Demonstrating the randomness of my corner of the universe, I twigged to working on the scalar and vector potentials recently. I'm not sure how much of my library covers them.
Feynman scatters his material across several chapters of Volume II
II-2 Differential Calculus of Vector Fields
II-15 Vector Potential
II-25 Electrodynamics in Relativistic Notation
Carver Mead also discusses it in Collective Electro Dynamics, and there is some information in Griffiths.
But I wasn't quite finding what I was looking for - a demonstration of the theorems from which the scalar and vector potential follow.
The derivation of the scalar and vector potentials rests on two theorems.
∇ ∙ B = 0 ↔ B = ∇ x A
∇ x F = 0 ↔ F = ∇ ψ
The drill in both cases is the same - right to left is completely straightforward, but left to right requires a process for discovering the potential.
The first of these is outlined as a problem in Griffiths (5.30). We can write out the vector B as ( Bx, By, Bz ).
∇ ∙ B = 0
therefore ∂Bx/∂x + ∂By/∂y + ∂Bz/∂z = 0
therefore ∂Bx/∂x = - ( ∂By/∂y + ∂Bz/∂z )
Now A is vector, and we can discover an A that satisfies the equation by discovering (Ax,Ay,Az). From the definition of the cross product, you can deduce that A ∇ (B+C) = A ∇ B + A ∇ C. Using the result from the second theorem, right to left, it follows that B = ∇ ⨯ A implies B = ∇ ⨯ ( A + ∇ ψ ), because the latter addend has no cross product of its own.
Therefore, there's a degree of freedom available, which is used at once to make life easier. Assume that Ax = 0 ! Now the equations of the cross product are reduce to:
Bx = ∂Ay/∂z - ∂Az/∂y
By = - ∂Az/∂x
Bz = ∂Ay/∂x
The latter two, from the Fundamental Theorem of calculus, produce
Ay = ( ∫ Bz(t,y,z) dt ) + C
Az = D - ( ∫ By(t,y,z) dt )
so now we just need to satisfy one more equation. We don't need both constants, so it's easier at this point to take C=0, and solve the first equation.
Bx(x,y,z)
= ∂Az/∂y - ∂Ay/∂z
= ∂D/∂y - ∂/∂y ∫ By(t,y,z) dt - ∂/∂z [∫ Bz(t,y,z) dt]
= ∂D/∂y - ∫ ∂Bz(t,y,z)/∂z dt - ∫ ∂By(t,y,z)/∂y dt
= ∂D/∂y - ∫ ( ∂Bz(t,y,z)/∂z + ∂By(t,y,z)/∂y ) dt
= ∂D/∂y + ∫ ∂Bx(t,y,z)/∂x dt
= ∂D/∂y + Bx(x,y,z) - Bx(0,y,z)
The last two steps relying on B being divergenceless and the Fundamental Theorem of Calculus.
Bx(0,y,z) = ∂D/∂y
D = ∫ Bx(0,t,z) dt
So our candidate solution vector potential is
A = ( 0, ∫ Bz(t,y,z) dt, ∫ Bx(0,t,z) dt - ∫ By(t,y,z) dt )
Calculating ∇ x A demonstrates the first theorem right to left. QED.
The other problem gave me fits - I'm hopeful that this indicates only that I'm terribly out of practice, rather than incapable of discovering these demonstrations on my own - until I tracked down this entry from MathWorld.
∇ x F = 0 ↔ ∂Fy/∂z = ∂Fz/∂y etc.
So let's discover ψ
Fx(x,y,z) = ∂ψ(x,y,z)/∂x
ψ(x,y,z) = ∫ Fx(t,y,z) dt + C(y,z)
now we plug that result into the next restriction
Fy(x,y,z)
= ∂ψ(x,y,z)/∂y
= ∂/∂y [ ψ(x,y,z) ]
= ∂/∂y [ ∫ Fx(t,y,z) dt + C(y,z) ]
= ∫ ∂/∂y Fx(t,y,z) dt + ∂/∂y C(y,z) ]
= ∫ ∂Fx(t,y,z)/∂y dt + ∂C(y,z)/∂y ]
= ∫ ∂Fy(t,y,z)/∂x dt + ∂C(y,z)/∂y ]
= Fy(x,y,z) - Fy(0,y,z) + ∂C(y,z)/∂y
Fy(0,y,z) = ∂C(y,z)/∂y
C(y,z) = ∫ Fy(0,t,z) dt + D(z)
Running through the same sequence starting from Fz(x,y,z) produces
D(z) = ∫ Fz(0,0,t) dt + K
ψ(x,y,z) = ∫ Fx(t,y,z) dt + ∫ Fy(0,t,z) dt + ∫ Fz(0,0,t) dt + K
K is a completely arbitrary constant, unless you've some additional boundary condition to satisfy
June 20, 2005 10:40 PM
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