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.
Definitions
- $$A(r)=\text{number of prime factors of r}$$
- 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)
Prologue
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:
$$A(x)+A(y)=A(xy)$$
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 $S=k_a A(\Omega)$
$$\therefore \beta= \frac{-1}{k_a T} \times \frac{\mathrm d \ln(\Omega)}{\mathrm d A(\Omega)} $$
Questions
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.
Related
asymptotic and monotonically increasing properties of prime factorization function?