In some of the answers and comments from this question people contended (not in so many words) that because entropy is parameterized by number of microstates $\Omega$, and the definition of $\Omega$ is up to the person doing the calculation, entropy is subjective. Such an argument might go thus:
"Consider a box containing $n$ particles. Suppose the particles do not interact and can be crammed as many as you like into as little space as you like. Subdivide the box into $m$ smaller boxes of equal dimensions. Then the number of box-contents microstates is $Ω=m^n$. $n$ is an observable, but whoever choses $m$ chooses $Ω$, and $S=k_blog(Ω)$, therefore whoever subdivides the box sets the entropy."
It seems to me that $\Omega$ (and hence $S$) isn't subjective, it's simply parameterized by degrees of freedom per particle. I couldn't suggest a formula, but as far as I can tell the relationship is such that if you use all the degrees of freedom each particle can have, you get $d S = 0$ and $S = 0$, and if you use only some of the degrees of freedom, you get $dS \ge \delta Q / T$ and $S = k_b log(\Omega)$ as normal.
(Of course if you introduce quantum mechanics you can't actually fully specify all the states of each particle, because you can't express a state in which the particle's momentum and position are both known to arbitrary precision, but I think that's not directly relevant to my question.)
It also seems to me that most if not all observables have a "degree of freedom" parameterization, so entropy is not a special case, it's just that most quantities' degree of freedom parameterization is obvious and implicit, whereas entropy's emerges from difficult math.
Consider quantities velocity and length and the parameter mass. If we give the rocket in free space a mass degree of freedom, the rocket accelerates and stays the same length. If we don't give the rocket a mass degree of freedom, the rocket stays at one velocity and gets longer (the expelled fuel still counts as part of the rocket). Length and velocity are hardly subjective, and yet they're parameterized by the mass degree of freedom of the rocket.
Is this right? If so, is there a mathematical expression for the parameterization of $\Omega$ as a function of particle degrees of freedom?
An addition:
Going back to our box of n particles: we can make reference to spatial degrees of freedom from this by starting by subdividing parallel to the xy plane, then further subdividing along the xz plane, and along the yz plane. If we make equal subidivisions for each dimension, we multiply m by the number of dimensions.
We can add additional complexity by gradating other variables than position (for instance, momentum) and asking what the average in each subdivision is closest to.
So perhaps our $\Omega$ might be something like,
$\Omega = f(n, m_1, ...m_i, D_1, ...., D_i) = (D_1m_1)^n \times (D_2m_2)^n\times ...\times (D_im_i)^n$
where $m_i$ is the number of subdivisions per dimension per observable and $D_i$ is the number of corresponding dimensions. This would seem to contradict my intuition that $S=0$ if you perfectly specify every possible parameter, since $S$ would trend in the opposite direction as you add parameters and precision.
Another addition:
Wikipedia suggests going around the question by applying H theorem or (if I understand correctly) by using quantum eigenstates as your "boxes" and then applying the thermodynamical limit to reduce first to a countable infinity and then to a finite number of maximum microstates, to the limits of the system's uncertainty. This probably is directly relevant since (I think) it implies that an appropriately chosen variable captures all the existent information, allowing the problem to be reduced to counting the number of microstates of that single variable. Is that a correct inference?
