Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

206 submissions , 164 unreviewed
5,106 questions , 2,251 unanswered
5,390 answers , 22,970 comments
1,470 users with positive rep
822 active unimported users
More ...

  What does it mean for thermodynamic entropy to be defined only up to a constant?

+ 5 like - 0 dislike
11 views

I'm reading the Wikipedia article on thermodynamic definition of entropy.

Because the energy of a particle in a classical thermodynamic system is a continuous variable, the number possible states is actually uncountably infinite. So, to be able to define entropy using the definition $S=k\ln(W)$, we need to somehow "coarse grain" the continuous distribution so that we end up with a finite number. We can do this by grouping together states, so that states with energies close to each other (within a, say, $dE$ from each other) are taken to be the same state.

According to Wikipedia, this means that entropy is actually "defined up to an additive constant". I read from somewhere else that this is analogous to potential energy, which is also defined up to an additive constant (by the choice of the zero point). It is also mentioned in Wikipedia that in the thermodynamic limit, the definition of entropy becomes independent of the choice of $dE$.

These points are where I am confused. Could somebody elaborate a bit more on what does it mean the entropy is defined up to a constant, and why the choice of $dE$ does not affect the definition of entropy in the "thermodynamic limit"?

This post imported from StackExchange Physics at 2025-01-21 21:52 (UTC), posted by SE-user S. Rotos
asked Feb 18, 2019 in Theoretical Physics by S. Rotos (25 points) [ no revision ]

2 Answers

+ 1 like - 0 dislike

That "entropy is defined up to an additive constant" – if this is taken as a universal statement valid for all thermodynamic systems – is actually a myth, sadly perpetuated by many textbooks in thermodynamics especially for undergraduates.

This statement is only valid for so-called elastic systems. These are systems for which any two states can in principle be connected by a reversible process (without excluding the possibility of irreversible processes too). For such an elastic system we can define an entropy function that can be used for various purposes, for example to determine equilibria or other function of states. It turns out that in making such a definition we are free to add a real constant to the entropy function: the new entropy function we obtain will still perfectly serve the same purposes.

Non-elastic systems, instead, are systems for which some pairs of states cannot be connected by any reversible process, but only by irreversible ones. Systems presenting hysteresis are an example, as are many viscous or plastic systems. This kind of systems is very common in everyday life. For such systems we can define several entropy functions, which cannot be obtained from one another simply by adding some constant. Yet all these inequivalent entropy functions are perfectly fine and can be again used for the usual purposes. This non-uniqueness is sometimes expressed by saying that for such systems there do not exist any "entropometer" (see eg Samohýl & Pekař §2.3 p. 52).

From a mathematical point of view the difference between these two situations boils down to the conditions for which Stokes's theorem holds. For elastic systems, we can draw a cyclic thermodynamic process in the space of thermodynamic states, and its closed curve can be decomposed into a collection of smaller closed curves with overlapping segments: these segments are traversed in one direction in one curve and in the opposite direction in the other curve. The possibility of traversing in both directions comes from the existence of reversible processes connecting any two states – a reversible process is one that can be traversed in both directions. This leads to the possibility of defining a global entropy function on the whole state space, up to a constant.

For non-elastic system, instead, an analogous decomposition of a cyclic process into smaller ones cannot be made, because some segments of overlapping curves can be physically traversed in one direction only. This leads to the possibility of defining several inequivalent entropy functions on the state space.

For further details see for example these references:

This post imported from StackExchange Physics at 2025-01-21 21:52 (UTC), posted by SE-user pglpm
answered Apr 26, 2023 by pglpm (510 points) [ no revision ]
+ 0 like - 0 dislike

I'm assuming this is just due to the fact that derivatives are unique up to addition of a constant. For example, if we have the simple relationship $\text d y=\text dx$, then we know that $y=x+C$.

This post imported from StackExchange Physics at 2025-01-21 21:52 (UTC), posted by SE-user BioPhysicist
answered Feb 18, 2019 by BioPhysicist (0 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ys$\varnothing$csOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...