Jackson proof of eq 6_5
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A brief Word note by Phil, dated 1.22.03, deriving equation (6.5) on page 172 of Jackson's electrodynamics text. It considers the total derivative of a surface integral of a vector field over a contour moving with velocity v, expands A(r+vdt) to first order, and applies the curl identity for constant v with div B = 0. Stokes' theorem then converts the area integral to the line integral term. Equations are missing from the extracted text.
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Proof of (6.5) Jackson PhL 1.22.03
Imagine a contour that is moving in a direction v. Imagine we are interested in the area integral over a surface bounded by this contour of some vector field A,
I =
If we want the total derivative of I, we get one contribution from dA/dt, but we get another one because the contour moves. The first term is obvious. The second term can be written like this:
[ - ] /dt
If you draw a picture of the contour moving in direction v, you see that the values of A that are picked up at time t+dt were the values that A had at t but at a distance shifted by v dt. This is just because the entire contour and its attached surface has shifted by v dt. Thus we can rewrite the above as:
[ - ] /dt =
where this A first term expansion used here is obvious if you just write out the components.
A(r+a) = A(r) + (a)A(r) + ...
Now we have this little vector identity,
x A x v = (v)A - v(A)
where v is a constant so all gradients on it vanish, killing two of the usual terms in this identity. Now, if A happens to be B, the magnetic field, we know that B = 0 so we ignore the second term. So we rewrite our term as follows:
=
But Stokes lets us relate an area integral like this with a curl to line integral
=
and this then is the rightmost term in Jackson's equation (6.5) on page 172.