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This is the text from Reif Statistical mechanics. In the screenshot he changes the summation to integral(Eq. 1.5.17) by saying that they are approximately continuous values. However, I don't see how. Can anyone justify this change?

enter image description here

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    $\begingroup$ What is the context? What does $W$, $n_1$ means and why do we sum? I guess that in general, if you approximate a summation variable as being quasicontinuous, you can always approximate by an integral. $\endgroup$ – fgoudra Mar 22 at 9:59
  • $\begingroup$ Here W(n) refers to probability of given problem(The random walk by drunken man). My question is how can you approximate the quasicontinuous variables which is W above by integrals? Can you justify it? $\endgroup$ – Abhi7731756 Mar 22 at 10:07
  • $\begingroup$ Well usually one does this by saying that the $n_1$ are "really close to each other" such that changing the summation by an integral introduces a negligible error. It's an approximation, thus it's justified depending on how large would you allow the error to be. $\endgroup$ – fgoudra Mar 22 at 12:33
  • $\begingroup$ I will apply mean value theorem. If we apply this I get $W_{n}\delta{n}$. Now,how do I tell that this and above sum yield approximately same result? $\endgroup$ – Abhi7731756 Mar 22 at 13:58
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I think Reif was being sloppy there. I don't have the book handy but that seems to be the case. A passage from discrete to continuous would go something like this:

Say we're talking about the normalization of a probability mass function: $$ 1= \sum_{n=0}^N W(n) \,. $$ You first note that $n$ is a counting number, so its increments are by 1. Namely, ${\rm d}n = 1$. So you can just insert one: $$ 1 = \sum_{n=0}^N {\rm d}n W(n) \,.$$ Then for some reason, you might be interested in the variable $r=n/N$. The ratio of rightward steps or something. You rewrite: $$ 1 = \sum_{n=0}^N \frac{{\rm d}n} {N} N W(n) \,.$$ Now you realize that as $N\to\infty$, the quantity ${\rm d}n/N$ approaches an infinitesimal ${\rm d}r$ . Now, thanks to the infinitesimal element, the summation looks like a Riemann sum, allowing the passage to an integral $$ 1= \sum_{n=0}^N \frac{{\rm d}n} {N} N W(n) \to \int_0^\infty {\rm d} r \, N W(Nr) =1 \,.$$ Then, one proclaims that a new function, say, $P(r)$ has to be the probability density for the continuous variable $r=n/N$ such that $$P(r) = N W(N r) \,.$$

From what I can see, Reif's treatment is not emphasising that the new variable is proportional to $1/N$, and also failing to identify that the probability density of the new variable is not the same $(W)$ as that of the old variable, even though they are very closely connected (equation above).

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  • $\begingroup$ You're welcome @Abhi7731756. If the answer was not only helpful but it actually did the trick for you, please don't neglect accepting the answer :-) If not, don't do it of course. $\endgroup$ – Cem Yolcu Mar 25 at 8:33

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