MATHEMATICAL METHODS III—CALCULUS OF VARIATIONS (Answers)
Q3 All of these correspond to horizontally stratified media. The integrand in the
expression for the time taken between points does not depend on the variable x. Hence we
have the first integral:
f − y′
1
∂f
=
= constant
′
∂y
v(y) (1 + y ′2 )1/2
3i) This is the same as Q5.
ii) Write the first integral as
2
′2
y(1 + y ) = κ
(κ2 − y)
dy
=
dx
y 1/2
or
1/2
Now use the substition y = κ2 sin2 θ, dy = 2κ2 sin θ cos θ dθ. Find
Z
sin 2θ
2
2
2
2 sin θ dθ = xc + κ θ −
x = xc + κ
2
2
2
y = κ sin θ.
Here xc is some constant. The answer is best left in this parametric form.
3iii) Here we can write
√
(1 +
y
y ′2 )1/2
=
1
κ
y ′ = κ2 y − 1
⇒
which can be integrated directly to give
1/2
2
2
κ
y
−
1
= x − xc
κ2
so
1/2
,
κ2 (x − xc )2
1
y=
+ 2
4
κ
p
Q5 Area of a ribbon is its width, (1 + r ′ (z)2 )dz times its length 2πr(z). The area of the
film is the integral given in the question. The integrand does not depend explicitly on z,
hence we have the first integral (c is a constant)
f − r′
∂f
2πr
= 2πc.
=
′
∂r
(1 + r ′ (z)2 )1/2
Rearranging gives r ′ = ((r/c)2 − 1)1/2 which can be integrated using the substitution
r = c cosh θ, dr = c sinh θdθ to give (z0 is a constant)
z − z0
z − z0
θ=
.
so r(z) = c cosh
c
c
Symmetry between z and −z means z0 = 0 for our problem and the boundary condition
gives a = c cosh(b/c) as required.
If we rewrite the boundary condition in terms of a′ = a/b and c′ = c/b we obtain the
condition
1
a′ = c′ cosh ′ ,
c
′
which has no solutions for real c for small enough a′ <∼ 1.5089. In fact for
a′ >∼ 1.5089 there are two solutions - one gives the minimal surface area while other
gives a critical point but not a minimum. We have not been worrying about how to classify
critical points in this module.
Q6 Normalising the trial wavefunction gives
Z a
Z a
2
2
2 2
1 =
A a − x dx =
A2 a4 − 2a2 x2 + x4 dx
−a
−a
1
2
= 2A2 a5 1 − +
3 5
2 5
8A2 a3
5
16A a
or
= 2.
1 =
15
3
2a
Then (after integrating once by parts and using the boundary condition as in the lectures)
we have
2 Z a
Z a
8
dφ
hEi =
dx
=
4A2 x2 dx = A2 a3
dx
3
−a
−a
5
=
,
2a2
which is very close to the exact value of (but, as it should be, slightly above) the lowest
eigenvalue: π 2 /4a2 .
Q7 Eigenvalue is 1 (substitute in and get the answer). This is the harmonic oscillator
problem. Normalisation of the trial wavefunction is as in question 6, A2 = 15a5 /16, as is
the expectation value of the kinetic energy. Then
!
Z a 2
dφ
2
hEi =
+ A2 x2 a2 − x2
dx
dx
−a
a
4 3
8A2 a3
2a2 x5 x7
2 a x
=
+A
−
+
3
3
5
7 −a
2 3
4
8A a
2a
=
1+
3
35
2
2a
5 1
.
+
=
2 a2
35
p
This expression is minimised by choosing a2 = 35/2 giving
r
10
.
hEi =
7
2
which is larger than one but it’s not too bad given the crudity of the approximation
(approximating a Gaussian by a chopped off parabola).
Q8 Euler-Lagrange equations for r(t) are (one eqn for each component of r):
d ∂L
∂L
= 0.
−
∂xi dt ∂ ẋi
So (using the summation convention)
∂Aj
∂φ
d
−q
− [mẋi + qAi ]
∂xi
∂xi dt
∂Aj
∂φ
∂Ai
∂
= q ẋj
Ai
−q
− mẍi − q
− q ẋj
∂xi
∂xi
∂t
∂xj
0 = q ẋj
Hence
∂Ai
∂Ai
∂φ
∂Aj
−q
− ẋj
mẍi = −q
+ q ẋj
∂xi
∂t
∂xi
∂xj
|
{z
} |
{z
}
qEi
(q ṙ∧B)i
To identify the last two terms with the Lorentz force, you need to show explicitly
ṙ ∧ B = ṙ ∧ (∇ ∧ A).
Q9 (Last Part) The dimension of the electric field can be found from the fact that in SI
units, we have
ǫ0 ESI2
Energy
=
.
2
Volume
Then
E ′2
Energy
ǫ0 ESI2 = ǫ0
⇒ E ′2 = (~c)3
= Energy4 .
3
ǫ0 (~c)
Volume
Hence E′ has the dimensions of energy-squared. A similar argument gives the same
dimensions for B′ .
Write the equation containing j in terms of the scaled variables
∂E
∂t
1 ∂ E′
= j+
~c ∂t′ (~c)3/2
∂E′
√
= j′ + ′
with j′ = (~c)5/2 j = (~c)5/2 µ0 jSI
∂t
∇∧B = j+
1 ′
B′
∇ ∧
~c
(~c)3/2
∇′ ∧ B ′
The dimension of j ′ is energy cubed (each time we differentiate with respect to scaled
position or time we increase the dimension of energy by 1). It follows that the scaled
conductivity has the dimension of energy and
r
µ0
′
σSI .
σ = ~c
ǫ0
3
Q10 Four independent (xµ ) and four dependent (Aµ ) variables so
∂L
∂L
− ∂ρ
0 =
∂Aσ
∂(∂ρ Aσ )
1
= −j σ + ∂ρ η µα η νβ [(δαρ δβσ − δβρ δασ ) (∂µ Aν − ∂ν Aµ )] ;
2
1
∂ρ [η µρ η νσ (∂µ Aν − ∂ν Aµ ) − η µσ η νρ (∂µ Aν − ∂ν Aµ )]
⇒ jσ =
2
= ∂ρ (∂ ρ Aσ − ∂ σ Aρ )
= ∂ρ F ρσ .
Q11
S=
Z
d4 x η µν ∂µ φ∂ν φ − m2 φ2
{z
}
|
f (xµ ,φ,∂µ φ)
The Euler-Lagrange equation is then
0 =
=
=
=
2
⇒ −∂t φ =
∂f
∂f
− ∂α
∂φ
∂ (∂α φ)
2
−2m φ − 2∂α η µν (δαµ ∂ν φ)
−2m2 φ − 2∂α (η αν ∂ν φ)
−2m2 φ − 2∂α ∂ α φ;
m2 φ − ∇2 φ.
This is consistent with classical result (c = 1): E 2 = m2 c4 + p2 c2 , if we make the usual
identifications E → i~∂t and p → −i~∇.
Q12 This questions looks daunting but is actually quite simple so long as you are accurate
with your notation and prepared to relabel dummy (summed over) indices at will.
Z
µ
S [x (λ)] = dλgαβ ẋα ẋβ .
Here λ is the independent variable and the xα are the four dependent variables. E-L gives
d ∂f
∂f
−
0 =
∂xµ dλ ∂ ẋµ
d gαβ δµα ẋβ + gαβ ẋα δ βµ
= gαβ,µ ẋα ẋβ −
dλ
= gαβ,µ ẋα ẋβ − gµβ ẍβ + gαµ ẍα + gµβ,ν ẋβ ẋν + gαµ,ν ẋα ẋν .
Here we have had to take account of the fact that the full derivative with respect to λ picks
up terms because g depends on the xν (λ) and I have written the Kronecker delta so that
the summation convention on Greek indices is respected (but note that δαβ = δαβ ). Now,
using the property gαβ = gβα and renaming some of the dummy indices, we obtain
0 = −2gµβ ẍβ − (gµγ,β + gβµ,γ − gγβ,µ ) ẋβ ẋγ .
4
Multiplying by −g αµ /2 gives
1
0 = g αµ gµβ ẍβ + g αµ (gµγ,β + gβµ,γ − gγβ,µ ) ẋβ ẋγ
2
α
α
β γ
≡ ẍ + Γ γβ ẋ ẋ .
5
MATHEMATICAL METHODS III—COMPLEX VARIABLES (Answers)
Q5 a) Limit exists and is equal to 1. b) Unique limit does not exist.
Q6 a) f (z) = −iz + ic, b) f (z) = ic + z1 , c) f (z) = i(c − log z), c real.
Q7
u = x2 − y 2 = (x + y)(x − y),
v = 2xy
Lines of constant u and constant v are hyperbolae which intersect at right angles. This
follows from
∇u.∇v = (2x, −2y).(2y, 2x) = 0.
(∇u and ∇v are perpendicular to lines of constant u and constant v respectively.)
For the general case, we can use the Cauchy-Riemann equations to obtain
∂u ∂u
∂v ∂v
∂v ∂v
∂v ∂v
∇u.∇v =
.
=
.
= 0.
,
,
,−
,
∂x ∂y
∂x ∂y
∂y ∂x
∂x ∂y
Q8 w(z) is well-defined, it is not zero and w ′ (z) (= −1/(z − 2)2 ) exists except at z = 2.
Hence the function is analytic and the mapping is conformal. (We did not go into this, but
conformal means that the local magnification M = |w ′ (z)| is independent of direction.
AF’s correspond to conformal mappings with local magnification of figures by |dw/dz|
and local rotation by arg dw/dz.) For z = 32 + iy
1
2
+ iy
+1 + i2y
,
=
−1 + i2y
+ iy
1 + 4y 2
=
= 1,
1 + 4y 2
w(z) =
|w|2
− 12
ie all points lie on a circle about the origin with radius 1.
R
Q9 Γ z dz = z 2 /2. Integral independent of path as z is analytic.
Then on Γ1 (let the end point be (x0 , y0 ))
Z
Z
∗
On 1st part of contour:
z dz =
On 2nd part of contour:
Z
z ∗ dz =
Z
(x0 ,0)
x dx =
(0,0)
(x0 ,y0 )
x20
2
(x − iy) idy = ix0 y0 +
(x0 ,0)
y02
2
Adding together gives (x20 + y02 )/2 + ix0 y0 .
On Γ2 , we have
On 1st part of contour:
On 2nd part of contour:
Z
Z
Z
∗
z dz =
(0,y0 )
(−iy) idy =
(0,0)
Z
z ∗ dz =
(x0 ,y0 )
(0,y0 )
6
y02
2
(x0 − iy0 ) dx =
x20
− ix0 y0
2
Adding together gives (x20 + y02 )/2 − ix0 y0 .
The two contours give different answers. We should expect this as f (z) = z ∗ is not an
analytic function.
Q10 Any contours not including the zeros α and β of the quadratic form will be OK. Here
(z − α)(z − β) = z 2 + z + 1. Also any contour which encloses both poles also gives zero
(but this is a special case).
Q11 Deform the contours to be circles centred on a. On the circular contour, C ′ , we have
z = a + reiθ and dz = ireiθ dθ giving
I
Z 2π
Z 2π
iθ m
iθ
m+1
=
re
ire dθ = ir
ei(m+1)θ dθ
C′
0
m6=−1
=
=
m=−1
0
0
2πi.
Q12 a)(i) f (z) = 1 + z + z 2 + z 3 . . .. Radius of convergence 1.
a)(ii)
2 3
z+1
z+1
z+1
+
+
+ ...,
f (z) = 1 +
2
2
2
which converges if |z + 1| < 2.
The radii of convergence in both parts (i) and (ii) are equal to the distance to the closest
singular point. This is at z = 1.
b)
−(3 + (z − 2))
1
(z − 2)
2 + (z − 2)
′
−(3 + z ) 1
where z ′ = z − 2,
=
z′
2 + z′
−1
−(3 + z ′ )
z′
=
1+
2z ′
2
!
′ ′ 2 ′ 3
′
−(3 + z )
−z
−z
−z
=
+
1+
+
+ ...
′
2z
2
2
2
f (z) =
=
∞
X
m=−1
Fm (z − 2)m .
Here Fm = −3/2 for m = −1 and Fm = (−1/2)m+2 for m ≥ 0.
13 (i) Simple poles at z = 0 and z = 2;
(ii) Simple poles at z = ±i and essential singularity at z = 0;
(iii) Singularities when ez = −1, ie when z = zn ≡ i(2n + 1)π. To classify the
7
′
′
singularities need to expand about the zn . Writing z = zn + z ′ and ez = ezn ez = −ez , we
have
1
1
′ =
′2
z
′
1−e
1 − 1 − z − z2! −
1
1
= − ′
′
z
z 1 − 2! − z3!′2 − . . .
1
≡ − ′ g(z ′ ),
z
f (z) =
z ′3
3!
−...
where g(z ′ ) is analytic at z ′ = 0. Hence poles at zn are simple.
iv) Branch points at z = 2 and z = −1
Q15 Add on semicircle in the upper half-plane (need this to ensure that eikz bounded for
large Imz) with radius |R| > 1. Then show that
Z ikz e dz lim lim O(1/R) → 0.
= R→∞
2
R→∞
⌢ z +1
Then take real parts to show that
I
Z ∞
cos kt
dt =
. . . = 2πi(residue at z = i) = πe−k .
2
C
−∞ t + 1
NB We needed the condition k > 0 to close in the upper half-plane. If we had k < 0 we
would have closed in the lower half-plane (as in the lectures).
√
Q16 Quadratic form has roots α± = 1 ± 3i, one of which is in the upper half-plane and
one in the lower half-plane. Here we can close in either the upper or lower half-plane.
Choosing the upper half-plane, the contribution from the semicircular bit of the closed
contour vanishes as R → ∞ by orders of magnitude. Hence the integral around the closed
contour in this limit is the one we want, which we can evaluate using the calculus of
residues:
Z ∞
I
dx
dz
π
= lim
= 2π(residue at z = α+ ) = √ .
2
R→∞ C (z − α+ )(z − α− )
3
−∞ x − 2x + 4
NdA
November 23, 2015
8