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I too was put off by the appeal to quantum mechanics.

Let $\Omega(n,V,E)=\Omega_q(n,V)\Omega_p(n,E)$. I assume the factor $1/n!$ stems from $\Omega_q$, the configurational source of entropy.

I'm sure it is not always possible to factor the $\Omega(n,V,E)$ neatly like this. But, in some simple cases it is possible.

The number of ways of putting $n$ indistinguishable objects into $k$ boxes is $$ N_{k,n}={n+k-1\choose k-1}=\frac{\Gamma(k+n)}{n!\Gamma(k)}.$$ We can make $k$ really big, like $k>n^2/(2\varepsilon)$ and why not? We have point particles and space is infinitely divisible - continuous. If we make $k$ this large, then the fraction of configurations with boxes containing more than one particle is at most $\varepsilon$. Anyway, as $k\to\infty$ $$ \frac{\Gamma(n+k)}{n!\Gamma(k)}\sim \frac{k^n}{n!}$$ (Note the similarities with the birthday paradox). So increasingly (with ever larger $k$) we have boxes with either zero or one objects. So, for a finite volume $V$, some reference volume $V_0$, maybe $V_0=V/k$ and some $\lambda>0$, we could argue that we should have $$ \Omega_q(n,V)=\frac{\lambda^n}{n!}\left(\frac{V}{V_0}\right)^n.$$

In the continuum case it is even easier to avoid particles having the same birthday (location, cell, sub volume). So, this explanation is already too complex, I think. It is just some normalized volume to the n-th power, taking into account particle relabeling.

The divider argument: $\Omega(n,.)\star \Omega(m,.)=\Omega(n+m,.)$, along with $\Omega(n,V)=c(n)V^n$ is pretty compelling. Basically the convolution identity $\frac{1}{n!}Y(x)x^{n-1}\star \frac{1}{m!}Y(x)x^{m-1}=\frac{1}{(n+m)!}Y(x)x^{n+m-1}$, implying $c(n)=\lambda^n/n!$. (Some off by one errors here and there. Shouldn't matter that much)

I too was put off by the appeal to quantum mechanics.

Let $\Omega(n,V,E)=\Omega_q(n,V)\Omega_p(n,E)$. I assume the factor $1/n!$ stems from $\Omega_q$, the configurational source of entropy.

I'm sure it is not always possible to factor $\Omega(n,V,E)$ as neatly as this. But, in some simple cases it is possible.

The number of ways of putting $n$ indistinguishable objects into $k$ boxes is $$ N_{k,n}={n+k-1\choose k-1}=\frac{\Gamma(k+n)}{n!\Gamma(k)}.$$ We can make $k$ really big, like $k>n^2/(2\varepsilon)$ and why not? We have point particles and space is infinitely divisible - continuous. If we make $k$ this large, then the fraction of configurations with boxes containing more than one particle is at most $\varepsilon$. Anyway, as $k\to\infty$ $$ \frac{\Gamma(n+k)}{n!\Gamma(k)}\sim \frac{k^n}{n!}$$ (Note the similarities with the birthday paradox). So increasingly (with ever larger $k$) we have boxes with either zero or one objects. So, for a finite volume $V$, some reference volume $V_0$, maybe $V_0=V/k$ and some $\lambda>0$, we could argue that we should have $$ \Omega_q(n,V)=\frac{\lambda^n}{n!}\left(\frac{V}{V_0}\right)^n.$$

In the continuum case it is even easier to avoid particles having the same birthday (location, cell, sub volume). So, this explanation is already too complex, I think. It is just some normalized volume to the n-th power, taking into account particle relabeling.

The divider argument: $\Omega(n,.)\star \Omega(m,.)=\Omega(n+m,.)$, along with $\Omega(n,V)=c(n)V^n$ is pretty compelling. Basically the convolution identity $\frac{1}{n!}Y(x)x^{n-1}\star \frac{1}{m!}Y(x)x^{m-1}=\frac{1}{(n+m)!}Y(x)x^{n+m-1}$, implying $c(n)=\lambda^n/n!$. (Some off by one errors here and there. Shouldn't matter that much)

I too was put off by the appeal to quantum mechanics.

Let $\Omega(n,V,E)=\Omega_q(n,V)\Omega_p(n,E)$. I assume the factor $1/n!$ stems from $\Omega_q$, the configurational source of entropy.

The number of ways of putting $n$ indistinguishable objects into $k$ boxes is $$ N_{k,n}={n+k-1\choose k-1}=\frac{\Gamma(k+n)}{n!\Gamma(k)}.$$ We can make $k$ really big, like $k>n^2/(2\varepsilon)$ and why not? We have point particles and space is infinitely divisible - continuous. If we make $k$ this large, then the fraction of configurations with boxes containing more than one particle is at most $\varepsilon$. Anyway, as $k\to\infty$ $$ \frac{\Gamma(n+k)}{n!\Gamma(k)}\sim \frac{k^n}{n!}$$ (Note the similarities with the birthday paradox). So increasingly (with ever larger $k$) we have boxes with either zero or one objects. So, for a finite volume $V$, some reference volume $V_0$, maybe $V_0=V/k$ and some $\lambda>0$, we could argue that we should have $$ \Omega_q(n,V)=\frac{\lambda^n}{n!}\left(\frac{V}{V_0}\right)^n.$$

In the continuum case it is even easier to avoid particles having the same birthday (location, cell, sub volume). So, this explanation is already too complex, I think. It is just some normalized volume to the n-th power, taking into account particle relabeling.

The divider argument: $\Omega(n,.)\star \Omega(m,.)=\Omega(n+m,.)$, along with $\Omega(n,V)=c(n)V^n$ is pretty compelling. Basically the convolution identity $\frac{1}{n!}Y(x)x^{n-1}\star \frac{1}{m!}Y(x)x^{m-1}=\frac{1}{(n+m)!}Y(x)x^{n+m-1}$, implying $c(n)=\lambda^n/n!$. (Some off by one errors here and there. Shouldn't matter that much)

I too was put off by the appeal to quantum mechanics.

Let $\Omega(n,V,E)=\Omega_q(n,V)\Omega_p(n,E)$. I assume the factor $1/n!$ stems from $\Omega_q$, the configurational source of entropy.

I'm sure it is not always possible to factor the $\Omega(n,V,E)$ neatly like this. But, in some simple cases it is possible.

The number of ways of putting $n$ indistinguishable objects into $k$ boxes is $$ N_{k,n}={n+k-1\choose k-1}=\frac{\Gamma(k+n)}{n!\Gamma(k)}.$$ We can make $k$ really big, like $k>n^2/(2\varepsilon)$ and why not? We have point particles and space is infinitely divisible - continuous. If we make $k$ this large, then the fraction of configurations with boxes containing more than one particle is at most $\varepsilon$. Anyway, as $k\to\infty$ $$ \frac{\Gamma(n+k)}{n!\Gamma(k)}\sim \frac{k^n}{n!}$$ (Note the similarities with the birthday paradox). So increasingly (with ever larger $k$) we have boxes with either zero or one objects. So, for a finite volume $V$, some reference volume $V_0$, maybe $V_0=V/k$ and some $\lambda>0$, we could argue that we should have $$ \Omega_q(n,V)=\frac{\lambda^n}{n!}\left(\frac{V}{V_0}\right)^n.$$

In the continuum case it is even easier to avoid particles having the same birthday (location, cell, sub volume). So, this explanation is already too complex, I think. It is just some normalized volume to the n-th power, taking into account particle relabeling.

The divider argument: $\Omega(n,.)\star \Omega(m,.)=\Omega(n+m,.)$, along with $\Omega(n,V)=c(n)V^n$ is pretty compelling. Basically the convolution identity $\frac{1}{n!}Y(x)x^{n-1}\star \frac{1}{m!}Y(x)x^{m-1}=\frac{1}{(n+m)!}Y(x)x^{n+m-1}$, implying $c(n)=\lambda^n/n!$. (Some off by one errors here and there. Shouldn't matter that much)

I too was put off by the appeal to quantum mechanics.

Let $\Omega(n,V,E)=\Omega_q(n,V)\Omega_p(n,E)$. I assume the factor $1/n!$ stems from $\Omega_q$, the configurational source of entropy.

The number of ways of putting $n$ identical objects into $k$ distinguishable boxes is $$ N_{k,n}={n+k-1\choose k-1}=\frac{\Gamma(k+n)}{n!\Gamma(k)}.$$ We can make $k$ really big, like $k>n^2/(2\varepsilon)$ and why not? We have point particles and space is infinitely divisible - continuous. If we make $k$ this large, then the fraction of configurations with boxes containing more than one particle is at most $\varepsilon$. Anyway, as $k\to\infty$ $$ \frac{\Gamma(n+k)}{n!\Gamma(k)}\sim \frac{k^n}{n!}$$ (Note the similarities with the birthday paradox). So increasingly (with ever larger $k$) we have boxes with either zero or one objects. So, for a finite volume $V$, some reference volume $V_0$, maybe $V_0=V/k$ and some $\lambda>0$, we could argue that we should have $$ \Omega_q(n,V)=\frac{\lambda^n}{n!}\left(\frac{V}{V_0}\right)^n.$$

In the continuum case it is even easier to avoid particles having the same birthday (location, cell, sub volume). So, this explanation is already too complex, I think. It is just some normalized volume to the n-th power, taking into account particle relabeling.

The divider argument: $\Omega(n,.)\star \Omega(m,.)=\Omega(n+m,.)$, along with $\Omega(n,V)=c(n)V^n$ is pretty compelling. Basically the convolution identity $\frac{1}{n!}Y(x)x^{n-1}\star \frac{1}{m!}Y(x)x^{m-1}=\frac{1}{(n+m)!}Y(x)x^{n+m-1}$, implying $c(n)=\lambda^n/n!$. (Some off by one errors here and there. Shouldn't matter that much)

I too was put off by the appeal to quantum mechanics.

Let $\Omega(n,V,E)=\Omega_q(n,V)\Omega_p(n,E)$. I assume the factor $1/n!$ stems from $\Omega_q$, the configurational source of entropy.

The number of ways of putting $n$ indistinguishable objects into $k$ boxes is $$ N_{k,n}={n+k-1\choose k-1}=\frac{\Gamma(k+n)}{n!\Gamma(k)}.$$ We can make $k$ really big, like $k>n^2/(2\varepsilon)$ and why not? We have point particles and space is infinitely divisible - continuous. If we make $k$ this large, then the fraction of configurations with boxes containing more than one particle is at most $\varepsilon$. Anyway, as $k\to\infty$ $$ \frac{\Gamma(n+k)}{n!\Gamma(k)}\sim \frac{k^n}{n!}$$ (Note the similarities with the birthday paradox). So increasingly (with ever larger $k$) we have boxes with either zero or one objects. So, for a finite volume $V$, some reference volume $V_0$, maybe $V_0=V/k$ and some $\lambda>0$, we could argue that we should have $$ \Omega_q(n,V)=\frac{\lambda^n}{n!}\left(\frac{V}{V_0}\right)^n.$$

In the continuum case it is even easier to avoid particles having the same birthday (location, cell, sub volume). So, this explanation is already too complex, I think. It is just some normalized volume to the n-th power, taking into account particle relabeling.

The divider argument: $\Omega(n,.)\star \Omega(m,.)=\Omega(n+m,.)$, along with $\Omega(n,V)=c(n)V^n$ is pretty compelling. Basically the convolution identity $\frac{1}{n!}Y(x)x^{n-1}\star \frac{1}{m!}Y(x)x^{m-1}=\frac{1}{(n+m)!}Y(x)x^{n+m-1}$, implying $c(n)=\lambda^n/n!$. (Some off by one errors here and there. Shouldn't matter that much)