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


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?


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

  • $\begingroup$ Maybe look into the radioactive decay of a thermal distribution of unstable particles? The particles' half lives should act as the clock. $\endgroup$
    – WAH
    Jun 19 '17 at 9:41
  • $\begingroup$ @WAH Thanks, that is actually where I had planned to head with this eventually $\endgroup$
    – R. Rankin
    Jun 19 '17 at 9:46
  • 2
    $\begingroup$ The particles (i.e. clocks) in your gas are assumed to exchange energy with each other rapidly enough for the system to overall be in equilbrium. That is they collide with each other and exchange momentum. So there is no one clock that remains stationary while other clocks move with some thermal velocity. If you time average then all the clocks have the same average speed, so all have the same average time dilation. $\endgroup$ Jun 19 '17 at 9:48
  • $\begingroup$ @JohnRennie If course there will be some average of the distribution, but even in thermal equilibrium, there is a distribution of velocities (or speeds here for simplicity. I could simply consider the "rest/inertial frame" clock to be a larger one following the center of mass motion of the ensemble or one of the small clocks kept very cold. I dont believe you could claim all the clocks would have the same time. $\endgroup$
    – R. Rankin
    Jun 19 '17 at 9:58
  • 1
    $\begingroup$ @R.Rankin: if your clocks are subject to thermal motion (which your questions says they are) then the clocks will change speed randomly every time they interact with another clock. If you take any one clock and graph a histogram of all its speeds in between the collisions then you'll get the Maxwell-Boltzmann distribution. And this is true for any clock provided that clock is in thermal equilibrium. So all clocks will have the same time averaged speed. $\endgroup$ Jun 19 '17 at 10:14

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?

  • $\begingroup$ I originally obtained something like this myself, but consider If we initially sync our clocks as stated and wait till MUCH later and measure them again they will be synced still according to the your answer, this is like starting the experiment over again. If we abruptly measure them yet again, according to the above they will have a distribution of (yes very small) fluctuations in time. But we could have skipped the second measurement in which case they should still be synced? This seems to indicate that there should always be a distribution $\endgroup$
    – R. Rankin
    Jun 20 '17 at 10:58
  • $\begingroup$ Either that or it's something classically akin to the quantum zeno effect @Peter Shor any thoughts? $\endgroup$
    – R. Rankin
    Jun 20 '17 at 11:09
  • $\begingroup$ +1 and accepted; however my above comment remains unresolved to my point of view (maybe I'm wrong about that). Ultimately, I'd planned to use the notion of the clocks remaining synced over large times to draw up a distribution of allowable worldlines from $t_1$ to $t_2$ as the syncing of the times places a tight constraint on possible paths traveled between the two events. I thought it was interesting in that some paths (that would otherwise seem reasonable) are strictly forbidden by such a constraint. This in turn made me wonder if there was indeed such a constraint $\endgroup$
    – R. Rankin
    Jun 21 '17 at 4:44
  • $\begingroup$ @R.Rankin Sorry I don't fully understand your comments, but that is because my knowledge of relativity is pretty basic. If we initially sync our clocks and wait for a long enough time, then they will not be exactly synced. There will be a distribution, whose standard deviation grows as $\sqrt{n}\sigma_\tau$. However the mean is growing much faster, equal to $n\mu_\tau$. Therefore ratio of standard deviation to mean is going to zero at the rate $1/\sqrt{n}$, which means that the probability distribution is becoming more peaked with time. But they will never be perfectly synced. $\endgroup$
    – Deep
    Jun 21 '17 at 7:41
  • $\begingroup$ Thank you again, I hadn't thought of it in terms of the coefficient of variation. I suppose more reading is in order $\endgroup$
    – R. Rankin
    Jun 21 '17 at 8:28

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.

  • $\begingroup$ If you're trying to say that the standard theory of statistical mechanics doesn't let you say anything about individual particles, I don't know where you get that idea. It predicts the distribution of velocities at any given time with very good accuracy. $\endgroup$ Jun 19 '17 at 14:58

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