Page 479 - Electromagnetics

into (A.61), we find that only the first two terms give non-zero contributions to the
integral and

∂F(r, t) F(r, t) · dS − F(r, t) · dS

ψ(t) = · dS + lim S 2 S 1 . (A.62)
S(t) ∂t t→0 t
The second term on the right can be evaluated with the help of Figure A.5. As the surface
moves through a displacement v t it sweeps out a volume region V that is bounded
on the back by S 1 , on the front by S 2 , and on the side by a surface S 3 = S. We can
thus compute the two surface integrals in (A.62) as the difference between contributions
from the surface enclosing V and the side surface S (remembering that the normal
to S 1 in (A.62) points into V ). Thus

∂F(r, t) S 1 +S 2 + S F(r, t) · dS − S F(r, t) · dS 3

ψ(t) = · dS + lim
∂t t→0 t
S(t)

∂F(r, t) V ∇· F(r, t) dV 3 − S F(r, t) · dS 3

= · dS + lim
S(t) ∂t t→0 t
by the divergence theorem. To compute the integrals over S and V we note from
Figure A.5 that the incremental surface and volume elements are just
dS 3 = dl × (v t), dV 3 = (v t) · dS.
Then, since F · [dl × (v t)] = t(v × F) · dl, we have

∂F(r, t) S(t) t [v × F(r, t)] · dl
t [v∇· F(r, t)] · dS
ψ(t) = · dS + lim − lim .
S(t) ∂t t→0 t t→0 t
Taking the limit and using Stokes’s theorem on the last integral we have finally
d ∂F
F · dS = + v∇· F −∇ × (v × F) · dS, (A.63)
dt S(t) S(t) ∂t
which is the Helmholtz transport theorem [190, 43].
In case the surface corresponds to a moving physical material, we may wish to write
the Helmholtz transport theorem in terms of the material derivative. We can set v = u
and use

∇× (u × F) = u(∇· F) − F(∇· u) + (F ·∇)u − (u ·∇)F
and (A.60) to obtain
d DF
F · dS = + F(∇· u) − (F ·∇)u · dS.
dt S(t) S(t) Dt
If S(t) in (A.63) is closed, enclosing a volume region V (t), then

[∇× (v × F)] · dS = ∇· [∇× (v × F)] dV = 0
S(t) V (t)
by the divergence theorem and (B.49). In this case the Helmholtz transport theorem
becomes

d ∂F
F · dS = + v∇· F · dS. (A.64)
dt S(t) S(t) ∂t

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