#14 (10/09/2023)

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 r2r

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 yy + C.
(12)
Example 4:
Compute




C 

3 dx + 6 y2 z3/2 dy + (3 y3z + 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 y3z + z) is an irrotational field (verify !).
Solving
3 = ∂ϕ

x
,    6 y2 z3/2 = ∂ϕ

y
,     3 y3z + 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 y3z + 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)

Leibniz integral rule

The Leibniz integral rule (the Leibniz rule) is also known as the Reynolds Transport Theorem and plays an important rule to derive important field equations in mechanics. It refers to the time rate of change of an integral where the integral range is a function of time.


Problem: Given
I(t) ≡
b(t)

a(t) 
f(x, t) d x,
(33)
find I′(t) also denoted as [(DI)/(Dt)].

Approaches

Method 1: (not discussed in class)
Let
F(x, t) ≡
x

0 
f(y, t) d y.
(34)
Note that the integral variable has been changed to y from x to avoid confusion with the upper bound so that

F(x, t)

x
= f(x, t).
(35)
Using F(x, t), I(t) can be written as

I(t) = F(x, t)|x = b(t)F(x,t)|x = a(t),
(36)
hence

I′(t)
=

F

x
|x = b(t) db(t)

dt
+F

t
|x = b(t)

F

x
|x = a(t) da(t)

dt
+F

t
|x = a(t)
=
f(x,t)|x = b(t) db(t)

dt
f(x,t)|x = a(t) da(t)

dt
+F

t
|x = b(t)|−F

t
|x = a(t)|
=
f(x,t)|x = b(t) db(t)

dt
f(x,t)|x = a(t) da(t)

dt
+
b(t)

0 
f

t
dy
a(t)

0 
f

t
dy
=
f(x,t)|x = b(t) db(t)

dt
f(x,t)|x = a(t) da(t)

dt
+
b(t)

a(t) 
f

t
dy
=
f(b(t), t) db(t)

dt
f(a(t), t) da(t)

dt
+
b(t)

a(t) 
f

t
dy.
(37)
Method 2:(discussed in class)

I(t) ≡
b

a 
f(x, t) dx.
(38)
liebnitz.gif
The quantity, I(t+∆t), is expressed as

I(t+∆t)
=

b+∆b

a+∆a 
f(x,t+∆t) dx
=

b

a 
f(x, t+∆t) dx +
b+∆b

b 
f(x,t+∆t) dx
a+∆

a 
f(x,t+∆t) dx
 ∼ 

b

a 
f(x,t+∆t) dx + f|bbf|aa.
(39)
Therefore, the rate of time change is

I(t+∆t)−I(t)

t
 ∼ 

b

a 
f(x,t+∆t)−f(x,t)

t
dx +f|b b

t
f|a a

t
.
(40)
By taking a limit of ∆t → 0, Eq. (40) becomes

DI

Dt
=

b

a 
f

t
dx +(−1) fv|a + (+1) fv|b
=

b

a 
f

t
dx +

boundary 
n f v
=

b

a 
f

t
dx +
b

a 
( f v)′dx
=

b

a 

f

t
+ (f v)′
dx.
(41)
This result can be immediately extended to 2-D as

I(t) ≡

f(x, t) dS.
(42)

DI

Dt
=


f

t
dS +
(⎜)

n · f  v dl
=


f

t
dS +

∇·( f v) dS
=



f

t
+ ∇·(f v)
dS,
(43)
and to 3-D as

I(t) ≡




 
f(v, t) dV.
(44)

DI

Dt
=





 
f

t
dV +
(⎜)

n· f v dS
=





 
f

t
dV +




 
∇·( f v) dV
=





 

f

t
+ ∇·(f v)
dV.
(45)
Example 1
I(t) ≡
t2

t 
sin (x2 t) dx.
Example 2 Evaluate
I
1

0 
x − 1

ln x
d x.
by starting with this integral:
I(t) ≡
1

0 
xt−1

ln x
dx.
Example 3 Evaluate

1

0 
xt ln x d x
by starting with the known integral

I(t) ≡
1

0 
xt d x.

Equation of continuity

Choose f as the mass density, ρ, as

f(r, t)=ρ(r, t),
then its integral is the total mass, m(t), as
m(t) =



 
ρ(r, t) dS.
The Lagrangian derivative of the total mass, then, is
Dm(t)

Dt
=




 

∂ρ

t
+ ∇·( ρv)
dS
=
0,
(46)
as the total mass is conserved in classical mechanics. Therefore, we have
∂ρ

t
+ ∇·(ρv) = 0,
(47)
which is known as the equation of continuity. If ρ = const, Equation (47) is reduced to
∇·v = 0.
(48)
The fluid is called incompressible.

Equation of motion

By choosing f = ρv (momentum), its integral defines the total momentum, M(t), as
M(t) ≡



 
ρ v  dS,
(49)
where Ω is a 2-D region occupied by the body and dS is an area element of the 2D integral. The Lagrangian derivative of the total momentum, therefore, is

DM

Dt
=



 


t
(ρ v) + ∇·( ρv v)
dS.2
(50)
According to the Newton's second law, the rate change of the total momentum is due to the total force applied to the body as

DM

Dt
=

(⎜)



∂Ω 
t dl+



 
b  dS
=




 
( ∇·σ+ b) dS,
(51)
where ∂Ω is the boundary of Ω and dl is an area element of the boundary integral.
By equating Eq. (51) with Eq. (50), we have
∇·σ+ b
=

t
(ρ v) + ∇·( ρv v)
=
∂ρ

t
v + ρv

t
+∇·(ρv) v + ρvv
=

∂ρ

t
+ ∇·(ρv)
v+
ρv

t
+ ρv ∇·v
=
ρ
v

t
+ ∇·v v
.
(52)
Thus the equation of motion is expressed as

ρ
v

t
+ ∇·v v
= ∇·σ+ b.
(53)
Equation (53) is known as the equation of motion and universal for any deformable bodies.

Isotropic fluids

For isotropic fluids, it can be shown that

σ = −p I + μ∇v 2

3
μ I  ∇·v.
(54)
Therefore,

∇·σ
=
−∇p + μ∇∇·v 2

3
μ∇∇·v
=
−∇p + μ∇∇·v + μ

3
∇∇·v,
(55)
so that the equation of motion becomes

ρ
v

t
+ ∇v v
= μ∆v+ μ

3
∇∇·v −∇p +b.
(56)
This is called the Navier-Stokes equation.

ρ
u

t
+ u

x
u +u

y
v+u

z
w
=
μ
2 u

x2
+2 u

y2
+2 u

z2

+ μ

3

x

u

x
+v

y
+w

z

p

x
+ bx,
ρ
v

t
+ v

x
u +v

y
v+v

z
w
=
μ
2 v

x2
+2 v

y2
+2 v

z2

+ μ

3

y

u

x
+v

y
+w

z

p

y
+ by,
ρ
w

t
+ w

x
u +w

y
v+w

z
w
=
μ
2 w

x2
+2 w

y2
+2 w

z2

+ μ

3

z

u

x
+v

y
+w

z

p

z
+ bz.
(57)


Footnotes:

1
∇·(∇×ψ)
=
(∇×∇)·ψ
=
0
2
∇·( ρv v) =








x
(ρ u u)+

y
(ρ u v)+

z
(ρ u w)

x
(ρ v u)+

y
(ρ v v)+

z
(ρ v w)

x
(ρ w u)+

y
(ρ w v)+

z
(ρ w w)











File translated from TEX by TTH, version 4.03.
On 09 Oct 2023, 13:07.