# Is the Boltzmann constant really that important?

I read a book in which one chapter gave a speech about the fundamental constants of the Universe, and I remember it stated this:

If the mass of an electron, the Planck constant, the speed of light, or the mass of a proton were even just slightly different (smaller or bigger) than what they actually are, then the whole Universe would not exist as we know it. Maybe we all wouldn't exist.

This speech works for all the fundamental known constants of the Universe but one: the Boltzmann constant. Its value is well known but even if its value were $10$ times bigger or if it were exactly $1$, or $45.90$ or $10^6$ well... the Universe would remain the same as it is now. The Boltzmann constant is not really fundamental to the existence of the Universe.

Maybe they weren't the exact words, but the concept is correct.

Now I ask: is that true, and why?

• @CuriousOne Beware mistaking a change of dimensions with a choice of units. They are not the same thing. – DanielSank Jan 22 '16 at 5:37
• @DanielSank: I was not talking about either dimensions or units but just the scale. The thermodynamic relation between energy and temperature is fixed by nature, independently of how we express it, we might just make it one on paper. – CuriousOne Jan 22 '16 at 5:47
• @CuriousOne I don't understand your comment, apologies. Temperature is defined (ignoring $k_b$) as $1/T \equiv \partial \ln W / \partial E$. – DanielSank Feb 5 '16 at 20:40
• @DanielSank: $kT$ is $kT$, isn't it? Unless one can find a physical phenomenon in which a term like $kT^2$ plays a role, one can set the constant to anything we like, can't we? You even gave the theoretical reason for why that is yourself. Nice post, by the way! – CuriousOne Feb 5 '16 at 21:15
• @CuriousOne Yes, we could not only set $k_b$ to a any value but we could even change its dimensions because we made up the dimensions of $k_b$ anyway in the first place. Of course, as I said in the answer, it makes the most sense to make $k_b$ dimensionless, and pick the value 1. – DanielSank Feb 5 '16 at 23:05

We can understand all of this business if we visit the statistical mechanics notion of temperature, and then connect it to experimental realities.

# Temperature is a Lagrange multiplier (and should have dimensions of energy)

First we consider the statistical mechanics way of defining temperature. Given a physical system with some degree of freedom $X$, denote the number of possible different states of that system when $X$ takes the value $x$ by the symbol $\Omega(x)$. From statistical considerations we can show that modestly large systems strongly tend to sit in states such that $\Omega(x)$ is maximized. In other words, to find the equilibrium state $x_\text{eq}$ of the system you would write $$\left. \left( \frac{d\Omega}{dx} \right) \right|_{x_\text{eq}} = 0$$ and solve for $x_\text{eq}$. It's actually more convenient to work with $\ln \Omega$ so we'll do that from now on.

Now suppose we add the constraint that the system has a certain amount of energy $E_0$. Denote the energy of the system when $X$ has value $x$ by $E(x)$. In order to find the equilibrium value $x_\text{eq}$, we now have to maximize $\ln \Omega$ with respect to $x$, but keeping the constraint $E(x)=E_0$. The method of Lagrange multipliers is the famous mathematical tool used to solve such problems. One constructs the function $$\mathcal{L}(x) \equiv \ln \Omega(x) + t (E_0 - E(x))$$ and minimizes $\mathcal{L}$ with respect to $x$ and $t$. The parameter $t$ is the Lagrange multiplier; note that it has dimensions of inverse energy. The condition $\partial \mathcal{L} / \partial x = 0$ leads to $$t \equiv \frac{\partial \ln \Omega}{\partial x} \frac{\partial x}{\partial E} \implies t = \frac{\partial \ln \Omega}{\partial E} \, .$$ Now remember the thermodynamic relation $$\frac{1}{T} = \frac{\partial S}{\partial E} \, .$$ Since the entropy $S$ is defined as $S \equiv k_b \ln \Omega$ we see that the temperature is actually $$T = \frac{1}{k_b t} \, .$$ In other words, the thing we call temperature is just the (reciprocal of the) Lagrange multiplier which comes from having fixed energy when you try to maximize the entropy of a system, but multiplied by a constant $k_b$.

# Logically, $k_b$ doesn't need to exist

If not for the $k_b$ then temperature would have dimensions of energy! You can see from the discussion above that $k_b$ is very much just an extra random constant that doesn't need to be there. Entropy could have been defined as a dimensionless quantity, i.e. $S \equiv \ln \Omega$ without the $k_b$ and everything would be fine. You'll notice in calculations that $k_b$ and $T$ almost always shows up together; it's no accident and it's basically because, as we said, $k_b$ is just a dummy factor which converts energy to temperature.

# But then there's history :(

Folks figured out thermodynamics before statistical mechanics. In particular, we had thermometers. People measured the "hotness" of stuff by looking at the height of a liquid in a thermometer. The height of a thermometer reading was the definition of temperature; no relation to energy. Entropy was defined as heat transfer divided by temperature. Therefore, entropy has dimensions of $[\text{energy}] / [\text{temperature}]$.$^{[a]}$

We measured the temperatures $T$, pressures $P$, volumes $V$, and number of particles $N$ of some gasses and found that they always obeyed the ideal gas law$^{[b,c]}$

$$P V = N k_b T \, .$$

This law was known from experiment for a long time before Boltzmann realized that entropy is actually proportional to the logarithm of the number of available microstates, a dimensionless quantity. However, since entropy was already defined and had this funny temperature dimensions, he had to inject a dimensioned quantity for "backwards compatibility". He was the first to write $$S = k_b \ln \Omega$$ and this equation is so important that it's on his tomb.

# Connecting temperature and energy

In practice, it is actually rather difficult to measure temperature and energy in the same system over many orders of magnitude. I think that it's for this reason that we still have independent temperature and energy standards and units.

# Summary

• Boltzmann's constant is just a conversion between energy and a made-up dimension we call "temperature". Logically, temperature should have dimensions of energy and Boltzmann's constant is just a dummy that converts between the two for historical reasons. Boltzmann's constant contains no physical meaning whatsoever. Note that the value of $k_b$ isn't the real issue; values of constants depend on the units system you use. The important point is that, unlike the speed of light or the mass of the proton, $k_b$ doesn't refer to any unit-independent physical thing in Nature.

• Temperature is the Langrange multiplier that comes from imposing fixed energy on the problem of maximizing entropy. As such, it logically has dimensions of energy.

• Boltzmann's constant $k_b$ only exists because people defined temperature and entropy before they understood statistical mechanics.

• You will always see $k_b$ and $T$ together because the only logically relevant parameter is $k_b T$, which has dimensions of energy.

# Notes

$[a]$: Note that if temperature had dimensions of energy then under this definition entropy would have been dimensionless (as it "should" be).

$[b]$: Actually, this law was originally written as $PV = n R T$ where $n$ is the number of moles of a substance and $R$ is the ideal gas constant. That's not really important though because you can group Avogadro's number in with $R$ to get $k_b$. $R$ and $k_b$ have equivalent "status".

$[c]$: Note again how $k_b$ and $T$ show up together.

• Isn't this true of speed of light too? It's in the question – innisfree Jan 22 '16 at 8:27
• @innisfree I think not. While the numerical value of the speed of light depends on the unit system, I think that physical quantity really does represent something in Nature. When theorists say that they "set $c$ to 1", they really mean that the letter $x$ means "length divided by the speed of light". It's an incredibly annoying bad habit that theorists don't say this correctly. Similarly, when people say they "set $\hbar$ to 1", they really mean that their Hamiltonians are actually $H/\hbar$, i.e. has dimensions of angular frequency. Does that answer your question? – DanielSank Jan 22 '16 at 8:54
• No, when I set c=1, I really set it to one. SR is telling that x and t are mixed by boost and ought to really have the same units. – innisfree Jan 22 '16 at 8:57
• Similarly, k=1 just reflects that T and E ought to meausred in the same units, I don't see a difference. – innisfree Jan 22 '16 at 8:59
• @DanielSank Amazing answer, you really enlighten me! I feel usually uncomfortable in writing those comments but I have to tell when someone made an excellent work! – Les Adieux Jan 22 '16 at 11:20

I think this question can be interpretted a few ways. I will frame the argument using the example of phase transitions.

1. Does temperature need to exist (or do we really need another constant)?:

No.

Let us consider what it means to have a phase transition. In the broadest terms, we are introducing energy into a system and approaching a critical point that either leads to long-range order (gas to solid) or disorder (solid to gas). We typically have defined these transitions to occur at a critical temperature. However, temperature is related to energy through the Boltzmann constant as

$$E = k_bT$$

Thus, we could also define a transition energy instead of temperature which would remove the need for the Boltzmann constant. Therefore, I would say that we are more or less arguing that we could define the entire universe without a temperature like parameter, which is true.

2. Can we redefine temperature (or fixing our energy scale)?

Absolutely. However, we can do this for any fundamental constant and there is nothing special about Boltzmann. This is just trivial unit conversion.

3. What would happen if the relationship between the microscopic and macroscopic world were different? (or fixing the energy AND temperature scale)

In this interpretation, we would fix our temperature and energy scales, but change the relationship between the two. This would mean that the amount of energy required to heat or cool would be different. Thus, we are changing the amount of energy required to phase transition for example. We could still eliminate temperature and just use energy, but the amount of energy required would be fundamentally different.

In Conclusion

Temperature is an unnecessary variable as all physics can be simply transcribed in terms of energy. Thus, the Boltzmann constant could be removed. However, if we consider the temperature and energy scales fixed, changing the Boltzmann constant amounts to changing the energy required for many physical processes (i.e. phase transitions).

So how do we interpret the question?

The questions original statement specifically indicated that changing the Boltzmann constant would have no effect on the universe. Based on this, I interpret this question to be related to points 2 and 3. Since 2 applies to any constant as well, I think it's fair to assume the author meant 3.

If I'm mistaken and the author mean point 1, I believe the question should be reworded.

• Which would mean we would live on Mercury or Neptune, rather than on Earth? It's not really answering the question, though I don't think it can be answered easily. – pfnuesel Jan 22 '16 at 3:30
• I guess it depends on what you mean by "The universe as we know it". Can't you make your same argument about the gravitational constant? – Greg Petersen Jan 22 '16 at 3:44
• As far as I understand, it's about how much can we change the universal constants and life would still be possible in the Universe. Impossible to answer, of course, since we don't know at what strange places life might be possible. But when changing the gravitational constant, either no galaxies/stars would form or stars would burn up too quickly so at least life as we know it would become impossible. Not sure if same holds for the Boltzmann constant. – pfnuesel Jan 22 '16 at 4:02
• Besides other issues in your answer, the Boltzmann constant is written as $k_\mathrm{B}$ not as $K_b$ and your expression $E = K_b/T$ is not even dimensionally correct, the correct expression is $E = k_\mathrm{B}T$. Funny to see that no one of the five upvoters mentioned this glaring mistake. – juanrga Aug 6 '16 at 12:35
• Oh my! Yes that is an embarrassing typo. Thanks for pointing it out! I've corrected it. – Greg Petersen Aug 8 '16 at 13:59

Perhaps the author is thinking that $k_B$ really serves as the rate of exchange between units that we use to measure energy and those we use to measure temperature (which are different more for historical reasons than anything else). From this point of view, if we doubled $k_B$, it would be the same as rescaling our definition of temperature, so things we now call 100 K are instead said to be at 50 K and so on. Of course, like any change of units this doesn't actually change anything physically.

This is fine, but it is not clear why the author thinks that changing the value of $c$ or another dimensionful constant is any different. The only type of constant whose absolute value clearly matters for the universe is some dimensionless parameter, like the fine structure constant $\alpha$ or the ratio of the proton to electron mass.

• Doubling $k_B$ redefines the relationship between energy and temperature. Doubling $c$ redefines the relationship between length and time. – Rococo Jan 23 '16 at 2:49
• Anyway, $k_B$ is, for example, the relationship (up to a constant 3/2) between the internal energies of two samples of monatomic ideal gas and to what extent they exchange energy when brought into contact with one another. This seems to me to be a well-defined physical property, independent of units, to the same extent as "the speed of light." I just don't see an ontological distinction there. It is, certainly, a physical property which is particularly natural to make dimensionless. – Rococo Jan 23 '16 at 3:54
• You'll see the ontological difference between $k_b$ and things like $c$ once you note that there is literally not a single equation in all of physics where $k_b$ and $T$ do not appear multiplied by each other. This is a clear indication that $k_b$ and $T$ ought to be grouped together and $k_b$ forgotten entirely. If you object by citing $S \equiv k_b \ln \Omega$ then I'd point out that this is the same mistake as putting $k_b$ in the definition of temperature, which becomes obvious if we go back to the thermodynamic definition $dS \equiv dQ / T$. – DanielSank Jan 23 '16 at 4:01
• @DanielSank Well sure there are; heat capacities end up with $k_B$ by itself, just as properties involving space divided by time tend to end up with a $c$. Maybe if we called $k_B$ something like the fundamental heat capacity instead the comparison is more apparent. – Rococo Jan 30 '16 at 17:19

If you look at the two forms of the ideal gas law, $PV = nRT$ and $PV = Nk_BT$, you'll note that $nR = Nk_B$, where units are conventionally defined so that $R$ and $n$ are reasonably sized quantities. That means that the hugeness of $N$ has to be cancelled out by $k_B$.

So I think what your quote is really saying is that a universe with a different $k_B$ means one with a different scale for Avogadro's number, i.e. one where we'd be made of 10 times as many atoms, or 10 times fewer. This doesn't matter so much for us, because in either case, we're much larger than atomic scales, which is what's important.

Taking $\mathcal{E}= k_\mathrm{B}T$ and pretending that Boltzmann constant is a "historical artifact" because we are not measuring temperatures in "units of energy" is like taking $E = mc^2$ and $E=\hbar \nu$ and pretending that the speed of light and Planck constant are also artifacts because we are not measuring masses and frequencies in units of energy. The physical nature of $c$ and $\hbar$ follows from analyzing other physical expressions. The same happens with $k_\mathrm{B}$; its physical meaning cannot be obtained from $\mathcal{E}= k_\mathrm{B}T$.

Niels Bohr was the first to suggest that $k_\mathrm{B}$ would have a role similar to $\hbar$. He suggested that temperature $T$ and energy $E$ would be complementary properties in a way analogous to the complementarity of $x$ and $p$ in quantum theory. He was partially wrong because the complementary quantity for energy is not $T$, but the inverse temperature $1/T$, but Bohr suggested analogy with quantum mechanics is complete as posterior authors have shown:

• The Boltzmann constant plays the role of 'quantum' of thermodynamic action.
• The Boltzmann constant appears in the thermodynamic 'quantization' rules for inverse temperature and rest of intensive parameters $Y$ $$\delta (1/T) = k_\mathrm{B} \frac{\partial}{\partial E}, \qquad \delta Y = k_\mathrm{B} \frac{\partial}{\partial \Theta}$$
• The Boltzmann constant appears in the thermodynamic commutators that yield the thermodynamic uncertainty relations $$\Delta (1/T) \Delta E \ge k_\mathrm{B}, \qquad \Delta Y \Delta \Theta \ge k_\mathrm{B}.$$

Above expressions and others like Schwartz inequalities or the commutators for the thermal quantities can be found in section "7.5.2 Thermodynamic complementarity" of Byung Chan Eu Nonequilibrium Statistical mechanics (Kluwer, 1998).

DanielSank's answer is 100% rigth about the temperature issue. The question deserves much more argumentation because the book you use, and many others, are plain wrong. I will use numbers and the laws of physics to show my point: it is not true that

If the mass of an electron, the Planck constant, the speed of light, or the mass of a proton were even just slightly different (smaller or bigger) than what they actually are, then the whole Universe would not exist as we know it. Maybe we all wouldn't exist.

I will reinforce the distinction between the 'amount' to be measured and the value of the measure. As an example: the size of my yard do not change if measured in yards or in meters.
All the systems of units are derived from, and proportional to, the size of the atoms, even those that set several units to $1$. In other words: there are no independent references. There is a loop in the definitions of the units: for instance the mass unit - $kg$ is the mass of a bunch of atoms (N for instance) as represented in the prototype in Paris and the electron's mass is a definite part of that mass.

Imagine a universe where, compared with ours, particles have half the mass and charge, atomic radii are halved, and field constants ($c, \varepsilon,G$) are the same; how would an inhabitant of such an universe describe it?
Because the units of mass, charge, length, and time are half of ours (using the same definitions of units), the values of the field constants in this “half universe” are the same as ours. The values of bodies’ mass, charge, or size measured in each world by the respective observer are the same (by definition of standard units). Earth’s gravitational acceleration is doubled (half mass, half radius, same G); however, the measuring unit is also doubled ($LT^{-1}$), so the measured value is the same. Whatever the quantity, all quantity values are the same as ours except for one: the distance between bodies that are not gravitationally linked is doubled because the length unit is halved, and this distance is not affected by the fact that matter has half the size.
Now, let us examine what happens with the spectral radiation. Atoms in the “half universe” are half-sized – the Bohr radius is halved - and so are the characteristics of particles, namely the associated wavelength and the energy. Spectral radiations have half the wavelength and half the energy because only in this way can the transformation be self-similar, i.e., invariant in standard units. This implies that Planck constant is four times ours. Of course, its value in the standard units of the “half universe” is the same as ours because the unit is also four times ours. Local constants, in order to have the same value for the observers of each universe, must be different – in accordance with their dimensional functions. Different when measured with the same units, but equal when measured with the units of their own universe.
Now, we can go one step further and consider that in this conceptual universe, matter (each particle) is decreasing in size, mass, and charge in relation to our universe. The inhabitant of the conceptual universe can detect a redshift in radiation from distant sources because the radiation was emitted when atoms were larger (therefore, radiation was emitted with proportionally larger wavelengths). As this is the only locally detectable consequence of the variation of matter, such inhabitant concludes that matter is invariant, field constants are invariant, and space exhibits uniform and isotropic expansion - as we observe. This result shows that cosmic observations may trace a self-similar evolution of the matter/space relationship, which appears as a dilation of space in standard units.

Inserting the numbers we see that Electromagnetism, Gravitation and Mechanics are invariant under a synchronous change of units, aka atomic 'sizes':

The halved charge and length units : Coulomb force law - $F_c=\frac{1}{4\pi\varepsilon_0}\frac{Q_1}{2}\frac{Q_2}{2}(\frac{2}{d})^{2}$
The halved mass : Universal Gravitation law - $F_g=G\frac{M_1}{2}\frac{M_2}{2}(\frac{2}{d})^{2}$
The halved time and length units : 2nd Newton's law - $F=M\cdot \frac{d(\frac{x}{2})^2}{d(\frac{t}{2})^2}$

The energy levels of the atomic spectra: (the Sommerfeld relation) $E_{j,n}=−m_e∗f(j,n,\alpha,c)$ will be blueshifted if the electron's mass is halved, but keep all the masses in proportion, for instance $\frac{m_e}{m_p}$. Conversely if the atoms in the past were larger than the ones around then the past radiation is redshifted, as we see.

( the model was presented here (arxiv) with a formal proof here (vixra pdf) and a new version is in preparation)

Note: I see no reason to say anything about the viability of a faster universe (greater $c$ =$\frac{1}{\sqrt{\varepsilon\mu}}$) without a proper study

Most of the time, I believe (as most people here) that Boltzmann's constant is "just" an arbitrary unit conversion factor (temperature $$\Leftrightarrow$$ energy), and that we could get rid of it. It's not a fundamental constant imposed by nature.

However, when I'm thinking more about it, I'm frequently having a doubt. Here's my - unsure - opinion that $$k_{\mathrm{B}}$$ may actually be a very deep fundamental constant, like $$\hbar$$ and $$c$$ (which themselves aren't simple unit conversion factors, as I'll try to show below).

We define statistical entropy as this (of course, $$p_n$$ are probabilities, but I'll not elaborate on this): $$$$\tag{1} S_{\text{stat}} = - k \sum_n p_n \ln{p_n},$$$$ where $$k$$ is an arbitrary positive constant with any units. It would be simpler to just use $$k \equiv 1$$ by definition (without any unit), or to choose $$k = 1/\ln{2}$$ in units of "bits" to make a contact with Shannon's information theory. By its definition, (1) is not something that we can actually measure in a laboratory.

But then, when we want to give a relation with this quantity and the things we can measure in a laboratory, using macroscopic bodies, we have to introduce a coupling constant which states that:

Extracting information from a macroscopic system has an energy cost.

This cost is imposed by nature. This implies that the coupling constant between a measurable quantity (energy, for example) and the non-measurable statistical entropy (1) is not 0. Boltzmann's constant is a measure of that cost.

In this context, it's natural to set $$k \equiv k_{\mathrm{B}}$$, so the coupling constant is included in the definition of the statistical entropy (which could be identified with the thermodynamic entropy, i.e the one which appears in the empirical/physical relations). Its small value in human units ($$k_{\mathrm{B}} \sim 10^{-23} \, \mathrm{J/K}$$) is a manifestation of the empirical fact that information is cheap in our universe. Humans can get a lot of information about Nature by doing measures which aren't removing much energy from them, or else they would die (the cost is low).

In principle, we could imagine an hypothetical universe in which extracting information is extremelly painfull. The cost is then very high and $$k_{\mathrm{B}} \sim 10^{12} \, \mathrm{J/K}$$ in this universe (using the same units as in our universe, which may not have any meaning actually!). We could define another universe in which the cost is infinite: $$k_{\mathrm{B}} \rightarrow \infty$$. Observers couldn't get any information at all by any measure in their lab. Life wouldn't be possible in that universe. On the opposite side, we could imagine another universe in which information is totally free: $$k_{\mathrm{B}} \rightarrow 0$$. In that case, any tiny measure could bring a lot of information and life would be easy (actually, way too easy. Life would probably destroy itself by overpopulation, since living organisms could be immortal!).

Once you get $$k_{\mathrm{B}} \ne 0$$, you could introduce a system of units which gives $$k_{\mathrm{B}} = 1$$. The "arbitrary" choice $$k_{\mathrm{B}} \sim 10^{-23} \, \mathrm{J/K}$$ is a way of stating our size (scale) in our universe.

I believe there's something similar with other constants of Nature which are usually interpreted as simple unit conversion factors, like $$\hbar \sim 10^{- 34} \, \mathrm{J \cdot s}$$ (unit of action) and $$c^{-1} \sim 10^{- 9} \, \mathrm{s / m}$$ (unit of time in spacetime).

The important point to notice isn't the particular (small) value of these constants (which depend on our scale in the universe), but the fact that they aren't 0 in our universe. Our universe isn't just a newtonian world, for which $$k_{\mathrm{B}} = 0$$, $$\hbar = 0$$ and $$c^{-1} = 0$$. From this point of view, the Boltzmann constant isn't just arbitrary: It's a fundamental property of our very large and complicated non-newtonian world, and that the living beings are defining a special scale, the only one at which Life is possible.

• Can’t you redefine the joule and/or kelvin to make $k_B$ as big or small as you want? – user76284 Aug 13 at 11:54
• @user76284, the joule is associated to the human scale : $K = \frac{1}{2} \, m v^2 \sim 2 \, \mathrm{J}$ for a typical mass $m \sim 1 \, \mathrm{kg}$ moving at a typical velocity $v \sim 2 \, \mathrm{m/s}$, on a typical human scale. Same for the kelvin (or Celsius) : temperature of about $20 \, \mathrm{^{\circ} C}$, and so on... – Cham Aug 13 at 12:11
• But you say “$k_B$ may actually be a very deep fundamental constant“. That doesn’t make sense because you can change these conversion factors even slightly while retaining the same “human scale” (order of magnitude). Hence no value is “fundamental”, just the rough order of magnitude (if you think there’s something fundamental about human scales). – user76284 Aug 14 at 20:34
• @user76284, fundamental constants (like $c$ and $\hbar$) have values that depend on your arbitrary units. Yet, they are still fundamental characteristics of Nature. This may be the same for $k_{\mathrm{B}}$. – Cham Aug 14 at 21:20
• I suspect they aren't, by themselves, fundamental characteristics of nature. It's as if someone decided to introduce an arbitrary conversion factor $\kappa$ between heat and work. The whole of physics can be done without sprinkling the equations with these additional, arbitrary conversion factors. – user76284 Aug 14 at 23:25

## protected by Qmechanic♦Jan 22 '16 at 8:51

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