# Why is this Taylor Expansion, Leading to the Boltzmann Distribution, Acceptable?

In Stephen Blundell's "Concepts in Thermal Physics" chapter 4 he derives the Boltzmann distribution. The equation that leads to the Taylor expansion is the following:

$$P_s(\epsilon) \propto \Omega(E-\epsilon)$$

where $$P_s(\epsilon)$$ is the probability of a system in thermal equilibrium with a reservoir being a microstate $$s$$ of energy $$\epsilon$$, and $$\Omega(E-\epsilon)$$ is the number of microstates associated with a reservoir of energy $$E-\epsilon$$. In order to get the Boltzmann distribution he performs a Taylor expansion of the following function:

$$\ln(\Omega(E-\epsilon))$$

where $$\Omega(E-\epsilon)$$ is the number of microstates associated with a reservoir of energy $$E-\epsilon$$, $$E$$ is the total energy of the reservoir and an attached system and is therefore constant, and $$\epsilon$$ is the energy of the attached system. Here $$E \gg \epsilon$$, and we take the Taylor expansion about $$\epsilon = 0$$ to get:

$$\ln(\Omega(E-\epsilon)) = \ln(\Omega(E)) -\frac{d\ln(\Omega(E))}{dE}\epsilon + \dots$$

I don't understand how he has arrived at this equation. It seems to me setting $$\epsilon = 0$$ is the same as setting $$a = E$$ in a standard Taylor expansion, so that part is fine, but surely as we are representing $$x$$ from the normal taylor expansion as $$x = E-\epsilon$$, the second term should be:

$$-\frac{d\ln(\Omega(E))}{d(E-\epsilon)}\epsilon$$

Why doesn't the second term have this form as its denominator? Surely I can't set the $$\epsilon$$ in the denominator to zero without setting the multiplying $$\epsilon$$ to zero also and wiping out the whole term?

Edit: As $$E$$ is the total energy of the reservoir plus the system it is my assumption it stays constant and cannot be a variable.

• You are evaluating the derivative at $\epsilon = 0$, so it can not depend on $\epsilon$. Your notation makes it look like it depends on $\epsilon$. Think of it like this: $\ln(\Omega(x))$ is a function of x. Call it f(x). Call the derivative of f with respect to x, g(x). You are interested in g(E). This is what the notation means.
– hft
Oct 22, 2021 at 22:41
• What do you mean it cannot depend on $\epsilon$? Surely in a normal Taylor expansion I am evaluating it at $x = a$ which could be equal to zero, and yet the differential depends on $x$? Oct 22, 2021 at 22:56
• I mean it cannot depend on epsilon. Not sure how else to say that. The epsilon dependence of the linear term is just linear, there's no other epsilon dependence.
– hft
Oct 23, 2021 at 0:10

I'm not sure what your objection is. A generic Taylor expansion might look like $$f(x-\epsilon) = f(x) + f'(x) (-\epsilon)+ \frac{1}{2} f''(x) (-\epsilon)^2 + \ldots$$

Perhaps the confusion is related to using the Liebniz notation rather than the prime notation for the derivative? Try defining $$f(E) = \ln(\Omega(E))$$ and then using the prime notation.

• But is it not the case that your Taylor expansion is derived from the $f(x)$ about $x = a$ version of the Taylor expansion? So surely if you define $x = x - \epsilon$ that would carry through to the denominator? Oct 22, 2021 at 22:55
• @Connor I’m afraid I don’t know what that means. It sounds like you are confused by Leibniz notation. $f$ is a function of one variable, and $f’$ is its derivative. Using $E-\epsion$ as a dummy variable of differentiation seems like a recipe for trouble to me. Oct 22, 2021 at 23:01
• @Connor $f$ is a function which eats an energy $E$ and spits out the log of the number of microstates accessible to the system at that energy. I don’t know why you don’t think you can differentiate it. Oct 22, 2021 at 23:09
• @Connor, the differential "dx" is a different type of animal than the variable "x". You should think of "d/dx" as a different kind of symbol. You don't really "substitute in" for every literal occurrence of the letter "x", only every occurrence of the variable "x". Alternatively you can just understand that "d/d(x-y)" is the same as "d/dx" for any constant "y". The differential notation like "d/dx" is kind of frustrating sometimes.
– hft
Oct 23, 2021 at 0:15
• For example, if instead of f(x) = f(a) + (df(a)/dx)*(x-a), I write f(x) = f(a) + f'(a)*(x - a), there is probably less confusion. Now when I substitute in for the variable x I get: f(E - e) = f(E) + f'(E)*(E- e - E)). No "E-e" needed in the "denominator".
– hft
Oct 23, 2021 at 0:19

I think you are just getting a little lost in notation; there is nothing particularly deep happening.

The way the Taylor expansion is often defined by mathematicians (eg see wikipedia) is to relate the value of a function at some point $$x$$, with the value of the function (and derivatives) at some other point $$a$$, like so $$\begin{equation} f(x) = f(a) + \frac{df}{dx}\Big|_{x=a} (x-a) + \frac{1}{2} \frac{d^2f}{dx^2}\Big|_{x=a} (x-a)^2 + \cdots \end{equation}$$ This point of view makes sense if you want to ask questions like "what is the radius of convergence of the Taylor series done at a point $$a$$" that a mathematician is interested in.

In physics, we are almost always interested in using the Taylor series to approximate some function. Therefore, defining $$|x - a| = \epsilon$$, we are almost always interested in the case where $$\epsilon$$ is small (in some units).

We therefore tend to use notation that makes it easy to truncate the series when $$\epsilon$$ is small. To do this, we replace $$x\rightarrow x-\epsilon$$ and $$a\rightarrow x$$ in the above expression. Note this requires a change in how we think about what $$x$$ means. Previously, we thought of $$a$$ as a fixed reference point for the expansion, and $$x$$ as a test point that we can vary; we want to know the value of $$f$$ at $$x$$ given knowledge of $$f$$ and its derivatives at $$a$$. Now, $$x$$ is the reference point, and we vary $$\epsilon$$; we want to know the value of $$f$$ at $$x-\epsilon$$ given the value of $$f$$ and its derivatives at $$x$$.

It's also important to know that the variable "in the denominator of the derivative" is a dummy variable. We can use any symbol for it, since ultimately we are going to evaluate the derivative at the reference point $$x$$. For clarity, I will use $$y$$ as the variable in the derivative.

With all this in mind, we make the substitutions described in words above:

1. $$x \rightarrow x-\epsilon$$
2. $$a \rightarrow x$$
3. $$\frac{d^n f}{d x^n}\Big|_{x=a} \rightarrow \frac{d^n f}{d y^n}\Big|_{y=x}$$

and we find $$\begin{equation} f(x-\epsilon) = f(x) + \frac{df}{dy}\Big|_{y=x} \left(-\epsilon\right) + \frac{1}{2}\frac{d^2 f}{dy^2}\Big|_{y=x} \left(-\epsilon\right)^2 + \cdots \end{equation}$$ As you can see, this is an exactly equivalent way of representing the Taylor expansion as defined on wikipedia -- all we have done is change variables. This form is the same as the one in your question -- all we need to do to fully make the connection is rename $$x$$ as $$E$$, and $$f(x)$$ as $$\ln \Omega(E)$$.

• Thanks, that's a really good explanation. My question then is, does the evaluation at the point $x=a$ make the denominator of the derivative a dummy variable? Or is it always the case that $dx$ could be replaced by anything in any standard derivative? My current intuition is that $\frac{df(x)}{dy}$ is meaningless/trivial if $y$ and $x$ aren't related in some way. Oct 23, 2021 at 11:31
• @Connor I think this is easiest to think about from the definition of the derivative. To evaluate the expression $\frac{df(y)}{dy}\Big|_{y=x}$, we have to consider $\frac{1}{h}(f(y+h) - f(y))$ and take the limit $h\rightarrow 0$. At the end of this process, we set $y=x$. So $y$ is a dummy variable since it drops out at the end. Oct 23, 2021 at 20:10
• Where you can get subtleties is when you need to chain rule, e.g. if you evaluate $\frac{df(y^2)}{dy}\Big|_{y=x}$. Another way you could do the calculation in your question (without introducing $y$) is to evaluate $\frac{df(E-\epsilon)}{dE}$ -- the argument of the function and the "denominator" are different, so you should use the chain rule, but when you do so this turns into $\frac{df(E)}{dE} \cdot 1$. So for a simple translation of the argument in a derivative, you don't need to worry too much about this. Note we are not translating the point at which the derivative is evaluated. Oct 23, 2021 at 20:15

Technically, you CAN take derivative with respect to $$E^{\prime} = E-\varepsilon$$. Expansion around $$E^{\prime} = E$$ should look like that then: $$\ln{\Omega\left( E^{\prime} \right)} \approx \ln{\Omega\left( E \right)} + \left. \frac{d \ln{\Omega\left( E^{\prime} \right)}}{ d E^{\prime} } \right|_{E^{\prime} = E} (E^{\prime} - E) = \ln{\Omega\left( E \right)} - \left. \frac{d \ln{\Omega\left( E^{\prime} \right)}}{ d E^{\prime} } \right|_{E^{\prime} = E} \varepsilon$$

Compare to the traditional expansion around $$x_0$$:

$$f(x) \approx f(x_0) + \left. \frac{d f(x)}{d x} \right|_{x=x_0}(x-x_0)$$

Basically the same. The trick is that you can differentiate function over its argument, and if the argument is constant then set it to the constant value. Even if $$E$$ is constant, you still can take derivative over that. It's like saying "IF the energy changed, what would be the derivative?"

Or, in other words, you want to describe the behavior of a function, not its variables. You can rename the variables as you wish: $$f(x) \approx f(x_0) + \left. \frac{d f(😀)}{d 😀} \right|_{😀=x_0}(x-x_0),$$ and that is still correct.

Sometimes, for convenience, to avoid drawing that ugly stick on the right, people write something like in your textbook. In conventional functions it can be written like that:

$$f(x) \approx f(x_0) + \frac{d f(x_0)}{d x_0}(x-x_0)$$

That is a little bit less traditional form, but still true. It's like you are saying at the very beginning at which point you take the derivative.

UPD: I was using partial derivatives initially. Sorry, got used to it.