Sources that discuss the derivation of the Maxwell-Boltzmann Statistics end up with two unknown constants ($\alpha$ and $\beta$) through the Lagrange Multipliers, of which $\alpha$ is derived by normalizing an integrand containing the derived Maxwell-Boltzmann probability distribution formula.

However, $\beta$ is approached differently by introducing a completely different information; saying that on average, a particle has $\frac{3}{2}kT$ energy (translational). Equalizing this with the average energy per particle according to the derived Maxwell-Boltzmann formula: $$\int^\infty_0 2 \cdot \sqrt{\frac{E}{\pi}} \cdot \beta^{1.5} \cdot e^{-\beta E} \cdot E \cdot dE = \frac{3}{2}kT$$ Which shows that $\beta = \frac{1}{kT}$.

However, it is not explained how $kT$ itself is derived. How did this value get associated with a degree of freedom in the first place? Is this derived experimentally by measuring the amount of energy needed for a system to reach a certain temperature $T$ and somehow knowing the number of moles and the number of degrees of freedom for a particle in that system?

  • $\begingroup$ I find this a strange way to define $\beta$. In the books that I have you define it using the thermodynamics identity $\frac{\partial S}{\partial U} = 1/T$. Solving for $\beta$ you find that $\beta = 1/(KT)$. $\endgroup$
    – Shamaz
    Commented Apr 1, 2019 at 16:17
  • $\begingroup$ @Shamaz Is there a way to explain how the association between $kT$ and a degree of freedom is discovered? Or is this experimentally deduced? $\endgroup$
    – Phy
    Commented Apr 1, 2019 at 16:45
  • $\begingroup$ @JohnnyGui at some point you need to introduce a definition of $T$ in terms of $\beta$. Where you do this is somewhat arbitrary so long as its consistent with what everyone else means by $T$. $\endgroup$
    – jacob1729
    Commented Apr 1, 2019 at 16:57
  • $\begingroup$ @jacob1729 Sorry, but I can not see from this how it is then deduced that a degree of freedom must contain the value $kT$. Could you give an example? $\endgroup$
    – Phy
    Commented Apr 1, 2019 at 17:05
  • $\begingroup$ @JohnnyGui I simply mean that if we don't introduce $T$ ever then the theory still makes sense. For a system with $D$ quadratic degrees of freedom the mean energy will be $(D/2) (1/\beta)$ (this itself is not a hard calculation). We are free to define $1/\beta$ as $kT$ if we wish to have a quantity that tends to increase with energy rather than decrease, otherwise $\beta$ alone is perfectly good. $\endgroup$
    – jacob1729
    Commented Apr 1, 2019 at 22:19

4 Answers 4


So presumably your derivation looks like this, that in the discrete case we have a set of states $A$ and a probability variable for each state $p_a$ such that $\sum_{a\in A}p_a = 1$ and an energy of each state $E_a$ such that the average energy is fixed, $\sum_{a\in A} p_a E_a = \eta$. The goal is to maximize $s(\{p_a\})=-\sum_a p_a \ln p_a$ subject to those constraints and then with Lagrange multipliers we institute two parameters, let's call them $\beta$ and $\gamma$, so that we instead minimize this constrained entropy, $$\bar s(\{p_a\}) = \sum_{a \in A} \big(-p_a \ln p_a - \beta p_a E_a - \gamma p_a\big).$$ We then would find that $$\frac{\partial \bar s}{\partial p_a} = -1 - \ln p_a - \beta E_a - \gamma = 0$$ and thus that $$p_a = \frac{1}{Z}~e^{-\beta E_a}$$ for some $Z = e^{1 + \gamma}$ that enforces the normalization of the distribution, $Z = \sum_a e^{-\beta E_a}.$

We call this function $Z = Z(\beta)$ the partition function and recognize it specially because, for example, if we were to want to calculate $\sum_a p_a E_a = \eta$ we can now do it simply by looking at $$\eta=-\frac\partial{\partial \beta}\ln Z = -\frac{1}{Z}\frac{\partial Z}{\partial \beta} = -\frac1Z\sum_a e^{-\beta E_a}\cdot(-E_a) = \sum_a p_a E_a,$$ so that in its functional form it contains a great deal more information than you might have expected out of a simple normalization constant. Similarly $$s = -\sum_a p_a \ln p_a = +\sum_a p_a~(\beta E_a + \ln Z) = \ln Z + \beta \eta = \left(1 - \beta \frac{\partial}{\partial \beta}\right)\ln Z.$$

Proof that $\beta$ determines temperature

Now you are asking about the details of why we say that this parameter $\beta = 1/k_\text B T,$ where $T$ is the absolute temperature. Suppose we give a little bit of energy $\delta\eta$ to the system and allow it to again return to equilibrium. Since $Z = Z(\beta)$ we know that this requires somehow changing $\beta$ with some $\delta\beta.$ We can then work out that $$\delta Z = \sum_a e^{-\beta E_a} (-E_a) \delta \beta = -Z~\eta~\delta\beta$$ and thus that $$\delta s = \delta(\ln Z + \eta\beta) = \frac{\delta Z}{Z} + \eta~\delta\beta + \beta~\delta\eta = \beta ~\delta \eta.$$ We don't need to work out the details about $\delta\beta$ because it simply cancels in this expression.

So now if you imagine two such systems trying to exchange one packet of energy $\delta \eta$ you can see that the energy will flow from system 1 into system 2 spontaneously if the total entropy increases, $$\delta s = \delta s_1 + \delta s_2 = \beta_1~(-\delta\eta) + \beta_2~\delta \eta = (\beta_2 - \beta_1)\delta \eta,$$so the criterion for this is simply $\beta_2 > \beta_1$, and you might therefore say that $\beta$ measures some "coldness" of a system, with energy spontaneously flowing into a colder system from a warmer system.

So it turns out that far from $\beta$ being some sort of instance-specific parameter, we can use thermal contact to compare the $\beta$ factors between two otherwise-thermalized objects and it therefore represents some sort of universal property akin to temperature that we can use to describe thermal heat flows.

The functional form revealed in one example

A direct implication of the last point is the potential existence of thermometers. A thermometer is just a well-known system which we can use to tell you a value for $\beta$ without disturbing that $\beta$ by all that much.

One such thermometer would simply be an ideal gas thermometer. If the energy is independent of position in such a thermometer (i.e. gravity is negligible here) then we basically wish to parcel out velocity space into a bunch of discrete chunks so that a given atom has the state $p_{(v_x,v_y,v_z)} = \frac1Z e^{-\beta\frac12 m (v_x^2 + v_y^2 + v_z^2)}.$ We can see that in the limit of small chunks we have some sort of Gaussian integral, $$Z_1 = \alpha \int_{-\infty}^\infty dv~e^{-\beta \frac 12 m v^2} = \sqrt{2\pi\alpha^2\over m\beta},\\Z = Z_1^3.$$

We thus have for the single molecule that $\eta = -Z'/Z = \frac32 \beta^{-1},$ and so for a bunch of molecules $N$ that $E = N \eta = \frac32 N \beta^{-1}.$

From the kinetic theory it is known that $E = \frac32 n R T.$ (This comes purely from considering the momentum imparted to a ceiling of a piston and the time between collisions with that ceiling giving $E= \frac32 PV$ where the 1/2 comes from a kinetic energy prefactor while the 3 comes from the 3 dimensions of space, see comments below.) Thus this ideal gas thermometer is measuring $\beta^{-1} = (n/N)~R T.$ If we simply define that $k_\text B = (n/N) R$ then we have your resulting expression, that $$\beta = 1/(k_\text B T).$$

Note that this is a stronger result than it first appears, because coldness and temperature are such broad properties. There is an amplification where if this holds for any thermometer it must hold for all thermometers. So simply connecting the kinetic theory of gases to the statistical interpretation of temperature means that we must have one of two results:

  1. The statistical mechanism of heat transport is not the only one typically at play in real physical systems coming to equilibrium, or
  2. $\beta = 1/k_\text B T$ for everything.

If there is an empirical side to this, then, it is the rejection of (1). (And, to a lesser extent, the fact that we suspect $n/N$ to be some constant is also an empirical observation.) In physicists’ experience with statistical mechanics they have never needed to introduce any other mechanism; it has always been sufficient that spontaneous heat flow can be understood as the result of an overall system transferring into a more likely state by a transfer of energy.

Sums over degrees of freedom

The general relationship can now be derived. Above the 3 comes from the exponent in $Z_1^3$ which is our major hint. We take two places where energy can live (degrees of freedom) $A+B$ with their own energy contributions $E^{A,B}$ and then discover that $$Z_{A+B}=\sum_a\sum_be^{-\beta E^A_a-\beta E^B_b}=Z_AZ_B,$$ but both $\eta$ and $s$ have been expressed as derivatives of $\ln Z$ so must be additive: energy and entropy must sum over these degrees of freedom.

Now take any one continuous degree of freedom $q$ and attempt to calculate the average value of $ q\frac{\partial H}{\partial q}, $ where H is a Hamiltonian function g giving a total energy, to find:$$ \begin{align} \left\langle q\frac{\partial H}{\partial q}\right\rangle &= \frac1{Z_Q} \int_{-\infty}^\infty dq~e^{-\beta H}~ q~\frac{\partial H}{\partial q}\\ &= \frac1{Z_Q} \left[q~\frac{-e^{-\beta H}}{\beta}\right]_{-\infty}^\infty + \frac1{Z_Q}\int_{-\infty}^\infty dq~\frac{e^{-\beta H}}{\beta} \end{align} $$ by integration by parts. The boundary term looks suspiciously ignorable for most nice Hamiltonians that grow faster than logarithmically in energy towards $+\infty$ or that have one fixed bound that can be placed at $q=0$ or so, while the latter integral is just $Z_Q/\beta$, so we derive that in general $$\left\langle q\frac{\partial H}{\partial q}\right\rangle = \beta^{-1} = k_\text B T.$$ If a term contributes quadratically to the Hamiltonian $H=\alpha q^2$ then the expression on the left will be twice that contribution $\left\langle q\frac{\partial H}{\partial q}\right\rangle=\langle 2\alpha q^2 \rangle= k_\text B T$ and so we see the average energy occupation being $\eta_Q =\frac1n k_\text B T$ for a Hamiltonian that goes like $q^n.$

  • $\begingroup$ Thanks a lot for your detailed explanation. I have a couple of questions regarding this post 1. I have difficulty understanding how $\sum_a e^{-\beta E_a}\cdot -E_a = \frac{\partial Z}{\partial \beta} = Z \cdot \frac\partial{\partial \beta}\ln Z$. How is this concluded? 2. Is the kinetic theory $E = \frac{3}{2}nRT$ purely based on experiments or derived from elsewhere? $\endgroup$
    – Phy
    Commented Apr 2, 2019 at 13:37
  • $\begingroup$ 1. For the first part, I mean, just take the derivative. What is $\frac\partial{\partial\beta}\sum_a e^{-\beta E_a}$? Use all the rules of calculus you know -- linearity, chain rule, derivative of exponentials. For the second part, the chain rule on $(\ln Z)' = Z'/Z$ does all the heavy lifting. $\endgroup$
    – CR Drost
    Commented Apr 2, 2019 at 16:23
  • $\begingroup$ 2. The kinetic theory looks at a piston of length $L,$ cross-section $A$, and finds that a particle with velocity $v_z$ impacts the top every $\tau=2L/v_z$ imparting a momentum $2m v_z$ thus contributing a force $f=m v_z^2/L.$ Then we sum over particles $\langle X\rangle=\sum_pX_p/N$, and note that $\langle v_z^2\rangle=\frac13\langle\|\vec v\|^2\rangle$ if velocity is isotropic, because $v^2=\|\vec v\|^2=v_x^2+v_y^2+v_z^2$ for each particle, so that $PV=Nm\frac13\langle v^2\rangle$ while $E=\frac12Nm\langle v^2\rangle=\frac32PV.$ The ideal gas law is then the experimental part: $PV=nRT.$ $\endgroup$
    – CR Drost
    Commented Apr 2, 2019 at 16:30
  • $\begingroup$ @JohnnyGui Sorry, forgot to ping you in the above two. See the above two comments :) $\endgroup$
    – CR Drost
    Commented Apr 2, 2019 at 16:31
  • $\begingroup$ This has helped me a lot. Thank you so much for your trouble. $\endgroup$
    – Phy
    Commented Apr 4, 2019 at 18:30

The relationship between the degrees of freedom of a system and $kT$ is defined by the equipartition theorem, which states, in its most general form:

Given a system with Hamiltonian $H$ and degrees of freedom $\{x_n\}$,$$\left\langle x_m\frac{\partial H}{\partial x_n}\right\rangle=\delta_{mn}kT$$ where $\langle\cdot\rangle$ denotes an ensemble average.

A special case of this theorem occurs when the Hamiltonian contains terms that are quadratic in the degrees of freedom; in this case, the statement simplifies to:

Given a system with Hamiltonian $H$ and degrees of freedom $\{x_n\}$, any term in the Hamiltonian that is quadratic in $x_m$ for some $m$ contributes $\frac{1}{2}kT$ to the total internal energy of the system.

Derivations of both of these statements can be found in any sufficiently advanced thermodynamics textbook, or on Wikipedia here: https://en.wikipedia.org/wiki/Equipartition_theorem#Derivations.

As an aside, one thing to keep in mind is that the theorem only holds if the system is explicitly in a classical regime (such that the degrees of freedom have access to a continuum of different states), and systems that exhibit quantum behavior (where the possible states of the degrees of freedom are restricted to specific values) violate the equipartition theorem.


A proper derivation of $U=\frac{3}{2}kT$ is given in a book by Schrödinger on statistical thermodynamics. First the book gives the partition function $$Z= \Sigma e^{-\mu \epsilon_{l}} $$ Where $\epsilon_l$ is energy Eigen values of the system and $\mu$ is a Lagrange multiplier ( I will be using $\mu$ instead of $\beta$) that comes out to be $\frac{1}{kT}$ (as is later shown in this answer). Since the partition function is completely multiplicative we define $$ \Psi = klogZ$$ Which would be completely additive. For an ideal monotonic gas we do not have to worry about quantisation. We consider a six dimensional phase space $x,y,z,p_x ,p_y ,p_z$. The energy of the system is $$ \frac{1}{2m}(p²_x+p²_y+p²_z)$$ The sum in the expression for $Z$ can be replaced by an integral over the phase space. Hence $$Z= V\int{\int{\int{e^{\frac{-\mu}{2m}p^i p_i} dp_x dp_y dp_z}}}$$ I am using Einstein summation convention above. It simply says that a repeated index is summed over. The integrals over $dx,dy,dz$ have become the volume. By making the transformation $$\zeta_i = \frac{2m}{\mu} p_i $$ The integral above has become $$Z= V(\frac{2m}{\mu})^{\frac{3}{2}}\int{e^{-\zeta^i \zeta_i} d³\zeta}$$ Now $\Psi$ can be written as $$\Psi = klogV + \frac{3}{2}klog(2mk)T +log(\int{e^{-\zeta_i \zeta^i}d³\zeta})$$ $$ \Psi = klogV + \frac{3}{2} k logT + const.$$ From the relation $$U = T² \partial_{T}\Psi$$ On solving the above expression you would get the internal energy of the system to be $$U= \frac{3}{2} kT$$ All these relations are only for a single particle. Since $\Psi$ is additive we can simply multiply by the number of moles. This is the derivation for the internal energy. I have used $\mu=\frac{1}{kT}$ for the derivation of this result. To find $\mu$ Schrödinger has taken a different approach. Consider the function $F$ defined as $$F= log(\Sigma e^{-\mu \epsilon_l})$$ Finding the change in this function $$d(F+\mu U) = \mu(dU-\frac{1}{N} \Sigma a_l d\epsilon_l) $$ Now quoting the book

When the above equation is applied to this process, $a_l d\epsilon_l$ is the work we have to do on the pistons,etc.,attached to these a_l systems in order to ‘lift them up’ from the old level $\epsilon$ to the altered level $\epsilon_l +d\epsilon_l$;$\Sigma a_l d\epsilon_l$ is the work done in this way on the assembly, $-\Sigma a_l d\epsilon_l$ is the work done by the system and $-\frac{1}{N}\Sigma a_l d\epsilon_l$ is work done by one of the members of the system. And hence, the round bracket to the right of the above equation must be the average heat supply $dQ$ supplied to it. $\mu$ is seen to be an integrating factor thereof. This alone really suffices to say that $\mu$ must be essentially $\frac{c}{T}$ because there is no further function of $T$ which has this property for every system. And so, $F+U\mu$ must be the entropy.

$c$ can be easily then shown to be $\frac{1}{k}$ by using first law and second law of thermodynamics.

I hope this answer helps.

  • $\begingroup$ Is this last equation related to the Gibbs-Helmholtz equation, for they look similar, is that a chemical potential in there? $\endgroup$ Commented Apr 2, 2019 at 2:08
  • $\begingroup$ @GarethMeredith No $\endgroup$ Commented Apr 2, 2019 at 4:31

At first, when thermodynamics and statistical mechanics were still being fleshed out, I am sure that they used the experimental fact that the average transitional energy of a particle is $3KT/2$ to help them build the theory.

But in the current framework, the derivation above does not make sense. The equipartition theorem states that for a system of non-interacting particles with quadratic degrees of freedom, the average kinetic energy of a particle is $3fKT/2$ where f is the number of degrees of freedom.

But the important thing to note is that the equipartition theorem itself is derived using the fact that $\beta = 1/KT$. Hence your derivation using Maxwell-Boltzmann distribution is circular.

If you want to make proper sense of it all, you need to start with the fundamental quantity: Entropy. Using the information definition of entropy, you minimize the entropy of the system subject to the constraint that the average energy of the system is constant. The Lagrange multiplier in this case is called $beta$.

Then you define the temperature as being $1/T = \frac{\partial S}{\partial U}$, which yields $\beta = 1/KT$.

  • $\begingroup$ Ok this helped me a bit. So does this mean that the deduction of the number of degrees of freedom of a particle is fundamentally based on experiments? I'm amazed that one was able to deduce experimentally that $kT$ has a link with the number of degrees of freedom. $\endgroup$
    – Phy
    Commented Apr 1, 2019 at 17:54
  • $\begingroup$ I believe they first studied the heat capacity at constant volume and at constant pressure of ideal gases and made the association with the degrees of freedom since some gases only displayed 3 degrees at low temperature but 5 degrees of freedom at higher temperature. At the time they used $nR$ instead of $Nk_b$ where $n$ is the number of moles and $N$ is Avocados Number. $\endgroup$ Commented Apr 2, 2019 at 2:52

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