Mathematics for Physics 3: Dynamics and Differential Equations
15
74
Lecture 15: Coupled oscillators: normal coordinates and energy
conservation
In the last lecture we considered the class of dynamical problems involving coupled oscillators. As an example we
took a system with three springs and two masses connected as in fig. 13. Regarding
the
�
� coordinates of the two
x1
particles with respect to equilibrium, x1 and x2 , as components of a vector x =
, led us to formulate the
x2
equations of motion in a matrix form:
�
�
D2 + G [x(t)] = 0
(234)
where the first term (D2 ) is proportional to the unit matrix in the x vector space, while G is non-diagonal. We now
consider the general solution for a set of linear differential equations of the form (234).
Normal coordinates
In the last lecture we have seen the normal mode ansatz of (227)
x(t) = p cos(ωt − φ)
(235)
lead to a solution to (234). Let us return to the more general problem and explain why this must be the case.
To this end we would like to describe the general procedure of diagonalizing and solving the homogeneous linear
system of differential equations in (234), recalling the relevant concepts from linear algebra. This discussion applies
also to systems with n components. The symmetric two-dimensional system (223) discussed above provides a simple
example.
If G is diagonalizable (note that real symmetric matrices are always diagonalizable, by orthogonal transformations
P T = P −1 ), namely, if there exists a matrix P such that
P −1 GP = Λ
where Λ is diagonal,
Λ=
⇐⇒
�
λ1
0
0
λ2
�
GP = P Λ
≡
�
ω12
0
0
ω22
�
(236)
,
then we can solve (234) as follows: let us multiply (234) from the left by P −1 and insert a unit matrix I = P P −1 to
the right of the operator there, obtaining:
�
�
P −1 D2 + G P P −1 [x] = 0
(237)
which can also be written as:
�
�
D2 + P −1 GP P −1 [x] = 0
(238)
where we used the fact that the differentiation operator commutes with P (P being time independent). Using now
the definition of the diagonalizing matrix (236) we get:
�
�
D2 + Λ [�
x(t)] = 0,
(239)
where we defined the normal coordinates by
x
� = P −1 x =
�
x
�(1)
x
�(2)
�
.
(240)
In this basis the equation is diagonal: the equation for x
�(j) with a particular j does not involve any of the other
coordinates. Each normal coordinate x
�(j) has a unique characteristic frequency ωj corresponding to the square root
of the j-th element on the diagonal of the diagonal matrix Λ. This is the j-th eigenvalue of G, a normal mode of
the system.
The equations in the new basis x
� are therefore decoupled. Each of them is a simple harmonic oscillator:
which as we know, is solved by
(D2 + ωj2 ) x
�(j) (t) = 0 ,
x
�(j) = β (j) cos(ωj t − φ(j) ) ,
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where β (j) and φ(j) are constants of integration (to be fixed eventually by the initial conditions). Recall now that
x
�(j) are known linear combinations of the original coordinates xi . We can therefore obtain the solution in the original
coordinates by
�
�
xi (t) =
Pij x
�(j) (t) =
Pij β (j) cos(ωj t − φ(j) )
j
j
or in a vector notation:
x(t) =
�
j
p(j) β (j) cos(ωj t − φ(j) )
where
p(j) =
�
P1j
P2j
(241)
�
is the j-th eigenvector of G, the one corresponding to the eigenvalue ωj2 .
We see here the general property that the columns of the diagonalizing matrix P are the eigenvectors {p(1) , p(2) }.
Indeed eq. (236) is equivalent to the statement that for any j, the eigenvector p(j) with the eigenvalue λj = ωj2 solve
the eigenvalue problem
Gp(j) = p(j) λj .
(242)
In the two dimensional system we simply have two solutions to the eigenvalue problem, thus two normal modes.
The general solution of (234) is then a linear combination of the two:
x(t) = p(1) β (1) cos(ω1 t − φ(1) ) + p(2) β (2) cos(ω2 t − φ(2) ) .
(243)
Similarly, in a system with n degrees of freedom, we will have n such modes (j will then go from 1 to n), and the
general solution will be a linear combination of all of them.
We note that the coefficients β (j) determine the overall magnitude of the oscillation in a given normal mode ωj .
For example, by preparing the initial conditions appropriately we can make all the β (j) but one, say βk , vanish, thus
putting the system into one of normal modes (the k-th mode). This would mean that all the components of the
system (mi ) will oscillate simultaneously with a common frequency ωk and common phase φk , but with different
amplitudes Pik , as prescribed by the eigenvector p(k) (of course some of the components mi may be at rest in this
particular mode, if Pik = 0). This generalises what we found above in the two-dimensional example.
Energy conservation
Our starting point in the previous lecture was the second Newton law. We analysed the system in fig. 13 writing
down the forces which the springs assert on the two masses:
�
�
F1 = −k1 y1 − l1
�
�
F2 = −k2 y2 − y1 − l2
(244)
�
�
F3 = −k3 L − y2 − l3 .
We will now analyse the same system starting from energy conservation. Instead of looking at the forces, we write
down the corresponding potential energy which is stored in each spring. we have:
�2
k1 �
y1 − l1
2
�2
k2 �
V2 =
y 2 − y1 − l 2
2
�2
k3 �
V3 =
L − y2 − l 3 .
2
(245)
�2 k �
�2 k �
�2
k1 �
2
3
y1 − l 1 +
y2 − y1 − l2 +
L − y2 − l3
2
2
2
(246)
V1 =
The total potential energy is
V (y1 , y2 ) = V1 + V2 + V3 =
Note that the total force (244) on a given body i can be obtained by differentiating V with respect to yi : F =
−∂V /∂yi . We can expand the potential around the minimum, which is the equilibrium point. At this point the
total force on each particle, and thus the first derivative of the potential vanish, F = −∂V /∂yi = 0. Because the
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76
potential is quadratic to begin with, the expansion will yields a constant plus a quadratic term in the displacements
from equilibrium, namely using y(t) = x(t) + η (where η is fixed such that F = −∂V /∂yi = 0) we get:
V (y1 , y2 ) = V (x1 , x2 ) + const
where
�2 k
k1 2 k2 �
3
x1 +
x2 − x1 + x22
(247)
2
2
2
Had we started with a more complicated potential, a quadratic form such as (247) would be obtained as a small
oscillation approximation.
Let us now turn our attention to the kinetic energy in this problem. It is
V (x1 , x2 ) =
T (ẋ1 , ẋ2 ) =
m1 2 m2 2
ẋ +
ẋ .
2 1
2 2
(248)
The total energy is therefore:
E =T +V =
�2 k
m1 2 m2 2
k1 2 k2 �
3
ẋ1 +
ẋ2 +
x1 +
x2 − x1 + x22 .
2
2
2
2
2
(249)
This equation can be written using matrices as follows:
E=
where
M=
�
m1
0
0
m2
�
1 T
1
ẋ M ẋ + xT Kx .
2
2
,
K=
�
k1 + k2
−k2
(250)
−k2
k2 + k3
�
.
At this point it is easy to make the connection with the second Newton law. Taking the derivative of E with respect
to time we get:
d
1
1
1
1
E = ẋT M ẍ + xT K ẋ + ẍT M ẋ + ẋT Kx
dt
2 �
2 �
2
2 �
�
1 T
1 T
= ẋ M ẍ + Kx + x K + ẍT M ẋ = 0 .
2
2
(251)
(252)
where we collected the terms such that it is easy to see that each square brackets vanishes owing to the equation
of motion M ẍ + Kx = 0 and its transpose. In the latter case one makes use of the fact that both M and K are
symmetric matrices.
√
Let us now consider (250). It is convenient to rescale the coordinates xi by the corresponding mi such that
√
zi = xi mi (note that here we use explicitly the fact that M is a diagonal matrix. If it were not, we would have to
diagonalise it first). This leads to the following expression for the energy:
E=
where
1 T
1
ż ż + z T Gz .
2
2
(253)
G = M −1/2 KM −1/2
where
M −1/2 =
�
√
1/ m1
0
0
√
1/ m2
�
G=
�
(k1 + k2 )/m1
√
−k2 / m1 m2
�
√
−k2 / m1 m2
.
(k2 + k3 )/m2
Note that rescaling the coordinates preserves the symmetric form of the matrix.
Now the kinetic energy part involves a unit matrix, so all the non-trivial structure is in the potential term. The
next step is to diagonalise G. Note that being a symmetric matrix it can always be diagonalised, and furthermore
the diagonalising matrix P is orthogonal P −1 = P T , so
�
�
λ1 0
T
P GP = Λ =
0
λ2
As in the previous lecture, the normal coordinates, where the equation of motion takes a diagonal form,
�
�
D2 + Λ z� = 0 ,
(254)
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77
are defined such that �
z = P T z, or z = P �
z . Using these coordinates now for the expression of the energy we get:
E=
1 ˙T ˙
1
1 T
1
z� z� + z�T P T GP z� = z�˙ z�˙ + z�T Λ z� .
2
2
2
2
(255)
The final expression is diagonal. In particular, using the normal coordinates the potential energy takes the form:
V =
�
�2
1�
λj z�(j) ,
2 j
so, given that the kinetic energy is T = 12 ż T ż, each normal mode contributes independently to the energy:
Ej = T j + V j =
1 � ˙ (j) �2 1 � (j) �2
z�
+ λj z�
.
2
2
The contributions from several the different modes contribute additively to the total energy E =
the energy in a given mode is constant in time, as can be deduced from the fact that
�
�
(j) (j)
(j)
(j) (j)
d
Ej = z�˙ z¨� + λj z�˙ z�(j) = z�˙
z¨� + λj z�(j) = 0 ,
dt
(256)
�
j
Ej . Moreover,
which vanishes owing to the equation of motion in (254). In other words, not only the total energy is conserved, the
total energy in each mode is conserved. This is an important property of normal modes.
Example: For example, in the case we analysed in the previous lecture, where m1 = m2 = 1 and k1 = k3 , the
eigenvalues are:
λ1 = k 1 ,
λ2 = k1 + 2k2 ,
the matrix P is
and the normal modes are:
1
P =√
2
�
1
1
√
z (1) = (z1 + z2 )/ 2,
1
−1
�
,
√
z (2) = (z1 − z2 )/ 2 .
The potential energies in the two modes, respectively, are:
V1 =
k1 (x1 + x2 )2
,
2
2
V2 =
k1 + 2k2 (x1 − x2 )2
2
2
where we used xi instead of the rescaled zi (they are the same given that we now take mi = 1). The kinetic energies
of the two normal modes, respectively, are,
T1 =
1 (ẋ1 + ẋ2 )2
,
2
2
T2 =
1 (ẋ1 − ẋ2 )2
2
2
For the total energy one then obtains, respectively:
E1 = T 1 + V 1 =
1 (ẋ1 + ẋ2 )2
k1 (x1 + x2 )2
+
2
2
2
2
and
1 (ẋ1 − ẋ2 )2
k1 + 2k2 (x1 − x2 )2
+
2
2
2
2
Let us now consider the time derivative of E1 is:
�
�
�
d 1 (ẋ1 + ẋ2 )2
k1 (x1 + x2 )2
1�
+
= (ẍ1 + ẍ2 ) + k1 (x1 + x2 ) (ẋ1 + ẋ2 )
dt 2
2
2
2
2
E2 = T 2 + V 2 =
where the square brackets vanishes due to the equation of motion in (223) – this is the sum of the two equations
there. Similarly one can see that the time derivative of E2 vanishes using the difference of the two equations in (223).
This completes the picture of normal modes: not only can we diagonalise the equations of motion and describe
the general solution as a sum of normal modes, also the energies in these modes are separately conserved.