PHZ 3113 Fall 2010
Exam 2
Name:
This is a closed-book exam. Some possible useful information:
(1 + x)p = 1 + p x +
∞
X
1
p!
1
1
p(p − 1) x2 + p(p − 1)(p − 2) x3 + · · · =
xn
2
6
n! (p − n)!
n=0
ex = 1 + x +
∞
X
1 n
1 2 1 3
x + x +··· =
x
2
6
n!
n=0
Z x
0
sec3 x0 dx0 =
1
[sec x tan x + log(sec x + tan x)]
2
r̂ = x̂ cos θ + ŷ sin θ
∇ · (vr r̂ + vθ θ̂) =
θ̂ = −x̂ sin θ + ŷ cos θ
1 ∂vθ
1 ∂
(rvr ) +
r ∂r
r ∂θ
h1 ∂
1 ∂vr i
(rvθ ) −
ẑ
∇ × (vr r̂ + vθ θ̂) =
r ∂r
r ∂θ
(∂/∂z = 0)
1. Let z = x + iy. Let f (z) = eiz .
(a) Write the real and imaginary parts of f (z) as real functions of the real variables x, y.
f (z) = eiz = ei(x+iy) = eix−y = (cos x + i sin x) e−y
= cos x e−y + i sin x e−y = u(x, y) + i v(x, y).
The real part is u(x, y) = cos x e−y , the imaginary part is v(x, y) = sin x e−y .
(b) Verify that the Cauchy-Riemann relations hold for your functions.
∂u
∂v
= − sin x e−y =
,
∂x
∂y
∂u
∂v
= − cos x e−y = − .
∂y
∂x
(c) Let the vector v have components vx = Re[f (z)] and vy = Im[f (z)]. Compute ∇ · v and
∇ × v.
The vector is v = x̂ cos x e−y + ŷ sin x e−y , and
∇·v =
∇ × v = ẑ
∂vx ∂vy
+
= − sin x e−y − sin x e−y = −2 sin x e−y ,
∂x
∂y
∂v
y
∂x
−
i
h
∂vx = ẑ cos x e−y − (− cos x e−y ) = 2 cos x e−y ẑ.
∂y
The first is a scalar, the second is a vector. As you can see from the Cauchy-Riemann
relations, in both cases the terms do not cancel, but reinforce.
(d) Compute ∇(∇ · v). Compute ∇ × (∇ × v). Compute ∇2 v.
∇(∇ · v) = x̂
∂
∂
(−2 sin x e−y ) + ŷ (−2 sin x e−y ) = x̂ (−2 cos x e−y ) + ŷ (2 sin x e−y ),
∂x
∂y
∂
∂ + ŷ
× (2 cos x e−y ẑ)
∇ × (∇ × v) = x̂
∂x
∂y
∂
∂
= x̂ (2 cos x e−y ) − ŷ
(2 cos x e−y ) = x̂ (−2 cos x e−y ) + ŷ (2 sin x e−y ).
∂y
∂x
Either directly, or from ∇(∇ · v) − ∇ × (∇ × v), ∇2 v = 0.
(e) Compute any two of the integrals
I
I1 = (vx dx + vy dy),
I
I3 = (vx dx − vy dy),
I
I2 =
I4 =
I
(vx dy + vy dx),
(vx dy − vy dx),
around the perimeter of the square 0 < x < π2 , 0 < y < π2 . (Some choices may be easier
than others.)
Recall Green’s Theorem,
I
Z ∂Q ∂P (P dx + Q dy) =
−
dx dy.
∂y
C
S ∂x
So, the pairs with ∂Q/∂x = ∂P/∂y are easy: from the Cauchy-Riemann relations, I2 = I3 =
0. These are also the real and imaginary parts of
Z
Z
Z
f (z) dz =
(u + iv)(dx + idy) =
[(u dx − v dy) + i(u dy + v dx)] = I2 + i I3 = 0.
C
C
C
The other two combinations amount to Stokes’s theorem and the divergence theorem,
I
I1 =
C
Z
v · dr =
I
I4 =
S
(∇ × v) · ẑ dx dy = 2
Z
C
v · n̂ ds =
S
(∇ · v) dx dy = −2
Z π/2
Z π/2
cos x dx
0
Z π/2
0
0
Z π/2
sin x dx
0
e−y dy = 2(1 − e−π/2 ),
e−y dy = −2(1 − e−π/2 ).
Or, do the integral around the circumference directly, but that has four different pieces and
you need to worry about directions and/or signs as well as integrating. Here is one example,
first across the bottom, then up the right side, back across the top, and down the left side:
Z π/2
I1 =
0
+
−y
cos x e
Z 0
π/2
Z π/2
dx y=0
cos x e−y dx +
0
y=π/2
−y
dy sin x e
x=π/2
Z 0
+
sin x e−y dy π/2
= 1 + (1 − e−π/2 ) − e−π/2 − 0 = 2(1 − e−π/2 ).
x=0
2. In solving quantum mechanical scattering of a particle with energy E = h̄2 k 2 /2m from a
hard sphere of radius a, one finds from matching boundary conditions in a spherical harmonic
expansion that an amplitude coefficient α` is given by
n − i j`
1 + α` = − `
,
n` + i j`
where j` (ka) and n` (ka) are spherical Bessel functions.
(a) Write the complex number 1 + α as a magnitude and a phase, 1 + α = r eiθ . Find the
magnitude r.
p
n` − i j` |n` − i j` |
j 2 + n2
=
p
|1 + α| = −
= 1.
=
n` + i j` |n` + i j` |
j 2 + n2
This does not imply |α| = 0.
(b) Write the phase as θ = 2δ` . Find tan δ` in terms of j` , n` . Write α` in terms of sin δ` .
Show that |α` |2 = 4 sin2 δ` .
Write
n − ij
(n − ij)(n − ij)
j 2 − n2 − 2ijn
−
=−
=
.
n + ij
(n + ij)(n − ij)
n2 + j 2
Then,
2jn
2 cos δ sin δ
sin 2δ
tan 2δ = 2
=
.
=
cos 2δ
j − n2
cos2 δ − sin2 δ
This is equivalent to
cos δ = p
j
j 2 + n2
,
sin δ = p
n
j 2 + n2
,
tan δ =
With unit amplitude, 1 + α` = e2iδ` ,
α` = e2iδ` − 1 = −1 + cos 2δ` + i sin 2δ`
= −1 + cos2 δ` − sin2 δ` ) + 2i sin δ` cos δ`
= −2 sin2 δ` + 2i sin δ` cos δ` = 2i sin δ` eiδ` ,
|α` |2 = 4 sin2 δ` .
j
.
n
(Bonus) The functions j` and n` are such that the phase is small [δ` ≈ (ka/`)2`+1] when
ka ` and the phase is large [tan δ` ≈ tan ka] when ka `. The total scattering cross
section is found from
∞
π X
σscatt = 2
(2` + 1) |α` |2 .
2k
`=0
Estimate σscatt when ka 1. (Hint: When angles are large, they are effectively random,
h sin θ i = h cos θ i = 0, h sin2 θ i = h cos2 θ i = 12 .)
“Estimate” is not quite the same as “compute.” The sum breaks into ` < ka and ` > ka.
For ` > ka,
[(2` + 3)(ka)/(` + 1)]`+1 ka ` `
ka/e
→
→
→ 0,
`
`+1 `+1
`+1
(2` + 1)(ka/`)
and the sum converges. Terms fall off rapidly, and the sum is of order π ka; (the coefficient
is 4.54823 for ka = 1, 3.31118 for ka = 10, 3.17835 for ka = 100, 3.16538 for ka = 1000,
3.16410 for ka = 10000), most of which comes from the first term), but the exact form of
the sum is not important.
For 0 ≤ ` ≤ ka, the factor |α` |2 is h 4 sin2 δ` i = 2, independent of `. The sum of 2` + 1 is 1
for ka = 0, 1 + 3 = 4 for ka = 1, 1 + 3 + 5 = 9 for ka = 2, and in general
ka
X
(2` + 1) = (ka + 1)2 .
`=0
Thus,
σscatt ≈
π
π
[2(ka + 1)2 + πka] → 2 2(ka)2 = πa2 .
2
2k
2k
3. It is suggested that an analytic function of a complex variable F (z) with z = z +iy = r eiθ
can describe the velocity potential and streamline function for two-dimensional flow of an
incompressible fluid. Write F = Φ + i Ψ, where Φ and Ψ are real, and let the flow velocity
v = ∇Φ.
a2 (a) Let F = v0 z +
. Find v. Compute ∇ · v. (An incompressible fluid should have
z
∇ · v = 0.) What is the radial component of v at r = a? (Solid obstacles with surface S are
allowed to occupy positions where vn = n̂ · v = 0 along S.) Compute ∇ × v. (A fluid with
∇ × v = 0 is called “irrotational.”) Does your velocity plausibly describe irrotational fluid
flow around an infinite cylinder of radius a?
It is easiest to work in the polar representation, z = r eiθ ,
a2 −iθ a2 a2 a2 iθ
F = v0 z +
= v0 r +
= v0 r e +
e
cos θ + iv0 r −
sin θ.
z
r
r
r
Thus,
a2 Φ = v0 r +
cos θ,
r
and
v = ∇Φ = r̂
a2 Ψ = v0 r −
sin θ,
r
∂Φ
1 ∂Φ
+ θ̂
= v0 r̂ 1 −
∂r
r ∂θ
a2 cos
θ
−
v
θ̂
1+
0
r2
a2 sin θ.
r2
In r-θ components, it is easy to see that at r = a, vr = 0. The divergence is
i 1 ∂h i
a2 a2 1 ∂h rv0 1 − 2 cos θ +
v0 1 + 2 sin θ
∇·v =
r ∂r
r ∂θ
r
r
2
2
a
a
= v0 1 + 2 cos θ + v0 1 + 2 (− cos θ) = 0;
r
r
incompressible it is. The curl is
i 1 ∂h i
1 ∂h a2 a2 ∇×v =
rv0 1 + 2 sin θ −
v 1 − 2 cos θ
r ∂r
r ∂θ 0
r
r
a2 a2 = v0 1 − 2 sin θ + v0 1 − 2 sin θ) = 0,
r
r
and irrotational it is too. It all works.
For the cartesian representation z = x + iy, the function F
h
a2 a2 (x − iy) i
F = v0 z +
,
= v0 x + iy + 2
z
x + y2
the potentials are
a2 Φ = v0 x 1 + 2
,
x + y2
a2 Ψ = v0 y 1 − 2
,
x + y2
and the velocity is
v = v0 x̂
(x2 + y 2 )2 + a2 (y 2 − x2 )
2a2 xy
ŷ
−
v
.
0
(x2 + y 2 )4
(x2 + y 2)2
The divergence of that does vanish, as does the curl. If you do x-y components, you can
extract
x2 + y 2 − a2
,
rvr = r · v = x vx + y vy = v0 x
x2 + y 2
which does vanish at x2 + y 2 = a2 .
(b) Show that Ψ is constant along v.
If Ψ is constant along v, then the derivative in the v-direction vanishes
v · ∇Ψ = vr
∂Ψ vθ ∂Ψ
+
= 0.
∂r
r ∂θ
This works for the explicit form provided, or in general from the Cauchy-Riemann relations,
v · ∇Ψ =
since
∂Φ ∂Ψ 1 ∂Φ 1 ∂Ψ
+
=0
∂r ∂r
r ∂θ r ∂θ
∂Φ
1 ∂Φ
=
,
∂r
r ∂θ
1 ∂Ψ
∂Ψ
=−
.
r ∂θ
∂r
a2 (c) Let F 0 = v0 z +
− i v1 ln z. What about the resulting flow v 0 = ∇Φ0 is different
z
from the result in part (a)?
This adds to F the function
iv1 ln z = iv1 (ln r + i θ) = −v1 θ + iv1 ln r.
So, in Φ there is an additional term −v1 θ, which adds an additional piece to the θ-velocity
v
vθ0 = − 1 .
r
This piece has no divergence, ∂vθ0 /∂θ = 0, and it has no curl, ∂/∂r(rvφ0 ) = 0, and it has no
radial component at r = a.
H
It does add a nonzero circulation, v · dr = 2πv1 , which means there is a nonzero curl
somewhere.