1 Home exercise sheet 1.1 Generalized coordinates

1
1.1
Home exercise sheet
Generalized coordinates
Exercise 1.1: The thingy
Four massless rods of length L are connected at their ends to form a rhombus. A particle
of mass M is attached to each vertex. The opposite corners are joined by springs of spring
constant k. In the square configuration, the strings are unstretched. The motion is confined
to a plane, and the particles move only along the diagonals of the rhombus. How many
degrees of freedom are there in the system? Find appropriate generalized coordinates and
describe the kinetic and potential energy in terms of these.
√
Let a be the equilibrium length of the strings, namely a = L 2. The masses are located at
the positions (X± , 0) and (0, ±Y ) where X = L cos φ and Y = L sin φ. Moreover, the spring
extensions are δX = 2X − a and δY = 2Y − a. Hence the kinetic and potential energies are:
a2
T = m Ẋ 2 + Ẏ 2 = m φ̇2
2
2 √
2 √
1
1
2
2
2
U =
k δX + δY = ka
2 cos φ − 1 +
2 sin φ − 1
2
2
1
1.2
Calculus of variation
Exercise 1.3: Geodesic on a sphere
A sphere of radius R is a set of points which satisfy the following:
x(θ, φ) = R sin θ cos φ, y(θ, φ) = R sin θ sin φ, z(θ) = R cos θ
(1)
Find the Geodesics on a sphere.
Note: a geodesic is a generalization of a straight line in curved space, namely geodesics are
(locally) the shortest path between points in the curved space.
Recalling that S =
following
R
ds and also that ds2 = dr2 + r2 dθ2 + r2 sin2 θdφ2 we derive the
r
r
Z
Z q
2
dφ
dφ2
R2 dθ2 + R2 sin2 θdφ2 = R dθ 1 + sin2 θ 2 → L = 1 + sin2 θ 2
S=
dθ
dθ
(2)
We can now use the Euler-Lagrange equations to obtain the equations of motion (the
geodesic)
d ∂L
∂L
=0
=
0
dθ ∂φ
∂φ
sin2 θφ0
∂L
p
=
∂φ0
1 + sin2 θφ02
(3)
(4)
(5)
Giving us the differential equation √ sin
2
θφ0
1+sin2 θφ02
φ0 =
= c which simplifies into
c
p
sin θ sin2 θ − c2
(6)
Using u = cot θ we solve the integral and receive
√
1 − c2
cos(φ − φ0 )
cot θ =
c
2
(7)
This is the geodesic of a sphere, where c and φ0 are dependent on the initial conditions.
Exercise 1.3: Shape of a bubble
Find the curve c passing through two given points A(x1 , y1 ) and B(x2 , y2 ) such that the
rotation of the curve c about the x-axis generates a surface of revolution having minimum
surface area.
Note: Start by thinking what you need to minimize and see how it relates to the infinitesimal
distance ds (which you know in Euclidean coordinates).
Note: No need to find the constants of integration
We start by looking at a infinitesimal area dA = 2πyds where y is the radius and ds is
the infinitesimal distance (In Euclidean coordinates).
Z
Z x2 p
(8)
I =
dA = 2π
y 1 + y 02 dx
x1
p
f = y 1 + y 02
(9)
All that is left now is to use E-L and do some algebra.
∂f
∂y
p
1 + y 02
∂f
yy 0
p
=
∂y 0
1 + y 02
yy 00
yy 02 y 00
y 02
+p
−
= p
3
q + y 02
1 + y 02 (1 + y 02 ) 2
=
p
1 + y 02
;
3
(10)
(11)
After some algebra we receive
1 + y 02 − yy 00 = 0
(12)
This equation is complicated to solve, you can use a program to solve this but there is a
simpler way.
p
We start with the same functional f = y 1 + y 02 . Using the Beltrami identity (if we haven’t
done it in class you can find the full derivation on Wikipedia) we have
d
∂f
0 ∂f
f −y 0 =
(13)
dx
∂y
∂x
Since our functional is independent of x we get that
f − y0
∂f
= c1
∂y 0
p
y 02 y
y q + y 02 − p
= c1
1 + y 02
(14)
s
⇒
y0 =
y 2 − c21
c21
(15)
Finally giving us the expression for y, where the constants of motion are found by using the
initial (boundary) conditions.
x − c2
y = c1 cosh
(16)
c1
4