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In questions that ask about Planck's constant entering into statistical mechanics, a common and accepted answer is that Planck's constant is an arbitrary normalization that falls out when calculating experimentally measurable quantities.

Specifically, it's said in the above questions that $h$ enters to de-dimensionalize the product $dp dq$ or to normalize $dp dq$ to count states, and the choice for such a constant with units of action is nearly arbitrary.

To the contrary, consider how one calculates quantities in the grand canonical ensemble with a variable number of particles - mentioned in this question.

In the grand canonical ensemble, the key potential is the grand potential, which is $$\Phi = U - TS-\mu N .$$ The link to statistical mechanics comes from $$\Phi = -kT \ln(\mathcal{Z}), $$

where $$\mathcal{Z} = \sum_N e^{\beta \mu N} Z(V,N,T).$$

For simplicity, consider a classical partition function $Z$ for N non-interacting particles. We have then that

$$Z = \frac{1}{h^{3N} N!} \Big(\int d^3p d^3q \, e^{-\beta E({p}, {q})}\Big)^N$$

This yields in turn that

$$\mathcal{Z} = e^{\frac{1}{h^3} \Big(\int d^3p d^3q \, e^{-\beta E({p}, {q})}\Big)e^{\beta \mu}}$$

and in turn that $\Phi$ is proportional to $\frac{1}{h^3}$:

$$\Phi = -kT \frac{1}{h^3} \Big(\int d^3p d^3q \, e^{- \frac{E({p}, {q})}{kT}}\Big)e^{\frac{\mu}{kT}}. $$

For example, for non-interacting massless particles with $E = cp$ in a box with two internal degrees of freedom with $\mu = 0$ (i.e. a naive picture of light, without any $h$ put into the energy or whatnot), we have

$$\Phi = -\frac{16 \pi k^4}{h^3 c^3} T^4 V.$$ This leads to a pressure

$$P = \frac{16 \pi k^4}{h^3 c^3} T^4$$ from the thermodynamic relation above.

As you can see, $h$ enters into an experimentally measurable quantity! Thus one could experimentally measure Planck's constant via the pressure of such a gas. I'll leave it to you to check that $h$ also enters into the pressure of a non-relativistic gas at a fixed $\mu$.

I will take from the above arguments that $h$ is not simply an arbitrary constant for de-dimensionalizing the product $dp dq$. Given this point of view that the de-dimensionalization of $dp dq$ has experimental ramifications, can we show that for every system in thermodynamic equilibrium that the de-dimensionalizing constant $h$ must be a universal constant? That is, since the de-dimensionalization is not arbitrary, can we show the de-dimensionalizing constant must be the same for all equilibrium systems in classical statistical mechanics?

user196574
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2 Answers2

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I do not think that the value of the normalization constant that de-dimensionalizes the phase space integral in the partition function is arbitrary. Already in 1912, Sackur and and Tetrode used vapor pressure data of mercury to determine its numerical value, and found that it was the same as the constant Planck discovered when studying blackbody radiation.

I do not know if at that time there was a fundamental reason to assume that this constant is universal, i.e., independent of the used material. Heisenberg's uncertainty principle, discovered 15 years later, provided that reason.

I highly recommend this paper for an insightful discussion and historical perspective of Sackur and Tetrode's work.

user8153
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    (h/t to Semiclassical who helped me figure this out) The Sackur and Tetrode equation only yields an experimentally measurable $h$ from vapor pressure if you have a good model for the heat capacity of the solid as a function of temperature. They both use variants of the Einstein model for this [see the bottom of page 10 in the paper you link], which is a quantum model, so this is not establishing the claim that $h$ is measurable purely in classical statistical physics. – ACuriousMind Sep 10 '18 at 17:26
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    @ACuriousMind Thank you for looking into this. Two comments: (1) As you point out, the cited paper used an Einstein-ish model to compute the heat capacity of solid mercury. But is that strictly necessary, or due to lack of experimental data? I would assume that at least in principle the heat capcaity could be measured, which would make the Sackur-Tetrode approach a viable way to determine $h$ wihout an underlying quantum-mechanical model. – user8153 Sep 10 '18 at 18:25
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    @ACuriousMind (2) Perhaps more importantly, I believe it is incorrect to say that the value of the quantity $h$ that appears in the classical partition function is "arbitrary", which to me sounds like "it doesn't matter which value you use, the results will always be the same". It seems to me that there is only one correct valuer (6.626E-34 m^2 kg / s), but this value is not predicted by classical theory, and must be obtained from experiment instead. – user8153 Sep 10 '18 at 18:31
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    @user8153 Maybe the following is a better way to say what $h$ is: In pure classical statistical mechanics, the value of $h$ is irrelevant - you can express all physically measurable quantities in terms that do not involve it. Not only is there no classical prediction, there is also no purely classical measurement prescription. Classical statistical mechanics is agnostic as to whether $h$ is a fundamental feature of the universe or a trick of the theorist. Our knowledge of quantum mechanics nowadays fixes its role firmly to be the former. – ACuriousMind Sep 10 '18 at 18:44
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    Note also that a measurement of the heat capacity at discrete points is not enough - the ST equation contains an integral over the heat capacity at constant pressure all the way from absolute zero. As good as we may be in figuring out experiments, I doubt that there is an experiment that allows you to collect the data needed for a good approximation of this integral. – ACuriousMind Sep 10 '18 at 18:48
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    Thank you for the discussion. The measurement of the heat capacity at discrete points can approximate an integral to high accuracy. I believe the use of cryostats should allow one to sample hundreds of temperatures and find the value of h to high precision. Even in the 1970s (pubs.acs.org/doi/pdf/10.1021/je60047a035) we could find decent heat capacity data from absolute zero – user196574 Sep 10 '18 at 20:07
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    @ACuriousMind Thank you for the clarification. I am not sure how to combine your comments with the work by Sackur-Tetrode. Isn't equation (41) in the cited retrospective a measurement prescription, based purely in classical statistical mechanics, which lets one determine the numerical value of $h$ from experimental observations (at least in principle - as you say, it might be difficult to obtain accurate heat capacity data)? It seems to me that different values of $h$ would yield different vapor pressures, so only one specific value of $h$ makes the classical theory consistent with experiment. – user8153 Sep 10 '18 at 20:30
  • I realized that the problem with the heat capacity is not experimental at all: A purely classical theory would have to use the classical theory of solids, i.e. have constant heat capacity as required by Dulong-Petit. The very idea of a temperature-dependent heat capacity of a solid is unclassical. That is, to even conceive of the experimental prescription in which you have to measure $c(T)$ you already need to have thrown away part of classical thermodynamics. – ACuriousMind Sep 10 '18 at 21:49
  • @ACuriousMind I agree that one cannot typically make a correct theoretical prediction for the heat capacity without quantum mechanics. For example, the heat capacity vanishing at absolute zero is typically a sign of quantized energy levels. However, ones heat capacity can certainly vary with temperature in classical statistical mechanics. One simply needs a Hamiltonian say with a slightly complicated dependence on momentum or position such that one cannot substitute out the temperature dependence in the requisite integrals for e.g. the partition function. – user196574 Sep 10 '18 at 22:25
  • cont'd. For example, including interactions between particles in a gas leads to a non-constant heat capacity with temperature and other effects like non-zero virial coefficients. The desire to experimentally measure a heat capacity arises out of the temperature dependence caused by inter-particle interactions, not just from quantization of energy levels (the admittedly typically larger effect near absolute zero). – user196574 Sep 10 '18 at 22:32
  • Though this answer doesn't completely tackle the universality aspect of the question, I found the article excellent and feel satisfied that a classical model paired with experimental data can allow one to measure Planck's constant. Thus I have selected this as the answer. I will keep thinking about the universality aspect and will add any updates as an answer of my own. – user196574 Apr 15 '19 at 06:15
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The derivation in the question is misleading because it prematurely sets the chemical potential to zero. Even in systems where it is zero, you must carry it with you until the final evaluation of expressions, since it is a free parameter of a macrostate in the grand canonical ensemble.

In particular, the chemical potential contains a $\ln(h^3/\text{stuff})$ factor (e.g. the classical ideal non-relativistic gas has $\mu \propto \ln\left(\frac{\lambda^3 N}{V}\right)$, with $\lambda$ the thermal de Broglie wavelength proportional to $h$), that, since it is the fugacity that enters into physical quantities, cancels against the $h^3$ in the denominator.

Therefore, your approach does not necessitate a universal classical value for $h$, since the results can be re-expressed in terms of quantities that do not involve $h$. Note, in particular, that you cannot experimentally directly measure the chemical potential $\mu$ - you need to compute it from indirect measurements, with formulae that in this case also will involve $h$.

ACuriousMind
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  • Thanks for the response. In the grand canonical ensemble, the fixed parameters are T, V, and $\mu$. The pressure and number of particles fluctuate because they are not fixed. I don't believe that $\mu$ is a free parameter. – user196574 Sep 10 '18 at 19:59
  • Also, there are cases where nature "fixes" $\mu$ and not number of particles, the classic example being a photon gas. However, we can even fix $\mu$ and not the number of particles by affixing our system to a "particle number" bath. This is how I think about it, please let me hear your thoughts: A heat bath can give heat without noticeably changing its temperature until an adjoined smaller system is at the same temperature. A particle number bath can give particles without noticeably changing its chemical potential until an adjoined smaller system is at the same chemical potential. – user196574 Sep 10 '18 at 20:03
  • @user196574 $\mu$ is fixed for a specific macrostate in the grand canonical example. It's a free parameter in the sense that derivations that are supposed to hold for a system must hold for all possible values of this parameter, just like they must hold for all values of $T$ and $V$ (that obey the equation of state). You can't just say "but $T = 1\mathrm{K}$ in my system". That's not a property of the system, but of a state of the system. In any experimental test, you would have to establish that indeed $\mu = 0$ for the state you're measuring, thus losing the ability to measure $h$. – ACuriousMind Sep 10 '18 at 20:42
  • I think I will need to refresh my knowledge on the chemical potential before returning to the role of $h$ in the grand canonical ensemble. Specifically, whether it is necessary to have some theoretical expression for $\mu$ (that depends on $h$) in order to be able to set $\mu$, or whether they may be ways to experimentally determine a chemical potential without knowledge of $h$, just as there are ways to experimentally determine a volume without knowledge of $h$. – user196574 Sep 10 '18 at 21:09