# Can Boltzmann's Constant be Calculated from Fundamental Constants?

Boltzmann's constant is not "fundamental" in the same sense as $c$ or $G$. Rather, it is an artifact of measuring temperature in units of kelvins rather than joules. Other non-fundamental constants can often be expressed compactly in terms of other more fundamental constants. (For example, the Rydberg constant can be expressed as $R_\infty=\frac{m_ee^4}{8\epsilon_o^2h^3c}$.)

So it would stand to reason that $k_B$ can be calculated from other more fundamental constants as well (even if not from a nice compact equation like the one given for $R_\infty$ above). In particular, $k_B$ essentially converts between units of joules and units of kelvins. The kelvin is defined as 1/273.16 the temperature difference between absolute zero and the temperature of the triple point of water. Water is made of H2O molecules, which obey the laws of quantum mechanics. So I would expect there to be a (perhaps ugly) expression for $k_B$ in terms of statistical properties of water, which in turn should reduce to fundamental constants like $h, c, m_e,$ etc.

1. Does such an expression for $k_B$ exist (at least in principle)?

2. Could $k_B$ be calculated theoretically from computer simulations of water, and if so, has anyone done it?

• @nicael I am asking about Boltzmann's constant $k_B$, not the Stefan–Boltzmann constant. Of course other less fundamental constants (like the Stefan–Boltzmann constant) can be expressed in terms of $k_B$, but that's not what I'm asking. I'm asking if $k_B$ can be expressed in terms of more fundamental constants. I don't think anyone would consider the Stefan–Boltzmann constant more fundamental than $k_B$. – WillG Apr 13 '18 at 4:55
• Oh sorry! I did somehow manage to misread it. – nicael Apr 13 '18 at 4:57

Simply by dimensional analysis, what you are asking for is not possible. $k_B$ is the fundamental unit of entropy, and no combination of other dimensionful universal constants can make one with units of entropy. A more rigorous treatment would use the Buckingham π theorem . Given the set of SI constants $c, G, \hbar, \frac1 {4 \pi \epsilon_0}, k_B$, we find that there are no dimensionless quantities that can be derived, and that every dimensionful quantity in SI can be expressed as a dimensionless quantity times some power of these units. That is the starting point in the definition of Planck units. The inclusion of $k_B$ in the set is crucial; the non-existence of any dimensionless quantities that can be constructed from these means that no expression of the form $k_B = f(c,G,\hbar, \frac1{4\pi \epsilon_0})$ can be found, because dividing the RHS by the LHS would result in such a dimensionless quantity.

What you could do would be to augment this set by one or more additional dimensionful constants $\kappa_i$, and decide that the $\kappa_i$ are more fundamental than $k_B$. One such example would be the temperature of the triple point of water (though few would claim this is "fundamental"). If you did this (and $\kappa_i$ has appropriate units) you could obtain a function $k_B = f(c,G,\hbar,\frac1{4 \pi \epsilon_0}, \kappa_i)$ which you could claim is a derivation of $k_B$ in terms of more fundamental constants, but really there is no such example of a quantity $\kappa _i$ which we would consider more fundamental than $k_B$ today, and any such function could be inverted to alternatively give $\kappa_i = g_i(c,G,\hbar,\frac1{4\pi \epsilon_0}, k_B)$ which we would think of as more "fundamental" (though ultimately what is "fundamental" is pretty arbitrary and mostly a matter of aesthetics).

One more thing, one could adopt an information-theoretic definition of entropy as in statistical mechanics. For example, $S = \log \Omega$, the logarithm of the number of microstates. If you adopt this convention then $k_B = 1$. But even still, there is nothing that stops us from instead defining $S = k_B \log \Omega$ and making entropy dimensionful. Or, even if you want a dimensionless entropy, you could make the logarithm base $b$ rather than a natural logarithm, in which case $k_B = \log b$.

• But isn't the triple point of water determined by $c$ $\hbar$ and $\epsilon_0$? Admittedly, this relation will be complicated but there should be one in principle. – sagittarius_a Apr 12 '18 at 6:02
• @sagittarius_a The temperature of the triple point cannot be determined in terms of just these three quantities (in the SI system), because there is no way to construct a quantity with units of temperature from them. It must include a factor of $k_B^{-1}$ to be a temperature. Allowing the use of additional physically relevant quantities like the masses and charges of the particles involved change that fact because none of them have units of temperature. Note that in other systems of units, this may not be true, and then it may be that $k_B$ takes some dimensionless value (e.g. $1$). – Logan M Apr 12 '18 at 6:20
• Ok! Ageed. Regarding your last point: the value of $k_B$ will probably be fixed in a future SI redefinition – sagittarius_a Apr 12 '18 at 6:59
• I'm not sure I agree. Kelvins are just a unit of energy in disguise, so I see no reason we shouldn't be able calculate the conversion factor from kelvins to joules from first principles. Your argument is similar to the following (false) argument: Let the average distance of an electron from the proton in hydrogen define the unit of "one bohr"; call the "bohr constant" the conversion factor from meters to bohrs; we cannot express this "bohr constant" in terms of fundamental constants because none of them involves units of "bohrs". But we can, and it is given by the Bohr radius formula. – WillG Apr 13 '18 at 4:52
• Relating the above back to my question: We should (in principle) be able to calculate the triple point of water in terms of joules. Statistically, this means the number of joules that must be added to water at the triple point in order to increase its multiplicity by a factor of $e$, based on the definition of temperature $T=\frac{\partial E}{\partial (\ln \Omega)}$. This number of joules is defined to be 273.16 kelvins, so divide this number of joules by 273.16 and you should have $k_B$—correct? – WillG Apr 13 '18 at 5:04

Molecular dynamics can only very roughly estimate the triple point of water. The melting temperature is in principle the same. These authors claim an accuracy of 10 kelvin: http://www.pnas.org/content/early/2016/07/07/1602375113

One can doubt that. I would regard 50 kelvin already as impressive.

An ab initio definition of temperature or coldness could be $\beta = \frac{1}{kT} = \frac{1}{\Omega} \frac{d\Omega}{dE},$ the fractional change of the multiplicity with internal energy. Room temperature would then be approximately 4 % per meV for any system.

Anyway, in the impending redefinition of the kelvin, the value of the Boltzmann constant will be exactly fixed, and the triple-point temperature will be determined experimentally.