Time distribution of many small clocks in thermal motion

Question

Suppose we make many very small clocks such that they're subject to thermal motion.

To make things simple suppose at some moment $t_0$ all of their times are synced (yes they're in close proximity). We then check on them at some later time and graph the distribution of the various time measurements.

It seems evident that a clock not moving at all (in our reference frame) will record the greatest time change, and there will be a distribution of others that record various lesser time intervals. Clearly such a distribution will be dependent upon the mass and temperature of the clocks, but I was curious about the general shape of such a distribution. Any thoughts or places I could look?

This is a classical system, I get that quantum theory would generally come into play here

EDIT:

This should be solvable utilizing the Maxwell-Boltzmann distribution:

$$f(v)=\left(1/2\pi a\right)^{3}4\pi v^{2}e^{\frac{-1}{2}\left(\frac{v}{a}\right)^{2}}$$

where $a=\sqrt{kT/m}$ (T and m being the temperature and mass of our clocks respectively). The time measured by a randomly chosen clock is then:

$$\Delta t=\int_{t_{2}}^{t_{1}}dt\gamma^{-1}$$

Where dt is the observers' (non moving clocks') proper time and the Lorentz factor gamma is going to depend upon the probablity of the particle having a particular speed at some particular time. I'm not sure how to proceed with this, but already it seems like an interesting problem if one considers that our consituent parts are continually getting “smeared” out in time. Surely there's a physically measureable consequence of this. Or maybe I should use the random walk?

EDIT 1

To address Rennie's comments below, it is claimed that the clocks will go toward the same time. ie. all of their individual time-average velocities will converge. The issue with this is that for Any arbitrarily large (but non-infinite) random walk in velocity space, the average need not fall on zero, and in fact there's always a finite probability of it falling quite far from zero (the more steps in the walk the further it could possibly be, but yes that probablity also goes down).

Furthermore, if the clocks were in equilibrium prior to syncing their times (a reasonable proposition), the origin of each clocks random walk (in velocity space) would vary by a probablity depending upon the initial (Maxwell-Boltzmann) velocity distribution, such that even after an infinite amount of time the clocks would be very out of sync.

I'm asking about the shape of such a distribution for a purely thermal system as it pertains to the time interval experienced by such a body relative to another having maintained an inertial frame consistently. What is the distribution of the clocks measurements in time. I have trouble believing they would all be synced as Rennie says, such behavior goes against a maximal entropy state

Answer 1

We shall make some assumptions. My knowledge of relativity theory is basic while that of quantum mechanics is negligible, so someone more knowledgeable about these topics can comment on how realistic the following assumptions are.

First, we shall take molecules of an ideal gas to be our clocks. Some periodic internal process within the molecule is supposed to act like a clock. We shall assume that whenever exchange of energy between molecules occurs due to a collision it manifests entirely as kinetic energy of the molecules involved. In what follows, whenever we speak of time we shall mean the time of an observer w.r.t. whom the mean motion of molecules is zero.

Let $g(s)$ be the probability density function such that $g(s)\delta s$ gives the probability that any given molecule travels for a distance lying in the interval $[s,s+\delta s]$ between consecutive collisions. Let $f(v)$ be the probability density function for molecular speed $v$. We shall assume that $f$ and $g$ are statistically independent. This means that knowledge that a molecule has speed $v$ does not alter the probability values for flight distance $s$ between collisions, and vice versa. Then the probability for a molecule having speed $v$ and flight distance $s$ (between consecutive collisions) is simply $f(v)g(s)$.

For a given speed $v$, a molecule in flight for a distance $s$ measures a proper time $\tau=(s/v)\sqrt{1-v^2}$, in $c=1$ units. For a given $v$, probability that the measured proper time of molecule is $\leq \tau$ is equal to the probability that $(s/v)\sqrt{1-v^2}\leq\tau$ i.e. $s\leq \tau v/\sqrt{1-v^2}$. Accounting for all possible values of $v$, the c.d.f. for $\tau$ is obtained: \begin{align} P(\tau)=\int_0^1dv~f(v)G(\tau v/\sqrt{1-v^2}) \end{align} where $G$ is the c.d.f. corresponding to the p.d.f. $g$. The p.d.f. for $\tau$ is: \begin{align} p(\tau)=\frac{dP}{d\tau}=\int_0^1dv~\frac{v}{\sqrt{1-v^2}}f(v)g(\tau v/\sqrt{1-v^2}) \end{align}

Using $p(\tau)$ we may calculate mean and variance of proper time $\tau$ if it exists: $\mu_\tau,\sigma^2_\tau$. This is for one collision. For $n$ collisions the total measured proper time of a molecule is $T_n=\tau_1+\tau_2+...+\tau_n$. If the variance $\sigma^2_\tau$ is finite, then assuming that $\tau_i$ are independent variables, by virtue of central limit theorem we have (for large $n$, which happens over a large enough observation time): \begin{align} z_n & \equiv \frac{T_n-n\mu_\tau}{\sqrt{n}\sigma_\tau} \\ \phi(z_n) & =\frac{1}{\sqrt{2\pi}}e^{-z_n^2/2} \end{align}

This shows that it is practically certain (for large $n$) that all the molecules will have measured the same proper time, equal to $n\mu_\tau$.

P.S. I could not find an expression for $g(s)$ in the links for kinetic theory of gases. Anybody knows?

Answer 2

If you are thinking in terms of a relativistic distribution then you probably ought to use the Maxwell–Jüttner distribution.

However, regarding the comments by @John-Rennie on time dilation averaging out, I think your confusion lies with the way you're thinking of the distribution of velocities.

The distribution just describes the probability for the entire ensemble and over a large enough timescale for averaging to take place that a give velocity will be found.

It does not say that a particle will be in a specific velocity or remain in it for any measurable period.

The whole principle of a statistical treatment of this type is that the motion of individual particles is not just random, but that interactions are small and that measurements take place over a time scale that averages out the effect of interactions.

You are, in effect, trying to track individual particles, which is meaningless in the context of a statistical distribution like this.

Every particle is "jostled" by the rest and will have many directions and velocities in the period of time required for the distribution to be valid.

If you try to measure a distribution of particles over a smaller time scale that would avoid averaging, then the result could be any distribution shape that preserves conserved quantities. For example, There could be a distribution with no particles at velocities in the mid range. The next snapshot you took could be completely different.

It's the time averaging of interactions and their effects that allows for such a distribution to have meaning.

Your idea would only be valid if the particles did not interact at all and were not constrained and simply flew off into space unhindered.