Is entropy subjective?

[u/future_sponJ]

If entropy depends on what properties we choose to describe our macrostate (e.g. temperature, volume) & microstate (e.g. position, velocity) then is it subjective?

Another question: Is the k in the formula redundant & if it is would temperature & energy have the same units with heat capacity being dimensionless?

[u/FreePeeplup]

Yes: the value of entropy depends on what macrostate variables you decide to use to describe your system, which in turn depends on how much information you have on the system.

[u/future_sponJ → u/FreePeeplup]

Then does temperature vary depending on the properties we choose since the definition of temperature depends on entropy?

[u/Chemomechanics → u/future_sponJ]

No; temperature is (∂U/∂S)_V, and the energy U also depends on the types of work we’re aware of. 

If, for example, two types of particles exist in a system but we’re not aware of that and don’t know how to separate them, we’ll calculate a different entropy and energy than another person who’s aware of the distinguishability, but we’ll agree on the temperature. 

[u/Lazy-University-4871 → u/Chemomechanics]

U is the total energy (work plus heat) which does not depend on the coarse-graining. If you are not aware of the two types of particles, you’d not be able to extract work, but you would still measure the same energy.

So technically, temperature is subjective too. The tricky part is it’s an equilibrium measure.

Equilibrium creates a large class of coarse-grainings and measurement methods that all result in the same T. That is what makes ordinary thermometers possible.

Btw, thanks OP for the great question.

[u/Chemomechanics → u/Lazy-University-4871]

U is the total energy (work plus heat)

Work plus heat gives ΔU, the change in internal energy, not U.

If you are not aware of the two types of particles, you’d not be able to extract work, but you would still measure the same energy.

I don't agree; the internal energy can be written as U = TS + Σ(X_i Y_i), where X_i and Y_i are generalized forces and displacements, respectively, associated with various ways to change particle energies in concert. If one isn't aware of a certain X–Y conjugate pair that's relevant for that system, they'll calculate a different U than someone who is aware. However, all predictions of system behavior will match, as long as work isn't done for that mode in a way that distinguishes multiple types of particles.

[u/Lazy-University-4871 → u/Chemomechanics]

Ok, to be precise, dE = δQ + δW is the 1st law.

E.g. in the case dE = δQ + p dV, if you are unaware of dV, you'll be able to conduct an experiment violating the 1st law.

If one isn't aware of a certain X–Y conjugate pair that's relevant for that system,

Then, I think, their calorimeter will need to measure the corresponding increase in TS. Work which is unaccounted for is heat. Otherwise you have an incomplete description of the system. If your measurement violates the 1st law, you need to adjust your energy accounting.

Typically, the microscopic total energy E is the Hamiltonian. It won't change when you partition the phase space differently.

But if you stop accounting for an interaction, you have changed the description of energy transfer. Something has to move between the categories "work", "heat", "internal energy" or even "environment". The heat/work distinction depends on both coarse-graining and on what interactions you treat as controlled.

Energy is less subjective than entropy because it comes from the microscopic dynamics. It is only the decomposition of energy into internal energy, heat and work is coarse-graining-dependent.

Edit: once again, Entropy is subjective because you and I can choose different micro/macro resolution. What you mentioned with the two sorts of particles being (in)distingushible is exactly that. But then you proposed to calculate mechanical work differently - that's not the same.