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Abstract

Two different combinatorial models of particles that are free to move on a finite number of distinguishable lattice points have been shown. Average pressures at the boundaries for the two models have been calculated using statistical mechanical principles. The difference in average pressures from the two models has been deemed significant and interesting results have been shown. The model has been interpreted as 'radiation pressure of a finite numbers of photons present in a 1-dimensional internally reflecting tube of various finite lengths.'


Introduction

What is pressure?

Pressure ($p$) of gas results from particles in motion colliding with the walls of a container. Pressure is a derived physical quantity with the SI units $kg\ m^{-1}s^{-2}=Nm^{-2} =1Pa$. 1 Pascal corresponds to a 1 Newton force applied uniformly over an area of $1m^2$.

Note: It is known that photons can also exert pressure.

In order to study this, let's model a finite number of photons in a 1-dimensional tube of finite length:

photons in a tube

This may be modelled from first principles of statistical mechanics and the calculations may be performed using Python3.7 using Spyder5 IDE. I tried searching for 'pressure of photons in a tube' and did not find any results. Perhaps we are doing something new here!


A simplified model in 1D

Consider $n$ point particles confined on integer points between the boundaries $x=0$ and $x=N$, making $N+1$ lattice points.

Say, whenever a particle is at the left boundary ($x=0$), a 'unit pressure' is applied at the boundary. Thus, pressure exerted when $j$ particles are present at the left boundary is $j$.

Labeled particles

illustration of labeled photons in an internally reflecting tube

  1. Particles are labeled.
  2. Multiple particles can exist at the same point.
  3. Particles cannot cross the boundaries.

Since each of the $n$ particles may be present at any one of the $N+1$ points ($0,1,2,\cdots,N$), there are $(N+1)^n$ possible arrangements.

Assumption: Each of these arrangements is equally likely.

The number of arrangements in which $j$ particles are presented at the left boundary ($N=0$) is $\binom{n}{j}N^{n-j}$: there are $\binom{n}{j}$ ways of choosing the $j$ particles present at the left boundary, and $N^{n-j}$ ways of arranging the remaining $n-j$ particles at the remaining $N$ lattice points $(1,2,\cdots,N)$.

Average pressure can be calculated as $$ \langle p\rangle =\dfrac{\Sigma_j j\binom{n}{j}(N^{n-j})}{(N+1)^n} $$

Unlabeled Particles

illustration of unlabeled photons in an internally reflecting tube

  1. Particles are unlabeled.
  2. Multiple particles can exist at the same point.
  3. Particles cannot cross the boundaries.

I found this useful formula for number of ways in which $k$ indistinguishable particles can be placed into $n$ distinguishable boxes:

$$ \binom {k+n-1}{n-1} $$

We have to place $n$ indistinguishable particles into $N+1$ boxes. Thus $k \mapsto n$ and $n \mapsto (N+1)$:

$$ \binom {n+(N+1)-1}{(N+1)-1} = \binom {n+N}{N} $$

Assumption: Each of these arrangements is equally likely.

The number of arrangements in which $j$ particles are presented at the left boundary $(N=0)$ is $\binom {(n-j)+N-1}{N-1}$: we are placing the remaining $n-j$ particles at the remaining $N$ lattice points $(1,2,\cdots,N)$.

Average pressure can be calculated as

$$ \langle p\rangle =\dfrac{\Sigma_j j\binom {(n-j)+N-1}{N-1})}{\binom {n+N}{N}} $$


Interesting results

Labeling the particles should have no effect on the pressure. However, when we take the same measurement of pressure (for same $n$ and $N$) for the cases of labeled and unlabeled particles, we find that the pressure values actually differ. Since $\langle p \rangle (n,-N)$ is a monotonically increasing function, we may also plot the differences of pressure for labeling and not labeling the particles against the pressure measurements arranged according to the average values of the pressures obtained from the two models:

difference in pressure values on labeling and not labeling arranged according to pressure

Note: Though the fluctuations are very small $(\approx 1e{-13}\%)$, it seems that the pressure difference fluctuates more (around zero) when the values of pressure are higher. It may just be numerical error, but, as far as I know, python values are reliable upto $1e{-16}$.

Here's why I don't think that this is a numerical error: Consider the ratio

$$ \dfrac{\langle p_{\text{l}} \rangle -\langle p_{\text{u}} \rangle}{\dfrac{\langle p_{\text{l}} \rangle +\langle p_{\text{u}} \rangle}{2}} = 2\dfrac{\dfrac{\Sigma_j j\binom{n}{j}(N^{n-j})}{{(N+1)^n}} - \dfrac{\Sigma_j j\binom {(n-j)+N-1}{N-1})}{\binom {n+N}{N}}}{\dfrac{\Sigma_j j\binom{n}{j}(N^{n-j})}{{(N+1)^n}} + \dfrac{\Sigma_j j\binom {(n-j)+N-1}{N-1})}{\binom {n+N}{N}}} $$

where $\langle p_{\text{l}} \rangle$ and $\langle p_{\text{u}} \rangle$ are the average values of pressure from the labeled and the unlabeled model, respectively.

If we plot this ratio of the difference and average of pressures from the two models, the graph undeniably shows regular patterns:

plot of the ratio difference and average of pressures from the two models arranged according to pressure

Note: This is a function of the type $z = 2\dfrac{x-y}{x+y}$


Question

Why does pressure change upon labeling the particles?

Further Reading

Statistical models for finite particles in a finite number of distinguishable boxes

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  • $\begingroup$ Related: Proof of pressure of ideal gas from first principles $\endgroup$
    – Roger V.
    Dec 6, 2022 at 9:36
  • $\begingroup$ I think this is a good question and can be pretty easily reworked into an acceptable format to this site. try to destill the question as much as you can to being about 1. rather then your specific model $\endgroup$ Dec 7, 2022 at 4:03
  • $\begingroup$ @ananta, suppose we are moving the particles one by one, then your scenario "labeled particles" and "unlabeled particles" would be exactly the same, would it? Because whether we label or not label them explicitly, we need some internal labeling to be able to move them one by one? $\endgroup$
    – James
    Dec 7, 2022 at 4:53
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    $\begingroup$ I am not convinced that this is a homework question. Actually I think it seems more original and looks like OP put a lot of effort into it. Voting to reopen. $\endgroup$ Dec 7, 2022 at 11:00
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    $\begingroup$ @doublefelix indeed, this is not a homework question. Though the initial approach I used was faulty as pointed out in the answers, the question is now more complete. Thank you for your support. $\endgroup$
    – ananta
    Dec 7, 2022 at 15:40

1 Answer 1

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You are being inconsistent with indistinguishability. If your $n$ particles are considered indistinguishable, then there are not $(N+1)^n$ possible configurations; this is instead the number of configurations of labelled particles. E.g. for two positions and two particles ($N=1,n=2$), the labelled configurations are "both on 0", "$1\mapsto0,2\mapsto1$", "$1\mapsto1,2\mapsto0$", and "both on 1," for $(1+1)^2=4$ configurations, but the unlabelled configurations are "both on 0", "different positions", and "both on 1," for only 3 configurations.

Dealing with unlabelled particles is an unnecessary complication. Just deal with distinguishable ones. Then your error is that the numerator in $$\langle p\rangle=\frac{\sum_jj\cdot N^{n-j}}{(N+1)^n}$$ is wrong. $N^{n-j}$ only counts how many ways you can distribute the "non-bunched" particles among the $N$ remaining positions. You also have the freedom to choose which $j$ particles are bunched $(\binom nj$). The corrected expression $$\langle p\rangle=\frac{\sum_jj\cdot\binom njN^{n-j}}{(N+1)^n}$$ produces a more reasonable graph.

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  • $\begingroup$ Thank you for the insight. I have added the corrected result to the question statement. Though I still wanted to keep the particles indistinguishable, so I used the formula $\binom{k+n-1}{n-1}$, as said in the question. $\endgroup$
    – ananta
    Dec 6, 2022 at 5:08

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