19.1
Chapter Nineteen
Some Physics
19.1 Fluid Mechanics
Suppose
)
,
,
,
(
t
z
y
x
v
is the velocity at
k
j
i
r
z
y
x
z
y
x
+
+
=
=
)
,
,
(
of a fluid flowing
smoothly through a region in space, and suppose
)
,
,
,
(
t
z
y
x
ρ
is the density at r at time t. If
S is an oriented surface, it is not hard to convince yourself that the flux integral
∫∫
⋅
S
dr
v
ρ
is the rate at which mass flows through the surface S. Now, if S is a closed surface, then
the mass in the region B bounded by S is, of course
∫∫∫
B
dV
ρ
.
The rate at which this mass is changing is simply
∫∫∫
∫∫∫
∂
∂
=
∂
∂
B
B
dV
t
dV
t
ρ
ρ
.
This is the same as the rate at which mass is flowing across S into B:
∫∫
⋅
−
S
dr
v
ρ
, where S
is given the outward pointing orientation. Thus,
∫∫∫
∫∫
⋅
−
=
∂
∂
B
S
d
dV
t
r
v
ρ
ρ
.
We now apply Gauss’s Theorem and get
.
)
(
dV
d
dV
t
B
B
S
v
r
v
ρ
ρ
ρ
∫∫∫
∫∫∫
∫∫
⋅
∇
−
=
⋅
−
=
∂
∂
Thus,
dV
t
B
∫∫∫
⋅
∇
+
∂
∂
)
( v
ρ
ρ
.
Meditate on this result. The region B is any region, and so it must be true that the
integrand itself is everywhere 0:
19.2
0
)
(
=
⋅
∇
+
∂
∂
v
ρ
ρ
t
.
This is one of the fundamental equations of fluid dynamics. It is called the equation of
continuity.
In case the fluid is incompressible, the continuity equation becomes quite simple.
Incompressible means simply that the density
ρ is constant. Thus
0
=
∂
∂
t
ρ
and so we have
,
0
)
(
)
(
=
⋅
∇
=
⋅
∇
=
⋅
∇
+
∂
∂
v
v
v
ρ
ρ
ρ
ρ
t
or
0
=
⋅
∇
v
.
Exercise
1. Consider a one dimensional flow in which the velocity of the fluid is given by
)
(x
f
=
v
, where
0
)
(
>
x
f
. Suppose further that the density
ρ of the fluid does not vary
with time t. Show that
)
(
)
(
x
f
k
x
=
ρ
,
where k is a constant.
19.2 Electrostatics
Suppose there is a point charge q fixed at the point s. Then the electric field
)
(r
E
q
due to q is given by
3
)
(
s
r
s
r
r
E
−
−
=
kq
q
.
It is easy to verify, as we have done in a previous chapter, that this field, or function, is
conservative, with a potential function
|
|
)
(
s
r
r
−
−
=
kq
P
q
;
so that
q
q
P
∇
=
E
. Physicists do not like to be bothered with the minus sign in
q
P , so they
define the electric potential
q
V to be -
q
P . Thus,
19.3
|
|
)
(
s
r
r
−
=
kq
V
q
,
and
)
(
)
(
r
r
E
q
q
V
−∇
=
.
We have already seen that the flux out of a closed surface S is
=
⋅
∫∫
origin
the
enclose
does
if
4
origin
the
enclose
not
does
if
0
S
kq
S
d
S
q
π
S
E
Some meditation will convince you there is nothing special here about the origin; that is, if
the point charge is at s, then
=
⋅
∫∫
s
s
S
E
enclose
does
if
4
enclose
not
does
if
0
S
kq
S
d
S
q
π
Next, suppose there are a finite number of point charges
,
,
at
,
at
2
2
1
1
K
s
s q
q
and
n
n
q
s
at
.
Suppose
j
E is the electric intensity due to
j
q . Then it should be clear that the
electric field due to these charges is simply the sum
∑
∑
=
=
−
−
=
=
n
j
j
j
j
n
j
j
q
k
1
3
1
|
|
)
(
s
r
s
r
E
r
E
.
Also,
∑
=
−
=
n
j
j
j
q
k
V
1
|
|
)
(
s
r
r
; and
)
(
)
(
r
r
E
V
−∇
=
.
Finally,
∫∫
∑
=
⋅
S
j
q
k
d
π
4
S
E
where the sum is over those charges enclosed by S.
19.4
Things become more exciting if instead of point charges, we have a charge
distribution in space with charge density
ρ . To find the electric field
)
(r
E
produced by
this distribution of charge in space, we need to integrate:
s
U
dV
k
∫∫∫
−
−
=
3
|
|
)
(
)
(
)
(
s
r
s
r
s
r
E
ρ
.
But this appears to be a serious breach of decorum. We are integrating over everything,
and at
r
s
=
we have the dreaded 0 in the denominator. Thus what we see above is an
improper integral—that is, it is actually a limit of integrals. Specifically, we integrate not
over everything but over everything outside a spherical solid region of radius a centered at
r. We then look at the limit as
0
→
a
of this integral. With the integral for the electric
field, this limit exists, and so there is no problem with 0 on the bottom of the integrand. In
the same way, we are safe in writing for the potential
s
U
dV
k
V
∫∫∫
−
=
|
|
)
(
)
(
s
r
s
r
ρ
.
Everything works nicely so that we also have
)
(
)
(
r
r
E
V
−∇
=
.
If R is a solid region bounded by a closed surface S, then we can also integrate to get
(*)
∫∫
∫∫∫
=
⋅
S
R
dV
k
d
.
)
(
4
s
S
E
ρ
π
The divergence of E is the troublesome item in extending matters to distributed
charge. If we simply try to calculate the divergence by
∫∫∫
∫∫∫
=
U
U
dV
div
dV
div
stuff)
(
stuff
,
then things go wrong because the improper integral of the divergence does not exist.
Gauss saves the day. Let R be any region and let S be the closed surface bounding R.
Then
∫∫
∫∫∫
⋅
∇
=
⋅
S
R
dV
d
E
S
E
.
But from equation (*) we have
∫∫∫
∫∫
∫∫∫
⋅
∇
=
=
⋅
R
S
R
dV
dV
k
d
)
(
4
E
s
S
E
ρ
π
.
This gives us
19.5
∫∫∫
∫∫∫
⋅
∇
=
R
R
dV
dV
k
4
E
ρ
π
, or
(
)
dV
k
R
∫∫∫
−
⋅
∇
ρ
π
4
E
.
But R is any region, and so it must be true that
ρ
πk
4
=
⋅
∇
E
for all r.
Finally, remembering that
V
−∇
=
E
, we get
ρ
πk
V
4
)
(
=
∇
⋅
−∇
=
⋅
∇
E
;
ρ
πk
V
4
2
−
=
∇
, or
ρ
πk
z
V
y
V
x
V
4
2
2
2
2
2
2
−
=
∂
∂
+
∂
∂
+
∂
∂
.
This is the celebrated Poisson’s Equation, a justly famous partial differential equation, the
study of which is beyond the scope of this course.
.