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This question came up because the ideal gas law says that for, say, a gas enclosed in a bottle the pressure is:

$ P \propto \frac{N \cdot T}{V}$

Where P: Pressure, N: Number of Gas particles in the bottle, T: Temperature and V: Volume of the bottle.

I am only investigating the dependence on volume:

The Pressure does not depend on the shape of the volume.

I find it intuitive that the pressure would be inversely proportional to the bottle's surface area. I also find it plausible that the pressure would be inversely proportional to the average distance of a point inside the volume to the volume's surface, as the particles inside the volume will take more time to impinge on the surface (and cause a momentum transfer that causes pressure) the farther the particles are away from the bottle's surface.

The average distance to the surface should be understood as the average over all points inside the volume and all solid angles subtending each point to the surface, and not averaged over the surface itself. This is because the average distance needed here should be a sum of the contributing distances weighed by the likelihood this distance will be traversed by a particle: This likelihood is identical for all directions. If one were to calculate the average by weighing the distances proportionally to the (infinitesimal) surface areas that are connected to the point whose average distance is being calculated, one would weigh those directions more that make a very small or large angle with the surface, as the solid angle cones intersect a larger piece of the surface there.

I checked the conjecture for a sphere, and it seems to work, but I do not see how to extend this result to all "well behaved" volumes.

Also, the proportionality factor between (Average Distance multiplied by Surface Area) and Volume would have to be identical for all body shapes....

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  • $\begingroup$ Do you mean "the mean distance from the center of mass to the body's surface"? If not, your conjecture is definitely false: take a point $P = (100,0,0)$ and the unit sphere $S$. The mean distance from $P$ to $S$ is about 100; the area of $S$ is $4\pi$, but the volume of $S$ is $\frac{4}{3}\pi$, which is far from $100 \cdot 4\pi$. $\endgroup$ – John Hughes Nov 29 '17 at 11:32
  • $\begingroup$ I mean the mean distance of all points within the volume to the volume's surface. For simplicity, I am only considering the shortest distance from a point in the volume to the surface, not all distances. $\endgroup$ – yippy_yay Nov 29 '17 at 11:57
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    $\begingroup$ So if $P$ is in a solid, $M$, and the boundary of $M$ is $S$, then you're talking about the min of $d(P, s)$, where $s$ is any point of $S$. In that case, the conjecture is almost certainly wrong. Consider, on the unit sphere, drilling a "well" from the north pole towards the south pole, but not quite reaching it. If the well is very narrow, then the volume is essentially unchanged, as is the surface area. But all points in the "middle" of the sphere (distance < 1/2 from the origin) are now "closer" to the surface than they were, by about an average of 1/4. $\endgroup$ – John Hughes Nov 29 '17 at 12:28
  • $\begingroup$ You're right! The conjecture is certainly false if one takes the minimum distance averaged over all points in the volume, as I naively suggested. The correct conjecture would involve the average distance to the surface averaged over all points in the volume - your example with the well bored into the sphere would change the minimum average distance, but not the averaged average distance (averaged over all points AND all distances). $\endgroup$ – yippy_yay Nov 29 '17 at 15:13
  • $\begingroup$ @JohnHughes, you should write that as an answer, as it clearly shows the conjecture is false. Yippy: your conjecture, even in its revised form is still based on rather naive intuition: in ideal gas, the particles don't "average" their distance to the surface, and in fact the mean free path (average distance between successive collisions) is typically much shorter than any such distance. What is even more problematic is your argument for pressure being inversely proportional to area - there is no meaning to be given here to pressure as some "total" force divided by total area. $\endgroup$ – Nick Pavlov Nov 29 '17 at 18:08
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The first conjecture is not quite right, but it can be modified into a second conjecture which is correct.

Conjecture 1:

The average distance-to-surface (over points in the interior and solid angles from them) times the surface area is proportional to the volume.

We can disprove this by contradiction. Consider two blobs $B_1$ and $B_2$ connected by a short (negligible) thread to form the full volume $B_3$. If the conjecture were true, we would have: \begin{align} V_1 & = k S_1 \hat d_1 \tag{1} \\ V_2 & = k S_2 \hat d_2 \tag{2} \\ V_3 & = k S_3 \hat d_3 \tag{3} \end{align} By construction, we know that: \begin{align} V_1 + V_2 & = V_3 \tag{4} \\ S_1 + S_2 & = S_3 \tag{5} \\ \frac{V_1}{V_3} \hat d_1 + \frac{V_2}{V_3} \hat d_2 & = \hat d_3 \tag{6} \end{align} and then we can use (5) and (6) to rewrite (3) as \begin{align} V_3 &= k (S_1 + S_2) \left( \frac{V_1}{V_3} \hat d_1 + \frac{V_2}{V_3} \hat d_2 \right) \tag{7} \\ V_3^2 &= k (S_1 + S_2) \left( V_1 \hat d_1 + V_2 \hat d_2 \right) \tag{8} \\ V_3^2 &= k S_1 V_1 \hat d_1 + k S_1 V_2 \hat d_2 + k S_2 V_1 \hat d_1 + k S_2 V_2 \hat d_2 \tag{9} \\ V_3^2 &= V_1^2 + k S_1 V_2 \hat d_2 + k S_2 V_1 \hat d_1 + V_2^2 \tag{10} \\ V_3^2 - V_1^2 - V_2^2 &= k ( S_1 V_2 \hat d_2 + S_2 V_1 \hat d_1 ) \tag{11} \\ \left( \frac{V_3^2 - V_1^2 - V_2^2}{k} \right) &= S_1 V_2 \hat d_2 + S_2 V_1 \hat d_1 \tag{12} \end{align} where we get from (9) to (10) by using (1) and (2).

Now we see that we are in a strange situation. If we squish the blobs around without changing their volumes, then the left side of (12) is constant, while the right side has strange, physically meaningless cross-terms. (These arose because the conjecture is multiplying quantities that essentially combine additively over independent subparts.)

In particular, if we use a low-distance, high-surface volume (e.g. a wide wrinkled crêpe or a very long wrinkled thread) for $B_2$, then we can hold $V_2$ constant while making $\hat d_2$ arbitrarily small, with $S_2$ arbitrarily large. If we leave $B_1$ alone while we do this, this will make the right hand side of (12) arbitrarily large, contradicting the intransigence of the left hand side.$\tag*{$\blacksquare$}$

However, your original intuition is correct. But particles hitting the surface many times due to a short distance between bounces contribute a lot to the pressure while particles that rarely hit the surface due to long distances are almost irrelevant, so instead of the average distance, we should use the average inverse total distance, where the total distance is the full distance traveled between bounces (i.e. the sum of the two distances in opposite directions, the width).

Conjecture 2:

The harmonic mean (over points in the interior and solid angles from them) of the width (at that point, in that direction), times the surface area, is equal to 4 times the volume.

\begin{align} \frac{1}{\mbox{harmonic mean }}\ =\ & \frac{1}{V} \int_{p \in B} \frac{1}{2\pi} \int_\phi \frac{1}{\mbox{width (through $p$ to surfaces in direction $\pm\phi$)}}\;\; dB d\Phi \\=\ & \frac{1}{2\pi V} \int_\phi \int_{p \in B} \frac{1}{\mbox{width}}\ dB d\Phi \\=\ & \frac{1}{2\pi V \cdot 2 \epsilon} \int_\phi \int_{p \in B} \left\{{ \begin{array}{ll} 1 & \mbox{if $p$ is within $\epsilon$ of a surface in direction $\pm \phi$} \\ 0 & \mbox{otherwise} \end{array}}\right\} dB d\Phi \\=\ & \frac{1}{2\pi V \cdot 2 \epsilon} \int_{p \ \in \ surface(B)} \int_{\phi\ pointing\ out\ of\ B} \epsilon \sin (\mbox{angle from $\phi$ to surface}) \ dS \\=\ & \frac{1}{2\pi V \cdot 2 \epsilon} \int_{p \ \in \ surface(B)} 2 \pi \cdot \frac{1}{2} \epsilon \ dS \\=\ & \frac{1}{2\pi V \cdot 2 \epsilon} \cdot S \cdot \pi \epsilon \\=\ & \frac{S}{4V} \tag*{$\blacksquare$} \end{align}

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  • $\begingroup$ This also shows that the surface area is 4 times the expected shadow size (for convex objects). This is because the large line above (containing the "if-then" expression with big curly braces) can be seen to be computing 1/V times the expected shadow size. This extends to non-convex objects if you count the shadow multiple times where the light path is occluded multiple times. I am surprised that I have never heard of these simple identities. $\endgroup$ – Matt Dec 2 '17 at 20:56
  • $\begingroup$ Brilliant counter example! Why doesn't it contradict the conjecture #2? The argument seems to stay the same... $\endgroup$ – yippy_yay Dec 2 '17 at 23:23
  • $\begingroup$ @yippy_yay Ah... I think I see: The "Widths" cross over to the other Blob, whereas the "Distances" remains inside their own Blob. $\endgroup$ – yippy_yay Dec 2 '17 at 23:26
  • $\begingroup$ In the second conjecture, you use the expression "harmonic mean" instead of average - what does harmonic mean here? $\endgroup$ – yippy_yay Dec 2 '17 at 23:41
  • $\begingroup$ @yippy_yay Yes, if $\hat d$ is the harmonic mean, then equation (6) would become $\frac{V_1}{V_3} \frac{1}{\hat d_1} + \frac{V_2}{V_3} \frac{1}{\hat d_2} = \frac{1}{\hat d_3}$, and so in equation (7) we would put the weighted $\hat d$ sum on the left hand side, and no cross terms would be produced. $\endgroup$ – Matt Dec 3 '17 at 7:17
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So if $P$ is in a solid, $M$, and the boundary of $M$ is $S$, then you're talking about the min of $d(P,s)$, where $s$ is any point of $S$. In that case, the conjecture is almost certainly wrong. Consider, on the unit sphere, drilling a "well" from the north pole towards the south pole, but not quite reaching it. If the well is very narrow, then the volume is essentially unchanged, as is the surface area. But all points in the "middle" of the sphere (distance < $\frac{1}{2}$ from the origin) are now "closer" to the surface than they were, by about an average of $\frac{1}{4}$.

I would add that the conjecture is also false if for each interior point, instead of minimum distance to the surface, an average is taken. For that, it suffices to consider very long, thin cylinders ($l \gg r$). For any point, the average distance to the surface is at least $l/4$ and no more than $l+2r$ (in fact, a much stricter bound can be put on that, but there is no need here). The average for all points is thus also within these bounds. But the ratio of volume to surface area can be made arbitrarily close to $(\pi r^2l)/(2\pi rl) = r/2$, so the two quantities cannot possibly be proportional to each other for cylinders of sufficiently different "aspect ratios" $r/l$.

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  • $\begingroup$ Could you point me towards a clearer understanding of why the average distance of a point to the surface is at least $l/4$? $\endgroup$ – yippy_yay Nov 30 '17 at 21:49
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    $\begingroup$ Suppose you have the line segment $0 \le x \le s$. For any point $p$ on this segment, compute $\frac{1}{s} \left( \int_0^p p-x ~ dx + \int_p^s x-p ~dx \right)$; this gives you the average distance form $p$ to all points of the segment, as a function of $p$. Observe that this function is minimize at $p = s/2$, where its value is $s/4$. Now for your problem: the distance form a point in a skinny cylinder to a surface point is very nearly the same as the axial distance between them (because it's skinny!) Hence the computation is well approximated by the 1D problem I just described. $\endgroup$ – John Hughes Dec 1 '17 at 2:02
  • $\begingroup$ There are two different methods of calculating average distance of a point to a surface - either you weigh the contributions to the average distance using the infinitesimal surface area to which the contributing distance belongs - or you weigh it by taking the size of the infinitesimal solid angle subtended by that surface (looking out from the point). The latter method would be suggested by the experiment that motivated the conjecture: All angles are equally likely to be traveled along by the gas atoms to reach the surface. The average you are using is calculated using the first method. $\endgroup$ – yippy_yay Dec 1 '17 at 10:07
  • $\begingroup$ I should add that these two methods for calculating the average distance are not equivalent: The first one obtains a result >0 for the skinny cylinder, no matter how skinny it gets. The "solid angle" method however, will make the average distance go to zero as the cylinder gets skinnier and skinnier. $\endgroup$ – yippy_yay Dec 1 '17 at 10:22
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    $\begingroup$ I'd say that the first of these is not a method of computing average distance in the usual sense "the expected value of the distance between $p$ and $q$, where $p$ is any point of the interior, and $q$ is any point of the surface, and they are chosen with respect to the volume and surface measure respectively." The approach described involves picking $q$ with respect to a projected-solid-angle measure; reasonable, but surely requiring some extra description (like "average wrt subtended solid angle"). As a computer graphics guy, this is my bread and butter, but I'd never say just "average". $\endgroup$ – John Hughes Dec 1 '17 at 13:01

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