When I've studied Thermodynamics I did so in Callen's book and there the author talks about temperature as a single thing, which mathematically is simply defined as:

$$T = \dfrac{\partial U}{\partial S}.$$

Now, currently I'm needing to learn a more conceptual approach to Thermodynamics and in some books I see the authors talk about "two kinds of temperature".

One is called the empirical temperature and the other is called the thermodynamic temperature. As far as I understood the thermodynamic temperature is the one I've always known, which can be defined by the equation I've mentioned.

Now there's this empirical temperature and I have no idea about what it is. The authors introduce it by talking about lots of experiments relating properties of systems to thermodynamic equilibrium. In one of the books it is said:

Let $X$ represent the value of any thermodynamic property such as the emf $\mathcal{E}$ of a thermocouple, the resistance $R$ of a resistance thermometer, or the pressure $P$ of a fixed mass of gas kept at constant volume, and $\theta$ the empirical temperature of the thermometer or of any system with which it is in thermal equilibrium. The ratio of two empirical temperatures $\theta_2$ and $\theta_1$, as determined by a particular thermometer, is defined as equal to the corresponding ratio of the values of $X$:


I don't really get what is going on here. What is this empirical temperature? What is its relation to the usual temperature? And why would anyone define something like that anyway?

  • $\begingroup$ "And why would anyone define something like that anyway?" History, and practical concerns, of course. The science of thermodynamics predates the thermodynamic notion of temperature, and even once a good abstract definition of temperature was available, finding instruments that provably represented that notion of temperature was not trivial. I'd like to write an answer to this, but I think I'm going to be to busy for a few days. $\endgroup$ Commented Mar 16, 2016 at 3:58
  • $\begingroup$ @dmckee: Absolutely nobody measures temperature like that, neither in physics nor in engineering, as far as I know. Gas thermometers and thermocouples come with calibration curves to correct for the deviations of the measured quantities from a linear relationship to thermodynamic temperature. While temperature was measured with ad-hoc methods, these definitions of temperature (melting butter, really?) have absolutely no life left outside of science history books. $\endgroup$
    – CuriousOne
    Commented Mar 16, 2016 at 4:17
  • $\begingroup$ @CuriousOne I know that nobody except pupils at school measure temperature like that but are we not allowed to answer such a question? The passage has turned up in a book and I think it valuable to discuss such points as it gives an idea as to why the current temperature scale is used. Otherwise it goes like this. "This the temperature scale that we use at present . . . . . . . . " which seems rather a blunt statement particularly since the practical realisation of the current scale in its present form was only agreed in 1954. $\endgroup$
    – Farcher
    Commented Mar 16, 2016 at 22:15
  • 1
    $\begingroup$ @Farcher: The problem is analogous to that of the definition of "force". It's defined by the acceleration of a mass, but hand a force gauge to a student and they will internalize that "force" is defined by the elongation of a spring, which is 100% false. $\endgroup$
    – CuriousOne
    Commented Mar 17, 2016 at 8:01
  • 1
    $\begingroup$ @Farcher: At what age do you give up the illusion that kids learn better when they are being told "lies to children"? I would say Kindergarten is a pretty good place to start. The problem isn't that kids can't handle the truth, the problem is that there are way too many incompetent teachers and, worse, people who prefer incompetent teachers over competent ones. $\endgroup$
    – CuriousOne
    Commented Mar 18, 2016 at 6:17

2 Answers 2


The definition of an empirical temperature is basically what the Zeroth Law of Thermodynamics does.

Let us suppose we do not have any prior knowledge about temperature. What we do know is that if we put two bodies in contact with each other they may change some thermodynamic properties -volume, for instance - of one another. When such a thing happens we say the bodies are in thermal contact. After a while the thermodynamic properties stop changing and we say the bodies are in thermal equilibrium. The Zeroth Law consists on the empirical fact that if $A$ is in thermal equilibrium with $B$ and $B$ is in thermal equilibrium with $C$, then $A$ is in thermal equilibrium with $C$. This is an equivalence relation which classify a set of bodies into subsets called equivalence classes. Each class is labeled by a number $T>0$ which we shall call temperature. The Zeroth Law allows us establish thermal equilibrium just in terms of temperature.

To understand the importance of this empirical temperature, defined through the Zeroth Law, imagine a substance - a portion of Mercury would do well - characterized by its volume. You put it in thermal contact with a body $A$, wait for thermal equilibrium and measure its volume $V_1$. Then we freely assign a temperature $T_1$ to this volume and consequently to the body A. If the substance is then put in thermal contact with body $B$ and reaches thermal equilibrium, we measure the volume $V_2$ and arbitrarily assign a temperature $T_2$. The next step is to make use of an interpolation to obtain a function $T(V)$. This is most easily done with a linear interpolation. In this case $$T=a+bV.$$ In this case $$\frac{T}{T'}=\frac{a+bV}{a+bV'}\equiv \frac{f(V)}{f(V')}.$$ Therefore this works as a thermometer.

The drawback of this empirical temperature is that it is not absolute. One can define different scales based on different physical properties, reference points or interpolations. This difficulty is overcame by the thermodynamic temperature which can be defined through a reversible thermal engine operating between two sources. Any of such engine has efficiency $$\eta_R=1-\frac{T_2}{T_1},$$ where $T_1$ and $T_2$ are the temperatures of the sources. Given the universality of this result one can for instance arbitrarily define the temperature of the cold source $T_2$, measure - mechanically - the efficiency of the engine and then the temperature $T_1$ is determined by $$T_1=\frac{T_2}{1-\eta_R}.$$ Note that there is no longer arbitrariness about the concept of temperature, except for the choice the the temperature of the cold source. Therefore it is appropriate to use as reference point which is highly reproducible anywhere. A standard choice is the triple point of water which is defined to be at $273.16\, \mathrm K$.


This is a supplement to @Diracology's answer.

If you have a perfectly calibrated thermometer that measures absolute temperature, it will give the same readings as the thermodynamic temperature under most circumstances--that is, an empirical temperature equal to the thermodynamic temperature. However, empirical temperatures are always positive (since they are proportional to volume, voltage, pressure, etc.), whereas thermodynamic temperatures can be negative--even on a Kelvin scale.

How? Let's look at the formula for thermodynamic temperature again: $$T = \frac{\partial U}{\partial S}.$$ The temperature can only be negative is adding energy to a system decreases its entropy. Seems weird, since adding energy usually increases the random motion of a system's atoms. One exception to this is inside the optical chamber of a laser.

In the simplest case, the active material of a laser is made of atoms with two energy levels: a ground state and an excited state. When the laser is off, almost all the atoms are in the ground state. This is a low entropy state since all the atoms are in the same state. When the laser turns on, energy is pumped in and atoms are elevated to the excited energy state, increasing entropy as atoms are more equally distributed between the states. Here the temperature is positive.

But, once the number of ground state atoms equals the excited atoms, entropy reaches a local maximum, so $$T = \frac{\partial U > 0}{\partial S = 0} = \textrm{undefined}$$ Odd, to say the least. Pump in more energy, and the atoms begin to concentrate in the excited state, actually lowering entropy. $$T = \frac{\partial U > 0}{\partial S < 0} < 0.$$ An extremely hot environment has a negative temperature.

This is almost a purely mathematical effect since no instrument can measure a negative temperature.


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