The Size of Phase Space 8/12/22

Science with Richard Bleil

A friend of mine and I did a very interesting calculation today.  We calculated the size of phase space for Helium.  It was an interesting calculation so I will walk through it here.  A couple of notes.  First, this will be a very mathematically heavy post.  Second, to the best of my knowledge, nobody, anywhere, ever has determined this value.  It’s a pretty simple idea, so maybe somebody does, but when I was studying statistical thermodynamics, I was told that this has never been done before. 

The first and obvious question, then, becomes what is phase space?  Phase space is the size of every possible distribution of a system in the order of the atoms, the velocities of the atoms (the kinetic energy distribution), and the distribution of electron excited states.  It’s every possible combination of every possible distribution and energies of the entire system.

First, a little bit of statistical thermodynamics.  We call the phase space “w” (for the sake of this post) whatever that might be.  In statistical thermodynamics, we learn that entropy, S, is related to the size of phase space through the deceptively simple equation S=k ln w, where k is Boltzmann’s constant (1.3807×10^-27 J s/mol).  You might be wondering how we can calculate entropy, S, if we can’t find phase space, and the answer is we cannot.  We can only calculate changes in entropy, Sf – Si where Sf is the final entropy, and Si is the initial entropy.  Thus, we measure relative entropy only.  But, the thought struck me that if we can find just one absolute entropy for Si, then we can calculate phase space, and entropy for everything else.

Is this important?  No, not really.  We will likely never need absolute entropy or to know the value of phase space, but honestly, we really don’t need to know the size of Avogadro’s number either (6.02×10^23).  We use it in first year chemistry courses in a (usually vain) attempt to make a connection in the mind of students that Avogadro’s number is just a number, just like a dozen is 12, but beyond freshman chemistry, I’ve never seen anybody actually use this value.  In the lab, we use grams because we can measure them, but we cannot measure the number of moles.  And on paper, we think in terms of moles, because it’s easier to think in such terms than grams.  It’s easier to think one mole of sulfur to six moles of fluorine than it is to think in terms of 32 grams of sulfur to 174 grams of fluorine.  So, honestly, this is a conceptual exercise with no real value.

So we can re-arrange our formula for change in entropy from S=k ln w to Sf – Si = k(ln wf – ln wi), or, with a little algebra, Sf – Si = k ln (wf/wi).  Now, I expect this will be easier to use base 10 log instead of natural log, so we get Sf – Si = (k/2.303) log (wf/wi).  So, we need an initial system where the absolute entropy is known.  That system is helium in a perfect crystal at absolute zero.

This is a conceptual system since absolute zero will never be reached, but helium can crystalize, and will be a solid crystal at absolute zero.  In a perfect crystal, all of the exact positions are known relative to the position of just three atoms.  Since we know the location of the atoms, we remove the unknown of position.  What’s more, at absolute zero, there is no kinetic energy.  All of the helium atoms are fixed in location, so there is no unknown in that.  Additionally, at absolute zero, all elements are at the ground electronic state, so there is no distribution of excited state atoms. 

Some of the better read of you might be wondering if this isn’t, in fact, a Bose-Einstein condensate.  Yes, in reality, it would be, but this is conceptual only, and since we will never actually reach absolute zero, we can ignore the Bose-Einstein condensate.

So, in our system, we have one, and only one, distribution, where the relative positions of all atoms are known, the velocities are all zero, and the electronic configurations are all ground state.  Thus, S=k ln (1) =0 since ln(1)=0.  Entropy is zero because we know all there is to know about this system.  So, in the equation Sf – Si = (k/2.303) ln (wf/wi), we substitute for Si and wi and have Sf = k/2.303 ln (wf).  As it turns out, we know the value of helium at STP (standard temperature and pressure, namely room temperature and 1 atm of pressure), S=126.153 J/mol K.  So, for Helium, 126.153=(1.3803×10^-23/2.303)log(w). 

To the best of my knowledge, this is the first and only time we have been able to solve for w.  I’ll not go through step by step, but with a little algebra, I am getting w=1.27×10^(2×10^27).  This number is so huge that I cannot even fathom it.  1×10^27 is one billion trillion.  To raise this to the power ten is just more than I can comprehend, so I believe that, at the very least, I’m on the right track. 

There are two important results here.  First, it means that phase space is, indeed, finite.  There are a fixed number of possible distributions, even if we include distribution and energies.  This is only possible because the number of energy levels at a quantum level are quantized, or finite.  The number of potential energy levels of even one atom of helium is beyond comprehension, but it is finite.  Second, it’s important that this number “looks” like infinity to us.  No, it’s not infinity, but the way Statistical Thermodynamics works, it is critical that the number is so large that to our limited comprehension, it looks like infinity.  And to me, it does.

And you may be one of the first people on earth to know this value!

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