Physics 5040
Spring 2009
Problem Set 2 Solutions
1. In plane polar coordinates we have
ds2 = dr2 + r2 dθ2
or
ds = (ṙ2 + r2 θ̇2 )1/2 dt .
The functional to be minimized is
Z 2
Z
ds
I=
dt = (ṙ2 + r2 θ̇2 ) dt .
dt
We have (where f denotes the integrand)
∂f
= 2rθ̇2
∂r
∂f
= 2ṙ
∂ ṙ
∂f
=0
∂θ
∂f
∂ θ̇
= 2r2 θ̇
Then the Euler-Lagrange equations for r and θ are
∂f
d ∂f
−
= r̈ − rθ̇2 = 0
dt ∂ ṙ
∂r
and
d ∂f
∂f
= 2rṙθ̇ + r2 θ̈ = 0 .
−
dt ∂ θ̇
∂θ
This last equation may be written as
2
θ̈ + ṙθ̇ = 0 .
r
2. The line element in plane polar coordinates is
2
ds2 = dr2 + r2 dθ2 = (1 + r2 θ′ ) dr2
where θ′ = dθ/dr. Minimizing the curve length means that we want to minimize the integral
Z
Z
2
I = ds = (1 + r2 θ′ )1/2 dr .
Here the independent variable is r, and since ∂f /∂θ = 0 (where again we let
f denote the integrand) we have ∂f /∂θ′ = const := c. Therefore
∂f
r2 θ ′
=c
=
′
∂θ
(1 + r2 θ′ 2 )1/2
1
so that
2
2
r4 θ′ = c2 (1 + r2 θ′ )
or
θ′ =
Integrating this we have
θ − θ0 =
c
.
r(r2 − c2 )1/2
Z
r(r2
c
dr .
− c2 )1/2
To evaluate this integral, let
r=
c
cos u
so that
dr =
c sin u
du .
cos2 u
Using (r2 − c2 )1/2 = c sin u/ cos u we see that
c
dr = du
r(r2 − c2 )1/2
and hence
θ − θ0 = u = cos−1
or
c
r
r cos(θ − θ0 ) = c .
One way to see that this is the equation of a straight line perpendicular to
the line θ = θ0 is to look at the figure below:
P = P (r, θ)
r
θ − θ0
θ
r=c
θ0
Note that any point P on the line shown has the property that r cos(θ−θ0 ) = c.
Another way to see this is as follows. First note that the relation between
rectangular and polar coordinates is given by x = r cos θ and y = r sin θ. Then
c = r cos(θ − θ0 ) = r cos θ cos θ0 + r sin θ sin θ0
= x cos θ0 + y sin θ0
2
or
cos θ0
c
x+
.
sin θ0
sin θ0
To see that this represents the line perpendicular to the line θ = θ0 , note that
the line θ = θ0 has slope tan θ0 , and the line perpendicular to this has slope
y=−
tan(90◦ + θ0 ) =
1
cos θ0
sin(90◦ + θ0 )
=−
.
=−
cos(90◦ + θ0 )
sin θ0
tan θ0
slope = − tan1 θ0
slope = tan θ0
r=c
θ0
At a larger scale that shows the y-intercept b, this is
b
θ0
c
θ0
From the figure we see that
c
b
which implies that the y-intercept is given by
sin θ0 =
b=
c
.
sin θ0
Therefore the line through the point (r = c, θ = θ0 ) and perpendicular to the
line θ = θ0 is just y = mx + b where
m=−
cos θ0
1
=−
sin θ0
tan θ0
3
and
b=
c
.
sin θ0
3. We have
ds =
where y ′ = dy/dx. Then
I=
q
p
dx2 + dy 2 = 1 + y ′ 2 dx
Z
ds
=
u(y)
Z p
1 + y′2
dx
u(y)
where the independent variable is x. Letting f denote the integrand as usual,
we see that ∂f /∂x = 0 so we may use the second form of Euler’s equation.
We have
y′
∂f
p
=
′
∂y
u(y) 1 + y ′ 2
and therefore Euler’s equation becomes
2
or
y′
∂f ′
p
y −f =
−
′
∂y
u(y) 1 + y ′ 2
p
1 + y′2
= const := −k
u(y)
−1
p
= −k .
u(y) 1 + y ′ 2
Referring to the figure below, we see that y ′ = tan ϕ and hence
q
1
1
1
1 + y ′ 2 = sec ϕ =
=
=
.
cos ϕ
cos(90◦ − θ)
sin θ
y
θ
ϕ
• (x2 , y2 )
y(x)
•
(x1 , y1 )
x
Therefore we have
1
sin θ
p
=
= k.
2
u(y)
u(y) 1 + y ′
4. Starting from the geodesic equation
d2 xl
dxi dxj
+ Γlij
= 0,
2
dt
dt dt
4
let t → σ = f (t). Then
and
dxl
dσ dxl
dxl
=
= f˙
dt
dt dσ
dσ
2 l
l
l
d dxl
d2 xl
˙ d f˙ dx = f˙2 d x + f˙ dx dt f¨ .
=
=
f
dt2
dt dt
dσ
dσ
dσ 2
dσ dσ
But
1
1
dt
=
=
dσ
dσ/dt
f˙
so
2 l
l
d2 xl
˙2 d x + f¨dx .
=
f
dt2
dσ 2
dσ
Alternatively, write
l
l
l
2 l
d2 xl
d ˙ dxl
¨dx + f˙ d dx = f¨dx + f˙2 d x .
=
f
f
=
dt2
dt
dσ
dσ
dt dσ
dσ
dσ 2
Either way, the geodesic equation then becomes
dxl
dxi dxj
d2 xl
+ f˙2 Γlij
=0
f˙2 2 + f¨
dσ
dσ
dσ dσ
or
j
i
f¨ dxl
d2 xl
l dx dx
+
Γ
=−
.
ij
2
dσ
dσ dσ
f˙2 dσ
Note the geodesic equation remains unchanged under affine transformations
t → σ = f (t) = at + b because then f¨ = 0.
5. We evaluate using d/dt = ẋi ∂/∂xi and the fact that gij = gji :
d
(gij ẋi ẋj ) = gij,k ẋk ẋi ẋj + gij ẍi ẋj + gij ẋi ẍj
dt
= gij,k ẋk ẋi ẋj + 2gij ẍi ẋj .
We substitute for ẍi using the geodesic equation in the form
1
ẍi = −Γilm ẋl ẋm = − g in (gnm,l + gnl,m − glm,n )ẋl ẋm
2
to obtain
d
(gij ẋi ẋj ) = gij,k ẋk ẋi ẋj − gij g in (gnm,l + gnl,m − glm,n )ẋj ẋl ẋm .
dt
But gij g in = δjn so gij g in ẋj = ẋn and we have
d
(gij ẋi ẋj ) = gij,k ẋk ẋi ẋj − (gnm,l + gnl,m − glm,n )ẋn ẋm ẋl .
dt
5
Noting that i, j, k and n, m, l are all just dummy indices, we see that all four
terms on the right hand side of this equation are the same (up to sign) and
hence
d
(gij ẋi ẋj ) = 0
along a geodesic.
dt
6. On the sphere we have ds2 = r2 dθ2 +r2 sin2 θ dϕ2 so the metric and its inverse
are given by
2
r
1/r2
ij
gij =
and
g
=
.
r2 sin2 θ
1/r2 sin2 θ
The Christoffel symbols are
Γlij =
1 lk
g
2
∂gkj
∂gki
∂gij
+
−
∂xi
∂xj
∂xk
and
1 ∂gkj
∂gki
∂gij
.
+
−
2 ∂xi
∂xj
∂xk
Note that here gij is diagonal, and r = const on the sphere, so the only nonvanishing term ∂gkj /∂xi is ∂gϕϕ /∂θ. Then the only nonvanishing Christoffel
symbols are
Γkij = gkl Γlij =
Γθϕϕ = −
1 ∂gϕϕ
= −r2 sin θ cos θ
2 ∂θ
1 ∂gϕϕ
= r2 sin θ cos θ .
2 ∂θ
Γϕθϕ = Γϕϕθ =
Therefore
Γθϕϕ = g θi Γiϕϕ = g θθ Γθϕϕ = − sin θ cos θ
ϕ
ϕi
ϕϕ
Γϕ
Γϕθϕ =
θϕ = Γϕθ = g Γiθϕ = g
cos θ
sin θ
and the geodesic equations become
θ̈ − sin θ cos θ ϕ̇2 = 0
and
ϕ̈ + 2
cos θ
θ̇ϕ̇ = 0 .
sin θ
A solution of these coupled equations is ϕ̇ = θ̈ = 0. Then ϕ̇ = 0 implies
ϕ = const which we might as well take to be ϕ = 0 (i.e., the xz-plane). And
θ̈ = 0 implies θ = at + b, so taking θ(t = 0) = 0 implies θ = at (i.e., start on
the z-axis).
A point on the sphere is given by x = (r sin θ cos ϕ, r sin θ sin ϕ, r cos θ), so
our geodesic becomes x = (r sin at, 0, r cos at) which is just the parametric
equation of a circle of radius r in the xz-plane, i.e., a great circle.
6
7. (a) To extremize I(ε) =
dI
=
dε
R
Z
f (e
x, x,
ė x,
ë t) dt we proceed as usual:
t1
t0
ė ∂f ∂ x
ë
∂f ∂e
x ∂f ∂ x
dt .
+
+
∂e
x ∂ε
∂x
ė ∂ε
∂x
ë ∂ε
For the second term in the integrand we have (as usual)
Z t1 Z t1
Z t1
x
d ∂f ∂e
∂f d ∂e
x
∂f ∂
dt = −
x
ė dt =
dt
∂ε
dt
∂ε
dt
∂ε
ė
ė
∂x
ė
t0
t0 ∂ x
t0 ∂ x
where the boundary terms vanish at the endpoints because
∂e
x ∂e
x =
= 0.
∂ε t0
∂ε t1
For the third term in the integrand we integrate by parts twice. First we
have
Z t1
Z t1
ė
ė
∂f d ∂ x
∂f ∂ dx
dt =
dt
∂
ẍ
∂ε
dt
dt
∂ε
ë
t0 ∂ x
t0
Z t1 d ∂f ∂ x
d ∂f ∂ x
ė
ė
−
dt .
=
dt
∂ε
dt
∂ε
∂
x
ë
∂
x
ë
t0
Since we are assuming that x
e(t, ε) has continuous second derivatives, we
also assume that
∂x
ė ∂x
ė =
=0
∂ε t0
∂ε t1
so the boundary term in this last integral vanishes and we have
Z t1 Z t1 Z t1
d ∂f ∂ x
d ∂f d ∂e
ė
x
∂f ∂ dx
ė
dt
dt = −
dt = −
∂ε
dt
dt
∂ε
dt
dt
∂ε
ë
∂x
ë
∂x
ë
t0
t0
t0 ∂ x
Z t1 Z t1 2 d d ∂f ∂e
x
x
d
∂f ∂e
=−
dt +
dt .
2
dt
dt
∂ε
dt
∂ε
∂
x
ë
∂
x
ë
t0
t0
The first integral on the right hand side again vanishes because of our
x/∂ε|t1 = 0. Putting these results
first boundary condition ∂e
x/∂ε|t0 = ∂e
together we have
Z t1 dI
∂f
d ∂f
d2 ∂f
∂e
x
=
−
dt .
+ 2
dε
∂e
x dt ∂ x
dt ∂ x
∂ε
ė
ë
t0
Since ∂e
x/∂ε|ε=0 is arbitrary, we see that dI/dε|ε=0 = 0 yields our desired
equation
∂f
d2 ∂f
d ∂f
+ 2
= 0.
−
∂x dt ∂ ẋ
dt ∂ ẍ
7
(b) Using
1
1
L = − mq q̈ − kq 2
2
2
we have
∂L
1
= − mq̈ − kq
∂q
2
∂L
=0
∂ q̇
∂L
1
= − mq .
∂ q̈
2
Therefore the equation derived in part (a) yields
1
1
− mq̈ − kq − mq̈ = 0
2
2
or simply
q̈ +
k
q = 0.
m
This is a simple harmonic oscillator of frequency ω =
p
k/m.
8. A point on the surface of a right circular cylinder has the coordinates
x = (ρ cos ϕ, ρ sin ϕ, z)
as shown on the figure below:
z
ρ
x
z
y
ϕ
x
Then dx = ρdϕϕ̂ + dzẑ so that ds2 = ρ2 dϕ2 + dz 2 and
2
ρ
gij =
.
1
But ρ is constant, so all of the Christoffel symbols vanish, and the geodesic
equations become simply
d2 z
d2 ϕ
=0= 2 .
2
dt
dt
8
These have the solutions ϕ = ωt + α and z = kt + β where α, β, ω and k are
all constants. If we choose out initial conditions to be ϕ(0) = z(0) = 0, then
we have simply ϕ = ωt and z = kt. Therefore the geodesic becomes
x = (ρ cos ωt, ρ sin ωt, kt)
which is the equation of a circular helix of radius ρ.
9. Our functional is
F [y] =
Z
sin y(x)
dx .
y(x)
From the definition of functional derivative we have (writing simply y instead
of y(x) for much of the calculation)
δF [y]
F [y(x) + εδ(x − x0 )] − F [y]
= lim
δy(x0 ) ε→0
ε
Z sin[y + εδ(x − x0 )] sin y
1
dx
−
= lim
ε→0 ε
y + εδ(x − x0 )
y
Z 1
y sin[y + εδ(x − x0 )] − [y + εδ(x − x0 )] sin y
dx
= lim
ε→0 ε
y(y + εδ(x − x0 ))
Z 1
y[sin y + εδ(x − x0 ) cos y] − [y + εδ(x − x0 )] sin y
= lim
dx
ε→0 ε
y[y + εδ(x − x0 )]
Z εδ(x − x0 )[y cos y − sin y]
1
dx
= lim
ε→0 ε
y[y + εδ(x − x0 )]
Z δ(x − x0 )[y cos y − sin y]
dx
= lim
ε→0
y[y + εδ(x − x0 )]
Z δ(x − x0 )[y(x) cos y(x) − sin y(x)]
dx
=
y(x)2
=
y(x0 ) cos y(x0 ) − sin y(x0 )
.
y(x0 )2
Therefore
cos y(x0 ) sin y(x)
δF [y]
.
=
−
δy(x0 )
y(x0 )
y(x0 )2
10. (a) Start from
Z
1
(xẏ − ẋy) dt .
2
Change variables to polar coordinates in the usual way with x = r cos θ
and y = r sin θ. Then
A=
ẋ = ṙ cos θ − rθ̇ sin θ
and
9
ẏ = ṙ sin θ + rθ̇ cos θ
so that
xẏ − ẋy = r cos θ(ṙ sin θ + rθ̇ cos θ) − (ṙ cos θ − rθ̇ sin θ)r sin θ
= r2 θ̇ .
Therefore
A=
1
2
Z
r2 θ̇ dt =
1
2
Z
r2 dθ .
(b) We want to minimize the length
Z
Z
I = ds = (ẋ2 + ẏ 2 )1/2 dt
subject to the constraint
1
J=
2
Z
(xẏ − ẋy) dt = const .
Applying equation (21) of the notes, we first form the function
h = f + λg = (ẋ2 + ẏ 2 )1/2 +
λ
(xẏ − ẋy)
2
and then substitute into the Euler-Lagrange equations:
ẋ
∂h
d ∂h
λ
d
λ
−
= ẏ −
0=
− y
∂x dt ∂ ẋ
2
dt (ẋ2 + ẏ 2 )1/2
2
ẋ
d
= λẏ −
2
dt (ẋ + ẏ 2 )1/2
and
ẏ
d ∂h
λ
d
∂h
λ
−
= − ẋ −
0=
+ x
∂y
dt ∂ ẏ
2
dt (ẋ2 + ẏ 2 )1/2
2
ẏ
d
= −λẋ −
.
dt (ẋ2 + ẏ 2 )1/2
We integrate both of these to write
λ(y − y0 ) =
ẋ
(ẋ2 + ẏ 2 )1/2
and
λ(x − x0 ) = −
ẏ
.
(ẋ2 + ẏ 2 )1/2
Squaring and adding we have
λ2 [(x − x0 )2 + (y − y0 )2 ] =
10
ẋ2 + ẏ 2
=1
ẋ2 + ẏ 2
or
1
.
λ2
This is the equation of a circle centered at (x0 , y0 ) (determined by some
given initial location of the curve) and with radius 1/λ. Since the area
A is given, it follows that
(x − x0 )2 + (y − y0 )2 =
A = πr2 =
or
π
λ2
1/2
π
λ=
.
A
11. Here is the general situation:
y
x
da = y dx
From ds2 = dx2 + dy 2 we have dx2 = ds2 − dy 2 or
q
p
2
2
dx = ds − dy = 1 − y ′ 2 ds
where y ′ = dy/ds. Then
A=
Z
y dx =
Z
l
0
Z l
q
2
′
y 1 − y ds =
f (y, y ′ , s) ds .
0
Since ∂f /∂s = 0 we may use the second form of Euler’s equation
∂f ′
y − f = const
∂y ′
which becomes
2
−yy ′
p
−y
1 − y′2
q
−y
=R
1 − y′2 = p
1 − y′2
2
where we have labeled the constant by R. Squaring we have y ′ = 1−y 2/R2 =
(R2 − y 2 )/R2 so that
p
dy
R2 − y 2
′
y =
=
ds
R
11
which implies
Z
ds = R
Z
dy
p
.
R2 − y 2
p
To integrate this, let y = R sin θ so dy = R cos θ dθ and R2 − y 2 = R cos θ.
Then s = Rθ + const = R sin−1 (y/R) + const. Since s = 0 at y = 0 we have
the constant of integration equal to zero, and hence
y
s
s
or
y = R sin
.
= sin
R
R
R
At the other end of the string we also have y = 0 at s = l, and therefore
l/R = nπ. In fact, we must have l/R = π because l is fixed, so if l = nπR,
then a larger n means a smaller R. As we will see below, this means a smaller
area, and we are looking for a maximum.
p
1 − y ′ 2 ds = sin(s/R) ds so that
Z
s
s
ds = −R cos
+ const .
x = sin
R
R
Next we have dx =
But x = 0 at s = 0 implies that the constant is equal to R, which in turn
implies that x = R − R cos(s/R). Together with y = R sin(s/R) we see that
(x − R)2 + y 2 = R2
so the string is a semi-circle of radius R centered at (R, 0). This also verifies
the claim made above that a smaller R means a smaller area.
12. Let the parallelepiped have sides a, b, c so its volume is V = abc. We want to
maximize this volume subject to the constraint
2 2 2
b
c
a
+
+
= R2 .
2
2
2
(By definition, the sides of the parallelepiped are parallel, and by symmetry
each side must be centered with respect to the center of the sphere.) Thus
our constraint equation becomes
g(a, b, c) = a2 + b2 + c2 − 4R2 = 0 .
Then h = V +λg = abc+λ(a2 +b2 +c2 −4R2 ) and we have ∇h = ∇(V +λg) = 0
or
∂h
= bc + 2λa =⇒ bc = −2λa
(1)
0=
∂a
0=
∂h
= ac + 2λb
∂b
=⇒
ac = −2λb
(2)
0=
∂h
= ab + 2λc
∂c
=⇒
ab = −2λc
(3)
12
Dividing equation (1) by equation (2) yields b/a = a/b or a = b. And dividing
(2) by (3) shows that b = c. Therefore a = b = c so that 3a2 = 4R2 or
2
a = b = c = √ R.
3
13