To talk about this right, we need a little physics. Most of you probably have heard of the principle of conservation of energy... If you haven't, or even if you have, make sure to read about how Hamiltonian Mechanics is Awesome. Anyway, so if you imagine little billiard balls of uniform mass $m$ freely zipping about, banging into stuff, each particle having position $\mathbf{q}_i$ and momentum $\mathbf{p}_i$, then since we know Hamiltonian Mechanics is Awesome, this says that the two quantities evolve in such a way that the energy of the $i$th particle \[

E_i = \frac{|\mathbf{p}_i|^2}{2m}

\]

stays constant. But if you imagine having a very, very large number of such billiard balls (several billiards of such (1 billiard = $10^{15}$ in most dialects of English, or such equivalent cognates) HAR HAR!), and compute the

*total energy*, summing them all up,\[

H = \sum_i E_i = \sum_i \frac{|\mathbf{p}_i|^2}{2m}.

\] To say that the

*total*energy is constant, then, is to say \[

\sum_i \frac{|\mathbf{p}_i|^2}{2m} = E

\] for some fixed number $E$. But what does this say? If we write out each \[

|\mathbf{p}_i|^2 = p_{ix}^2 + p_{iy}^2 + p_{iz}^2,

\] then \[

2mE=\sum_i p_{ix}^2 + p_{iy}^2 + p_{iz}^2,

\] meaning that if we sum over all the particles, say $N$ of them, where $N \approx 10^{24}$ (a milliard billiard of billiard balls), this means

*we're summing the squares of a $3N$-dimensional vector and equating it to a constant*. So if we ask ourselves:

*What is the collection of all possible momenta of $N$ particles that have one constant energy level $E$,*

this is the

*same*as saying,

*What are all the points on the $(3N-1)$-dimensional sphere of radius $\sqrt{2mE}$?*

*(the -1 from the fact that you're expressing the surface as the solution set to one equation: the boundary of the $k$-ball is the $(k-1)$-sphere)*

*Interesting*, you might say, but is there any occasion which actually requires you to

*truly*do calculations treating it precisely as a geometrical $(3N-1)$-dimensional object?

Well, what's one of the first things you think about when talking geometry? Areas and volumes, of course. There are situations in which you need to

*integrate*over such energy "surface", or within the $N$-ball contained within, for $N \sim 10^{24}$. Basically, to find quantities like entropy, one needs to "count some number of cells" inside the $3N$-ball . This yields the total "number" of energy states with level

*less than or equal to*$E$. "Number" in quotations because classically, of course, the actual number of states is the number of points on a sphere, which is of course uncountably infinite. We instead mean, per unit $3N$-dimensional volume, which actually leads to the

*density*of states. The distinction is a little blurry, because, with a little help of quantum mechanics, we can think of each state as a cell of volume $(2\pi \hbar)^{3N}$, and each distinct state of momentum, which must be quantized, ultimately arises from boundary conditions in solving Schrödinger's equation, or eigenstuff.

Suffice to say, however, let's talk as if calculating the volume of the $3N$-ball does indeed count the total number of momentum states with energy less than or equal to $E$, i.e., let's calculate \[

\frac{1}{(2\pi \hbar)^{3N}}\int_{|\mathbf{p}|^2 \leq 2mE} d\mathbf{q}d\mathbf{p} =\frac{V^N}{(2\pi \hbar)^{3N}}\int_{|\mathbf{p}|^2 \leq 2mE} d\mathbf{p} .

\]

where $V$ is the ordinary 3D volume of the spatial region where all our particles live. For the momentum integral, we need the formula (whose derivation is fodder for another post!) for the volume of an $K$-ball \[

V_K(R) = \frac{\pi^{K/2}}{\Gamma\left(1+\frac{K}{2}\right)}R^K

\]

where $\Gamma(1+ K/2)$ is $(K/2)!$, the factorial of $K/2$, if $K$ is even, and $\pi^{1/2} K!!/2^{K/2}$ where $K!!$ is the double factorial $1\cdot 3 \cdot 5 \cdots k$, for $k$ odd.

Plugging this in with $R = \sqrt{2mE}$ and $K=3N$ we have the total volume of a $3N$-ball in momentum space: \[

V_{3N}(\sqrt{2mE}) = \frac{\pi^{3N/2}}{\Gamma\left(1+\frac{3N}{2}\right)} (2mE)^{3N/2}

\]

To make headway here, we can use Stirling's Approximation for the Gamma function:\[

\Gamma\left(1+\frac{3N}{2}\right) \approx \sqrt{3N\pi} \left(\frac{3N}{2e}\right)^{3N/2}

\] This gives \[

V_{3N}(\sqrt{2mE}) \approx \frac{1}{\sqrt{3\pi N}}\left(\frac{4\pi meE}{3N} \right)^{3N/2}

\] or, the total number of states is \[

\#\text{states} = \frac{V^N V_{3N}(\sqrt{2mE})}{(2\pi \hbar)^{3N}} \approx \frac{1}{\sqrt{3\pi N}}\left(\frac{4\pi meEV^{2/3}}{3N (2\pi \hbar)^2} \right)^{3N/2}

\]

Considering the smallness of $\hbar$ ($10^{-34}$ in everyday units) and the exponent $3N/2$ being on the order of $10^{24}$, this could be a staggeringly huge number (but that's what you expect: for lots of particles, they really can occupy even more states. For example, two particles behaving independently is the

*square*the total number of states of for one particle). Thus to measure entropy, we take the natural log of all of this (and add a factor of the Boltzmann constant, $k$ which is on the order of $10^{-23}$ in ordinary units), we get \[

k\ln(\#\text{states}) = \frac{3Nk}{2} \left( \ln \left(\frac{4\pi mEV^{2/3}}{3N(2\pi \hbar)^2}\right) +1\right) -\frac{1}{2}\ln(3\pi N)k.

\]

Since those last terms are small potatoes (not being multiplied by $N$), we usually just drop it to arrive at the

**entropy**of our system of particles as

\[

S = \frac{3Nk}{2} \left( \ln \left(\frac{4\pi mEV^{2/3}}{3N(2\pi \hbar)^2}\right)\right)+\frac{3Nk}{2}.

\] And for you chemists, $Nk = nR$, where $n$ is the number of moles and $R$ is the gas constant:

\[

S = \frac{3nR}{2}\left( \ln \left(\frac{4\pi mEV^{2/3}}{3N(2\pi \hbar)^2}\right)\right)+\frac{3nR}{2}.

\]

That's really nasty. How about we try to plug numbers into this and see what it is? We'll do that more concretely next time. But suffice to say, let's get back to what's interesting: we calculated the volume of some high-dimensional stuff!

Cool stuff you have got and you keep update all of us. patronato genova

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