PHZ 3113 Fall 2010
Homework #4, Due Monday, October 4
1. Define elliptical coordinates u, v by x =
√
u2 + c2 cos v,
y = u sin v .
(a) Write the area element dA = dx dy in terms of elliptical u, v.
The element of area is dx dy = |J| du dv, where the Jacobian determinant is
 u cos v

√
2
2
√
−
u
+
c
sin
v
∂x/∂u ∂x/∂v

J = det
= det  u2 + c2
∂y/∂u ∂y/∂v
sin v
u cos v
=√
u2
u 2 + c2
cos2 v +
p
u2 + c2 sin2 v =
u2 + c2 sin2 v
√
.
u 2 + c2
(b) Using elliptical coordinates, compute the area of an ellipse with semimajor axis a, semiminor axis b, where a2 = b2 + c2 .
It is easy to see that curves of constant u are ellipses,
y2
x2
+
= cos2 v + sin2 v = 1,
2
2
2
u +c
u
and 0 < v < 2π traces the entire curve. The area is
Z b Z 2π
Z b
u2 + c2 sin2 v
2u2 + c2
A=
du
dv √
=π
du √
.
u 2 + c2
u 2 + c2
0
0
0
Since
d p 2
2u2 + c2
2
√
,
u u +c =
du
u 2 + c2
this is
h p
ib
p
A = πu u2 + c2 = πb b2 + c2 = πab.
0
2
1/3
x − y2
2. Let the scalar field φ be given by φ(x, y) =
.
x2 + y 2
(a) Compute the components of the gradient ∇φ. Where is the derivative troublesome?
The derivatives are
∂φ
4xy 2
,
=
∂x
3(x2 + y 2 )4/3(x2 − y 2 )2/3
and
∇φ =
∂φ
4x2 y
,
=−
∂y
3(x2 + y 2 )4/3 (x2 − y 2)2/3
4xy (y x̂ − x ŷ)
3(x2 + y 2)4/3 (x2 − y 2 )2/3
.
This is well-behaved except as x2 + y 2 → 0 and at x = y.
(b) Compute the directional derivative of φ in the direction n̂ = x̂ cos α + ŷ sin α. Write
your answer in terms of polar coordinates ρ, φ, where x = ρ cos φ, y = ρ sin φ. What is the
directional derivative along a radial ray (fixed φ) as ρ → 0?
Write the function as Φ so as not to confuse with the coordinate φ (sorry ’bout that). The
directional derivative is
∂Φ
4xy (y cos α − x sin α)
= n̂ · ∇φ =
.
∂n
3(x2 + y 2 )4/3 (x2 − y 2)2/3
In terms of ρ and φ, this is
∂Φ
4ρ3 sin φ cos φ(sin φ cos α − cos φ sin α)
2 sin 2φ sin(φ − α)
=
.
=
2
4/3
2/3
∂n
3ρ
3(ρ2 ) [ρ2 (cos2 φ − sin φ) ]
(cos 2φ)2/3
This is perhaps easier to obtain by writing
Φ = (cos 2φ)1/3,
∇Φ =
1 ∂Φ
φ̂,
ρ ∂φ
∂Φ
2
sin 2φ
=−
φ̂ · n̂.
∂n
3ρ (cos 2φ)2/3
with
φ̂ · n̂ = − sin φ cos α + cos φ sin α = sin(α − φ).
3. Let the vector field v be
v=
where r =
p
x2 + y 2 + z 2 .
y(y 2 + z 2 )
x(x2 + z 2 )
xyz
x̂
+
ŷ − 3 ẑ ,
3
3
r
r
r
(a) Compute ∇ · v. Compute ∇ × v.
All calculations can make good use of the relation
∂ 1
1
2x
∂
p
px
=
=−
= − p+2 ,
p
p/2
(p+2)/2
2
2
2
2
2
2
∂x r
∂x (x + y + z )
2 (x + y + z )
r
and similarly for y and z.
The derivatives that appear in the divergence are
∂vx
3xy(y 2 + z 2 )
=−
∂x
r5
∂vy
3xy(x2 + z 2 )
=−
∂y
r5
xy(x2 + y 2 − 2z 2 )
∂vz
=−
,
∂z
r5
and the divergence is
∇·v =
∂vx ∂vy ∂vz
4xy
+
+
=− 3 .
∂x
∂y
∂z
r
The derivatives that appear in the curl are
∂vx
3x2 y 2 + x2 z 2 + y 2 z 2 + z 4
=
∂y
r5
∂vy
3x2 y 2 + x2 z 2 + y 2 z 2 + z 4
=
∂x
r5
∂vz
yz (2x2 − y 2 − z 2 )
=
∂x
r5
and
∇ × v = x̂
∂v
z
∂y
−
yz (2x2 − y 2 − z 2 )
∂vx
=
∂z
r5
∂vy
xz (2y 2 − x2 − z 2 )
=
∂z
r5
2
∂vz
xz (2y − x2 − z 2 )
=
,
∂y
r5
∂v
∂v
∂vy ∂vz ∂vx y
x
−
−
+ ŷ
+ ẑ
= 0.
∂z
∂z
∂x
∂x
∂y
(b) Compute ∇(∇ · v) and ∇2 v. Compute ∇ × (∇ × v) two different ways.
Using (∂/∂x)(1/r 3 ) = −3x/r 5 , etc., it is straightforward to obtain
∇(∇ · v) =
4x (3y 2 − r 2 )
12xyz
4y (3x2 − r 2 )
x̂
+
ŷ +
ẑ.
5
5
r
r
r5
The Laplacian of the z-component
∂ 2 vx
3y(y 2 + z 2 )(r 2 − 5x2 )
=
,
∂x2
r7
∂ 2 vx
3y(2x4 − 3x2 y 2 + x2 z 2 − y 2 z 2 − z 4 )
=
,
∂y 2
r7
∂ 2 vx
y(2x4 + x2 y 2 − y 4 − 11x2 z 2 + y 2 z 2 + 2z 4 )
=
,
∂z 2
r7
4y (r 2 − 3x2 )
.
r5
The y-component follows from exchanging x and y,
∇2 vx =
∇2 vz =
12xyz
.
r5
The curl of the curl directly is easy,
∇ × (∇ × v) = ∇ × (0) = 0.
which is the same as ∇(∇ · v) − ∇2 v.
If you think of it, ∇ × v = 0 means that v = ∇φ for some φ; and from vz it is apparent that
this is
xy
φ=
.
r
Then,
4xy 2
2
2
∇ v = ∇ (∇φ) = ∇ (∇ φ) = ∇ − 3 .
r
4. In analogy with electrostatics, where the electric field E can be obtained from an electrostatic potential E = −∇V where V satisfies Poisson’s equation ∇2 V = −ρe /0 , the
gravitational field g can be obtained from a gravitational potential g = −∇φ that satisfies
∇2 φ = 4πGρg . In numerical simulations of large-scale cosmological structure, it is sometimes
useful to “soften” the Newtonian gravitational potential φg = −GM/r of a point particle of
mass M as
GM
φa = −
.
r+a
(a) What is the gravitational field g derived from φa ? How does g behave as r → ∞? Show
that the softening introduces a maximum on the gravitational force between particles.
g = −∇φ = −
GM
r̂.
(r + a)2
This behaves as GM/r 2 as r → ∞, but as r → 0 has magnitude at most GM/a2 .
(b) The softened φa can be interpreted as the potential of a slightlyR fuzzy particle with a
mass density ρa (r). What is ρa (r)? What is the total mass, Ma = d3 v ρa ? What radius
r1/2 contains half the mass?
From (7.10)
∇2 φ =
2aGM
∂ 2 φ 2 ∂φ
=
+
= 4πGρ,
2
r ∂r
∂r
r(r + a)3
ρa (r) =
aM
.
2πr(r + a)3
The mass within radius r is
Z r
Ma (r) =
0
4πr 2 ρa (r) =
Mr 2
,
(r + a)2
and so M(r → ∞) = M, independent of a. The half-mass radius is
2
r1/2
1
= ,
2
2
2
r1/2 + a
r1/2 = a(1 +
√
2).
As a → 0, this is a mass distribution that has zero density for r 6= 0, has infinite density
as r → 0, has extent of order a, and has constant integral total mass. Thus, as a → 0 this
returns to the point mass, or
ρa → M δ (3) (r).