Entropy is a measure of statistical nature. Thus, knowing the entropy value of a system requires knowing something of its internal structure, the hierarchy of microstates, and the values of their energies. Probably, for these reasons, it is not possible to construct an universal "entropometer," i.e. a measuring device/instrument able to determine differences in the value of the entropy for any two states of an arbitrary closed system.

On the other hand, temperature also is a magnitude which is essentially statistical in nature. But, indeed it is possible to construct a nearly universal "thermometer," able to determine the temperature of any system made of ordinary [baryonic] matter.

How can we simply explain that thermometers can be built but not entropometers?

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    $\begingroup$ It doesn't seem totally impossible, certainly if you have an ideal thermometer it seems like you could add heat slowly starting from absolute zero and get a curve $T(Q)$ and integrate, and you'd have something $\endgroup$
    – CR Drost
    Feb 17, 2022 at 23:37
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    $\begingroup$ I believe the reason for the nonexistence of an "entropometer" is because entropy is an "anthropomorphic concept" in other words it is a subjective measure; see this physics.stackexchange.com/questions/193677/… $\endgroup$
    – hyportnex
    Feb 17, 2022 at 23:46
  • $\begingroup$ And by the way there is no "forcemeter", either; directly you only measure displacement or strain never force, see physics.stackexchange.com/questions/395813/… $\endgroup$
    – hyportnex
    Feb 17, 2022 at 23:52
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    $\begingroup$ Entropy is measurable quantity. At least for the equilibrium state in a gas... $\endgroup$
    – K.R.Park
    Feb 18, 2022 at 0:55

1 Answer 1


If your system is ergodic (meaning it randomly occupies all possible microstates over long time periods) and has a discrete number of microstates, you could measure entropy by repeatedly measuring which microstate it's in and computing $S = \sum_i p_i \log(p_i)$. This is horribly impractical, because it requires knowledge of and access to a system's microstates, and enough time to build up a good guess for the probability distribution $p_i$. But at least it's in principle a passive operation; we don't have to dramatically change the system to get the quantity that we want.

Without the above expression, we're forced to course-grain over microstates and take the perspective of thermodynamics. All other conventional relations that involve entropy, like $dS=\delta Q_\text{rev}/T$, are ultimately expressions for change in entropy, so you can't use them to compute absolute entropy without doing drastic things to your system. To measure entropy using relations like $dS=(dU+PdV)/T$, we have to compute changes in entropy until we get the system to a state of known entropy, like absolute zero (which is usually zero but may not be for some tricky systems like glasses). But this is far from a passive operation! So while this means that there is an experiment you can do to measure entropy, it will be far more invasive than using a thermometer.

So why isn't there a simple thermodynamic expression for total entropy? The following isn't a proof, but I think it captures the essential reason. The operational, and therefore easily measurable thermodynamic quantities like temperature and chemical potential are derived by maximizing entropy, and to do that we take the derivative of entropy and set it to zero. For example, to derive temperature as a concept, we consider two systems that are allowed to exchange energy and compute how total entropy changes as they do so

$$\frac{\partial S_{tot}}{\partial E_{1\rightarrow 2}} = \frac{\partial S_1}{\partial E_{1\rightarrow 2}} + \frac{\partial S_2}{E_{1\rightarrow 2}} = -\frac{\partial S_1}{\partial E_{2\rightarrow 1}} + \frac{\partial S_2}{E_{1\rightarrow 2}}$$

The second equality follows because energy flowing into system 2 has to equal negative the energy flowing into system 1. In equilibrium, $S_{tot}$ is maximized and $\partial S_{tot}/\partial E_{1\rightarrow 2} = 0$, so we get $\partial S_1/\partial E_1 = \partial S_2/\partial E_2$, which we define to be $1/T$. The concept of entropy leads to the operational quantities like temperature when we compute how entropy changes. Therefore the resulting quantities are insensitive to the absolute magnitude of $S$.

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    $\begingroup$ Is the first part of your answer truly correct though? Discrete microstates are quantum in nature fundamentally, and doing this repeated observation is not possible without changing the density matrix of the system. I think the question is more subtly about whether the entropy is measurable without needing the full density matrix $\endgroup$
    – KF Gauss
    Feb 18, 2022 at 19:16
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    $\begingroup$ You can often get away with discrete, classical toy models like a bunch of spins that can only point up or down, but yeah I agree that things get even harder when you consider quantum mechanics (you would have to perform quantum state tomography on many copies of the system). But even then, you're left with the question of why thermodynamic entropy can't be measured easily/passively. So no matter what perspective you take on microstates, you still need the second half of this answer. $\endgroup$
    – user34722
    Feb 18, 2022 at 19:51

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