# Spheres in a Space with Trillions of Dimensions

I don’t venture into speculative science writing – this is just about classical statistical mechanics; actually about a special mathematical aspect. It was one of the things I found particularly intriguing in my first encounters with statistical mechanics and thermodynamics a long time ago – a curious feature of volumes.

I was mulling upon how to ‘briefly motivate’ the calculation below in a comprehensible way, a task I might have failed at years ago already, when I tried to use illustrations and metaphors (Here and here). When introducing the ‘kinetic theory’ in thermodynamics often the pressure of an ideal gas is calculated first, by considering averages over momenta transferred from particles hitting the wall of a container. This is rather easy to understand but still sort of an intermediate view – between phenomenological thermodynamics that does not explain the microscopic origin of properties like energy, and ‘true’ statistical mechanics. The latter makes use of a phase space with with dimensions the number of particles. One cubic meter of gas contains ~1025 molecules. Each possible state of the system is depicted as a point in so-called phase space: A point in this abstract space represents one possible system state. For each (point-like) particle 6 numbers are added to a gigantic vector – 3 for its position and 3 for its momentum (mass times velocity), so the space has ~6 x 1025 dimensions. Thermodynamic properties are averages taken over the state of one system watched for a long time or over a lot of ‘comparable’ systems starting from different initial conditions. At the heart of statistical mechanics are distributions functions that describe how a set of systems described by such gigantic vectors evolves. This function is like a density of an incompressible fluid in hydrodynamics. I resorted to using the metaphor of a jelly in hyperspace before.

Taking averages means to multiply the ‘mechanical’ property by the density function and integrate it over the space where these functions live. The volume of interest is a  generalized N-ball defined as the volume within a generalized sphere. A ‘sphere’ is the surface of all points in a certain distance (‘radius’ R) from an origin

$x_1^2 + x_2^2 + ... + x_ {N}^2 = R^2$

($x_n$ being the co-ordinates in phase space and assuming that all co-ordinates of the origin are zero). Why a sphere? Because states are ordered or defined by energy, and larger energy means a greater ‘radius’ in phase space. It’s all about rounded surfaces enclosing each other. The simplest example for this is the ellipse of the phase diagram of the harmonic oscillator – more energy means a larger amplitude and a larger maximum velocity.

And here is finally the curious fact I actually want to talk about: Nearly all the volume of an N-ball with so many dimensions is concentrated in an extremely thin shell beneath its surface. Then an integral over a thin shell can be extended over the full volume of the sphere without adding much, while making integration simpler.

This can be seen immediately from plotting the volume of a sphere over radius: The volume of an N-ball is always equal to some numerical factor, times the radius to the power of the number of dimensions. In three dimensions the volume is the traditional, honest volume proportional to r3, in two dimensions the ‘ball’ is a circle, and its ‘volume’ is its area. In a realistic thermodynamic system, the volume is then proportional to rN with a very large N.

The power function rN turn more and more into an L-shaped function with increasing exponent N. The volume increases enormously just by adding a small additional layer to the ball. In order to compare the function for different exponents, both ‘radius’ and ‘volume’ are shown in relation to the respective maximum value, R and RN.

The interesting layer ‘with all the volume’ is certainly much smaller than the radius R, but of course it must not be too small to contain something. How thick the substantial shell has to be can be found by investigating the volume in more detail – using a ‘trick’ that is needed often in statistical mechanics: Taylor expanding in the exponent.

A function can be replaced by its tangent if it is sufficiently ‘straight’ at this point. Mathematically it means: If dx is added to the argument x, then the function at the new target is f(x + dx), which can be approximated by f(x) + [the slope df/dx] * dx. The next – higher-order term would be proportional to the curvature, the second derivation – then the function is replaced by a 2nd order polynomial. Joseph Nebus has recently published a more comprehensible and detailed post about how this works.

So the first terms of this so-called Taylor expansion are:

$f(x + dx) = f(x) + dx{\frac{df}{dx}} + {\frac{dx^2}{2}}{\frac{d^2f}{dx^2}} + ...$

If dx is small higher-order terms can be neglected.

In the curious case of the ball in hyperspace we are interested in the ‘remaining volume’ V(r – dr). This should be small compared to V(r) = arN (a being the uninteresting constant numerical factor) after we remove a layer of thickness dr with the substantial ‘bulk of the volume’.

However, trying to expand the volume V(r – dr) = a(r – dr)N, we get:

$V(r - dr) = V(r) - adrNr^{N-1} + a{\frac{dr^2}{2}}N(N-1)r^{N-2} + ...$
$= ar^N(1 - N{\frac{dr}{r}} + {\frac{N(N-1)}{2}}({\frac{dr}{r}})^2) + ...$

But this is not exactly what we want: It is finally not an expansion, a polynomial, in (the small) ratio of dr/r, but in Ndr/r, and N is enormous.

So here’s the trick: 1) Apply the definition of the natural logarithm ln:

$V(r - dr) = ae^{N\ln(r - dr)} = ae^{N\ln(r(1 - {\frac{dr}{r}}))}$
$= ae^{N(\ln(r) + ln(1 - {\frac{dr}{r}}))}$
$= ar^Ne^{\ln(1 - {\frac{dr}{r}}))} = V(r)e^{N(\ln(1 - {\frac{dr}{r}}))}$

2) Spot a function that can be safely expanded in the exponent: The natural logarithm of 1 plus something small, dr/r. So we can expand near 1: The derivative of ln(x) is 1/x (thus equal to 1/1 near x=1) and ln(1) = 0. So ln(1 – x) is about -x for small x:

$V(r - dr) = V(r)e^{N(0 - 1{\frac{dr}{r})}} \simeq V(r)e^{-N{\frac{dr}{r}}}$

3) Re-arrange fractions …

$V(r - dr) = V(r)e^{-\frac{dr}{(\frac{r}{N})}}$

This is now the remaining volume, after the thin layer dr has been removed. It is small in comparison with V(r) if the exponential function is small, thus if ${\frac{dr}{(\frac{r}{N})}}$ is large or if:

$dr \gg \frac{r}{N}$

Summarizing: The volume of the N-dimensional hyperball is contained mainly in a shell dr below the surface if the following inequalities hold:

${\frac{r}{N}} \ll dr \ll r$

The second one is needed to state that the shell is thin – and allow for expansion in the exponent, the first one is needed to make the shell thick enough so that it contains something.

This might help to ‘visualize’ a closely related non-intuitive fact about large numbers, like eN: If you multiply such a number by a factor ‘it does not get that much bigger’ in a sense – even if the factor is itself a large number:

Assuming N is about 1025  then its natural logarithm is about 58 and…

$Ne^N = e^{\ln(N)+N} = e^{58+10^{25}}$

… 58 can be neglected compared to N itself. So a multiplicative factor becomes something to be neglected in a sum!

I used a plain number – base e – deliberately as I am obsessed with units. ‘r’ in phase space would be associated with a unit incorporating lots of lengths and momenta. Note that I use the term ‘dimensions’ in two slightly different, but related ways here: One is the mathematical dimension of (an abstract) space, the other is about cross-checking the physical units in case a ‘number’ is something that can be measured – like meters. The co-ordinate  numbers in the vector refer to measurable physical quantities. Applying the definition of the logarithm just to rN would result in dimensionless number N side-by-side with something that has dimensions of a logarithm of the unit.

Using r – a number with dimensions of length – as base, it has to be expressed as a plain number, a multiple of the unit length $R_0$ (like ‘1 meter’). So comparing the original volume of the ball $a{(\frac{r}{R_0})}^N$ to one a factor of N bigger …

$aN{(\frac{r}{R_0})}^N = ae^{\ln{(N)} + N\ln{(\frac{r}{R_0})}}$

… then ln(N) can be neglected as long as $\frac{r}{R_0}$ is not extreeeemely tiny. Using the same argument as for base e above, we are on the safe side (and can neglect factors) if r is of about the same order of magnitude as the ‘unit length’ $R_0$. The argument about negligible factors is an argument about plain numbers – and those ‘don’t exist’ in the real world as one could always decide to measure the ‘radius’ in a units of, say, 10-30 ‘meters’, which would make the original absolute number small and thus the additional factor non-negligible. One might save the argument by saying that we would always use units that sort of match the typical dimensions (size) of a system.

Saying everything in another way: If the volume of a hyperball ~rN is multiplied by a factor, this corresponds to multiplying the radius r by a factor very, very close to 1 – the Nth root of the factor for the volume. Only because the number of dimensions is so large, the volume is increased so much by such a small increase in radius.

As the ‘bulk of the volume’ is contained in a thin shell, the total volume is about the product of the surface area and the thickness of the shell dr. The N-ball is bounded by a ‘sphere’ with one dimension less than the ball. Increasing the volume by a factor means that the surface area and/or the thickness have to be increased by factors so that the product of these factors yield the volume increase factor. dr scales with r, and does thus not change much – the two inequalities derived above do still hold. Most of the volume factor ‘goes into’ the factor for increasing the surface. ‘The surface becomes the volume’.

This was long-winded. My excuse: Also Richard Feynman took great pleasure in explaining the same phenomenon in different ways. In his lectures you can hear him speak to himself when he says something along the lines of: Now let’s see if we really understood this – let’s try to derive it in another way…

And above all, he says (in a lecture that is more about math than about physics)

Now you may ask, “What is mathematics doing in a physics lecture?” We have several possible excuses: first, of course, mathematics is an important tool, but that would only excuse us for giving the formula in two minutes. On the other hand, in theoretical physics we discover that all our laws can be written in mathematical form; and that this has a certain simplicity and beauty about it. So, ultimately, in order to understand nature it may be necessary to have a deeper understanding of mathematical relationships. But the real reason is that the subject is enjoyable, and although we humans cut nature up in different ways, and we have different courses in different departments, such compartmentalization is really artificial, and we should take our intellectual pleasures where we find them.

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Further reading / sources: Any theoretical physics textbook on classical thermodynamics / statistical mechanics. I am just re-reading mine.