2
$\begingroup$

I could be doing something quite stupid here; this is a tad over my head, so I apologize if this is a stupid question, and if requested to delete it, I will do so.

I know that Boltzmann's formulation of entropy is $S = k \log W$ where $k$ is Boltzmann's constant, $S$ is the entropy of the given system, and $W$ is the number of microstates within the given macrostate - or, to put it another, simpler way, the number of ways you can arrange the pieces of a system such that the overall picture of they system is the same (in, say, a steam cloud, one could say the cloud-ish look is the macrostate and the position of each specific molecule is a particular microstate). Yet another way, this is configurations in phase space.

I read the log relation

$$\log a^b = b \log a$$

Which made me think that Boltzmann's formulation of entropy could be written as $$S = \log W^k$$

I have two questions about this:

  1. Is this right? If not, where did I go wrong?
  2. If this is right, what are the advantages are writing it the first way versus the other?

Edit:

In light of CuriousOne's comment:

Gibb's formula for entropy is $$S=-k_{{\text{B}}}\,\sum _{i}p_{i}\ln \,p_{i}$$

Which, by the same log rule gives

$$S = \log p_i^{-k_{{\text{B}}}\,\sum _{i}p_{i}}$$

At which point the same two questions apply. Also in terms of CuriousOne's comment, how is the second equation (in the two cases) physically meaningless?

$\endgroup$
  • $\begingroup$ While the formal manipulation you are doing there is mathematically correct, it's physically meaningless since the Boltzmann constant is merely a unit dependent proportionality between energy and temperature. In "natural units" we can set it to one. Boltzmann entropy is, by the way, only valid for a very limited case. See Gibbs entropy for the more general formula. Please also keep in mind that by configurations we mean configuration in phase space, not just the possible locations. $\endgroup$ – CuriousOne Aug 15 '16 at 23:45
4
$\begingroup$

Both $S=k_B\ln W$ and $S=\ln W^{k_B}$ are mathematically equivalent and equally valid, so you're not wrong in shifting the constant $k_B$ into the logarithm. You can even express the microstates $W$ as a function of the entropy $S$ if you want, i.e. \begin{align}W=e^{S/k_B}~, \end{align} as it is all the same.

Regarding the advantages of writing one form over the other, the regular expression $S=k_B\ln W$ might be easier to manipulate, since this expression is linear in $k_B$ as opposed to working out an expression which depends on the power of $k_B$. For example, if you have an expression for the microstates $W$ which contains factorials, (say an $N$-particle, two-state system, and $p$ particles in one of the states) \begin{align} W=\frac{N!}{p!(N-p)!}~, \end{align} you'll realize that keeping the $k_B$ as a linear factor in the entropy \begin{align} S = k_B \left[\ln N! - \ln p! - \ln (N-p)! \right]~, \end{align} allows you to easily make the Stirling's approximation for large numbers $x$, where \begin{align} \ln x! \approx x\ln x - x~. \end{align}

It might be harder to evaluate $S$ if you tried to brute force $\ln (x!)^{k_B}$ by keeping it in the second form which which you have described.

Reponse to Edit: Likewise for the Gibbs entropy, linearity of sums is easier to work with than with a large collection of terms with some multiplicative powers. Your expression with the Gibbs entropy is also incorrect because you have to sum over the individual $\ln p_i$'s first.

To show how complicated the expression becomes when you express it in terms of a single logarithm, we expand the geometric series in the entropy $S$ \begin{align} S &= -k_B \sum_{i} p_i \ln p_i \\ &= -k_B(p_1 \ln p_1 + p_2 \ln p_2 + p_3 \ln p_3 + \ldots) \end{align} and then raise the expressions within the logarithm to the power of the prefactors, as you would have done \begin{align} S = \ln p_1 ^{-k_Bp_1} + \ln p_2 ^{-k_Bp_2} + \ln p_3 ^{-k_Bp_3} + \ldots \end{align} Now we group all the terms into a single logarithm \begin{align} S = \ln \left( p_1 ^{-k_Bp_1}\cdot p_2 ^{-k_Bp_2} \cdot p_3 ^{-k_Bp_3}\ldots\right) \end{align} and if we were to express this in terms of $p_i$'s we would have \begin{align} S = \ln \left( \prod_i p_i^{-k_B p_i} \right)~, \end{align} which would be a nightmare to evaluate in practice.

Finally, with regards to CuriousOne's comment, you'll realize that $S=k_B \ln W = \ln W^{k_B}$ is the same expression when you set $k_B=1$, and this is possible when we change our system of measurements to a set of units that allows it to be so. (i.e. $k_B= 1.38\times 10^{-23}$ m$^{-2}$ kg s$^{-2}$ K$^{-1}$ only when you use SI units). Since the physics of the entropy equation should remain the same regardless of any system of units that we are using, you can think of $k_B$ as merely a bookkeeping factor that makes sure that the numbers you get for the entropy turns out correct for the system of units that you are using, and as such shifting the constant around does not attach any physical meaning to it.

$\endgroup$
1
$\begingroup$

The expression $W^k$, where $k$ is Boltzmann's constant, does not make sense mathematically because $k$ is not a number but has dimension of Energy/Temperature. So its numerical value depends on the units you choose. For example if you measure it in Joule/Kelvin then its value is about $1.38 \times 10^{-23}$ but if you measure it in eV/Kelvin it is about $8.62 \times 10^{-5}$. So it is not clear to what power you want to raise the number $W$ when you write $W^k$.

$\endgroup$
  • $\begingroup$ Phrased differently, it's because $k_B=1$ :) $\endgroup$ – user12029 Jun 10 '17 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.