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If we set the Boltzmann constant to $1$, then entropy would just be $\ln \Omega$, temperature would be measured in $\text{joules}$ ($\,\text{J}\,$), and average kinetic energy would be an integer times $\frac{T}{2}$. Why do we need separate units for temperature and energy?

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Better yet, we could express it in terms of the Planck temperature. Then it would be unitless. –  Ben Crowell Apr 12 '13 at 0:59
    
Because it is like that. –  ABC Apr 12 '13 at 1:45
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You could reason the same for everything. Mass in terms of the Planck mass ($m_P\frac{\hbar}{\ell_P c_0^2}$) is unitless. Action in terms of the Reduced Planck's constant $\hbar$ or the Planck's constant i s unitless. Energy in terms of the Planck Energy is unitless. Speed iin terms of the speed of light ($\frac{\ell_P}{t_P}$) is unitless, distance/length in terms of the planck length/string length is unitless, time in terms of the planck time/string length divided by the speed of light is unitless, entropy in terms of the boltzmann's constant $k_B$ (this is what yoou are doing) etc... . –  Dimensio1n0 Jun 22 '13 at 15:23

6 Answers 6

One reason you might think $T$ should be measured in Joules is the idea that temperature is the average energy per degree of freedom in a system. However, this is only an approximation. That definition would correspond to something proportional to $\frac{U}{S}$ rather than $\frac{\partial U}{\partial S}$, which is the real definition. The approximation holds in cases where the number of degrees of freedom doesn't depend much on the amount of energy in the system, but for quantum systems, particularly at low temperatures, there can be quite a bit of dependence.

If you accept that $T$ is defined as $\frac{\partial U}{\partial S}$ then the question is about whether we should treat entropy as a dimensionless quantity. This is certainly possible, as you say.

But for me there's a very good practical reason not to do that: temperature is not an energy, in the sense that it doesn't, in general, make sense to add the temperature to the internal energy of a system or set them equal. Units are a useful tool for preventing you from accidentally trying to do such a thing.

In special relativity, for example, it makes sense to set $c=1$ because then it does make sense to set a distance equal to a time. By doing that, you're simply saying that the path between two points is light-like.

But $T=\frac{\partial U}{\partial S}$ measures the change in energy with respect to entropy. Entropy and energy are extensive quantities, whereas temperature is an intensive one. This means that it doesn't very often make sense to equate them without also including some non-constant factor relating to the system's size. For this reason, it's very useful to keep Boltzmann's constant around.

My personal favorite way to do it is to measure entropy in bits, so that $k_B = \frac{1}{\ln 2} \,\mathrm{bits}$ and the units of temperature are $\mathrm{J\cdot bits^{-1}}$. Having entropy rather than temperature as the quantity with the fundamental unit tends to make it much clearer what's going on, and bits are a pretty convenient unit in terms of building an intuition about the relationship to probability theory.

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So the conversion is $1\ \mathrm{K}\approx2\times10^{-23}\ \mathrm{J/bit}$. Neat! –  Michael Brown Apr 12 '13 at 4:41
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@MichaelBrown sorry, I made a slight mistake, $k_B$ is $1/\ln 2$ (because then $S=\frac{\ln \Omega}{\ln 2}\,\mathrm{bits} = \log_2 \Omega\,\, \mathrm{bits}$), so the conversion factor is $k_B\ln 2 \approx 8\times 10^{-24}$. –  Nathaniel Apr 12 '13 at 8:00
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Additionally, for quantities like $\Delta S$ in chemistry, $1\,\mathrm{J/K/mole} \approx 5 \,\mathrm{bits/molecule}$. –  Nathaniel Apr 12 '13 at 8:00
    
The average of a unit cannot be expressed in the same units, if the average is actually a density measure! The average mass of a sack of potatoes is still measured in kilograms, right? But density is also a kind of "average" mass: the average mass of a unit of volume. And that cannot be measured in kilograms any more, because then we have to assume a hidden volume unit, which will then be terribly misleading in all the calculations involving density. The arithmetic of units will no longer follow the manipulations of formulas. –  Kaz Apr 12 '13 at 17:30

Temperature cannot be measured in units reserved for energy because, for instance, a grain of sand heated to the temperature as the Sun does not contain the same amount of energy as the Sun.

Temperature is the property that, when two bodies in thermal contact have the same value of it, no net heat flows from one body to the other: they are in thermal equilibrium.

If you try to equate this property with energy, and thus assign it Joule units, it isn't physically correct.

Two bodies that do not exchange heat due to being in thermal equilibrium are not isoenergetic. One could contain way more energy than the other.

But in fact, temperature is related to energy. In a (very simplified, idealized) particle model of a body of matter, temperature indicates the average kinetic energy of a particle of that body. That also directly tells you that it cannot have energy units: a phenomenon which corresponds to a density measure of energy (energy per unit of mass or volume, or per particle) cannot be measured in energy units.

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+1 clean and to the point –  Mr.Mindor Apr 12 '13 at 14:36
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I agree in principle, but keep in mind that if equipartition holds, measuring temperature in units of energy does make sense: it corresponds to average energy per degree of freedom, and as the latter is a unit-less cardinal number, the densities end up with the same unit –  Christoph Apr 12 '13 at 17:31
    
I disagree. You imply that if two things are measured in the same unit, then they have to be exactly the same thing. Therefore, you say, a unit system that measures temperature in J is not "correct". Well, I think that a knowledgeable person can understand that temperature is not exactly the same thing as total energy content, but can still measure both in J. (The advantage of using kelvin IMHO is not that it's more "correct", but to reduce the frequency of stupid confused mistakes and miscommunications -- which is still worthwhile!!) –  Steve B Apr 12 '13 at 21:12
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You can define temperature as "The temperature of a body is the energy of an ideal 1D classical simple harmonic oscillator in thermal equilibrium with that body." Then there is no "per degree of freedom" in the definition -- we're just talking about an energy. –  Steve B Apr 12 '13 at 21:15
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@Kaz: degree of freedom is not used as a unit - we just count them and divide the energy by that number –  Christoph Apr 12 '13 at 22:20

I've seen temperature being expressed in electron volts (eV) in Plasma Physics. Basically, you can equate $k_B T = e y$, where $y$ is the temperature in electron volts, and $T$ is the "thermal" temperature in Kelvin. $e$ is the quantum of charge and $k_B$ is the Boltzmann factor. So $1\mbox{ eV temperature} \approx 11600 \mbox{ K}$. ($y$ was set to 1 to obtain the expression). I guess it is convenient in Plasma Physics because of the energy scales involved.

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In practical life it boils down to tradition and convenience. In fundamental physics you are most certainly free to measure temperature in energy units. In fact it's better that way.

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SI units existed before statistical thermodynamics came along. Not a bad set of units, not the best set of units. Simply entrenched. (I have a weakness for FFF units)

But imagine listening to the radio: "it's a sunny four point one times ten to the negative 21 Joules in the metro area today". Bit wordy, wouldn't you say?

But Joules, and not eV? Frankly, I like books that set c=1 and drop many constants, but then it becomes hell trying to remember where the constants would be if I need the equation in other units.

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Suppose you want to use dimensional analysis to form a hypothesis about a relationship between two physical quantities. Let's say you are trying to find a formula relating unknowns $A$ and $B$ and you know there is a temperature T and an energy E involved. Suppose you use dimensions to hypothesis $A = F(T, E, B)$. If temperature and energy have the same dimensions then then the relation $A = g(E/T)F(T, E, B)$ will also be dimensionally valid for any mathematical function $g$. If you use SI dimensions then $A = g(E/kT)F(T, E, B)$ is also possible. However if you have a priori reasons for believing Boltzmann's constant isn't involved (eg. you're working with purely macroscopic quantities) then you can eliminate this possibility. You can't easily eliminate this possibility if temperature and energy have the same dimensions because you can't see where $k$ would have been involved.

Dimensions are a powerful tool. They allow you to detect errors and propose plausible hypotheses. They tell you about scaling properties and occasionally they even allow you to do better than get plausible hypotheses and allow you to prove useful theorems. Often the dimensions of a quantity can give helpful intuition about it. Not everyone wants to throw away this tool by setting all of their constants to 1. But sometimes it's more convenient to do eliminate a dimension. It all depends on what you are doing.

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