Circulation around a small circle
Let Cε be the circle which
◦ is centered on ~r0
◦ has radius ε
◦ lies in the plane through ~r0 perpendicular to n̂
◦ is oriented in the standard way with respect to n̂. Imagine standing on the circle
with your feet on the plane through ~r0 perpendicular to n̂, with the vector from your
feet to your head in the same direction as n̂ and with your left arm point towards
~r0 . Then your are facing in the positive direction for Cε .
We shall show that
I
~ × v(~r0 ) · n̂ + O(ε3 )
~v (~r) · d~r = πε2 ∇
Cε
To do so, pick any three vectors ıˆ′ , ˆ′ , k̂′ such that
◦ k̂′ = n̂
◦ ıˆ′ ⊥ n̂, ˆ′ ⊥ n̂
◦ ıˆ′ × ˆ′ = k̂′
Then
~r(t) = ~r0 + ε cos t ıˆ′ + ε sin t ˆ′
is a parametrization of Cε . Note in particular that, for all t, ~r(t) lies in the plane through ~r0
perpendicular to n̂ and k~r(t) − ~r0 k = ε. So
I
~v (~r) · d~r =
Cε
=ε
Z
2π
0
h
Z
2π
0
~v ~r0 + ε cos t ıˆ′ + ε sin t ˆ′ · − ε sin t ıˆ′ + ε cos t ˆ′ dt
i
− sin t ıˆ′ · ~v ~r0 + ε cos t ıˆ′ + ε sin t ˆ′ + cos t ˆ′ · ~v ~r0 + ε cos t ıˆ′ + ε sin t ˆ′ dt
Denote f (~r) = ıˆ′ · ~v (~r). Now Taylor expand F (ε) = f ~r0 + ε cos t ıˆ′ + ε sin t ˆ′ about ε = 0.
F (ε) = F (0) + F ′ (0)ε + O(ε2 )
~ ~r0 + ε sin t ˆ′ · ∇f
~ ~r0 + O(ε2 )
= f ~r0 + ε cos t ıˆ′ · ∇f
~ ıˆ′ · ~v ) ~r0 + ε sin t ˆ′ · ∇(
~ ıˆ′ · ~v) ~r0 + O(ε2 )
= ıˆ′ · ~v ~r0 + ε cos t ıˆ′ · ∇(
Similarly, if G(ε) = ˆ′ · ~v ~r0 + ε cos t ıˆ′ + ε sin t ˆ′ ,
~ ˆ′ · ~v ) ~r0 + ε sin t ˆ′ · ∇(
~ ˆ′ · ~v ) ~r0 + O(ε2 )
G(ε) = ˆ′ · ~v ~r0 + ε cos t ıˆ′ · ∇(
c Joel Feldman.
2003. All rights reserved.
1
Hence, the integrand
− sin t ıˆ′ · ~v ~r0 + ε cos t ıˆ′ + ε sin t ˆ′ + cos t ˆ′ · ~v ~r0 + ε cos t ıˆ′ + ε sin t ˆ′
~ ıˆ′ · ~v ) ~r0 − ε sin2 t ˆ′ · ∇(
~ ıˆ′ · ~v ) ~r0
= − sin t ıˆ′ · ~v ~r0 − ε sin t cos t ıˆ′ · ∇(
~ ˆ′ · ~v ) ~r0 + ε sin t cos t ˆ′ · ∇(
~ ˆ′ · ~v ) ~r0 + O(ε2 )
+ cos t ˆ′ · ~v ~r0 + ε cos2 t ıˆ′ · ∇(
Since
Z 2π
sin t dt =
2π
cos t dt
0
0
we have
Z
I
~v (~r) · d~r = πε
2π
Z
2π
0
h
i
′
′
′
′
ˆ
~
ˆ
ˆ
~
ˆ
−  · ∇(ı · ~v ) ~r0 + ı · ∇( · ~v ) ~r0 + O(ε3 )
2
Cε
2π
Z
Z
sin t cos t dt = 0
2
sin t dt =
ıˆ′ = −k̂′ × ˆ′
ˆ′ = k̂′ × ıˆ′
0
0
Sub in
~ =
∇
cos2 t dt = π
3
X
∂
ı̂ın ∂x
n
n=1
where I have renamed the standard basis for IR3 from ı̂ı, ̂, k̂ to ı̂ı1 ,ı̂ı2 ı̂ı3 and the standard
coordinates on IR3 from x, y, z to x1 , x2 , x3 . Then
3
3
X
X
ˆ′ ·~
v
∂ ıˆ′ ·~
v
′
′
′
′
′
ˆ
~
ˆ
ˆ
~
ˆ
ˆ
~
− · ∇(ı · ~v ) ~r0 + ı · ∇( · ~v) ~r0 =
− · ı̂ın ∂xn ~r0 +
r
ıˆ′ · ı̂ın ∂∂x
0
n
n=1
=
=
=
n=1
3
X
ıˆ′ ·~
v
~
−(k̂′ × ıˆ′ ) · ı̂ın ∂∂x
r
+
0
n
n=1
ıˆ′ ·~
v
~r0 +
−k̂′ · (ıˆ′ × ı̂ın ) ∂∂x
n
3
X
3
X
′
ıˆ′ ·~
v
ˆ′ ) ∂ˆ′ ·~v ~r0
~
k̂
·
(ı̂
ı
×

r
+
k̂′ · (ı̂ın × ıˆ′ ) ∂∂x
n
0
∂x
n
n
n=1
3
X
n=1
3
X
n=1
3
X
ˆ′ ·~
v
~r0
−(k̂′ × ˆ′ ) · ı̂ın ∂∂x
n
ˆ′ ·~
v
~r0
−k̂′ · (ˆ′ × ı̂ın ) ∂∂x
n
n=1
n=1
Since k̂′ ⊥ ı̂ın × k̂′ ), k̂′ · (ı̂ın × k̂′ ) = 0 and
3
X
′
′
′
′
~
~
ˆ
ˆ
ˆ
ˆ
k̂′ · ı̂ın ×
− · ∇(ı · ~v ) ~r0 + ı · ∇( · ~v) ~r0 =
∂
∂xn
n=1
ıˆ′ (ıˆ′ · ~v) + ˆ′ (ˆ′ · ~v ) + k̂′ (k̂′ · ~v ) ~r0
For any orthonormal basis, ıˆ′ , ˆ′ , k̂′ and any vector ~v ,
ıˆ′ (ıˆ′ · ~v ) + ˆ′ (ˆ′ · ~v ) + k̂′ (k̂′ · ~v ) = ~v
so
3
X
′
′
′
′
~
~
ˆ
ˆ
ˆ
ˆ
k̂′ · ı̂ın ×
− · ∇(ı · ~v ) ~r0 + ı · ∇( · ~v ) ~r0 =
∂~
v
(~r )
∂xn 0
~ × ~v (~r0 )
= k̂′ · ∇
n=1
and
c Joel Feldman.
I
~ × v(~r0 ) · n̂ + O(ε3 )
~v (~r) · d~r = πε2 ∇
Cε
2003. All rights reserved.
2