General Mechanics/The Continuum Limit

Introduction
In principle, we could use the methods described so far to predict the behavior of matter by simply keeping track of each atom.

In practice, this is not a useful approach. Instead, we treat matter as a continuum.

In this section we will see how this is done, and how it leads us back to waves.

For an example, we will consider a set of N+1 identical springs and masses, arranged as in the previous section, with spring 0 attached to the wall, and mass N free.

We are interested in what happens for large N, the continuum limit.

The system has three other parameters; the spring constant k, the particle mass &Delta;m, and the spring rest spacing &Delta;a, where variable names have been chosen for future convenience. These can be combined with N to give a similar set of parameters for the system.

Suppose all the masses are displaced by d from rest, with total displacement D=dN of the system end, then the total potential energy is kNd2/2 = kD2/(2N), so


 * The system's spring constant is K=k/N
 * The system mass is m=&Delta;mN
 * The system rest length is a=&Delta;aN

If, as we increase N we change the other three parameters to keep K, m, and a constant then in the large N limit this discrete system will look like a spring with mass continuously distributed along its length.

For coordinates, we will use the displacements of each mass, xn. For large N the displacement will vary approximately continuously with distance. We can regard it as a continuous function, x, with


 * $$x(n\Delta a)=x_n \,$$

The kinetic energy of the system is then simply


 * $$T=\frac{1}{2} \Delta m \sum_0^N \dot{x}_n^2 $$

The total potential energy is similarly


 * $$V=\frac{1}{2}k x_0^2 + \frac{1}{2}k \sum_1^N (x_n-x_{n-1})^2 $$

From this we may deduce the equations of motion in two different ways.

Equations of motion: First approach
If we take the large N limit first, these sums become integrals.


 * $$\begin{matrix}

T & = & \frac{m}{2a}\sum_0^N \dot{x}_n^2 \Delta a \\ \lim_{N \rarr \infty} T & = & \frac{m}{2a} \int_0^a \dot{x}^2 ds \end{matrix}$$

where s is the distance from the wall, and


 * $$\begin{matrix}

V & = & \frac{1}{2}k x_0^2 + \frac{1}{2}k \sum_0^{N-1} (x_n-x_{n+1})^2 \\ & = & \frac{1}{2}k x_0^2 + \frac{1}{2}K \sum \left( \frac{ x(n\Delta a+\Delta a)- x(n\Delta a)}{\Delta a} \right)^2 a \Delta a\\ \lim_{N \rarr \infty} V & = & \frac{1}{2}k x(0)^2 + \frac{1}{2}Ka \int \left( \frac{dx}{ds} \right)^2 ds \end{matrix}$$

Since the springs always remain attached to the wall, x(0)=0.

The integrand for T is the product of the density and the square of the velocity, just as we might naively expect. Simalarly, V is the integral of the potential energy of a infinitesimal spring over the length of the system.

Using the Lagrangian will let us get equations of motion from these integrals.

It is


 * $$\begin{matrix}

L & = & \frac{m}{2a} \int_0^a \dot{x}^2 ds - \frac{1}{2}Ka\int_0^a \left( \frac{dx}{ds} \right)^2 ds \\ & = & \frac{m}{2a} \int_0^a (\dot{x}^2 - \frac{Ka^2}{m} x^{\prime 2}) ds\\ \end{matrix}$$

The action of the system is


 * $$\int L dt = \int \int \mathcal{L}(\dot{x},x^\prime) dx dt \quad

\mbox{ where } \mathcal{L}=\frac{m}{2a}(\dot{x}^2 - \frac{Ka^2}{m} x^{\prime 2}) $$

Here, we are integrating over both space and time, rather than just space, but this is still very similar in form to the action for a single particle.

We can expect that the principle of least action will lead us to the natural extension of Lagrange's equations to this action,


 * $$\frac{d}{ds}\frac{\partial \mathcal{L}}{\partial x^\prime}+

\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{x}} = \frac{\partial \mathcal{L}}{\partial x} $$

This equation can be proven, using the calculus of variations. Using it for this particular Lagrangian gives


 * $$\frac{\partial^2 x}{\partial t^2} - \frac{Ka^2}{m}\frac{\partial^2 x}{\partial s^2} = 0$$

This is a partial differential equation. We will not be discussing its solution in detail, but we will see that it describes waves.

First, though, we will confirm that these equations are the same as we get if we find the equations of motion first, then take the large N limit.

Equations of motion: Second approach
First, we will look at the potential energy to see how it depends on the displacements.

$$\begin{matrix} V & = & \frac{1}{2}k x_0^2 + \frac{1}{2}k \sum_0^{N-1} (x_n-x_{n+1})^2 \\ & = & \frac{1}{2}k \left( x_0^2 \cdots (x_{n-1}-x_n)^2 + (x_n-x_{n+1})^2 \cdots (x_N-x_{N-1})^2 \right) \\ & = & \frac{1}{2}k \left( 2x_0^2 -2x_0 x_1 \cdots -2x_n x_{n+1} + 2x^2_n -2x_n x_{n+1} \cdots x^2_N - 2x_N x_{N-1} \right) \\ & = & \frac{1}{2}k \left( 2x_0^2 -2x_0 x_1 \cdots 2x_n (x_n-x_{n+1}-x_{n-1}) \cdots x^2_N - 2x_N x_{N-1} \right) \\ \end{matrix}$$

Notice that we need to treat the displacements of the first and last masses differently from the other coordinates, because the dependence of the Langragian on them is different.

The kinetic energy is symmetric in the coordinates,


 * $$T=\frac{1}{2} \Delta m \sum_0^N \dot{x}_n^2$$

Using Lagrange's equations, we get that, for x0


 * $$\Delta m \ddot{x}_0 + 2k x_0 = k x_1,$$

for xN,


 * $$\Delta m \ddot{x}_N + k x_N = k x_{N-1},$$

and, for all the other xn


 * $$\Delta m \ddot{x}_n + 2k x_n = k (x_{n+1}+x_{n-1}),$$

We can replace k and &Delta;m by the limiting values K and m using


 * $$\frac{k}{\Delta m} = \frac{K a^2}{m \Delta a^2} $$

giving us


 * $$\begin{matrix}

\ddot{x}_0 & = & \frac{K a^2}{m} \frac{x_1-2x_0}{\Delta a^2}\\ \ddot{x}_n & = & \frac{K a^2}{m} \frac{x_{n-1}+x_{n+1}-2x_n}{\Delta a^2}\\ \ddot{x}_N & = & \frac{K a^2}{m} \frac{x_{N-1}-x_N}{\Delta a^2} \end{matrix}$$

Looking at the general structure of the right hand side, we see difference between the displacement at nearby points divided by the distance between those points, so we expect that in the limit we will get differentials with respect to s, the distance from the wall.

As N tends to infinity both x0 and x1 tend to x(0), which is always zero, so the equation of motion for x0 is always true.

In the continuum limit, the equation for xN becomes


 * $$ \ddot{x}(a) = -\frac{K a^2}{m} \frac{1}{\Delta a} \frac{dx}{ds} $$

Since &Delta;a tends to zero, for this to be true we must have x'=0 at a.

For the other displacements,


 * $$\begin{matrix}

\lim_{\Delta a \rarr 0} \frac{x_{n-1}+x_{n+1}-2x_n}{\Delta a^2} & = & \lim_{\Delta a \rarr 0} & \frac{1}{\Delta a} \left( \frac{x_{n+1}-x_n}{\Delta a} - \frac{x_n-x_{n-1}}{\Delta a} \right) \\ & = & \lim_{\Delta a \rarr 0} & \frac{x^\prime((n+1)\Delta a)-x^\prime(n\Delta a)}{\Delta a} \\ & = & \left. \frac{d^2 x}{ds^2} \right|_{x=a} & \\ \end{matrix}$$

so we get the equation


 * $$\frac{\partial^2 x}{\partial t^2} - \frac{Ka^2}{m}\frac{\partial^2 x}{\partial s^2} = 0$$

just as with the other approach.

Waves
It is intuitively obvious that if we flick one of the masses in this system, vibrations will propagate down the springs, like waves, so we look for solutions of that form.

A generic travelling wave is


 * $$x = A \sin (\omega t - \kappa x + \alpha) \,$$

Substituting this informed guess into the equation gives


 * $$ -A \omega^2 \sin (\omega t - \kappa x + \alpha) =

-\frac{Ka^2}{m} \kappa^2 \sin (\omega t - \kappa x + \alpha) $$

so this wave is a solution provide the frequency and wavenumber are related by


 * $$ \omega^2 = \frac{Ka^2}{m} \kappa^2 $$

The speed of these waves, c, is


 * $$c = \frac{d\omega}{d\kappa} = \pm \sqrt{\frac{Ka^2}{m}}$$

Thus, we've gone from Newton's laws to waves.

We can do the same starting with a three-dimensional array of particles, and deduce the equations for longitudinal and transverse waves in a solid. Everything we said about waves earlier will be true for these systems.

This particular system has two boundary conditions: the displacement is zero at the wall, and a local extrema at the free end. This is typical of all such problems.

When we take account of the boundary conditions we find that the correct solutions is a combination of standing waves, of the form


 * $$x = A_m \sin \left( \frac{2b+1}{2}\frac{\pi s}{a} \right)

\sin \left( \sqrt{\frac{Ka^2}{m}}\frac{2b+1}{2}\frac{\pi}{a}t +\alpha_m \right) $$

where b is any integer.

If we also knew the initial displacement we could use Fourier series to obtain the exact solution for all time.

In practice, N is typically large but finite, so the continuum limit is only approximately true. Allowing for this would give us a power series in 1/N, describing small corrections to the approximation. These corrections can produce interesting effects, including solutions, but we will not calculate them here.

The continuum limit also fails for small wavelengths, comparable with the particle spacing.

Fields
In the continuum limit, the spring is described by a variable which is a function of both position and time. Variable such as this are commonly referred to as fields.

At first sight, classical fields look quite different to classical particles. In one case position is the dependent variable; in the other, it is an independent variable. However, as the above calculations suggest, fields and particles have an underlying unity, if we take a Lagrangian approach.

We can deal with both using essentially the same mathematical techniques, extracting information about both the field and the particles in that field from the same source.

E.g., once we know the Lagrangian for electromagnetism, we can deduce both the partial differential equations for the EM fields, and the forces on charged particles in those fields, from it. We will see precisely how later, when we come to study electromagnetism.

In the example above, the field Lagrangian was the continuum limit of a Lagrangian for the discrete system. It did not have to be. We can investigate the fields described by any Lagrangian we like, whether or not there is an underlying mechanical system.

So far, we've looked at waves and movement under Newton's law, and seen how the study of movement can lead us back to waves. Next, we will look at special relativity, and see how Einstein's insights affect all this.