A bit about the post

I apologize for the title. I know it sounds crazy but I could not think of an alternative one which was relevant. I know this is "wild idea" but please read the entire post.

Also, I wasn't sure if I should post this in the physics or mathematics community.

The referenced book is "Statistical Physics" Tony Guenault.


  • $$A(r)=\text{number of prime factors of r}$$
  • A micro-state by definition is a quantum state of the whole assembly. (Page 4)
  • Distribution of states : This is a set of numbers $(n_1,n_2,\dots,n_j,\dots)$ is defined as the number of particles in state $j$, which has energy $\epsilon_j$. Often, but does not always, this distribution will be an infinite set; the label $j$ must run over all the possible states for one particle. A useful shorthand for the whole set of distribution numbers $(n_1,n_2,\dots,n_j,\dots)$ is simply $\{ n_j \}$ (Page 6)
  • Usually and especially for a large system, each distribution $\{n_j \}$ will be associated with a very large number of micro-states. This we call $t(\{ n_j \})$ (page 9)
  • $ t(\{ n_j \})= \frac{N!}{\prod_j n_j !}$ (page 10)
  • $ S= k_b \ln(\Omega)$ (page 12)
  • $\Omega = \sum t( \{ n_j \}) \approx t( \{ n_j^* \}) =t^* $ where $t( \{ n_j^* \})$ is the maximum term (Page 16)


On page 13 the following was written:

  • .. For an isolated system a natural process ... is precisely the one in which thermodynamic entropy increases ... Hence a direct relation between $S$ and $\Omega$ is suggested, and moreover a monotonically increasing one ...
  • For a composite assembly, made up with two sub assemblies $1$ and $2$ say, we know the is whole assembly $\Omega$ is given by $\Omega=\Omega_1 \Omega_2$. This is consistent with relation ... ($ S= k_b \ln(\Omega)$)
  • ... $ \Omega=1$ corresponding to $S=0$, a natural zero for entropy.

If the above were the "rigorous" requirement to show that there $S=k_b \ln(\Omega)$

Then I believe I have found another function which satisfies the above criteria:

$$ S= k_a A(\Omega)$$

Where $k_a$ is an arbitrary constant. For example $ A(12)= A(2^2 \times 3 ) = 3$

To address bulletin point 1:


For example:

$$A(3)+ A(33)= A(99)=3 $$

We also note

$$ A(1) = 0 $$

About it monotonically increasing we note allowed values of $\Omega =\frac{N!}{\prod_j n_j !} $. Hence, for allowed values of $\Omega$:

$$\Omega_1 > \Omega_2 \implies A(\Omega_1) > A(\Omega_2) $$

Hence, we can use (as an alternative definition):

$$ S = k_a A(\Omega)$$

  • Logically perhaps ($S=k_b \ln(\Omega)$) is a derived result of statistical physics (page 13)

Rigorous Treatment

We can derive the Boltzmann distribution in the usual way with a few modifications ... We recognize the constraints:

$$ \sum_{j} n_j = N$$

$$ \sum_{j} n_j \epsilon_{j}= U $$

$$ \min (\Omega) = \min \ln( \Omega) = \min A( \Omega) \implies n^*_j = \exp(\alpha + \beta \epsilon_j)$$

Using the condition that $$ \sum_{j} n_j = N \implies \min (\Omega) \implies n_j = \frac{N}{Z} \exp{\beta \epsilon_j}$$

where $Z = \sum_j \exp{\beta \epsilon_j} $

However this does not give the usual kind of $\beta$

$$\begin{align} \mathrm d (\ln(\Omega))& = \mathrm d(\ln t^*) \\&= -\sum_{j} \ln {n^*_j} \,\mathrm dn_j\\ &= -\sum_j (\alpha + \beta \epsilon_j) \,\mathrm dn_j \qquad [\textrm{Using Boltzmann distribution}]\\ & = -\beta \sum_j \epsilon_j \,\mathrm dn_j\qquad \qquad[\textrm{ $\because \;N$ is fixed}] \\&= -\beta (\mathrm dU)\\ &= -\beta (T\,\mathrm dS)\end{align}$$ Inserting the new definition of $S=k_a A(\Omega)$

$$\therefore \beta= \frac{-1}{k_a T} \times \frac{\mathrm d \ln(\Omega)}{\mathrm d A(\Omega)} $$


Is this work correct? Has someone already worked on it? (If so, reference please) and does number theoretic loophole allow an alternative definition of entropy?

P.S: Ideally I would have liked to ask many spin-off questions but I think I need to first know if this is correct.


asymptotic and monotonically increasing properties of prime factorization function?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Feb 5 '16 at 14:41

While a funny-looking coincidence, this is not a valid alternative expression for entropy in general, since the entropy of a probability distribution (which are what rigorously hides behind the strange word "macrostate") is more generally given by $$ S = - k_B \sum_i p_i\ln(p_i) \tag{1}$$ and becomes only $$S = k_B \ln(\Omega) \tag{2}$$ in the case of a uniform distribution for which every state within $\Omega$ is equiprobable. It is precisely the fundamental postulate of thermodynamics that systems in equilibrium are described by such uniform distributions, but it is $(1)$, and not $(2)$, which gives the entropy for all statistical systems equilibrium or not.

Furthermore, it is $(1)$ that generalizes to the quantum case, where then entropy is given by $$ S(\rho) = k_B \mathrm{Tr}(\rho\ln(\rho))\tag{3}$$ for $\rho$ the density matrix, the quantum version of a statistical probability distribution. In particular, trying to apply $A$ to the density matrix doesn't make sense.

Let me also note that $A(r)$ and $\ln(r)$ differ in their growth behaviour: Let $x = \prod_i p_i^{r_i}$ be an integer with prime factors $p_i$. Then \begin{align} A(x) & = \sum_i r_i \\ \ln(x) & = \sum_i r_i \ln(p_i) \end{align} and while the logarithm is a monotonically increasing function, the prime-counting function $A$ is not: $A(3) = 1,A(4) =2,A(5)=1$. So your definition is markedly different from the usual definition - your entropy does not increase monotonically with the accessible microstates. In particular, since there are infinitely many primes, the function $A$ behaves more and more strangely at high values of $\Omega$ - and in particular will drop to its second-lowest value, $1$ for arbitrarily high $\Omega$. It should be evident that this does not model usual thermodynamics, and, in particular, cannot be a special case of $(1)$. You say you want to only allow special values for $\Omega$ on which $A$ is monotonically increasing, but then you've again restricted the validity of your formula to systems where $\Omega$ behaves indeed like that (I think you're having some "ideal gas" or "all particles are indistinguishable" assumption there).

All that this example shows is that the entropy as a function of $\Omega$ is not uniquely fixed by the functional equation $f(x_1x_2) = f(x_1)+f(x_2)$ and $f(1) = 0$ if one requires this to only hold on the integers. The solution $\ln$ becomes only unique if one requires the solution to be a continuous function on the positive real line.

  • $\begingroup$ Agreed $\ln(x)$ is not a monotonically increasing function ... $\ln(\Omega)$ is (kinda) monotonically increasing function ... Whats the difference? Recall allowed values of $\Omega= \frac{N!}{\prod n_j!}$ ... Keeping this in mind ... $ln(\Omega) = \ln(N!) - \sum \ln(n_j!)$ We not if we confine ourselves to this set of $\Omega$ then $\Omega_1 > \Omega_2 \implies A(\Omega_1)>A(\Omega_2)$ $\endgroup$ – drewdles Feb 4 '16 at 1:30
  • $\begingroup$ @AnantSaxena: I'm not sure what you mean. I address your restriction to special values of $\Omega$ - that's restricting to certain systems (admittedly those most often encountered in thermodynamics). I give several indications why using $A$ might work for those systems, but fails to generalize to the larger picture - that of non-equilibrium mechanics, non-ergodic systems, or quantum systems, while the $\ln$ generalizes effortlessly. What do you think is invalidated about that? $\endgroup$ – ACuriousMind Feb 4 '16 at 1:43
  • $\begingroup$ Sorry it seems you were right ... "You say you want to only allow special values for $\Omega$ on which $A$ is monotonically increasing, but then you've again restricted the validity of your formula to systems where $\Omega$ behaves indeed like that" ... Do you have any idea of physical special cases in mind? (If so would you care to elaborate more?) $\endgroup$ – drewdles Feb 4 '16 at 4:57

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