3.4: * Normal Coordinates and Initial Values

There is another way of looking at the solutions of (3.14). We can find linear combinations of the original coordinates that oscillate only with a single frequency, no matter what else is going on. This construction is also useful. It allows us to use the form of the normal modes to simplify the solution to the initial value problem.

To see how this works, let us return to the simple example of two identical pendulums, (3.78)-(3.93). The most general possible motion of this system looks like \[X(t)=b A^{1} \cos \left(\omega_{1} t-\theta_{1}\right)+c A^{2} \cos \left(\omega_{2} t-\theta_{2}\right) ,\]

or, using (3.88) and (3.93) \[\begin{aligned}
&x_{1}(t)=b \cos \left(\omega_{1} t-\theta_{1}\right)+c \cos \left(\omega_{2} t-\theta_{2}\right), \\
&x_{2}(t)=b \cos \left(\omega_{1} t-\theta_{1}\right)-c \cos \left(\omega_{2} t-\theta_{2}\right) .
\end{aligned}\]

The motion of each block is nonharmonic, involving two different frequencies and four constants that must be determined by solving the initial value problem for both blocks.

But consider the linear combination \[X^{1}(t) \equiv x_{1}(t)+x_{2}(t) .\]

In this combination, all dependence on \(c\) and \(\theta_{2}\) goes away, \[X^{1}(t)=2 b \cos \left(\omega_{1} t-\theta_{1}\right) .\]

This combination oscillates with the single frequency, \(\omega_{1}\), and depends on only two constants, \(b\) and \(\theta_{1}\), no matter what the initial conditions are. Likewise, \[X^{2}(t) \equiv x_{1}(t)-x_{2}(t)\]

oscillates with the frequency, \(\omega_{2}\), \[X^{2}(t)=2 c \cos \left(\omega_{2} t-\theta_{2}\right) .\]

\(X^{1}\) and \(X^{2}\) are called “normal coordinates.” We can just as well describe the motion of the system in terms of \(X^{1}\) and \(X^{2}\) as in terms of \(x_{1}\) and \(x_{2}\). We can go back and forth using the definitions, (3.103) and (3.105). While \(x_{1}\) and \(x_{2}\) are more natural from the point of view of the physical setup of the system, Figure \( 3.1\), \(X^{1}\) and \(X^{2}\) are more convenient for understanding the solution. As we will see below, by going back and forth from physical coordinates to normal coordinates, we can simplify the analysis of the initial value problem.

It turns out that it is possible to construct normal coordinates for any system of normal modes. Consider a normal mode \(A^{\alpha}\) corresponding to a frequency \(\omega_{\alpha}\). Construct the row vector \[B^{\alpha}=A^{\alpha T} M\]

where \(A^{\alpha T}\) is the transpose of \(A^{\alpha}\), a row vector with \(a_{j}^{\alpha}\) in the \(j\)th column.

The row vector \(B^{\alpha}\) is also an eigenvector of the matrix \(M^{-1}K\), but this time from the left. That is \[B^{\alpha} M^{-1} K=\omega_{\alpha}^{2} B^{\alpha} .\]

To derive (3.108), note that (3.68) can be transposed to give \[A^{\alpha T} K M^{-1}=\omega_{\alpha}^{2} A^{\alpha T}\]

because \(M^{-1}\) and \(K\) are both symmetric (see (3.18) and notice that the order of \(M^{-1}\) and \(K\) are reversed by the transposition). Then \[B^{\alpha} M^{-1} K=A^{\alpha T} M M^{-1} K=A^{\alpha T} K M^{-1} M\]

\[=\omega_{\alpha}^{2} A^{\alpha T} M=\omega_{\alpha}^{2} B^{\alpha}.\]

Given a row vector satisfying (3.108), we can form the linear combination of coordinates \[X^{\alpha}=B^{\alpha} \cdot X=\sum_{j} b_{j}^{\alpha} x_{j} .\]

Then \(X^{\alpha}\) is the normal coordinate that oscillates with angular frequency \(\omega_{\alpha}\) because \[\frac{d^{2} X^{\alpha}}{d t^{2}}=B^{\alpha} \cdot \frac{d^{2} X}{d t^{2}}=-B^{\alpha} M^{-1} K X=-\omega_{\alpha}^{2} B^{\alpha} \cdot X=-\omega_{\alpha}^{2} X^{\alpha} .\]

Thus each normal coordinate behaves just like the coordinate in a system with only one degree of freedom. The \(B^{\alpha}\) vectors from which the normal coordinates are constructed carry the same amount of information as the normal modes. Indeed, we can go back and forth using (3.107).

More on the Initial Value Problem

Here we show how to use normal modes and normal coordinates to simplify the solution of the initial value problem for systems of coupled oscillators. At the same time, we can use our physical insight to learn something about the mathematics of the eigenvalue problem. We would like to find the constants \(b_{\alpha}\) and \(c_{\alpha}\) determined by (3.99) and (3.100) without actually solving these linear equations. Indeed there is an easy way. We can make use of the special properties of the normal coordinates. Consider the combination \[B^{\beta} A^{\alpha} .\]

This combination is just a number, because it is a row vector times a column vector on the right. We know, from (3.112), that \(X^{\beta}=B^{\beta} X\) is the normal coordinate that oscillates with frequency \(\omega_{\beta}\), that is: \[B^{\beta} X(t) \propto e^{\pm i \omega_{\beta} t} .\]

On the other hand, the only terms in (3.98) that oscillate with this frequency are those for which \(\omega_{\alpha}=\omega_{\beta}\). Thus if \(\omega_{\beta}\) is not equal to \(\omega_{\alpha}\), then BβAα must vanish to give consistency with (3.115).

If the system has two or more normal modes with different \(A\) vectors, but the same frequency, we cannot use (3.115) to distinguish them. In this situation, we say that the modes are “degenerate.” Suppose that \(A^{1}\) and \(A^{2}\) are two different modes with the same frequency, \[M^{-1} K A^{1}=\omega^{2} A^{1}, \quad M^{-1} K A^{2}=\omega^{2} A^{2} .\]

Because the eigenvalues are the same, any linear combination of the two mode vectors is still a normal mode with the same frequency, \[M^{-1} K\left(\beta_{1} A^{1}+\beta_{2} A^{2}\right)=\omega^{2}\left(\beta_{1} A^{1}+\beta_{2} A^{2}\right) ,\]

for any constants, \(\beta_{1}\) and \(\beta_{2}\).

Now if \(A^{1 T} M A^{2} \neq 0\), we can use (3.117) to choose a new \(A^{2}\) as follows: \[A^{2} \rightarrow A^{2}-\frac{A^{1 T} M A^{2}}{A^{1 T} M A^{1}} A^{1} .\]

This new normal mode satisfies \[A^{1 T} M A^{2}=0 .\]

The construction in (3.118) can be extended to any number of normal modes of the same frequency. Thus even if we have several normal modes with the same frequency, we can still use the linearity of the system to choose the normal modes to satisfy \[B^{\beta} A^{\alpha}=A^{\beta^{T}} M A^{\alpha}=0 \text { for } \beta \neq \alpha .\]

We will almost always assume that we have done this.

We can use (3.120) to simplify the initial value problem. Consider (3.99). If we multiply this vector equation on both sides by the row vector \(B^{\beta}\), we get \[B^{\beta} X(0)=B^{\beta} \sum_{\alpha} b_{\alpha} A^{\alpha}=\sum_{\alpha} b_{\alpha} B^{\beta} A^{\alpha}=b_{\beta} B^{\beta} A^{\beta} .\]

where the last step follows because of (3.120), which implies that the sum over \(\alpha\) only contributes for \(\alpha = \beta\). Thus we can calculate \(b_{\alpha}\) directly from the normal modes and \(X(0)\), \[b_{\alpha}=\frac{B^{\alpha} X(0)}{B^{\alpha} A^{\alpha}} .\]

Similarly \[\omega_{\alpha} c_{\alpha}=\left.\frac{1}{B^{\alpha} A^{\alpha}} B^{N} \frac{d X(t)}{d t}\right|_{t=0} .\]

The point is that we have already solved simultaneous linear equations like (3.99) in finding the eigenvectors of \(M^{-1}K\) so it is not necessary to do it again in solving for \(b_{\alpha}\) and \(c_{\alpha}\). Physically, we know that the normal coordinate \(X^{\alpha}\) must be proportional to the coefficient of the normal mode \(A^{\alpha}\) in the motion. The precise statement of this is (3.122).

Matrices from Vectors

We can also use (3.120) and the physical requirement of linear independence of the normal modes to write \(M^{-1}K\) and the identity matrix in terms of the normal modes.

First consider the identity matrix. One can think of the identity matrix as a machine that takes any vector and returns the same vector. But, using (3.120), we can construct such a machine out of the normal modes. Consider the matrix \(H\), defined as follows: \[H=\sum_{\alpha} \frac{A^{\alpha} B^{\alpha}}{B^{\alpha} A^{\alpha}} .\]

Note that \(H\) is a matrix because \(A^{\alpha}B^{\alpha}\) in the numerator is the product of a column vector times a row vector on the right, rather than on the left. If we let \(H\) act on one of the normal mode vectors \(A^{\beta}\), and use (3.120), it is easy to see that only the term \(\alpha = \beta\) in the sum contributes and \(H \cdot A^{\beta}=A^{\beta}\). But because the normal modes are a complete set of \(N\) linearly independent vectors, that implies that \(H \cdot V=V\) for any vector, \(V\). Thus \(H\) is the identity matrix, \[H=I .\]

We can use this form for \(I\) to get an expression for \(M^{-1}K\) in terms of a sum over normal modes. Consider the product \(M^{-1} K \cdot H=M^{-1} K\), and use the eigenvalue condition \(M^{-1} K A^{\alpha}=\omega_{\alpha}^{2} A^{\alpha}\) to obtain \[M^{-1} K=\sum_{\alpha} \frac{\omega_{\alpha}^{2} A^{\alpha} B^{\alpha}}{B^{\alpha} A^{\alpha}} .\]

In mathematical language, what is going on in (3.124) and (3.126) is a change of the basis in which we describe the matrices acting on our vector space from the original basis of some obvious set of independent displacements of the degrees of freedom to the less obvious but more useful basis of the normal modes.

\(\omega^{2}\) is Real

We can use (3.120) to show that all the eigenvalues of the \(M^{-1}K\) are real. This is a particular example of an important general mathematical theorem. You will use it frequently when you study quantum mechanics. To prove it, let us assume the contrary and derive a contradiction. If \(\omega^{2}\) is a complex eigenvalue with eigenvector, \(A\), then then the complex conjugate, \(\omega^{2^{*}}\), is also an eigenvalue with eigenvector, \(A^{*}\). This must be so because the \(M^{-1}K\) matrix is real, which implies that we can take the complex conjugate of the eigenvalue equation, \[M^{-1} K A=\omega^{2} A ,\]

to obtain \[M^{-1} K A^{*}=\omega^{2^{*}} A^{*} .\]

Then if \(\omega^{2}\) is complex, \(\omega^{2}\) and \(\omega^{2^{*}}\) are different and (3.120) implies \[A^{* T} M A=0 .\]

But (3.129) is impossible unless \(A = 0\) or at least one of the masses in \(M\) is negative. To see this, let us expand it in the components of \(A\). \[A^{* T} M A=\sum_{j=1}^{n} a_{j}^{*} m_{j} a_{j}=\sum_{j=1}^{n} m_{j}\left|a_{j}\right|^{2} .\]

Each of the terms in (3.130) is positive or zero. Thus the only solutions of the eigenvalue equation, (3.127), for complex \(\omega^{2}\) are the trivial ones in which \(A = 0\) on both sides. All the normal modes have real \(\omega^{2}\).

Thus there are only three possibilities. \(\omega^{2} > 0\) corresponds to stable equilibrium and harmonic oscillation. \(\omega^{2} < 0\), in which case \(\omega\) is pure imaginary, occurs when the equilibrium is unstable. \(\omega^{2} = 0\) is the situation in which the equilibrium is neutral and we can deform the system with no restoring force.