If entropy is just a measure of the "randomness" or "chaos" of a system, then why does it have specific units. What is the reason behind the units of entropy being J/K?

  • 1
    $\begingroup$ Entropy is not a measure of the randomness. It is equal to the constant of gases R multiplied by the logarithm of the probability of a given state. It has the same unit as R. $\endgroup$ – Maurice Feb 27 '20 at 10:37
  • 1
    $\begingroup$ Wikipedia has what seems to me like a good explanation of the relation between the information-theoretic definition of entropy and the thermodynamic definition: en.wikipedia.org/wiki/… $\endgroup$ – padd13ear Feb 27 '20 at 16:58

The units of of energy over temperature (e.g. J/K) used for entropy in the thermodynamic definition follow from a historical association with heat transfer under temperature gradients, in other words, the definitions of temperature and entropy are intertwined, with entropy being the more fundamental property. Entropy can be thought of as a potential and temperature (or its inverse, rather) as a generalized force associated with displacements along energy dimensions in the entropy potential. The (inverse) temperature is a measure of the effect of changes in the amount of energy on the entropy of the system, as the following definition suggests:

$$\frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)_V$$

However if you start from a statistical mechanical definition of the entropy,


then it is easy to see that a unitless definition might be just as well suited:


However, in the absence of the Boltzmann constant you need some other way to relate entropy and energy (e.g. heat transfer) in our conventional system of units. You can of course subsume $k_\mathrm{B}$ into the definition of a new temperature scale:

$$\frac{1}{T_\mathrm{new}}=\left(\frac{\partial S_\mathrm{unitless}}{\partial U}\right)_V$$

where the old and new temperature scales are related as $T_\mathrm{new}=k_\mathrm{B}T$. The new temperature scale then has units of energy.

  • 3
    $\begingroup$ The "unitless entropy" should have some sort of unit even if it's not mathematically required. Like how an angle is 1 radian, not just 1 (even though we do not use radian units in the context of pure calculation, you still need to specify radians in order to understand what the number 1 actually means). $\endgroup$ – user253751 Feb 27 '20 at 11:16
  • 1
    $\begingroup$ @user253751 In the "unitless" case the entropy is the logarithm of the number of states, so in a mathematical sense it does not have units. I wouldn't know what to call the units in this case. $\endgroup$ – Buck Thorn Feb 27 '20 at 15:10
  • 2
    $\begingroup$ Entropy does have units, and the unit is based on the logarithmic base, e.g.: en.wikipedia.org/wiki/Nat_(unit) $\endgroup$ – gardenhead Feb 27 '20 at 15:33
  • 2
    $\begingroup$ ...and I guess most of the readers have heard about the unit of entropy if the logarithm is base-2, which is called bit. $\endgroup$ – JiK Feb 27 '20 at 17:13

Entropy isn't "just" a measure of randomness. It is the only physical property that gives the universe a temporal direction, i.e., that provides an "arrow of time".

Also, a "measure of randomness" a crude way to characterize entropy. Rather, the entropy of a system is proportional (through Boltzmann's constant) to the log of the number of possible macroscopically indistinguishable microstates that a system in a given state can sample, weighted by the relative probabilities of those microstates.

And what determines ($k_B$ times the log of) how many microstates a system at a given temperature can sample? It is how much heat we have let flow (reversibly) into the system divided by the temperature at each point.

And that is why the units of entropy are energy/temperature, because it is the integral of energy flow/temperature that determines the number of available states, and it is the number of available states that in turn determine the entropy.

More precisely, the entropy of a system is the integral of the inverse temperature times amount of reversible heat flow needed to bring that system from absolute zero to its current temperature. I.e.:

$$S(T') =\int_{0}^{T'} dS= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$$


Not the answer you're looking for? Browse other questions tagged or ask your own question.