# Notes on string theory #1: The non-relativistic string

The study of waves has far reaching applications throughout physics (and across the sciences). Fundamental physics is no exception. Essential for understanding waves is understanding oscillation, and as one will recall from classical mechanics a concept of fundamental importance in this regard is the harmonic oscillator. A simple harmonic oscillator can then also be generalised to quantum mechanical systems. Of course, coupled oscillators is another important generalisation (i.e., oscillators interacting with each other). Think, for example, of atoms in a crystal and the eventual discovery of the Debye theory of heat capacity.

The importance of understanding oscillation is true for the study of many other phenomena in nature: for sound waves, water waves, and even gravitational waves. In sound waves air molecules oscillate in the longitudinal direction (i.e., the coordinate direction the sound is travelling). In gravitation waves, the sort of waves predicted by General Relativity and other theories of gravity, we may think of oscillations in the very fabric of spacetime, similar to how electromagnetic waves are oscillations of the electromagnetic field that propagate through spacetime. In string theory, we will similarly study waves and their oscillations (for instance, we will find a general solution for the fields describing the centre of mass motion and oscillations of the string as it propagates through spacetime), and from this analysis we will discover some incredibly exciting results.

But in this note we begin with a much simpler story: the study of transverse waves travelling along a classical, non-relativistic string. The value in a such a review is, in my opinion, to refresh the connection with classical concepts in what is a lengthy journey of generalisation toward higher concepts. Or, at least this is how I like to approach and motivate string theory. (The same emphasis will become apparent in the next note on the relativistic point particle and p-branes). It will also help lay some groundwork intuition before studying the first-principle action of the bosonic string: namely the Nambu-Goto action, which of course will be relativistic in nature. Much of what is reviewed below will be generalised in the relativistic setting. What is nice is how, upon constructing the Nambu-Goto action for the relativistic string (as studied in the first pages of Polchinski’s textbook), we will show that in the non-relativistic limit we can recover (up to a total derivative) the action for the ordinary, classical vibrating string discussed in this note.

As this is a lightning review of a small selection of topics from classical and wave mechanics, for further reading see Taylor [1] – or, in the context of an introduction to string theory, see Chapter 4 in Zwiebach [2]. David Morin’s lecture notes as well as these course notes by Matt Jarvis are particularly good, to name a few.

## A stretched string with transverse oscillations

Consider a string stretched with fixed endpoints. For simplicity, we may use the $(x, y)$ plane and pin the string endpoints at $(0, 0)$ and $(a, 0)$ such that the string is stretched along the x-axis. The direction along the string is longitudinal (i.e., the x-coordinate direction), while the direction orthogonal to the string is transverse (i.e., the y-coordinate direction). When describing a transverse oscillation, the x-coordinate of any point on the string will not vary in time. Rather, the transverse displacements at any point will be described by the y-coordinate.

Two pieces of information are required in order to describe the classical mechanics of the homogeneous string: tension and mass per unit length.

Tension has units of force. This means that we can first write the tension in terms of [Energy / Length]. But energy has units of mass times velocity squared, so we can write the following equation

$T_0 : [T_0] = [\text{Force}] = [\text{Energy / Length}] = \frac{M}{L}[v^2]. \ \ (1)$

Denote the Mass / Length as $\mu_0$ such that (1) may be updated as $T_0 \approx \mu_0 v^2$, where the natural velocity $v = \sqrt{T_0 / \mu_0}$. The tension $T_0$ and the mass per unit length $\mu_0$ are dynamical parameters. The velocity will eventually prove to be the velocity of transverse waves.

As a continuation of the above reasoning, consider the next simplifying assumptions. Assume that the string is infinitesimally thin and completely flexible. Since we have not yet considered boundary conditions, let us change our previous assumption and now consider that this stretched homogeneous string extends infinitely in both directions. Next, consider exciting the string such that, for two nearby points separated by a displacement dx in the longitudinal direction, there is small transverse displacement dy.

It is worth noting that, in the non-relativistic case, if the string is stretched an infinitesimal amount dx, its tension will remain approximately constant. Whatever change in energy will be equal to the work done $T_0 dx$. For the total mass, there will be no change. But in the relativistic case, any increase in energy will of course correspond to a larger rest mass. Furthermore, equation (1) suggests for a relativistic string $T_0$ and $\mu_0$ may be expressed by the relation $T_0 = \mu_0 c^2$, where c is the canonical velocity in relativity. When we eventually study the relativistic string in detail, we find this relation to indeed be correct.

For a small dy (which is to say, if the slope dy / dx is small), then we can approximate that all points in the string move only in the transverse direction. This means we consider there to be no longitudinal motion. To make sense of this statement, consider a point on the string in transverse displacement.

Notice that the length of the hypotenuse equals

$\sqrt{dx^2 + dy^2} = dx \sqrt{1 + (\frac{dy}{dx})^2}$

$\approx dx(1 + \frac{1}{2}(\frac{dy}{dx})^2) = dx + dy\frac{1}{2}(\frac{dy}{dx}). \ \ (2)$

The length element that defines the farthest a given point can move longitudinally is generally different (indeed, smaller) than the length of the long side in the dx direction by dy(dy/dx)/2, which is only (dy/dx)/2 times as large as the transverse displacement dy. With the assumption that dy/dx is small, which is to say $\mid dy/dx \mid << 1$, the longitudinal motion can therefore be neglected. In fact, all points along this segment of string move only in the transverse direction. Hence, each point is considered to be labelled with a unique value of x. Again, given the string will stretch slightly, we can safely assume that the amount of mass in any given horizontal span stays essentially constant.

## Equations of motion

We want to calculate the equations of motion. The strategy is to invoke Newton’s law and write down the transverse $F = ma$ equation for the little piece of string in the span from x to x + dx (ignoring gravity for simplicity). To do this, we must first consider the forces acting on the string.

Let $T_1$ and $T_2$ be the tensions in the string at the end points. We define the angle $\theta_1$ at x and $\theta_2$ at x + dx. As shown in Figure 3, this infinitesimal segment of string is subject to opposing tension forces at its two ends. These tension forces act along the local tangent line to the string. As the working assumption is that the string displacement remains sufficiently small such that the tension does not vary in magnitude along the string, we may suppose the local tangent line to the string subtends angles $\theta_1$ and $\theta_2$ with the x-axis. Note that these angles are written as infinitesimal quantities because the string displacement is assumed to be infinitesimally small, which implies that the string is everywhere almost parallel with the x-axis (i.e., the string displacement is greatly exaggerated in Figure 3 for purpose of illustration). Precisely because the slope dy/dx is small, it is essentially equal to the $\theta$ angles.

The slope of the string is different at the points x and x + dx. A consequence of this difference in slope (however small) is that the tension changes direction, so there is a net force $F_{net}$.

Consider the net longitudinal or x-component of the tension force. We don’t need calculus here, just a simple reading of the force diagram shows

$F_{x} = T_2 \cos (\theta_1 + \theta_2) - T_1 \cos \theta_1. \ \ (3)$

Again, assuming the change in the angle is small at x and x + dx, then $F_x \approx 0$. Another way to look at this is to use the small angle approximation $\cos \theta \approx 1 - \theta^2 / 2$ that tells us that the longitudinal components of the tensions are equal to the tensions themselves (up to small corrections of order $\theta^2 \approx (dy / dx)^2$). And because there is no longitudinal motion (or because it is so small we can neglect it), the acceleration component in the longitudinal direction is zero and, as a consequence of this reasoning, the longitudinal forces must cancel. Therefore, we find $T_1 = T_2$.

Analysis of the transverse or y-components is obviously very different. The transverse components differ by a quantity that is first order in dy/dx. This difference cannot simply be neglected, because it causes the transverse acceleration.

For the y-component of the tension force, first recognise that the transverse force at x + dx is $T sin \ \theta_2$, which is approximately equal to T times the slope. So the upward force can be written as $Ty^{\prime}(x + dx)$. The downward force at x is then simply $-Ty^{\prime}(x)$. After a bit of calculus the net transverse force is shown to be

$F_{y} = T(y^{\prime}(x + dx) - y^{\prime}(x)) = T \frac{y^{\prime}(x + dx) - y^{\prime}(x)}{dx} \equiv T dx \frac{d^2 y(x)}{dx^2}, \ \ (4)$

where the definition of the derivative is used to obtain the equivalence on the right-hand side. Indeed, the difference in the first derivatives yields the second derivative.

The mass dx of this piece of string, originally stretched from x to x + dx, is given by the mass density $\mu_0$ times dx. By Newton’s law, the net vertical force equals mass times vertical acceleration. So we can simply write

$F_y = ma \implies T \frac{d^2y(x)}{dx^2} dx = (\mu_0 dx)\frac{d^2y(x)}{dt^2}. \ \ (5)$

Cancel dx on both sides and rearrange terms to give

$\frac{d^2y(x)}{dx^2} - \frac{\mu_0}{T}\frac{d^2 y(x)}{dt^2} = 0. \ \ (6)$

Since y is a function of x and t, we can explicitly include this dependence and write y as y(x, t). Then the standard derivatives become partial derivatives and we arrive at the expected wave equation.

$\frac{\partial^2 y(x,t)}{\partial x^2} - \frac{\mu_0}{T}\frac{\partial^2 y(x,t)}{\partial t^2}. \ \ (7)$

Now, compare (7) with the standard wave equation below

$\frac{\partial^2 y}{\partial x^2} - \frac{1}{c^{2}} \frac{\partial^2 y}{\partial t^2} = 0, \ \ (8)$

with c the parameter for the velocity of the waves. In the case for transverse waves on the classical stretched string, we find the velocity of the waves is

$c = \sqrt{T_0 / \mu_0}. \ \ (9)$

Physically, we see that the lighter the string or the higher the tension, the faster the wave moves. This makes a lot of sense.

## General solution

The wave equation provides a general equation for the propagation of waves, linking the displacements of a wave in the y-coordinate direction with the time and also the displacement along the perpendicular x-axis. We therefore require solutions that link together x and t dependences.

One approach to deriving such a solution follows d’Alembert. In this approach, the displacement in y is defined as a function of two new variables, u and v, such that

$u = x -ct, \ \ v = x + ct. \ \ (10)$

To relate these solutions to the wave equation, we first differentiate both u and v with respect to x and t. Using the chain rule,

$\frac{\partial y}{\partial x} = \frac{\partial y}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial x} = \frac{\partial y}{\partial u} + \frac{\partial y}{\partial v},$

$\frac{\partial y}{\partial t} = \frac{\partial y}{\partial u}\frac{\partial u}{\partial t} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial t} = -c \frac{\partial y}{\partial u} + c \frac{\partial y}{\partial v}. \ \ (11)$

For the second derivatives we obtain,

$\frac{\partial^2 y}{\partial x^2} = \frac{\partial}{\partial x}(\frac{\partial y}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial x})$

$= 2 (\frac{\partial^2 u}{\partial x \partial u} + \frac{\partial^2 v}{\partial x \partial v}) \frac{\partial y}{\partial x},$

and

$\frac{\partial^2 y}{\partial t^2} = \frac{\partial}{\partial t}(\frac{\partial y}{\partial u}\frac{\partial u}{\partial t} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial t})$

$= 2(\frac{\partial^2 u}{\partial t \partial u} + \frac{\partial^2 v}{\partial t \partial v}) \frac{\partial y}{\partial t}. \ \ (12)$

Using the equation for the respective first derivative in (11) we find,

$\frac{\partial^2 y}{\partial x^2} = (\frac{\partial u}{\partial x}\frac{\partial}{\partial u} + \frac{\partial v}{\partial x}\frac{\partial}{\partial v})(\frac{\partial y}{\partial u} + \frac{\partial y}{\partial v}), \ \ (13)$

and

$\frac{\partial^2 y}{\partial t^2} = (\frac{\partial u}{\partial t}\frac{\partial}{\partial u} + \frac{\partial v}{\partial t}\frac{\partial}{\partial v})(-c (\frac{\partial y}{\partial u} - \frac{\partial y}{\partial v})). \ \ (14)$

Rearranging terms and substituting the fact from (10)

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial x} = 1, \ \ -\frac{\partial u}{\partial t} = \frac{\partial v}{\partial t} = c, \ \ (15)$

we find

$\frac{\partial^2 y}{\partial x^2} = \frac{\partial^2 y}{\partial u^2} + 2 \frac{\partial^2 y}{\partial u \partial v} + \frac{\partial^2 y}{\partial v^2}$

and

$\frac{\partial^2 y}{\partial t^2} = c^2 (\frac{\partial^2 y}{\partial u^2} - 2 \frac{\partial^2 y}{\partial u \partial v} + \frac{\partial^2 y}{\partial v^2}). \ \ (16)$

Substituting this result into the wave equation (7) we obtain

$\frac{\partial^2 y}{\partial u \partial v} = 0. \ \ (17)$

Observe that y is separable into functions of u and v such that

$y(u,v) = g(u) + h(v). \ \ (18)$

Therefore, the general solution to the wave equation takes the form

$y(x,t) = g(x - ct) + h(x + ct), \ \ (19)$

where g and h are single variable arbitrary functions, representing the string’s initial shape and velocity. The values of these functions are to be determined by the initial conditions of the system.

## Boundary conditions and initial conditions

In order to find unique solutions to the partial differential equation (7) and (8), which involves space and time derivatives, we are required to apply both initial conditions and boundary conditions. The initial conditions we apply constrain the solution at some time, while the boundary conditions constrain the solution at the boundary of the system.

There are primarily two types of boundary conditions common when discussing the dynamics of a string: Dirichlet and Neumann boundary conditions. This is also true in string theory, and we will see in the context of the relativistic string that application of boundary conditions imply the existence of some interesting new objects. We will therefore return to this topic in a future note. But for now, there is an intuitive way to think about what happens at the endpoints of our stretched classical string.

Dirichlet boundary conditions specify the positions of the string endpoints. For example, imagine screwing each endpoint of a string into a wall or a wooden block. The endpoint of the string is fastened such that it cannot move up and down the wall in what we previously described as the y-coordinate direction. Therefore, when imposing Dirichlet boundary conditions we have

$y(t, x = 0) = y(t,x=a) = 0. \ \ (20)$

On the other hand, Neumann boundary conditions are akin to attaching a massless loop to each end of the string, with the loops allowed to slide along two frictionless poles. In this case, the Neumann boundary conditions mean that the endpoints are free to move along the y-coordinate direction. We specify the values of $\partial y / \partial x$ at the endpoints.

$\frac{\partial y}{\partial x}(t,x = 0) = \frac{\partial y}{\partial x}(t, x = a) = 0. \ \ (21)$

## Finite string with fixed endpoints

There are a number of particular solutions to the wave equation that we may derive for various boundary configurations and initial conditions. For instance, we can have a fixed end at $x = 0$ and a free end at $x = a$, or two free ends, or both ends fixed, and so on. First, we consider the case of an infinite string with one fixed endpoint. Then we’ll consider the case of a finite string with both endpoints fixed.

Consider a leftward-moving single sinusoidal wave that is incident on a wall located at $x = 0$. The most general form of a leftward-moving sinusoidal wave is given by

$y (x,t) = A \cos (kx - \omega t + \phi), \ \ (22)$

where $\omega / k = c = \sqrt{T / \rho}$, $\phi$ is arbitrary and depends on the location of the wave at $t = 0$. We model the wall as equivalent to a system of infinite impedance with reflection coefficient $r = -1$. This can, again, be reviewed in most classical mechanics texts. This means that at the wall we obtain a reflected wave with amplitude of the same magnitude as the incident wave but with the opposite sign and travelling in the opposite direction $y_r$ such that

$y_r (x,t) = -A \cos (kx + \omega t + \phi). \ \ (23)$

If we were to observe this system, we would see a summation of these two waves (using familiar trigonometric functions)

$y(x,t) = A \cos (kx - \omega t + \phi) - A\cos (kx + \omega t + \phi), \ \ (24)$

which, using familiar trig identities, we obtain an expression in terms of the $\sin$ function

$y(x,t) = -2A \sin (\omega t + \phi) \sin kx. \ \ (25)$

Alternatively, we can write

$y(x,t) = 2A \sin (\frac{2\pi x}{\lambda})\sin(\frac{2\pi t}{T} + \phi). \ \ (26)$

Of course, rather than assigning $r =-1$ for a wall, we could instead derive the change in sign given that at the wall where $x = 0$ it follows $y = 0$. Then for all t and working from the general solution of the wave equation we find

$y(x,t) = A_1 \sin(kx - \omega t) + A_2 \cos(kx - \omega t) + A_3 \sin (kx + \omega t) + A_4 \cos (kx + \omega t)$

$\implies y(x,t) = B_1 \cos kn \cos \omega t + B_2 \sin kx \sin \omega t + B_3 \sin kx \cos \omega t + B_4 \cos kn \sin \omega t \ \ (27)$

at $y(0,t) = 0$. Therefore, we should only have $\sin kx$ terms

$y(x,t) = B_2 \sin kn \sin \omega t + B_3 \sin kx \cos \omega t$

$= (B_2 \sin \omega t + B_3 \cos \omega t) \sin kx$

$= B \sin (\omega t + \phi) \sin kx. \ \ (28)$

If the coefficients $B_2 = B_3$ then $B = 2B_2 = 2B_3$.

For a system in which the string is fixed at both ends, i.e. at x = 0 and x = a, the boundary conditions we impose require both $y(0, t) = 0$ and $y(a, t) = 0$. Therefore, we see from the previous example that the only way to have $y(a, t) = 0$ for all t is to ensure that $\sin ka = 0$. This means that ka must be an integer number of $\pi$ such that

$k_n = \frac{n\pi}{a}, \ \ (29)$

where $n \in \mathbb{Z}$ describes which string mode is excited.

In the present configuration, each endpoint represents a node. This implies that we can only have wavelengths which are related to the length of the string by n such that

$\lambda_n = \frac{2\pi}{k_n} = \frac{2a}{n}. \ \ (30)$

Therefore, we can now write a solution of the form

$y(x,t) = -2A \sin (\omega t + \phi)\sin (\frac{n\pi}{a}) = -2A \sin (\omega t + \phi) \sin (\frac{n\pi}{a}). \ \ (31)$

Observe that the allowed wavelengths on the string are all integer divisors of twice the length of the string. For the mode $n = 0$, the string can be seen to be at rest and in equilibrium.

From what has been derived above, one can proceed with an analysis of the angular frequency $\omega$ to find that the frequency of oscillations of the string are in fact all integer multiples of the fundamental frequency. But, most importantly here, it is noted that since the wave equation (7) is linear, the most general motion of a string with both endpoints fixed is simply a linear combination of the solution (25), where k can only $k = n\pi/L$ and $\omega / k = v$. Therefore, the general expression for y(x, t) may be written as a summation over all n,

$y(x,t) = \sum\limits^{\infty}_{n = 0} F_n \sin (\omega_n t + \phi_n) \sin(k_n x). \ \ (32)$

We have obtained a sum of all possible solutions with the coefficients $F_n$ given by the initial displacement.

## Constructing the Lagrangian

The remaining space will be focused on a review of how to construct the Lagrangian for the classical non-relativistic string, and then we will calculate its equations of motion within the Lagrangian formalism. It should be familiar that, in general, an action will take the form $S = \int \ L \ dt$, where L is the Lagrangian. The Lagrangian will have both kinetic (T) and potential energy (V) such that schematically we have something of the form $L = T - V$.

Returning to the picture of the string with constant mass density $\mu_0$, constant tension $T_0$, and with its endpoints located at $x = 0$ and $x = a$, the first step is to see that we should integrate the kinetic energy over the infinitesimal dx pieces along the string. In other words, the kinetic energy will be the sum of the kinetic energies of all the infinitesimal segments along the string. We also consider the rate at which each segment along the string is moving. Following this reasoning we obtain for the kinetic part of the Lagrangian

$T = \int_{0}^{a} \frac{1}{2}\mu_{0}(\frac{\partial y}{\partial t})^{2}. \ \ (33)$

The potential energy relates the work done when each segment along the string is stretched. Similar to the picture earlier, when a single infinitesimal segment of string is stretched from (x, y) to (x + dx, y + dy) this individual segment will change by some $\Delta \ L$. To derive an expression for the potential V, we assume small oscillation such that $\mid \frac{dy}{dx} \mid << 1$. That is to say, we take $\delta L = \sqrt{dx^2 + dy^2} - dx$.

As an expression of the work done by deformation, the potential energy V may be written as

$V = T(ds -dx), \ \ (34)$

where

$(ds)^2 = (dx)^2 + (dy)^2 = (dx)^2 (1 + (\frac{\partial y}{\partial x})^2). \ \ (35)$

Again, as found earlier, using the binomial series expansion (and because we are invoking the small angle approximation, we ignore the higher order terms in the expansion),

$ds \approx dx (1 + \frac{1}{2}(\frac{\partial y}{\partial x})^2 + ...). \ \ (36)$

We take this result and account for the fact that the work done relates to the stretching of each infinitesimal segment such that $T_{0} ds$.

$V = \int_{0}^{a} \frac{1}{2}T_{0}(\frac{\partial y}{\partial x})^{2} dx. \ \ (37)$

Bringing together our expressions for T and V gives

$L = \int_{0}^{a} [\frac{1}{2} \mu_0 (\frac{\partial y}{\partial t})^2 - \frac{1}{2} T_0 (\frac{\partial y}{\partial x})^2] dx \\ \equiv \int_{0}^{a} \mathcal{L} dx, \ \ (38)$

where $\mathcal{L}$ is the Lagrangian density

$\mathcal{L}(\frac{\partial y}{\partial t}, \frac{\partial y}{\partial x}) = \frac{1}{2}\mu_0 (\frac{\partial y}{\partial t})^2 - \frac{1}{2}T_0 (\frac{\partial y}{\partial x})^2. \ \ (39)$

We may simplify notation with $\partial y / \partial t = \dot{y}$ and $\partial y / \partial x = y^{\prime}.$ The action for the non-relativistic string therefore takes the form,

$S_{NR} = \int_{t_{i}}^{t_{f}} L(t) dt \int_{0}^{a} dx [\frac{1}{2} \mu_0 \dot{y}^2 - \frac{1}{2} T_0 y^{\prime 2}]. \ \ (40)$

Recall from a previous section that we need to include a time component in addition to a spacial component, because in the action the path is the function y(t, x). So we see, again, that we are integrating from some initial time to some final time, and from an initial point to some final point in a region of (t,x) space.

## Equations of motion

To find the equations of motion, we need to vary the action (37). But before that, let’s introduce some notation that will allow us to achieve a more general derivation. The simplification has to do with the momentum density of the string. The momentum density will be denoted as $\mathcal{P}^t$ and $\mathcal{P}^x$. We compute these terms as follows,

$\mathcal{P}^t = \frac{\partial \mathcal{L}}{\partial \dot{y}} = \mu_0 \frac{\partial y}{\partial t}$

$\mathcal{P}^x = \frac{\partial \mathcal{L}}{\partial y\prime} = - T_0 \frac{\partial y}{\partial x}. \ \ (41)$

As the function y(t, x) represents the path of our string, in the variation $y (t, x) \rightarrow y(t, x) + \delta y (t, x)$ we may compute,

$\delta S_{NR} = \int_{t_i}^{t_f} dt \int_{0}^{a} dx (\frac{\partial \mathcal{L}}{\partial \dot{y}} \delta \dot{y} + \frac{\partial \mathcal{L}}{\partial y \prime} \delta y\prime). \ \ (42)$

Substituting for $\mathcal{P}^t$ and $\mathcal{P}^x$ in (40) we obtain,

$= \int_{t_i}^{t_f} dt \int_{0}^{a} dx [\mathcal{P}^t \delta \dot{y} + \mathcal{P}^x \delta y^{\prime}]$

$= \int_{t_i}^{t_f} dt \int_{0}^{a} dx (\mathcal{P}^t \frac{\partial}{\partial t} \delta y + \mathcal{P}^x \frac{\partial}{\partial x} \delta y)$

$= \int dt dx [\frac{\partial}{\partial t} (\mathcal{P}^t \delta y) + \frac{\partial}{\partial x} (\mathcal{P}^x \delta y) - \delta y (\frac{\partial \mathcal{P}^t}{\partial t} + \frac{\partial \mathcal{P}^x}{\partial x})]. \ \ (43)$

Following conventional procedure in the calculus of variations, we integrate by parts. Doing so gives

$\delta S =\int_{0}^{a} dx (\mathcal{P}^t \delta y)]_{t_i}^{t_f} + \int_{t_i}^{t_f } dt (\mathcal{P}^x \delta y)]_{0}^{a} - \int_{t_i}^{t_f} dt \int_{0}^{a} dx \ \delta y (\frac{\partial \mathcal{P}^t}{\partial t} + \frac{\partial \mathcal{P}^x}{\partial x}). \ \ (44)$

This final expression for the varied action contains three terms. Each term must vanish independently. Notice that the first term is determined by the string’s arrangement at $t_{i}$ and $t_{f}$. In the present case, our interest is not with these initial and final conditions (unless we were interested in the Hamilton-Jacobi), so we can just specify initial and final configurations which results in setting the variation to zero. We could even just set the times to infinity and forget about it entirely, the choice is ours.

Instead, our interest begins with the second term. Notice that it describes the motion of the string endpoints between 0 and a during the time interval from $t_i$ to $t_f$. We may expand this term as follows,

$\int_{t_i}^{t_f } [\mathcal{P}^x \delta y]_{0}^{a} = \int_{t_i}^{t_f } dt [\mathcal{P}^x(t, x=a) \delta y(t, x=a) - \mathcal{P}^t (t, x=0) \delta y (t, x=0)]. \ \ (45)$

What we require are boundary conditions for both terms.

Once again, putting the choice of mixed conditions to the side, we can either invoke Dirichlet or Neumann boundary conditions. As discussed in a previous section, there are physical implications dependent on either choice. Dirichlet boundary conditions means the string endpoints are fixed in time. Hence, consider some x coordinate at the endpoints, and with the choice of Dirichlet boundary conditions the variation $\delta y(t, x)$ must vanish for both terms. In this case, too, momenta along the string will not be conserved. Neumann boundary conditions, on the other hand, means that the endpoints are free to move, and thus under this choice $\delta y(t, x)$ would be unconstrained. If we were to do the calculations, we would see that momenta along the string is conserved.

Let us now focus on the remaining term in (42). This term is determined by the motion of the string for $x \in (0, a)$ and $t \in (t_{i}, t_{f})$. It gives us the equation of motion,

$\frac{\partial \mathcal{P}^{t}}{\partial t} + \frac{\partial \mathcal{P}^{x}}{\partial x} = 0. \ \ (46)$

Now let us ask: what is $\frac{\partial \mathcal{P}^t}{\partial t}$ and $\frac{\partial \mathcal{P}^x}{\partial x}$? Let’s ask another question: what is $\mathcal{P}^t$ and $\mathcal{P}^x$? We already computed these above. All we need to do is take their partial derivative respectively.

$\frac{\partial \mathcal{P}^t}{\partial t} + \frac{\partial \mathcal{P}^x}{\partial x} = 0$

$\implies \mu_0 \frac{\partial^2 y}{\partial t^2} - T_0 \frac{\partial^2 y}{\partial x^2} = 0. \ \ (47)$

Rearranging and remembering that $v^2 = \frac{T_0}{\mu_0}$ we find,

$\frac{\partial^2 y}{\partial x^2} - \frac{1}{\frac{T_0}{\mu_0}}\frac{\partial^2 y}{\partial t^2} = 0. \ \ (48)$

Once again, we have found the wave equation.

## Energy

In a final note, a vibrating string will of course carry energy. A natural question has to do with the quantity of energy. To answer this we may look at the relation between kinetic energy and potential energy as the string oscillates.

As before, we consider a small segment of string and study the linear density $\mu &fg=000000 &s=2$ between x and x + dx, displaced in the y-coordinate direction. Once again, assuming the displacement is small then we can calculate the kinetic energy density (i.e., the K.E. per unit length) and the potential energy density.

We know the kinetic energy (33) and the potential energy (37). We also know that the solutions to the wave equation take the form

$y(x,t) = f(x \pm ct). \ \ (49)$

Therefore,

$\frac{dK}{dx} = \frac{1}{2}\mu c^2 [f^{\prime}(x \pm ct)]^2, \ \frac{dV}{dx} = \frac{1}{2}T[f^{\prime}(x \pm ct)]^2. \ \ (50)$

Recall $c = \sqrt{T / \mu}$, and so we find the kinetic energy density to be equal to the potential energy density. Another way to observe this equality is by substituting a solution $y = A \sin (kx - \omega t)$ for the wave equation into these equations for the kinetic energy and potential energy density. We can then evaluate the energy over n wavelengths.

For the kinetic energy density,

$K = \frac{1}{2}\mu \int_{x}^{x + n\lambda} A^2 \omega^2 \cos^2 9kx - \omega t) dx$

$= \frac{1}{2} \mu A^2 \omega^2 \int_{x}^{x + n \lambda} \frac{1}{2}(1 + \cos[2(kx - \omega t)]) dx. \ \ (51)$

And for the potential energy density we find,

$V = \frac{1}{2}T \int_{x}^{x + n\lambda} A^2 k^2 \cos^2 (kx - \omega t) dx$

$= \frac{1}{2} T A^2 k^2 \int_{x}^{x + n\lambda} \frac{1}{2} (\cos [2(kx - \omega t)]) dx. \ \ (52)$

And so we see that

$K = \frac{1}{2} \mu A^2 \omega^2 \frac{n\lambda}{2}, \ V = \frac{1}{2} A^2 k^2 \frac{n\lambda}{2}. \ \ (53)$

But as $c = \sqrt{T / \mu} = \omega / k \rightarrow \mu \omega^2 = T k^2$, these expressions for K and V are once again equal. We can also deduce therefore that the total energy per unit length is $1/2 \mu A^2 \omega^2$. From this analysis it is possible to go on and study the energy flow per unit time and the power to generate the wave.

References

[1] Taylor, J., Classical mechanics. University Science Books, 2004.

[2] Zwiebach, B., A first course in string theory (2nd edition). Cambridge University Press, 2009.