Several Coupled Oscillators

For the quantum version, see quantum harmonic oscillator#Two coupled quantum harmonic oscillators.

Finite chain

Consider two beads of mass $m$ attached to springs and coupled together. Each mass is connected to a fixed wall by springs with constants $k$ , and the two masses are coupled to each other by a spring with constant $k_{c}$ . Let $q_{1}$ and $q_{2}$ represent the displacements of the masses from their equilibrium positions.
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The total energy of the system is described by the Hamiltonian $H$ , which consists of kinetic and potential energy terms:

H = \frac{p_{1}^{2}}{2 m} + \frac{p_{2}^{2}}{2 m} + \frac{1}{2} k q_{1}^{2} + \frac{1}{2} k q_{2}^{2} + \frac{1}{2} k_{c} (q_{2} - q_{1})^{2} .

Here, $p_{1}$ and $p_{2}$ are the conjugate momenta of $q_{1}$ and $q_{2}$ , respectively. The first two terms represent the kinetic energy of the masses, while the remaining terms describe the potential energy, including the coupling between the two oscillators.

The dynamics of the system are governed by Hamilton's equations:

{\dot{q}}_{j} = \frac{\partial H}{\partial p_{j}}, {\dot{p}}_{j} = - \frac{\partial H}{\partial q_{j}} .

For this system, these equations yield:

{\dot{q}}_{1} = \frac{p_{1}}{m}, {\dot{q}}_{2} = \frac{p_{2}}{m},

and

{\dot{p}}_{1} = - k q_{1} + k_{c} (q_{2} - q_{1}), {\dot{p}}_{2} = - k q_{2} - k_{c} (q_{2} - q_{1}) .

This Hamiltonian framework provides a complete description of the system's energy and dynamics. To simplify the analysis, we can transform to normal coordinates, which decouple the oscillators and reveal the system's normal modes of vibration. These modes describe the independent oscillatory patterns of the system, offering deeper insight into its behavior.

Normal Modes Transformation

Suppose we have $N = 4$ coupled harmonic oscillators with masses $m_{j}$ and spring constants $k_{i j}$ . The equations of motion for the displacements $q_{j} (t)$ can generally be written as:

m_{j} {\ddot{q}}_{j} + \sum_{i} k_{j i} q_{i} = 0,

where the coupling between oscillators is represented by the off-diagonal terms of the stiffness matrix $k_{i j}$ .

The equations of motion can be expressed in matrix form as:

M \ddot{q} + K q = 0,

where $M$ is the mass matrix (diagonal for point masses), $K$ is the stiffness matrix (symmetric for conservative systems), and $q$ is the vector of displacements.

To decouple the oscillators, we solve the eigenvalue problem:

- M^{- 1} K v_{k} = ω_{k}^{2} v_{k},

where $v_{k}$ are the eigenvectors (normal modes) and $ω_{k}^{2}$ are the eigenvalues (squared angular frequencies of the normal modes). These eigenvectors form a basis in which the motion of the system can be expressed.

Using the eigenvectors, we introduce new coordinates $Q_{k} (t)$ , called the normal coordinates, defined by:

q (t) = \sum_{k} Q_{k} (t) v_{k},

or in matrix form:

q (t) = V Q (t),

where $V$ is the matrix of eigenvectors.

Decoupled Equations

Substituting $q (t) = V Q (t)$ into the equations of motion and using the orthogonality properties of the eigenvectors, the system transforms into a set of uncoupled equations:

{\ddot{Q}}_{k} + ω_{k}^{2} Q_{k} = 0.

Each normal coordinate $Q_{k}$ behaves like an independent harmonic oscillator with angular frequency $ω_{k}$ . Thus, the normal modes represent a decomposition of the coupled system into $N$ uncoupled harmonic oscillators.

Transformation to Normal Coordinates via Discrete Fourier Transform

The transition from the original coordinates of several coupled oscillators to their normal coordinates is achieved using the discrete Fourier transform. This transformation diagonalizes the system, simplifying the equations of motion by expressing the coupled oscillations as a sum of independent normal modes.

The transformation is given by:

Q_{k} = \frac{1}{\sqrt{N}} \sum_{j} e^{- i \frac{2 π}{N} j k} q_{j}

where:

$q_{j}$ represents the original coordinates of the oscillators,
$Q_{k}$ are the transformed normal coordinates,
$N$ is the total number of oscillators.

Compare to the usual Fourier transform.

First-order system approach

(Notation must be adapted.)
Consider the system

\ddot{X} = - M^{- 1} K X

where $M$ and $K$ are symmetric matrices ( $M$ comes from kinetic energy in the Lagrangian and $K$ is the Hessian matrix of a potential function $V$ also in the exact Lagrangian).
To solve it, we construct the associated first order system:

Y = (\begin{matrix} X \\ \dot{X} \end{matrix})

and then

\dot{Y} = (\begin{array}{cc} 0 & I \\ - M^{- 1} K & 0 \end{array}) \cdot Y

\dot{Y} = A \cdot Y

As usual, we find the stationary states of this first-order system (see also system of first order linear ODEs- real case). Observe that:

All the eigenvalues of $- M^{- 1} K$ are reals, since is a symmetric matrix and they always can be diagonalized. Moreover, they are all negative numbers, because this matrix reflect a stable equilibrium point, i.e., from it potential energy increases in whatever direction we take.
$λ$ is eigenvalue of $A$ if and only if $λ^{2}$ is eigenvalue of $- M^{- 1} K$ . This is so because of the block matrix determinant formula

d e t (\begin{array}{cc} H & B \\ C & D \end{array}) = d e t (H - B D^{- 1} C) \cdot d e t (D)

applied to $λ I - A$ .

Therefore, all the eigenvalues of $A$ are pure imaginary. There is no $e^{α t}$ terms.
If

ξ = (\begin{matrix} ψ \\ η \end{matrix})

is an eigenvector of $A$ associated to $λ$ then $ψ$ is an eigenvector of $- M^{- 1} K$ associated to $λ^{2}$ . This is so because of

A ξ = (\begin{array}{cc} 0 & I \\ - M^{- 1} K & 0 \end{array}) \cdot (\begin{matrix} ψ \\ η \end{matrix}) = (\begin{matrix} η \\ - M^{- 1} K ψ \end{matrix}) = (\begin{matrix} λ ψ \\ λ η \end{matrix}) .

Thus,

A^{2} ξ = (\begin{array}{cc} - M^{- 1} K & 0 \\ 0 & - M^{- 1} K \end{array}) \cdot (\begin{matrix} ψ \\ η \end{matrix}) = (\begin{matrix} - M^{- 1} K ψ \\ - M^{- 1} K η \end{matrix}) = (\begin{matrix} λ^{2} ψ \\ λ^{2} η \end{matrix})

And therefore

- M^{- 1} K ψ = λ^{2} ψ

Reciprocally, if $(δ, ψ)$ is a pair such that $- M^{- 1} K ψ = δ ψ$ then it is easy to check that we have two eigenvalues with two eigenvectors for

(\begin{array}{cc} 0 & I \\ - M^{- 1} K & 0 \end{array}),

the pair

(\sqrt{δ}, (\begin{matrix} ψ \\ \sqrt{δ} ψ \end{matrix}))

and the pair

(- \sqrt{δ}, (\begin{matrix} ψ \\ - \sqrt{δ} ψ \end{matrix}) .

For every eigenvalue of $A$ it can be chosen an eigenvector

ψ = (\begin{matrix} ξ \\ η \end{matrix})

such that $ξ$ is real (because is an eigenvector of $- M^{- 1} K$ and their eigenvalues are all real). So

I m (ψ) = (\begin{matrix} 0 \\ I m (η) \end{matrix}) .

So for every eigenvalue-eigenvector of $- M^{- 1} K$ , $(λ_{n}, ξ^{n})$ , we have two solutions:

Ψ^{n} = e^{i ω_{n} t} ξ^{n}

{\tilde{Ψ}}^{n} = e^{- i ω_{n} t} ξ^{n}

where $ω_{n} = \frac{\sqrt{λ_{n}}}{i} \in R$ (remember $λ_{n}$ is negative). It may seem that both solutions have the same initial value, which is a contradiction, but they have different initial velocities!
Moreover, observe that

({\tilde{Ψ}}^{n})^{*} = Ψ^{n},

because we can choose $ξ^{n}$ to be real.

So if we are looking for real solutions we have that

Ψ (t) = \sum_{n} C_{n} Ψ^{n} + {\tilde{C}}_{n} {\tilde{Ψ}}^{n}

(Ψ (t))^{*} = \sum_{n} C_{n}^{*} (Ψ^{n})^{*} + ({\tilde{C}}_{n})^{*} ({\tilde{Ψ}}^{n})^{*}

must be equal, and since the solutions form a vector space we conclude that $C_{n}^{*} = {\tilde{C}}_{n}$ , and so

Ψ (t) = \sum_{n} C_{n} Ψ^{n} + C_{n}^{*} (Ψ^{n})^{*} = \sum_{n} 2 R e [C_{n} Ψ^{n}] = \sum_{n} A_{n} c o s (ω_{n} t + ϕ_{n}) ξ^{n}

where $A_{n}$ and $ϕ_{n}$ , determined by initial conditions, are reals and have absorbed the constant to leave $ξ^{n}$ as a real vector.

These fundamental solutions are called normal modes. The constants $ω_{k}$ are called fundamental frequencies.

Infinite discrete chain of oscillators

Consider now an infinite discrete chain of coupled oscillators with equal mass. This system is a classic example in physics, often used to model phenomena such as phonons in solids or to inspire the leap to quantum fields. Assume that each oscillator is connected to its nearest neighbors by springs with constant $k$ . Let $q_{j}$ represent the displacement of the $j$ -th oscillator from its equilibrium position. The equations of motion for the system are given by:

m {\ddot{q}}_{j} = - k (q_{j} - q_{j - 1}) - k (q_{j} - q_{j + 1}),

which simplifies to:

m {\ddot{q}}_{j} = k (q_{j - 1} - 2 q_{j} + q_{j + 1}) .

The system can be represented using infinite matrices. Define the mass matrix $M$

M = m I,

where $I$ is the infinite identity matrix.
and the stiffness matrix $K$ :

K = k (\begin{matrix} ⋱ & ⋮ & ⋮ & ⋮ & ⋮ & ⋱ \\ \dots & 2 & - 1 & 0 & 0 & \dots \\ \dots & - 1 & 2 & - 1 & 0 & \dots \\ \dots & 0 & - 1 & 2 & - 1 & \dots \\ \dots & 0 & 0 & - 1 & 2 & \dots \\ ⋱ & ⋮ & ⋮ & ⋮ & ⋮ & ⋱ \end{matrix}) .

The equations of motion can then be written in matrix form as:

M \ddot{q} + K q = 0.

To find the normal modes of the system, we solve the eigenvalue problem:

- M^{- 1} K v_{k} = ω_{k}^{2} v_{k},

where $v_{k}$ are the eigenvectors (normal modes) and $ω_{k}^{2}$ are the eigenvalues (squared angular frequencies of the normal modes).
The system exhibits translational symmetry, meaning it is invariant under shifts by an integer number of lattice sites. This symmetry can be exploited to simplify the eigenvalue problem. Define the shift operator $S$ , which acts on the displacement vector $q$ as:

S q = (\begin{matrix} ⋮ \\ q_{j - 1} \\ q_{j} \\ q_{j + 1} \\ ⋮ \end{matrix}),

and can be represented as

S = (\begin{array}{cccccc} ⋱ & ⋮ & ⋮ & ⋮ & ⋮ & ⋱ \\ \dots & 0 & 1 & 0 & 0 & \dots \\ \dots & 0 & 0 & 1 & 0 & \dots \\ \dots & 0 & 0 & 0 & 1 & \dots \\ \dots & 0 & 0 & 0 & 0 & \dots \\ ⋱ & ⋮ & ⋮ & ⋮ & ⋮ & ⋱ \end{array})

The shift operator $S$ has eigenvectors of the form:

v_{k} = (\begin{matrix} ⋮ \\ e^{- i k (j - 1)} \\ e^{- i k j} \\ e^{- i k (j + 1)} \\ ⋮ \end{matrix}),

where $k \in [- π, π)$ is a wave number (related to the spatial frequency of the mode).
Since $S$ commutes with the stiffness matrix $K^{- 1} M$ , the eigenvectors of $S$ are also eigenvectors of $K^{- 1} M$ (symmetry trick for eigenvectors). This allows us to write the normal modes as:

v_{k} = (e^{- i k j})_{j},

where $j$ is the lattice site index.
Substituting the normal mode $v_{k}$ into the eigenvalue problem, we obtain the dispersion relation:

- M^{- 1} K e^{- i k j} = ω_{k}^{2} e^{- i k j},

\dots

ω_{k}^{2} = \frac{2 k}{m} (1 - \cos (k)) .

Normal Coordinates

Normal modes correspond to special solutions of the system in which all oscillators move in a synchronized way. They are not indexed by a position $j$ , but by a wave number $k$ , which describes a spatial oscillation pattern across the original chain of oscillators, a kind of spatial frequency. Each normal mode is associated with a temporal frequency $ω_{k}$ (dispersion relation). Because these modes evolve independently in time (they are decoupled), they provide a particularly simple description of the system's dynamics. Any general motion of the system given by the displacement vector $q (t)$ can be represented as a superposition of these normal modes (they form a complete basis for describing the system’s behavior):

q (t) = \int_{- π}^{π} Q_{k} (t) v_{k} d k .

The components $Q_{k} (t)$ are called normal coordinates. In terms of them, the equations of motion decouple into independent harmonic oscillators:

{\ddot{Q}}_{k} + ω_{k}^{2} Q_{k} = 0.

Each normal coordinate $Q_{k}$ oscillates independently with angular frequency $ω_{k}$ , as given by the dispersion relation.

We could think that the normal coordinates could be obtained in terms of $q_{j} (t)$ by projecting the displacement vector $q (t)$ onto the normal modes $v_{k}$ , but this is not mathematically rigorous.
Instead, observe that the $j$ th component of $q (t)$ is given by

q_{j} (t) = \int_{- π}^{π} Q_{k} (t) e^{- i j k} d k,

which can be interpreted as if $q_{j} (t)$ are the components of the Fourier series expansion of the function $Q_{k} (t)$ (keep an eye: with respect to the $k$ variable, not $t$ !). Therefore

Q_{k} (t) = \sum_{j \in Z} q_{j} (t) e^{i k j} .

Sometimes the normal modes are equivalently defined as

v_{k} = (e^{i k j})_{j},

and then we would have the pair of equations

q_{j} (t) = \int_{- π}^{π} Q_{k} (t) e^{i j k} d k, Q_{k} (t) = \sum_{j \in Z} q_{j} (t) e^{- i k j} .

To ensure that $q_{j} (t)$ is real-valued, i.e., $q_{j} (t) = q_{j} (t)^{*}$ , we must impose the reality condition on $Q_{k} (t)$ :

Q_{- k} (t) = Q_{k}^{*} (t) .

Continuous limit

Source: ChatGPT.
We consider $N$ oscillators on a 1D lattice with positions $q_{i} (t)$ , where $i = 1, 2, \dots, N$ . The oscillators are coupled to their nearest neighbors, and the Newtonian equation of motion for the $i$ -th oscillator is:

m {\ddot{q}}_{i} = - k (q_{i} - q_{i - 1}) - k (q_{i} - q_{i + 1}) - m ω^{2} q_{i},

where:

$k$ is the spring constant between adjacent oscillators,
$m ω^{2}$ represents a restoring force on each oscillator (like a harmonic potential).

Rewriting, we have:

m {\ddot{q}}_{i} = k (q_{i - 1} - 2 q_{i} + q_{i + 1}) - m ω^{2} q_{i} .

In the continuum limit, the positions $q_{i} (t)$ are replaced by a continuous field $ϕ (x, t)$ , where $x = i a$ and $a$ is the lattice spacing. We make the following approximations:

The displacement difference $q_{i + 1} - q_{i}$ becomes $a \partial ϕ / \partial x$ ,
The discrete second difference $q_{i - 1} - 2 q_{i} + q_{i + 1}$ becomes $a^{2} \partial^{2} ϕ / \partial x^{2}$ .

Substituting into the equation of motion:

m \frac{\partial^{2} ϕ}{\partial t^{2}} = k a^{2} \frac{\partial^{2} ϕ}{\partial x^{2}} - m ω^{2} ϕ .

Divide through by $m$ and define:

c^{2} = \frac{k a^{2}}{m}, m^{2} = \frac{ω^{2}}{c^{2}} .

This gives the equation:

\frac{\partial^{2} ϕ}{\partial t^{2}} = c^{2} \frac{\partial^{2} ϕ}{\partial x^{2}} - m^{2} c^{4} ϕ .

Rewriting the equation in standard form:

\frac{\partial^{2} ϕ}{\partial t^{2}} - c^{2} \frac{\partial^{2} ϕ}{\partial x^{2}} + m^{2} c^{4} ϕ = 0,

which is nothing but Klein--Gordon equation.

Observe that if we had considered a null restoring force we would have obtained the wave equation.

From the point of view of Hamiltonian field theory we have this derivation.

Old stuff, to be incorporated above, maybe.

See the source in anotacioneslatex.tex.

A chain of a huge amount of coupled oscillators: passing to continuous

\begin

\includegraphics[width=15cm]

\end

We are going to study $N$ masses connected to a string with tension. We leave open ends, for the moment.

Let $Ψ_{j} (t)$ be the displacement from the equilibrium of every mass $j$ . Let the mass be $m$ and the distance between the mass be $d$ . Sometimes we will take $x = j d$ and write

Ψ (x, t) = Ψ_{x / d} (t)

Analyzing every mass individually we obtain

{\ddot{Ψ}}_{1} = - \frac{2 T}{m d} Ψ_{1} + \frac{T}{m d} Ψ_{2}

{\ddot{Ψ}}_{2} = \frac{T}{m d} Ψ_{1} - \frac{2 T}{m d} Ψ_{2} + \frac{T}{m d} Ψ_{3}

\dots

In general:

\ddot{Ψ} = \frac{1}{m d} (\begin{array}{cccc} - T & T & 0 & 0 \\ T & - 2 T & T & 0 \\ . . . & . . . & . . . & . . . \\ 0 & 0 & T & - T \end{array}) \cdot Ψ

or in short

\ddot{Ψ} = - M^{- 1} K \cdot Ψ

Observe that the matrix $- M^{- 1} K$ is symmetric and all their eigenvalues are negatives, since it comes from a stable equilibrium point (see section \ref{coupledoscillators}). Moreover, from this section and the previous ones we know that a basis for solutions of this differential equation is

A c o s (ω t + ϕ) ψ

where $- ω^{2}$ is an eigenvalue of $- M^{- 1} K$ ,

$ψ$ is an eigenvector for this eigenvalue, and $A$ and $ϕ$ are real constants determined by the initial conditions: the $A$ 's by the initial positions and the $ϕ$ 's by the initial velocities.

\bigbreak

\textbf{When $N$ is really big} is very difficult to find the eigenvalues by linear algebra, so we can proceed in a different way.

Also, the vector $ψ$ is a very large vector, since the number of masses $N$ will tend to infinity, and we want to study the shape of their components. In fact, when $N$ goes to infinity $ψ$ will be a function of position, giving the displacement of the masses in the initial time.

As usual, we will follow with the complex solution for $Ψ$ and take the real part in the final step. We think in solutions like

Ψ_{j} (t) = e^{i ω t} ξ (j)

where $ξ$ is a complex eigenvector of infinite length (keep an eye: in section \ref{coupledoscillators} we saw that with finite masses we can force $ξ$ to be real, but with infinite masses, a priori, we cannot).

To find infinite length eigenvectors we use the trick of \ref{symmetrytrick}. Since this infinite chain of oscillating masses has translation symmetry, we check that

- M^{- 1} K S = S (- M^{- 1} K)

where

- M^{- 1} K = (\begin{array}{cccccc} ⋱ & ⋮ & ⋮ & ⋮ & ⋮ & ⋱ \\ \dots & - 2 B & C & 0 & 0 & \dots \\ \dots & C & - 2 B & C & 0 & \dots \\ \dots & 0 & C & - 2 B & C & \dots \\ \dots & 0 & 0 & C & - 2 B & \dots \\ ⋱ & ⋮ & ⋮ & ⋮ & ⋮ & ⋱ \end{array})

and

S = (\begin{array}{cccccc} ⋱ & ⋮ & ⋮ & ⋮ & ⋮ & ⋱ \\ \dots & 0 & 1 & 0 & 0 & \dots \\ \dots & 0 & 0 & 1 & 0 & \dots \\ \dots & 0 & 0 & 0 & 1 & \dots \\ \dots & 0 & 0 & 0 & 0 & \dots \\ ⋱ & ⋮ & ⋮ & ⋮ & ⋮ & ⋱ \end{array})

What are the eigenvectors for $S$ ? We can take any $β \in \RR$ , and then produce an eigenvector $ξ$ . Observe that

S ξ (j) = ξ (j + 1) = β ξ (j)

so if we take $ξ (0) = 1$ we conclude $ξ (j) = β^{j}$ .

Moreover, since $j \in \ZZ$ we conclude that $| β | = 1$ and therefore must be $α \in [- π, π]$ such that $β = e^{i α}$ . In conclussion, for every $α \in [- π, π]$ we have the eigenpair $(e^{i α}, ξ^{α})$ where $ξ^{α} (j) = e^{i α j}$ , respect to $S$ .

But, what is the eigenvalue respect to $- M^{- 1} K$ ?

(- M^{- 1} K ξ^{α}) (j) = - 2 B e^{i α j} + C e^{i α (j - 1)} + C e^{i α (j + 1)} = (- 2 B + C e^{- i α} + C e^{i α}) e^{i α j}

= (- 2 B + C (e^{- i α} + e^{i α})) e^{i α j} = (2 C \cos (α) - 2 B) e^{i α j}

So $ξ^{α}$ is an eigenvector of $- M^{- 1} K$ associated to the eigenvalue $2 C \cos (α) - 2 B$ , which must be negative (probably $C = B$ , and we would get that).

Now, think that the pair $(2 C \cos (α) - 2 B, ξ^{α})$ produces two solutions for the associated first order system (and then ``truncated''):

Ψ^{α} (t) = e^{i ω_{α} t} ξ^{α}

{\tilde{Ψ}}^{α} (t) = e^{- i ω_{α} t} ξ^{α}

where $ω_{α} = \frac{\sqrt{2 C \cos (α) - 2 B}}{i} \in \RR$ . In the particular case where $B = C = \frac{T}{m d}$ ,

ω_{α} = 2 \sqrt{\frac{T}{m d}} | \sin \frac{α}{2} |

which is known as \textit

\bigbreak

A few remarks:

\begin

\item The general solution is a linear combination of $Ψ^{α}$ 's and ${\tilde{Ψ}}^{α}$ 's. But since $α$ has no restriction, beyond $α \in [- π, π]$ , because we don't have boundary conditions, $α$ varies continuously. So, instead of a linear combination we get

Ψ = \int_{- π}^{π} C (α) Ψ^{α} + \tilde{C} (α) {\tilde{Ψ}}^{α} d α

Observe that this has been developed with a trick: we forgot the initial and final rows of the matrix $- M^{- 1} K$ since we are dealing with a big $N$ . This is the reason why we are finding more than $N$ eigenvalues (infinite, actually): since we are no taking into account the initial and final restriction, it is totally as if we had infinite masses.

\item Let's look for the general \textbf{real} solution. First, observe that:

(Ψ^{α})^{*} = e^{- i ω_{α} t} (ξ^{α})^{*} = e^{- i ω_{α} t} ξ^{- α} = {\tilde{Ψ}}^{- α}

({\tilde{Ψ}}^{α})^{*} = e^{i ω_{α} t} (ξ^{α})^{*} = e^{i ω_{α} t} ξ^{- α} = Ψ^{- α}

Then

Ψ^{*} = \int_{- π}^{π} C^{*} (α) (Ψ^{α})^{*} + {\tilde{C}}^{*} (α) ({\tilde{Ψ}}^{α})^{*} d α = \int_{- π}^{π} C^{*} (α) {\tilde{Ψ}}^{- α} + {\tilde{C}}^{*} (α) Ψ^{- α} d α =

= \int_{- π}^{π} C^{*} (- α) {\tilde{Ψ}}^{α} + {\tilde{C}}^{*} (- α) Ψ^{α} d α

Since it must be $Ψ = Ψ^{*}$ , and so :

C^{*} (- α) = \tilde{C} (α)

(I don't know how to prove this!!, but the idea is that $Ψ^{α}$ and ${\tilde{Ψ}}^{α}$ constitute a basis for the vector space of solutions) and therefore:

Ψ = \int_{- π}^{π} C (α) Ψ^{α} + \tilde{C} (α) {\tilde{Ψ}}^{α} d α = \int_{0}^{π} 2 R e [C (α) Ψ^{α} + \tilde{C} (α) {\tilde{Ψ}}^{α}] d α

\item The parameter $α$ is the seed for the wave number. Let $d$ be the distance between the masses, we can write $x = d \cdot j$ to be the position in the horizontal direction. If we try to write everything int terms of $x$ instead $j$ is useful to choose $κ = \frac{α}{d}$ for the trial solution for the eigenvector. The normal modes would be:

Ψ^{κ} (x, t) = a \cdot c o s (ω_{κ} t + κ x + ϕ)

\item Dispersion relation is a name for the functional relation between wave number $κ$ and frecuency $ω_{κ}$ . In this case is

ω_{κ} = 2 \sqrt{\frac{T}{m d}} s i n (\frac{κ d}{2})

\item The idea for choosing $ξ (j) = A e^{i α j}$ can be seen from other point of view.

The eigenvectors satisfy the relation

- m d ω^{2} ξ (j) = - 2 T ξ (j) + T ξ (j - 1) + T ξ (j + 1)

where we forget the first and the last relation because we assume $N$ very very large

One can observe that if $d$ is very small and taking into account the name change $x = j d$ , equation \ref{eigenvectorrelation} can be interpreted (approximately) as:

- m ω^{2} ξ (x) = T [\frac{ξ (x + d) - ξ (x)}{d} - \frac{ξ (x) - ξ (x - d)}{d}] \approx T [ξ^{'} (x + d / 2)] - ξ^{'} (x - d / 2)]

and so

- m ω^{2} ξ (x) \approx T d ξ^{″} (x)

But since $m = d μ$ , where $μ$ is the density and is supposed to be constant

- \frac{μ ω^{2}}{T} ξ (x) \approx ξ^{″} (x)

We have got an equation quite similar but in the variable $x$ or $j$ if you prefer, so is natural to choice $ξ (j) = A e^{i α j}$ .

\end

\section

We have studied an infinite system. If we want to come back to a finite system we impose boundary conditions, so we reduce the number of fundamental solutions that are allowed.

Particular case: $N = 4$ . We have:

\ddot{Ψ} = \frac{1}{m d} (\begin{array}{cccc} - T & T & 0 & 0 \\ T & - 2 T & T & 0 \\ 0 & T & - 2 T & T \\ 0 & 0 & T & - T \end{array}) \cdot Ψ

We solve using Mathematica and find, evidently, four eigenvalues:

- ω_{1}^{2} = \frac{(- 2 - \sqrt{2}) T}{d m}, - ω_{2}^{2} = - \frac{2 T}{d m}, - ω_{3}^{2} = \frac{(- 2 + \sqrt{2}) T}{d m}, - ω_{4}^{2} = 0

and four eigenvectors

ξ_{1} = (- 1, 1 + \sqrt{2}, - 1 - \sqrt{2}, 1), ξ_{2} = (1, - 1, - 1, 1),

ξ_{3} = (- 1, 1 - \sqrt{2}, - 1 + \sqrt{2}, 1), ξ_{4} = (1, 1, 1, 1)

With pictures ( $T = 2; m = 1; d = 0.5$ ):

\begin{tabular}

\includegraphics[width=60mm]{imagenes/sol1.png} &

\includegraphics[width=60mm]{imagenes/sol2.png} \

\includegraphics[width=60mm]{imagenes/sol3.png} &

\includegraphics[width=60mm]{imagenes/sol4.png} \

\end

Could we obtain the same with the big-N-technique? Let's see. We start with infinite vibrating masses with displacement $Ψ_{j} (t)$ , and so infinite fundamental frequencies

ω_{α} = 2 \sqrt{\frac{T}{m d}} \sin (α / 2)

one for every $α \in [0, π]$ .

If we want to restrict to this particular case, we have to impose some conditions: \textbf{boundary conditions}. For example, if we force $Ψ_{0} = Ψ_{1}$ and $Ψ_{4} = Ψ_{5}$ we are removing the effect of masses 0 and 5 over the masses ranging from 1 to 4 (as desired). Let's study this two conditions: we are going to impose $Ψ_{0} = Ψ_{1}$ and $Ψ_{4} = Ψ_{5}$ and watch what $α$ 's survive.

e^{i ω_{α} t} [C (α) + C (- α)] = e^{i ω_{α} t} [C (α) e^{i α} + C (- α) e^{- i α}]

C (α) + C (- α) = C (α) e^{i α} + C (- α) e^{- i α}

and together with

e^{i ω_{α} t} [C (α) e^{i α 4} + C (- α) e^{- i α 4}] = e^{i ω_{α} t} [C (α) e^{i α 5} + C (- α) e^{- i α 5}]

we arrive to

α = \frac{π n}{4}

Let's study this values one by one:

\begin

\item n=0. Then $α = 0$ , $ω_{α}^{2} = 0$ and $ξ = (1, 1, 1, 1)$ .

Correspond to the case $ω_{4}$ following Mathematica.

\item $n = 1$ . Then $α = \frac{π}{4}$ , and $ω_{α}^{2} = \frac{T}{m d} (2 - \sqrt{2})$ . We are in case 3 from Mathematica. The eigenvector obtained is $ξ = (e^{i \frac{π}{4}}, e^{i \frac{π}{2}}, e^{i \frac{3 π}{4}}, e^{i π})$ , which is complex and do not coincide with the one obtained by Mathematica. But since $- α$ produces the same eigenvalue, we can mix the previous eigenvector with $(e^{- i \frac{π}{4}}, e^{- i \frac{π}{2}}, e^{- i \frac{3 π}{4}}, e^{- i π})$ to make new eigenvectors for this eigenvalue. In fact is a (long) computation to check that the linear combination of this complex vectors that produces the same that the real above eigenvector is

{(x \to - \frac{1}{2} + i (- \frac{1}{2} + \frac{1}{\sqrt{2}}), y \to - \frac{1}{2} + i (\frac{1}{2} - \frac{1}{\sqrt{2}})}

I found it by using Mathematica.

\item $n = 2$ . Then $α = \frac{π}{2}$ , and $ω_{α}^{2} = 2 \frac{T}{m d}$ . This correspond with case $ω_{2}$ above. The eigenvector is not the same, but the phenomenom is the same that for $n = 1$ : we can recover by mixing both eigenvectors corresponding to $α$ and $- α$ .

\item $n = 3$ . Then $α = \frac{3 π}{4}$ , and $ω_{α}^{2} = \frac{T}{m d} (2 + \sqrt{2})$ . Idem.

\item $n = 4$ . Then $α = π$ . Condition $Ψ_{0}^{π} = Ψ_{1}^{π}$ implies $C (π) + C (- π) = 0$ . Moreover, since $e^{i π j} = e^{- i π j}$ , we deduce that $C (π) Ψ^{π} + C (- π) Ψ^{- π} = 0$ , so we can ignore this normal mode.

\item $n = 5$ . Then $α = π + \frac{π}{4}$ . As we reasoned above, this would be the same case as $π + \frac{π}{4} - 2 π = - \frac{3 π}{4}$ which, in fact, is paired with $α = \frac{3 π}{4}$ (case $n = 3$ )

\item All the remaining cases are repeated

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