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Physics 200A Homework 6.4
Mark Derdzinski
December 2, 2013
Problem 4 Solution
Recall the Lagrangian density for a continuous string of length L with constant density µ and
tension τ clamped at both ends:
µ 2 τ 2
y − yx
(1)
2 t
2
where yt and yx are the derivatives of the position y(x, t) with respect to t or x. We want to
express the Lagrangian, and eventually the Hamiltonian, in terms of fourier coefficients. We will
expand y(x, t) in the (complete) basis of spatial eigenfunctions
L=
y(x, t) =
∞
X
Cn ρn (x) cos(ωn t + φn ) =
n=1
∞
X
An (t)ρn (x)
(2)
n=1
Where the spatial eigenfunctions ρn (x) are given by
ρn (x) =
∞
X
2 1
(
) 2 sin(kn x)
Lµ
(3)
n=1
Note our spatial eigenfunctions satisfy the orthonormality condition
Z
L
ρn (x)ρm (x)σdx = δnm
(4)
0
We are now equipped to describe the system in terms of the time-dependent fourier coefficients.
Suppressing x and t for clarity, the Lagrangian density expanded in fourier series becomes
"∞
#" ∞
#
"∞
#" ∞
#
X
X
µ X
τ X
dρn
dρm
L=
Ȧn ρn
Ȧm ρm −
An
Am
(5)
2
2
dx
dx
n=1
m=1
n=1
m=1
If we integrate over x to find the full Lagrangian, we can exploit the orthonormality of the ρn to
simplify the product of sums:
L(An , A˙n , t) =
Z
L
0
2
τ kn
µ
Ldx =
∞ ∞
i
1 X 2 τ kn2 2
1 Xh 2
Ȧn −
An =
Ȧn − ωn2 A2n
2
µ
2
n=1
(6)
n=1
Where we have used the fact
= c2 kn2 = wn2 . We are now equipped to find the Hamiltonian in
terms of the fourier coefficients. The generalized momentum is
πn =
dL
= A˙n
dA˙n
1
(7)
And so the Hamiltonian is given by
H(An , Ȧn , t) =
∞
X
n
πn Ȧn − L =
∞
1 X 2
πn + ωn2 A2n
2
(8)
n=1
The Hamiltonian EOM for the fourier coefficients is now simply
dH
dAn
=⇒ Än = −ωn2 An
π̇n = −
(9)
Note the Hamiltonian can be expressed in terms of
a+
n = πn + ıωn An
a−
n = πn − ıωn An
∞
1X + −
a a
=⇒ H =
2 n n n
(10)
−
The a+
n and an are analogous to creation and annihilation operators for our purely classical
system, where the zero-point energy is vanishing due to the classical commutator [πn , An ] = 0 .
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