussian pillbox and an Amperian loop straddling the boundary shows that the B field has a discontinuity at a surface current just as the electric field has a discontinuity at a surface charge. However in the case of B the tangential component is discontinuous which is perpendicular to the direction of the surface current. The normal component of B is continuous. As the vector potential is chosen to have ∇.A=0 and the magntic flux through an Amperian loop of vanishing width is zero, the vector potential A is continuous across the boundary for both its normal and tangential components. However the normal derivative of the vector potential inherits the discontinuity of B ( note that B comes from a derivative of A which is Curl A).

Magnetic Multipole Expansion: As the Vector Potential A is given by an integral of the current density ( line , surface or volume) similar to the scalar potential V ( which is given by an intergral of the charge density) it is obvious that the Vector potential A will also have a multipole expansion just like the scalar potential by expanding the denominator in a power series. As in the electrostatic case the multipole expansion of the Vector potential is also a series in powers of 1/r over simple current configurations. However as there is no free magnetic poles ( no magnetic monopoles in clasical electromagnetism) the magnetic multipole expansion starts from the pure magnetic dipole term. From the expansion it is possible to determine the vector potential due to a pure magnetic dipole in terms of the magnetic dipole moment m of a pure dipole where m=I a and a is the area of the corresponding current loop. The pure dipole is when the area of the loop goes to zero and the current to infinity keeping m=I a fixed. It is now possible to find the magnetic vector potential due to a small loop at the origin and has a similar form to that of the electric field of a pure electric dipole. Check Griffiths for the field line picture of a pure and a physical magnetic dipole.