#13 (10/07/2024)

Scalar potential

As shown in class, an irrotational field, v, has a scalar potential ϕ, such that
v = ∇ϕ,
(1)
i.e. if v is rotation-free, ∇×v = 0, then, v can be derived from a scalar function, ϕ, which simplifies the analysis (three components of v vs. a single component of ϕ).
Example 1:
v = r     (position vector)
(2)
As
∇×r = 0,
r has a scalar potential. Verify that
ϕ = 1
2
 (x2 + y2 + z2).
(3)
Example 2: (Gravitational field)
F = G m M r
r3
 .
(4)

∇×F
=
G m M ∇×⎛
⎝ 		r
r3
⎞
⎠
=
G m M ⎛
⎝ 		∇⎛
⎝ 		1
r3
⎞
⎠ 		×r + 1
r3
 ∇×r⎞
⎠
=
−Gm M ⎛
⎝ 		∇r3
r6
 ×r⎞
⎠
=
−G m M 3 r2 ∇r
r6
 ×r
=
−G m M 3
r5
 r×r
=
0.
(5)
Verify that
ϕ = − G m M
r
 .
(6)
Example 3: The vector field, v = (2 x y, x2 − 1, 0) is an irrotational field (verify that ∇×v = 0) so there must exist a scalar potential, ϕ, such that v = ∇ϕ, i.e.

2 x y
=
∂ϕ
∂x
 ,
(7)
x2 − 1
=
∂ϕ
∂y
 ,
(8)
0
=
∂ϕ
∂z
 .
(9)
By integrating Eq.(7) with respect to x, one obtains

ϕ = x2 y + f(y),
(10)
where f(y) is a function of y alone. Equation (10) is substituted into Eq.(8) as

x2 − 1 = x2 + f′(y),
(11)
which is solved for f(y) as f(y) = −y + c where c is a constant so the scalar potential for v is
ϕ(x, y, z) = x2 y − y + C.
(12)
Example 4:
Compute

⌠
⌡

C 
 ⎛
⎝ 		3 dx + 6 y2 z3/2 dy + (3 y3 √z + z) dz ⎞
⎠
where C is a curve x = sin t, y = cos 3 t, z = 2 t as t goes from 0 to 2 π.
(Solution)
The vector field, v = ( 3, 6 y2 z3/2, 3 y3 √z + z) is an irrotational field (verify !).
Solving
3 = ∂ϕ
∂x
 ,    6 y2 z3/2 = ∂ϕ
∂y
 ,     3 y3 √z + z = ∂ϕ
∂z
 ,
(13)
one gets
ϕ = 3 x + 2 y3 z3/2 + z2
2
 .
(14)
Therefore,

⌠
⌡

C 
 ( 3 dx + 6 y2 z3/2 dy + (3 y3 √z + z) dz )
=
⌠
⌡

C 
 d ⎛
⎝ 		3 x + 2 y3 z3/2 + z2
2
⎞
⎠
=
[3 x + 2 y3 z3/2 + z2
2
 ](0, 1, 0)(0, 1, 4 π)
=
16 π3/2 + 8 π2.
(15)


Note that
dϕ
=
∂ϕ
∂x
 dx+ ∂ϕ
∂y
 dy+ ∂ϕ
∂z
 dz
=
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
∂ϕ
∂x
∂ϕ
∂y
∂ϕ
∂z
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠ 		·⎛
⎜
⎜
⎜
⎝
dx
dy
dz
⎞
⎟
⎟
⎟
⎠
=
∇ϕ·dr
=
v·dr.
(16)
So that a line integral, ∫v·dr, (see an example in the last lecture) can be evaluated as

⌠
⌡ 		v ·dr
=
⌠
⌡ 		dϕ
=
[ ϕ]AB
=
ϕB − ϕA.
(17)
Note also that scalar potentials are not unique. If an irrotational field, v, is derived from different scalar potentials, ϕ1 and ϕ2, i.e.

v
=
∇ϕ1,
(18)
v
=
∇ϕ2.
(19)
It follows
0
=
∇ϕ1 − ∇ϕ2
=
∇(ϕ1 − ϕ2),
(20)
so
ϕ1 − ϕ2 = C,
(21)
where C is a constant, i.e. the difference of two scalar potentials which yield the same irrotational field is at most a constant.

Vector potential

When a vector field, v, is divergence-free, i.e.
∇·v = 0,
(22)
in a simply connected region, there exists a vector, ψ, whose rotation is v, i.e.

v = ∇×ψ.
(23)
The vector, ψ, is called a vector potential. The necessary condition for the existence of a vector potential is ∇·v=0 as

∇·(∇×ψ) = 0.1
(24)
The converse (the sufficient condition) is more difficult to prove and is not presented here.
Just like scalar potentials, vector potentials are not unique, i.e. if a vector v is derived from two different vector potentials,

v
=
∇×ψ1
=
∇×ψ2,
(25)
it follows
0 = ∇×(ψ1 −ψ2),
(26)
which implies that the vector, ψ1−ψ2, is irrotational. Thereby,
ψ1 − ψ2 = ∇ϕ,
(27)
i.e. the difference between vector potentials that yield the same vector filed is at most ∇ϕ.
Unlike scalar potentials, vector potentials are more difficult to derive and less useful as three components of ψ are needed to yield three components of v.

Helmholtz theorem

If a vector field, v, is neither rotation free nor divergence free, it can be still expressed by both a vector potential and a scalar potential, i.e.

v = ∇ϕ+ ∇×ψ.
(28)
Proof: Consider the Poisson equation,
∆ϕ = ∇·v.
(29)
Provided this equation is solved for ϕ, Eq.(29) can be written as
∇·(v − ∇ϕ) = 0,
(30)
which implies that v − ∇ϕ is divergence free. Hence, it has a vector potential, i.e

v − ∇ϕ = ∇×ψ,
(31)
or

v = ∇ϕ+ ∇×ψ.
(32)

Gauss theorem in multiply connected bodies

multi_1.jpg
The Gauss theorem needs to be modified when applied to multiply connected bodies. The integral path can be selected as
multi_2.jpg
so that the contour integral becomes

⌠
(⎜)
⌡


 
 n ·v dl
=
⌠
⌡

A→ B 
 n·v dl+⌠
⌡

B→ C 
 n·v dl+⌠
⌡

C→ D 
 n·v dl+⌠
⌡

D→ A 
 n·v dl
(33)
=
⌠
⌡

A→ B 
 n·v dl+⌠
⌡

C→ D 
 n·v dl.
(34)
Note that
⌠
⌡

C→ D 
 n·v dl
=
− ⌠
⌡

D→ C 
 −n′·v(− dl)
(35)
⇒
− ⌠
(⎜)
⌡


C2 
 n′·v dl,
(36)
where
n′ = − n.
(37)
Hence, the Gauss theorem for a multiply connected body can be written as

⌠
(⎜)
⌡


C1 
 n·v d l−⌠
(⎜)
⌡


C2 
 n·v d l = ⌠
⌡ 		⌠
⌡

D1−D2 
 ∇·v d S.
(38)
gausshole.jpg
Example: Verify Eq.(38) for v=(x,y).
gausshole2.jpg
Note that along C1, v=(x, y) = (2 cos θ, 2sin θ), and along C2, v=(x, y)=( cos θ, sin θ).

LHS
=
⌠
⌡ 		2π

0 
 ⎛
⎜
⎜
⎝
cos θ
sin θ
⎞
⎟
⎟
⎠ 		·⎛
⎜
⎜
⎝
2cos θ
2sin θ
⎞
⎟
⎟
⎠ 		2 dθ−⌠
⌡ 		2π

0 
 ⎛
⎜
⎜
⎝
cos θ
sin θ
⎞
⎟
⎟
⎠ 		·⎛
⎜
⎜
⎝
cos θ
sin θ
⎞
⎟
⎟
⎠ 		dθ
(39)
=
8π−2π
(40)
=
6π.
(41)
RHS
=
⌠
⌡ 		⌠
⌡

D1−D2 
 ⎛
⎜
⎜
⎜
⎜
⎜
⎝
∂
∂x
∂
∂y
⎞
⎟
⎟
⎟
⎟
⎟
⎠ 		·⎛
⎜
⎜
⎝
x
y
⎞
⎟
⎟
⎠ 		dS
(42)
=
⌠
⌡ 		⌠
⌡

D1−D2 
 2 dS
(43)
=
2 ×( π22−π12)
(44)
=
6π.
(45)

Poisson's equations (not covered in class, for your reference)

  1. Steady state heat conduction with heat generation

    ∇·( k ∇T ) = −f.
  2. Electrostatic field
    ϵo ∆ϕ = − ρ,     (E = −∇ϕ).
  3. Torsion
    −∆ϕ = 2 G θ,     (σxz= ∂ϕ
    ∂y
    		 , σyz=− ∂ϕ
    ∂x
    		 ).
Derivation
We start with Green's second identity applied to a multiply connected body.

⌠
(⎜)
⌡


C 
 ⎛
⎝ 		u ∂v
∂n
 − v ∂u
∂n
⎞
⎠ 		dl = ⌠
⌡ 		⌠
⌡

D 
 (u ∆v − v ∆u) dS.
(46)
poisson.jpg
When the above identity is applied to a multiply-connected body with a circular hole centered at x with the radius, ϵ, Eq.(46) becomes

⌠
(⎜)
⌡


C1 
 ⎛
⎝ 		u ∂v
∂n
 − v ∂u
∂n
⎞
⎠ 		dl−⌠
(⎜)
⌡


C2 
 ⎛
⎝ 		u ∂v
∂n
 − v ∂u
∂n
⎞
⎠ 		dl = ⌠
⌡ 		⌠
⌡

D1−D2 
 (u ∆v − v ∆u) dS,
(47)
where D1 is the total area and D2 represents the hole inside D. Note that n in the left hand side of Eq.(47) is both outward from the inside of the bodies.
It will be shown that by choosing u and v such that ∆u = −ρ and ∆v=0, Eq.(47) becomes something like

⌠
(⎜)
⌡


C1 
 (boundary values of u) dl− 2πu = ⌠
⌡ 		⌠
⌡

D 
 v ×ρdS
(48)
thus, the solution can be written as
u = 1
2π
 ⎛
⎝ 		⌠
⌡ 		⌠
⌡

D 
 (− v ×ρ) d S + ⌠
(⎜)
⌡


C1 
 (boundary value of u) dl⎞
⎠ 		.
(49)
If we choose v as

v = ln r,
(50)
then, v satisfies

∆v = 02.
(51)
Noting that

∆u = −ρ,
(52)
Eq.(47) becomes

⌠
(⎜)
⌡


C1 
 ⎛
⎝ 		u ∂
∂n
 ln r − ln r ∂u
∂n
⎞
⎠ 		dl− ⌠
(⎜)
⌡


C2 
 ⎛
⎝ 		u ∂
∂n
 ln r − ln r ∂u
∂n
⎞
⎠ 		dl = ⌠
⌡ 		⌠
⌡

D1−D2 
 ( − ln r ∆u) dS,
(53)
where ∆ln r = 0 was used.
If C2 is chosen as a circle centered at r0 with a radius of ϵ, the left hand side of Eq.(53) becomes

LHS
=
⌠
(⎜)
⌡


C1 
 ⎛
⎝ 		u ∂
∂n
 ln r− ln r ∂u
∂n
⎞
⎠ 		dl−⌠
(⎜)
⌡


C2 
 ⎛
⎝ 		u ∂
∂r
 (ln r)− ln r ∂u
∂r
⎞
⎠ 		dl
(54)
=
⌠
(⎜)
⌡


C1 
 ⎛
⎝ 		u ∂
∂n
 ln r− ln r ∂u
∂n
⎞
⎠ 		dl− ⌠
(⎜)
⌡


C2 
 ⎛
⎝ 		u
r
 − ln r ∂u
∂r
⎞
⎠ 		dl.
(55)
When ϵ→ 0, the integrations over C2 in Eq.(55) become

⌠
(⎜)
⌡


C2 
 u
r
 dl
=
⌠
⌡ 		2π

0 
 u
ϵ
 ϵd θ⇒ 2πu,
(56)
⌠
(⎜)
⌡


C2 
 ln r ∂u
∂r
 dl
=
⌠
⌡ 		2π

0 
 ln ϵ ∂u
∂r
 ϵdθ⇒ 2π ∂u
∂r
 ϵln ϵ.
(57)
As ϵ→ 0, Eq.(57) tends to 0 (ϵln ϵ→ 0 3) so Eq.(53) becomes

⌠
(⎜)
⌡


C1 
 ⎛
⎝ 		uA ∂
∂n
 ln r− ln r ∂uA
∂n
⎞
⎠ 		dl− 2πu = ⌠
⌡ 		⌠
⌡

D 
 ln r ρdS,
(58)
which can be solved for u as
u = 1
2π
 ⌠
(⎜)
⌡


C1 
 ⎛
⎝ 		uA ∂
∂n
 ln r− ln r ∂uA
∂n
⎞
⎠ 		dl− 1
2π
 ⌠
⌡ 		⌠
⌡

D 
 ln r ρdS.
(59)
Example 1: Infinitely extended body and a pointwise heat source:
In this case, both uA and ∂uA/∂n vanish as x→ 0. The point source can be set as

ρ(x,y) = ρ0 δ(x,y),
(60)
where δ(x,y) is the two-dimensional Dirac delta function.
Thus, Eq.(59) becomes

u(x,y)
=
− 1
2π
 ⌠
⌡ 		∞

−∞ 
 ⌠
⌡ 		∞

−∞ 
 ln 
√
 
(x−x′)2+(y−y′)2
 
 ρ0 δ(x′,y′) dx′dy′
(61)
=
− 1
2π
 ρ0 ln 
√
 
x2+y2
 
 .
(62)
Example 2: Same with Example 1 but the heat source is distributed over −1 ≤ x ≤ 1 and y=0.

u(x,y)
=
− 1
2π
 ⌠
⌡ 		∞

−∞ 
 ⌠
⌡ 		∞

−∞ 
 ln 
√
 
(x−x′)2+(y−y′)2
 
 ρ(x′,y′) dx′dy′
(63)
=
− 1
2π
 ⌠
⌡ 		1

−1 
 ln 
√
 
(x−x′)2+y2
 
 ρ0 dx′
(64)
=
…
(65)
=
(can be carried out by computer algebra)
(66)

Footnotes:

1
∇·(∇×ψ)
=
(∇×∇)·ψ
=
0
2
∆ln r
=
∇·∇ln r
=
∇· ∇r
r
=
∇· r
r2
=
∇·r r2 − r ∇r2
r4
=
2r2− r ·2 r ∇r
r4
=
2 r2− r ·2r r
r
r4
=
2r2−2r2
r4
=
0.
3

lim
ϵ→ 0 
 ϵln ϵ =
lim
ϵ→ 0 
 ln ϵ
( 1
ϵ
 )
 =
lim
ϵ→ 0 
( 1
ϵ
 )
( −1
ϵ2
 )
 =
lim
ϵ→ 0 
 (−ϵ) = 0.


File translated from TEX by TTH, version 4.03.
On 07 Oct 2024, 14:07.