#13 (10/04/2023)

Line integrals

Examples:

    (1)
    lineint_1.jpg

    W = ⌠
    ⌡     		(1,2)

    (0,0) 
    		 F·dr,
    where

    F
    ≡
    (x y, x2),
    		(1)
    dr
    =
    (dx, dy).
    		(2)

    W
    =
    ⌠
    ⌡     		x y dx + x2 dy
    =
    ⌠
    ⌡     		x (2x) dx + x2 2 dx
    =
    ⌠
    ⌡     		1

    0 
    		 (2x2 + 2x2) dx
    =
    ⌠
    ⌡     		1

    0 
    		 4x2 dx
    =
    4
    3
    		 .
    		(3)

    (2)
    lineint_2.jpg

    W
    =
    ⌠
    ⌡     		x y dx + x2 dy
    =
    ⌠
    ⌡     		x (2 x2) dx + x2 (4x dx)
    =
    ⌠
    ⌡     		1

    0 
    		 (2 x3 + 4 x3) dx
    =
    ⌠
    ⌡     		1

    0 
    		 6 x3 dx
    =
    3
    2
    		 .
    		(4)
    It can be seen that the values of line integrals generally depend on the integral paths.
    (3)
    lineint_3.jpg

    F ≡ (2 x y, x2 − 1).

    W
    =
    ⌠
    ⌡     		2 x y dx + (x2 − 1) dy
    =
    ⌠
    ⌡     		5/2π

    0 
    		 { 2 (rcosθ)(rsinθ) (dr cosθ− r sinθdθ)
    + ((rcosθ)2 − 1) (dr sin θ+ r cos θdθ) }
    =
    …
    =
    −5/2 π,
    		(5)
    where
    x
    =
    r cos θ,
    		(6)
    y
    =
    r sin θ,
    		(7)
    dx
    =
    dr cos θ− r sin θdθ,
    		(8)
    dy
    =
    dr sin θ+ r cos θdθ,
    		(9)
    was used.
    Alternatively, if one defines
    G(x, y) ≡ x2 y − y,1

    ⌠
    ⌡     		F·dr
    =
    ⌠
    ⌡     		dG
    =
    [G](x,y)=(0, 0)(x,y)=(0, 5/2π)
    =
    − 5
    2
    		 π.
    While the line integrals of Examples 1 and 2 depend on the integral path, the line integral of Example 3 is independent of the path. This difference will be clarified in the next section.

Stokes' theorem


⌠
(⎜)
⌡


∂S 
 F·dr = ⌠
⌡ 		⌠
⌡

S 
 m·(∇×F)dS.
(10)
stokes.jpg
Proof
From the figure above,
dr = m×n dr,
so it follows that
⌠
(⎜)
⌡


∂S 
 F ·dr
=
⌠
(⎜)
⌡


∂S 
 F ·(m ×n) dr
=
(scalar triple product)
=
⌠
(⎜)
⌡


∂S 
 m ·(n ×F) dr
=
(Gauss theorem)
=
⌠
⌡ 		⌠
⌡

S 
 m ·(∇×F) dS.
(11)


Example The velocity field of a vortex that rotates with the angular velocity of ω is given as
v
=
r ωeθ
=
r ω⎛
⎜
⎜
⎜
⎝
−sinθ
cosθ
0
⎞
⎟
⎟
⎟
⎠
=
⎛
⎜
⎜
⎜
⎝
−ωy
ωx
0
⎞
⎟
⎟
⎟
⎠ 		,
(12)
so
∇×v = 2 ωk.
(13)
If F=(P, Q, 0), m=(0,0,1) and the Stokes' theorem is rewritten as
⌠
(⎜)
⌡


∂S 
 Pdx + Qdy = ⌠
⌡ 		⌠
⌡

S 
 ⎛
⎝ 		∂Q
∂x
 − ∂P
∂y
⎞
⎠ 		dA.
(14)

Irrotational Field

Definition: If the vector field, v, satisfies ∇×v=0, v is called an irrotational field.
Theorem: The following four statements are all equivalent.

    (1)
    v = ∇ϕ.

    (2)
    ∇×v = 0.

    (3)
    ⌠
    (⎜)
    ⌡
    		v·dr = 0.

    (4)
    ⌠
    ⌡     		v·dr,
    is independent of the integral path.
  1. 1 → 2.
    ∇×(∇ϕ)=0
  2. 2 → 3
    ⌠
    (⎜)
    ⌡


    ∂S 
    		 v·dr = ⌠
    ⌡     		⌠
    ⌡

    S 
    		 m·(∇×v) dA = 0
  3. 3 → 4 (shown in class)
  4. 4 → 1
    Define ϕ as
    ϕ(r) ≡ ⌠
    ⌡     		r

    a 
    		 v·dr,
    then

    ϕ(r+dr)−ϕ(r) ≈ ∇ϕ·dr2
    		(15)
    On the other hand,

    ϕ(r+dr)−ϕ(r)
    =
    ⌠
    ⌡     		r+dr

    a 
    		 v·dr − ⌠
    ⌡     		r

    a 
    		 v·dr
    ≈
    v ·dr,
    		(16)
    so
    v = ∇ϕ.

Scalar potential

As shown in class, an irrotational field, v, has a scalar potential ϕ, such that
v = ∇ϕ,
(17)
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)
(18)
As
∇×r = 0,
r has a scalar potential. Verify that
ϕ = 1
2
 (x2 + y2 + z2).
(19)
Example 2: (Gravitational field)
F = G m M r
r3
 .
(20)

∇×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.
(21)
Verify that
ϕ = − G m M
r
 .
(22)
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
 ,
(23)
x2 − 1
=
∂ϕ
∂y
 ,
(24)
0
=
∂ϕ
∂z
 .
(25)
By integrating Eq.(23) with respect to x, one obtains

ϕ = x2 y + f(y),
(26)
where f(y) is a function of y alone. Equation (26) is substituted into Eq.(24) as

x2 − 1 = x2 + f′(y),
(27)
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.
(28)
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
 ,
(29)
one gets
ϕ = 3 x + 2 y3 z3/2 + z2
2
 .
(30)
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.
(31)


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.
(32)
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.
(33)
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,
(34)
v
=
∇ϕ2.
(35)
It follows
0
=
∇ϕ1 − ∇ϕ2
=
∇(ϕ1 − ϕ2),
(36)
so
ϕ1 − ϕ2 = C,
(37)
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,
(38)
in a simply connected region, there exists a vector, ψ, whose rotation is v, i.e.

v = ∇×ψ.
(39)
The vector, ψ, is called a vector potential. The necessary condition for the existence of a vector potential is proven easily as

∇·(∇×ψ) = 0.3
(40)
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,
(41)
it follows
0 = ∇×(ψ1 −ψ2),
(42)
which implies that the vector, ψ1−ψ2, is irrotational. Thereby,
ψ1 − ψ2 = ∇ϕ,
(43)
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.
(44)
Proof: Consider the Poisson equation,
∆ϕ = ∇·v.
(45)
Provided this equation is solved for ϕ, Eq.(45) can be written as
∇·(v − ∇ϕ) = 0,
(46)
which implies that v − ∇ϕ is divergence free. Hence, it has a vector potential, i.e

v − ∇ϕ = ∇×ψ,
(47)
or

v = ∇ϕ+ ∇×ψ.
(48)

Footnotes:

1
dG = ∂G
∂x
 dx + ∂G
∂y
 dy = 2 x y dx + (x2 − 1) dy.
2
ϕ(x+dx, y+dy, z+dz)−ϕ(x, y, z)
 ∼ 
ϕ(x, y, z)+ ∂ϕ
∂x
 dx+ ∂ϕ
∂y
 dy+ ∂ϕ
∂z
 dz − ϕ(x, y, z)
=
∂ϕ
∂x
 dx+ ∂ϕ
∂y
 dy+ ∂ϕ
∂z
 dz
=
∇ϕ·dr.
3
∇·(∇×ψ)
=
(∇×∇)·ψ
=
0


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On 03 Oct 2023, 21:07.