Comparing the variance of the position of particles of a sample of matter with the sample's temperature

Question

As the title suggests i have attempted comparing the variance of the position of particles of a sample of matter with the sample's temperature.
Now firstly we need to clarify the basics:

1. method for averaging a continuous function:
   $$ M\{f\}(a,b)=\frac{1}{b-a} \int_{a}^{b} f(x) \, dx $$
   

2. method for computing the variance of a continuous function:
   we have $$\tag{*} \sigma^2=\sum_{i=1}^N \frac{(x_i - \bar{x})^2}{N} = (\sum_{i=1}^N \frac{x_i^2}{N}) - \bar{x}^2 $$
   Now notice that $$(\sum_{i=1}^N \frac{x_i^2}{N}) $$
   is the same as the average of the function $f^2(x)$ (assuming $\forall x_i\Rightarrow x_i \in R_f$) so with a bit of substitution and simplification we will have:
   $$\sigma^2\{f\}(a,b)=M\{f^2\}(a,b) - (M\{f\}(a,b))^2$$
   Okay!Now that we have our tools;we can continue to compute the desired variance. Notice the "variance" of the function of position of a moving particle gives us a scalar that indicates that how much the particle -sort of-"wobbles".
   now if we have a sample consisting of 500 water molecules and we have the function of position of each particle with respect to time-and assume they are all moving in a simple harmonic manner (i.e $x(t) = A\cos{\omega t}$)-we can theoretically compute the total energy of the system:
   $$Q = \sum{K} = \frac{1}{2}m_{molecule}\sum_{i=1}^N(\frac{dx_i}{dt})^2 = m_{total}cT = cT\sum m_{molecule}$$ note that the total heat energy of a system can be computed like so:$$Q = mc\Delta T = mc(T_1-T_0)=mc(T_2-0) = mcT_2$$(because the heat energy of a system is zero at zero kelvin) and let's define the "variance" of a system as follows:
   $$\sigma^2[S]_{0}^{t}= \frac{1}{N}\sum_{i=1}^N \sigma^2\{x_i(t)\}(0,t)$$where $x_i(t)$ is the equation of motion of the i-th particle.
   now if plot the variance of a system against its temperature using a code I wrote you can see it here.

now it takes about ten minutes to completely render and do all of the needed computations - my code is inefficient and my computer slow:( - and the final plot (the $y$ axis is variance of the system and the x axis is the temperature of the system) and the computed image is like follows:

now if we had more continuous data i suspect it would strangely resemble this figure here.(And I have ran it several times and it was not an accident.)

Now my question is if this connection really exists and if so why? Please note that I am currently in high school and please try not to get things too complicated.

Answer 1

With $x(t)=A\cos(\omega t)$ we have the variance, if I did the integrals correctly, as \begin{align} \sigma^2\{x\}(0,t)&=\textstyle\frac{1}{t}\int_0^tx^2(s)\,ds-\frac{1}{t^2}\big(\int_0^tx(s)\,ds\big)^2\\ &=\textstyle \frac{A^2}{2\omega} +\frac{A^2\sin(2\omega t)}{4\omega t}-\frac{A^2\sin^2(\omega t)}{\omega^2t^2}\,. \end{align} It looks like you draw the amplitudes $A_i$ randomly from a uniform distribution on $\{0,1,...,10\}$ for an ensemble of $N$ particles. The frequencies you simulate by drawing uniformly a period $T_i$ from $\{1,...,10\}$ and setting $\omega_i=2\pi/T_i$. Then you take the average of the variance $$\tag{1} \frac{1}{N}\sum_{i=1}^N\sigma^2\{x_i\}(0,t)\,. $$ The temperature is proportional to the total kinetic energy and is therefore proportional to the sum $$\tag{2} \sum_{i=1}^N\dot x^2_i(t)=\sum_{i=1}^NA_i^2\omega_i^2\sin^2(\omega_i t)\,. $$ Your $N$ is 1000. A single point in your plot is (2) on the $y$-axis vs. (1) on the $x$-axis. You repeat this simulation with 500 different seeds.

So far so good. This python program can probably be made useful, however:

I find it extremely unlikely that the scatter plot of 500 points with five outliers has anything to do with Planck's radiation law.

• That law describes the intensity of black body ratiation, not the temperature distribution of water molecules.

• Even for the study of water molecules, the way you draw the random parameters $A_i$ and $\omega_i$ looks a bit too simplistic.

• With five outliers, the scatter plot is too insignificant.