I am a high school student. So, while studying about thermodynamics, I got a little curious about entropy. As I read, entropy is the rate of change of chaos. So, if the entropy change of a system is $\delta$S and the primary and final entropy is $S_1$ and $S_2$, then, $\Delta$$S$ = $S_2$ - $S_1$

Now, come to the mathematical expression of $\Delta$$S$ I am really disgusted by the very brief answer of the mathematical expression in my book. It says that, in a reversible process, the amount of heat or $\Delta$$Qrev$ a system is emitting or producing, divided by the temperature at which the heat is emitted or produced of $T$, or $\Delta$$Qrev$$/$$T$, is equal to $\Delta$$S$.

So, $\Delta$$S$ = $\Delta$$Qrev$$/$$T$

Ok, so my book doesn't tell how does this equation arise, and that is the irritating thing. Also, there is another question regarding this topic, when we are talking about the rate of change of chaos: what is the relation of temperature and heat in here?

Although I know that this is a complex part of physics, I am asking for your help so that I can be able to have a clear idea about the matter. Thank you.


5 Answers 5


For reversible processes, we may derive that $$ d S = \frac{\delta Q}{T} $$ (note that it's $dS$, not $\delta S$, on the left hand side because $S$ itself is actually a well-defined property of the state, namely the amount of disorder in the system, and not the "rate of chaos", as you misleadingly wrote; on the other hand, there is no unique $Q$ "overall accumulated heat" that would define a current state of a system because we don't know "from when" this heat would be counted and only the changes of $Q$ are well-defined which is why we use $\delta$ and not $d$) up to some changes of conventions or redefinitions of variables.

Why is it possible to choose conventions in which the equation above holds? Well, we may always transfer heat into a system or from the system at (nearly) the same temperature which is reversible. When we reverse such a heating by a cooling, there must be a function of the state $S$ that returns to the original value; if the processes were irreversible, it should increase to a higher level.

Now, the question is how such an $S$ and its changes have to be defined so that it has the properties demanded in the previous paragraph. Clearly, it has to be an extensive quantity. So the more heat we answer, the greater is $dS$. Because the heat may always go through an intermediate system of the same temperature, it doesn't matter where the heat is flowing from and into.

So it must be that $dS=\delta Q / B(T)$ where $B(T)$ is some universal function of temperature (and nothing else). How this function $B(T)$ may be determined using any system, for example the perfect gas which is really simple.

The gas obeys $pV=nRT$, the differential for the heat absorbed in the reversible expansion is $$\delta Q = C_v dT + p\, dV. $$ This differential is not exact which means that it cannot be written as $dM$ for some function of state (effectively, a function of $p,V,T$ etc.) $M$.

We want $$dS = \frac{\delta Q}{B(T)} = \frac{C_v dT}{B(T)} + \frac{p\,dV}{B(T)}$$ to be an exact differential of something. The first term is $C_v$ times the differential of some function of $T$, whatever $B(T)$ is. However, the second term $p\,dV/B(T)$ is only an exact differential if the non-constant factor $p$ in the numerator, in front of $dV$, cancels. So $B(T)$ must be equal to $p$ times a function of $V$. But because it's a function of $pV/nR=T$ at the same moment, it must be equal to $pV/nR=T$, up to an overall scaling.

So up to the freedom to rescale the temperature or entropy by a factor (a choice of units), we may choose $B(T)=T$ i.e. equal to the absolute temperature of the ideal gas. No nonlinear function such as $B(T)=T^n$ for $n\neq 1$ would be acceptable. For this example, we would have $B(T)=p^n$ times a power of the volume (times a constant) so in the denominator, we would get $p^n$ which wouldn't cancel $p$ in the numerator and the ratio would still depend on $p$, leaving us with $f(p)dV$ which is not an exact differential.

The ideal gas calculation above was simpler than others but with a working knowledge of the microscopic model of any physical system, we could use any physical system to derive the same thing. In fact, in statistical physics, which is the explanation of thermal phenomena in terms of the statistical properties of many atoms and their jiggling, it is possible to derive $\delta Q = T\,dS$ for all reversible processes in general. But I am afraid that statistical physics is OK material for the college level audience only – but I surely don't want to underestimate you.


First of all, entropy as "rate of change of chaos" is a piss poor definition and the source of your confusion. Forget it. Second, the definition of your book is the correct one sofar as thermodynamics is concerned.

What happened though is that in the 19th century, theoreticians were trying to give a mechanistic explanation for the laws of thermodynamics. Ludwig Boltzmann proposed a microscopic definition of entropy as counting the number of microstates realizing a certain macrostate. It is then possible to show for some idealized systems this indeed corresponds to the usual thermodynamic definition.

Boltzmann's interpretation then leads to the realization that the second law of thermodynamics is really describing an evolution from macrostates that have few microrealizations to macrostates that have lots of microrealizations which means they are more likely. In other words, the second law of thermodynamics is understood as a statistical law.

Now, in some of Boltzmann's work there is mention of chaos, particularly molecular chaos. This is a hypothesis Boltzmann made to derive a certain equation, nowadays bearing his name. It was later proved that this assumption was not necessary but it has nevertheless lead to much confusion about what Boltzmann was trying to tell.


Irritating things and deep concepts

Basically, what you call "the irritating thing" has irritated scientists for the latter half of the 19th century. Clausius had found that there was a quantity (later called entropy $S$) such that, for any transformation $\delta S\ge \frac T{\delta S}$, with equality only for reversible transformations. But it took time and many heated discussions to arrive on an agreement about what this entropy is. Nowadays, I don't think entropy is mysterious to physicists today, but this deep concept still leads to heated discussions, mainly about the relevant metaphors used to teach it, the words used to describe it without equations, or philosophical questions. If you want to learn more about it, I can advise you to read Wikipedia's entry about entropy, and the links therein. But be careful, the reading can last for hours ;-)

About the relation between "chaos" or "disorder" and temperature and heat.

You can read Luboš Motl's answer to have some ideas on the equations linking temperature and heat to entropy. I will try to give you a more qualitative answer. The following answer is much of course much less precise and rigorous than Luboš's, but I hope it will help you to build some intuition about these concepts.

The entropy $S$ of a system is now understood as a (logarithmic) measure of the number of microscopic states of a system corresponding to a single macroscopic state. For example, if you describe a pebble, you will not care about the individual state of each of its approximately 10$^{20}$ atoms. You have many possible such "microstates" corresponding to the same macroscopic state (the pebble is in your left shoe, at this position, at rest and has a temperature of 27°C = 300 K).This variety of "invisible" states is what people refer to when the speak about "chaos" and "disorder".

Now, all these microstates can be described statistically. For example, the temperature is (almost) nothing else than the average energy of the invisible "degrees" of freedom of the atoms in the pebble. For example, the atoms are vibrating, and these vibrations carry some energy. Of course, the exact amplitude and phase of the vibration of the 12049847193587392384th atom doesn't influence the macrostate of the pebble. But the fact that, on the average, all the atoms vibrate quickly or not can be felt at the macroscopic level. It's what we call temperature. The fact that this energy is spread among many atoms make it in some sense more difficult to manipulate than the macroscopic potential or kinetic energy of the pebble, and that is this difficulty which is measured by the second law of thermodynamics.

Let say you put the pebble in contact with another body at a different temperature (let say your toe, at 37°C = 310 K). If you look at the atomic scale, you will see many atoms and molecule vibrating at different speed and coupling to each other with energy transfer in all possible directions. Now, if you look at it statistically, you will see that the atoms of the colder body will have increased there average energy (aka Temperature) while the atoms of the hotter body will have decreased there average energy. In the long run, all the atoms will have the same average energy. This "disordered" energy transfer is basically what we call heat.

This three concepts (entropy, temperature and heat) are, by there very definition closely related to the statistical nature of the macroscopic world, and that's why they are closely related. Historically, the statistical nature of temperature and heat was unknown, and that's why entropy was so irritating.

  • 1
    $\begingroup$ >heated discussions $\endgroup$
    – Nikolaj-K
    Jun 3, 2013 at 13:10
  • $\begingroup$ @NickKidman : Sorry... I couldn't resist ;-) $\endgroup$ Jun 4, 2013 at 9:21

As an addition to the other answers I want to point out that a macroscopic and correct definition of entropy is quite young: Entropy from adiabatic accessibility by Lieb and Yngvason in 2003.

The less mathematical version of their publication is written by Thess in The Entropy principle, which does not require prior knowledge in statistical physics.


The OP asks two questions: where does the definition of entropy come from, and what is the relation of the formula for entropy with disorder. A simple physical example will answer the second question, which is what I will add to what the other posters have written.

To recapitulate a little from Prof. Mottl's answer. We start with thermodynamics, our study of how $T$ and $Q$ vary between different equilibrium states of, in this example, ice, water, and steam. We find that as $T$ and $Q$ vary, $1/T$ is an "integrating factor" for $\delta Q$. This formality led to the definition of the quantity $S$ of entropy via the formula $\Delta S = \int {\delta Q \over T }$. So now we have a formula, but no interpretation.

Consider a block of ice at exactly $0^\circ$C (and one atmosphere pressure). It is in a state of equilibrium with itself, but now we are going to steadily add heat to it. It remains at the same temperature even though it possesses more and more heat, but it melts. The amount of liquid water produced is exactly proportional to the added heat $\Delta Q$ up to the point at which all of it has melted, and the temperature does not increase until all the melting is finished. (Also, each state, part ice and part liquid, is at equilibrium with itself: if we stopped adding heat it would just stay there, at $0^\circ$C, without any more melting, but without any re-freezing either).

But clearly the molecules in crystal ice are in a relatively ordered state, while the molecules meandering around in the liquid are in a less ordered state, since the situation has become fluid instead of crystal clear. It is clear that the amount of increase in disorder is exactly proportional to the amount of fluid produced, which we just saw was proportional to $\Delta Q$. But by the definition of the change in entropy $S$, $\Delta S$ is also proportional to $\Delta Q$ since $T$ is constant during this transformation. Therefore it is intuitively clear that $\Delta S$ is proportional to the increase in disorder.

The exact same conclusion follows if we consider the special case of boiling water. As we pour more heat into the system of boiling water, its temperature does not increase, but more steam is produced, which is more disordered. So again, the increase in entropy is exactly proportional to the increase in disorder.

These two special situations show that after one has defined entropy from the formula, it is proportional to the increase of disorder at least in such cases. But since there is no precise mathematical definition of the "quantity" of disorder, we may as well make $S$ the precise definition of the amount of disorder in general: we extrapolate it from these two situations, using the formula we already have, $$\Delta S = \int {\delta Q \over T }.$$

Later, scientists like Maxwell and Boltzmann wanted to connect thermodynamics with the dynamics of the atoms making up the ice. In the more advanced theory of Statistical Mechanics which they created, a quantitative measure of disorder can be proposed, $S = k\log \Omega$, and then one can prove that this agrees with our formula for $\Delta S$.


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.