prove that divergence of a curl of a vector is zero using stoke’s theorem
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Consider (figure on the left) the volume V enclosed by surface S. Apply the divergence theorem to
function F~ (~r) = ∇ × ~ A~(~r), giving
Z V ∇ · ~ (∇ × ~ A~(~r)) d 3 r = Z S of V (∇ × ~ A~(~r)) · n dA.
ˆ Now slice out and remove from the surface a tiny sliver (figure on the right). Technically we’ve altered S, but this tiny alteration will not affect the value of the surface integral. The edge E of the altered S is the edge of the sliver. Apply the circulation theorem to F~ (~r) = A~(~r), giving
Z S (∇ × ~ A~(~r)) · n dA ˆ = Z E of S A~(~r) · d ~`. But Z E of S A~ · d ~`
= Z P1 to P2 A~ · d ~` + Z P2 to P1 A~ · d ~` = Z P1 to P2 A~ · d ~` − Z P1 to
P2 A~ · d ~` = 0. So for any volume V, Z V ∇ · ~ (∇ × ~ A~(~r)) d 3 r = 0.
Because this holds for any volume, ∇ · ~ (∇ × ~ A~(~r)) = 0.
Z V ∇ · ~ (∇ × ~ A~(~r)) d 3 r = Z S of V (∇ × ~ A~(~r)) · n dA.
ˆ Now slice out and remove from the surface a tiny sliver (figure on the right). Technically we’ve altered S, but this tiny alteration will not affect the value of the surface integral. The edge E of the altered S is the edge of the sliver. Apply the circulation theorem to F~ (~r) = A~(~r), giving
Z S (∇ × ~ A~(~r)) · n dA ˆ = Z E of S A~(~r) · d ~`. But Z E of S A~ · d ~`
= Z P1 to P2 A~ · d ~` + Z P2 to P1 A~ · d ~` = Z P1 to P2 A~ · d ~` − Z P1 to
P2 A~ · d ~` = 0. So for any volume V, Z V ∇ · ~ (∇ × ~ A~(~r)) d 3 r = 0.
Because this holds for any volume, ∇ · ~ (∇ × ~ A~(~r)) = 0.
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