How to turn a sum into an integral?

Question

In, An Introduction to Thermal Physics, page 235, Schroder wants to evaluate the partition function

$$Z_{tot}=\sum_0^\infty (2j+1)e^{-j(j+1)\epsilon/kT}$$

in the limit that $kT\gg\epsilon$, thus he writes

$$Z_{tot}\approx\int_0^\infty (2j+1)e^{-j(j+1)\epsilon/kT}\,dj$$

But how is this correct? There was no factor of $j$ in the sum that could be replaced with $dj$. Also, it is good that $j$ is just a number, otherwise even the dimensions of $Z_{tot}$ would be wrong.

Answer 1

So the trapezoid rule states that, if we partition an interval $(a, b)$ into $N + 1$ equally spaced points, $$\Delta x = \frac{b - a}{N},~~~x_k = a + k~\Delta x,$$ then $$ \int_a^b \mathrm dx ~f(x) = \Delta x\sum_{k=0}^{N} f(x_k) - \Delta x~\frac{f(a) + f(b)}{2} - \frac{(b - a) (\Delta x)^2}{12} f''(\xi)$$ for some $\xi$ on $(a, b).$

Normally we use this “forwards” to assess $\int \mathrm dx ~f(x)$ but your textbook author is using this “backwards” with $\Delta x$ fixed to $1$ to instead find that, $$\sum_{k=0}^{N} f(k) = \int_a^b \mathrm dx ~f(x) + \frac{f(0) + f(N)}{2} + \frac{N}{12} f''(n)$$ for some $0 < n < N.$

If one is doing this, one needs to be a little bit careful to make sure that the integral on the right hand side is growing much faster than the $N~f''(n)$ error term is, so that the relative error is not too great. But it's a perfectly mathematically valid way to use that equation. It is somewhat easy to forget, so worth underlining, that in mathematics, equals signs generally don't have a direction. If $a = b$, then you can always also say that $b = a$. The two expressions are just different names for, or ways to calculate, the same resultant number.

Answer 2

$$Z_{tot}=\sum_0^\infty (2j+1)e^{-j(j+1)\epsilon/kT}$$

in the limit that $kT\gg\epsilon$, thus he writes

$$Z_{tot}\approx\int_0^\infty (2j+1)e^{-j(j+1)\epsilon/kT}\,dj$$

But how is this correct? There was no factor of $j$ in the sum that could be replaced with $dj$.

Factors of $j$ in the sum translate to factors of $j$ in the integral, not factors of $dj$. The $dj$ terms basically is just there to remind us that the dummy integration variable is $j$.

If it makes you more comfortable, you are free to define $\Delta j = 1$ and write: $$ Z_{tot}=\sum_0^\infty (2j+1)e^{-j(j+1)\epsilon/kT} $$ $$ =\sum_0^\infty 1(2j+1)e^{-j(j+1)\epsilon/kT} $$ $$ =\sum_0^\infty \Delta j(2j+1)e^{-j(j+1)\epsilon/kT} $$

Also, it is good that $j$ is just a number, otherwise even the dimensions of $Z_{tot}$ would be wrong.

Yes. That is good, isn't it?

If the summation variable of interest happened to have dimensions, then you would need to introduce some constant with the same dimensions in order to proceed from sum to integral. But, your summation/integration variable already is dimensionless, so you don't have to worry about that in this case.

Answer 3

Here is an answer as a mathematician.

The situation you have is that you want to compare $$ \sum_{j=0}^\infty f_s(j)\qquad\text{ and} \qquad \int_0^\infty f_s(t)\,dt, $$ where $f_s(t)$ is a function that decreases on $t$, with $f_s(0)=1$ for all $s$, and such that $\lim_{s\to0}f_s(t)=0$ for all $t>0$. In this situation you have $$\tag1 \lim_{s\to0}\sum_{j=0}^\infty f_s(j)=-1+\lim_{s\to0} \int_0^\infty f_s(t)\,dt $$ That is, in the original notation $$\tag2 \lim_{\epsilon/kT\to0}\sum_0^\infty (2j+1)e^{-j(j+1)\epsilon/kT} =-1+\lim_{\epsilon/kT\to0}\int_0^\infty (2j+1)e^{-j(j+1)\epsilon/kT}\,dj, $$ and the $-1$ term would disappear if you started both the sum and the integral from $j=1$ instead of $j=0$.

To see why $(1)$ is true, note first that $$ \int_j^{j+1}f_s(t)\,dt\leq\int_j^{j+1} f_s(j)\,dt=f_s(j), $$ so $$ \int_1^\infty f_s(t)\,dt=\sum_{j=1}^\infty \int_j^{j+1}f_(t)\,dt\leq\sum_{j=1}^\infty f_s(j). $$ Hence \begin{align} \bigg|\sum_{j=1}^\infty f_s(j)-\int_1^\infty f_s(t)\,dt\bigg| &=\sum_{j=1}^\infty f_s(j)-\int_1^\infty f_s(t)\,dt\\[0.2cm] &=\sum_{j=1}^\infty \int_j^{j+1} [f_s(j)-f_s(t)]\,dt\\[0.2cm] &\leq\sum_{j=1}^\infty \int_j^{j+1} [f_s(j)-f_s(j+1)]\,dt\\[0.2cm] &=\sum_{j=1}^\infty [f_s(j)-f_s(j+1)]=f_s(1).\\[0.2cm] \end{align} As $f_s(1)\to0$ as $s\to0$, we have $$\tag3 \lim_{s\to0}\sum_{j=1}^\infty f_s(j)=\lim_{s\to0} \int_1^\infty f_s(t)\,dt. $$ It remains to see that $$ \lim_{s\to0} -1+\int_0^1f_s(t)\,dt=1. $$ Here $f_s(t)=(2t+1)e^{-t(t+1)s}$. Then, with the substitution $v=2t+1$, \begin{align} \int_0^1f_s(t)\,dt &=\int_0^1(2t+1)e^{-(t^2+t)s}\,dt =\int_0^2 e^{-vs}\,dv=\frac{1-e^{-2s}}s\xrightarrow[s\to0]{}2. \end{align}

Answer 4

Really, why $j$ , not any other variable like $\epsilon$ and even $T$ .

See , as we have summation variables in integrals , there is always some variable in pure sum ( $\Sigma$ ) also. They are generally told at the time of forming the sum .

Not about the specific partition function you have mentioned, but almost all of them are summed over microstates in the system .And most probably,I could say that the $j$ here would be some sort of microstate variable , right ?

If you want to know in detail , how they are summed over microstates, there is a very basic example (somewhat long) here.

Answer 5

Since $j$ increases by $1$ from term to term in the series, trivially, its increment is also $\Delta j= \left(j+1\right) - j = 1$, and it is correct to write

$$ \begin{aligned} Z_\mathrm{tot} & = \sum_{j=0}^\infty(2j+1)e^{-j\left(j+1\right)\epsilon/kT} = \sum_{j=0}^\infty(2j+1)e^{-j\left(j+1\right)\epsilon/kT}\Delta j \text{ .} \end{aligned} $$

Notice how this sum corresponds to a "left" rectangle integration rule of the function

$$ f(j)=(2j + 1)e^{-j\left(j+1\right)\epsilon/kT}\text{ ,} $$ using subinvervals in $j$ of width $\Delta j$.

(Look at the yellow figure in the appended link, where the function value at the left limit of each subinterval is used as approximation for the function in each whole respective subinterval, as if the function was in fact rectangular.)

In the limit $\Delta j\to 0$, the sum would exactly match the respective integral, that is,

$$ \lim_{\Delta j\to 0} \sum_{j=0}^\infty(2j+1)e^{-j\left(j+1\right)\epsilon/kT} \Delta j= \int_{j=0}^\infty(2j+1)e^{-j\left(j+1\right)\epsilon/kT}\,dj\text{ .} $$

However, that is not the case here since $\Delta j=1$. Still, the sum may be a valid approximation to the integral if function $f$ does not vary too much in each interval $[j,\, j + \Delta j]$ where the function magnitude is significant (since where the function magnitude is not, that is, where $f(j)\approx 0$ regardless of the variation, the contribution of the respective terms in the sum would be close to null just like in the integral). In a circumstance of significant magnitude, one would need to have

$$ f(l)\approx f(j),\quad \text{for } l\in[j,\,j+\Delta j]\text{ .} $$

Using a first order Tayler expansion around $l = j$ as an approximation to $f(l)$, one gets

$$ \begin{aligned} f(l) & \approx f(j) + f'(j)\cdot(l - j) \\ & \approx f(j) + (l - j)\cdot \left(2 - \frac{\epsilon}{kT}\left(2j+1\right)^2\right)e^{-j\left(j+1\right)\epsilon/kT}\text{ .} \end{aligned} $$

Thus,

$$ \begin{aligned} \frac{\left|f(l)-f(j)\right|}{f(j)} &\le \left|\frac{2}{2j + 1} - \frac{\epsilon}{kT}\left(2j+1\right)\right|\underbrace{\Delta j}_{=1} \\ & \le \left|\frac{2}{2j + 1} - \frac{\epsilon}{kT}\left(2j+1\right)\right| \end{aligned} $$

From this, one gets $\frac{\left|f(l)-f(j)\right|}{f(j)} \approx 0$ if $j\ll kT/\epsilon$, but $\frac{\left|f(l)-f(j)\right|}{f(j)} \gtrsim 1$ if $j\gtrsim kT/\epsilon$, meaning that the varition of function $f$ when compared to the function magnitude starts to be significant at $j\gtrsim kT/\epsilon$. Nevertheless, if the magnitude is negligible for such region, the approximation sum-to-integral would still be valid.

Well,

$$ f(j\sim \frac{kT}{\epsilon})\sim\frac{kT}{\epsilon}e^{-kT/\epsilon}\text{ .} $$ Since $e^{-kT/\epsilon}$ grows much faster to $0$ than $\frac{kT}{\epsilon}$ to $\infty$, one has $f(j\sim \frac{kT}{\epsilon})\approx 0$. The function magnitude at $j\gtrsim \frac{kT}{\epsilon}$ would be much smaller than for previous values. Indeed, the maximum magnitude is mugh grater than 1:

$$ f_{\mathrm{max}}=f\left(\frac{1}{2}\sqrt{2\frac{kT}{\epsilon}}-\frac{1}{2}\right)=\underbrace{\sqrt{2 \frac{kT}{\epsilon}}}_{\gg 1}\cdot \underbrace{e^{\frac{1}{4}\frac{\epsilon}{kT} - \frac{1}{2}}}_{\sim 1}\gg 1\text{ .} $$

One then concludes that under the condition $kT\gg\epsilon$, the approximation sum-to-integral is valid.