The local time is 3 o'clock

Classical Mechanics Level 5

There are some things in physics that we are so used to taking for granted that we don't even think about them. Consider, for example, the statement that classical physics is local - physical systems involve interactions that occur at the same points in space and at the same times. Does this have to be so? Let's look at a simple spring to see what happens to physics if we make things a little non-local.

The equation of motion for the position of a mass $m$ at the end of a spring with spring constant $k$ is

$\frac{d^2 x(t)}{dt^2}=-\frac{k}{m} x(t)$

if the other end of the spring is fixed in space. Consider solutions to this equation of the form $x = e^{iEt}$ where $E$ is a constant (which is a priori possibly complex). Let $N$ be the number of possible values for $E$ that satisfy this equation.

We'll now make the spring equation slightly non-local in time. We do this by letting the acceleration of the mass be determined by not just $x(t)$ , but by a combination of the position at two different times, $x(t-\Delta t)$ and $x(t+\Delta t)$ , where $\Delta t$ is a constant. The new equation of motion is

$\frac{d^2 x(t)}{dt^2}=-\frac{k}{2m}( x(t+\Delta t) + x(t-\Delta t))$

so that as $\Delta t \rightarrow 0$ we return to the usual equation for a spring. This second equation with $\Delta t \neq 0$ also has solutions of the form $x=e^{i E t}$ for certain values of $E$ . Let $M$ be the number of possible values of $E$ that satisfy this second equation of motion. What is $N/M$ ?

Hint: don't make any extra assumptions about $E$ than you absolutely need to.

1/4 1

\infty

2 solutions

Aleksandr V Sorokin
Sep 25, 2014

The local equation is the 2nd order ODE, which implies it has two linearly independent solutions. Thus $N=2$ .

The non-local equation can be transformed to the local one by assuming that $x(t)$ is an infinitely differentiable function and as such can be expanded into Taylor series around $t$ with displacement $\Delta t$ :

$x(t+\Delta t)=\sum_{n=0}^\infty{\frac{\mathrm{d}^n x}{\mathrm{d} t^n}\!\!(t)\frac{\Delta t^n}{n!}}$ ,

which, after substitution, leads to the ODE of order infinity. As before, this implies that the equation has infinitely many linearly independent solutions, and $M=\infty$ .

The result is then $\frac{N}{M}=\frac{2}{\infty}=0$ .

Andre Franca
Sep 25, 2014

If you plug the ansatz at the ODE, it's clear that the solutions satisfy

$E^2 = k/m \cos(E \Delta t)$

So the number of solutions actually depends on the ratio of $\delta t$ and the proper timescale of the oscillator $t_p = (m/k)^{1/2}.$ Naturally for $\Delta t$ going to infinity, you have infinitely many possible solutions, but for small $\Delta t / t_{p}$ , we still have only two possible solutions. Am I missing something?

Andre, I totally agree with you.

Jordi-Lluis Figueras - 6 years, 8 months ago

The local time is 3 o'clock

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