Taylor Series Expansion Confusion

[laser1]

Homework Statement

description

Relevant Equations

.

For context, this is when deriving the Boltzmann distribution by using a canonical ensemble (thermodynamics).

omega is a function to represent number of microstates. According to wikipedia...

is the first order expansion around 0 (Maclaurin series).

My confusion: What are even differentiating with respect to? It seems like we are differentiating with respect to E, but then the epsilon looks like the "x" in the Maclaurin formula. I have more questions, but I'll leave it at this to avoid confusion for now. Thanks!

 

[fresh_42]

Let's analyze it from behind. The Taylor series of a function $f(x)$ at a point $a$ is given by
$$
T(f,x,a)=\sum_{n\ge 0} \dfrac{f^{(n)}(a)}{n!}(x-a)^n=f(a)+f'(a)(x-a)+O((x-a)^2)
$$
For $f(a)$ we have $f(a)=\ln \Omega(E),$ i.e. $a=E.$ Then
$$
f'(a)(x-a)=f'(E)(x-E)=\dfrac{d\ln \Omega(E)}{dx}(x-E)
$$
which means the variable is $x=-\varepsilon + E$ and the second term of the series becomes
$$
f'(E)\cdot((-\varepsilon+E)-E)=\dfrac{d\ln \Omega(E)}{d(E-\varepsilon)}(-\varepsilon+E-E)=-\dfrac{d\ln \Omega(E)}{d(E-\varepsilon)}\varepsilon
$$
We have $d(E-\varepsilon)=dE - d\varepsilon$ and with $\varepsilon$ being a constant we have $d\varepsilon = 0.$
Proof:
\begin{align*}
\dfrac{df(E)}{d(E-\varepsilon)}&=\dfrac{df((E-\varepsilon)+\varepsilon)}{d(E-\varepsilon)}=\dfrac{df(y+\varepsilon)}{dy}\\[6pt]
&=\dfrac{df(y+\varepsilon)}{d(y+\varepsilon)}\cdot \underbrace{\dfrac{d(y+\varepsilon)}{dy}}_{=1}\\[6pt]
&=\dfrac{df(E)}{dE}
\end{align*}
In total we get
$$
\ln \Omega (E-\varepsilon)=\ln \Omega(E)- \dfrac{d\ln \Omega(E)}{dE}\varepsilon + O(\varepsilon^2)
$$

I have more questions, ...

Which?

 

[FactChecker]

According to wikipedia...

View attachment 351220
is the first order expansion around 0 (Maclaurin series).

It looks like they are really expanding around $E$, not around 0. The function and derivative are evaluated at $E$. The Maclaurin series is too specialized (centered at $x_0=0$) to give a good match to your equation. You should compare it to the more general Taylor series. $f(a)+\frac {f'(a)}{1!}(x-a)+...$. Then $E=a$, $E-\epsilon=x$, and $-\epsilon=x-a$.

My confusion: What are even differentiating with respect to? It seems like we are differentiating with respect to E,

That is correct.

 

[laser1]

@fresh_42 Thank you, that makes sense. I also tried differentiating it with respect to epsilon around 0 and got this, but I am stuck at this step:

 

[pasmith]

@fresh_42 Thank you, that makes sense. I also tried differentiating it with respect to epsilon around 0 and got this, but I am stuck at this step:

Apply the chain rule.

 

[laser1]

Apply the chain rule.

I did it a few days ago and was happy, but upon closer examination: I suspect dE/d(epsilon) = -1 only if the total energy of the ensemble is a constant.

Any idea where I have gone wrong with my second method?

 

[fresh_42]

View attachment 351397

I did it a few days ago and was happy, but upon closer examination: I suspect dE/d(epsilon) = -1 only if the total energy of the ensemble is a constant.

Any idea where I have gone wrong with my second method?

I don't think that you made a mistake.
\begin{align*}
\left.\dfrac{d\Omega(E-\varepsilon)}{d\varepsilon}\right|_{\varepsilon=0}&=\left.\dfrac{d\Omega(E-\varepsilon)}{d(E-\varepsilon)}\right|_{E-\varepsilon=E}\cdot \left.\dfrac{d(E-\varepsilon)}{d\varepsilon}\right|_{\varepsilon=0}\\
&=\left.\dfrac{d\Omega(E-\varepsilon)}{d(E-\varepsilon)}\right|_{E-\varepsilon=E}\cdot\left\{\underbrace{\left.\dfrac{dE}{d\varepsilon}\right|_{\varepsilon=0}}_{=0}+\underbrace{\left.\dfrac{d(-\varepsilon)}{d\varepsilon}\right|_{\varepsilon=0}}_{=-1}\right\}\\
&=-\left.\dfrac{d\Omega(E-\varepsilon)}{d(E-\varepsilon)}\right|_{E-\varepsilon=E}
\end{align*}
The trick is now that it doesn't matter how you call your variables. We simply have
$$
-\left.\dfrac{d\Omega(E-\varepsilon)}{d(E-\varepsilon)}\right|_{E-\varepsilon=E}=-\left.\dfrac{d\Omega(x)}{dx}\right|_{x=E}=-\left.\dfrac{d\Omega(E)}{dE}\right|_{E}=-\dfrac{d\Omega(E)}{dE}
$$