# Why isn't temperature measured in units of energy?

Temperature is the average of the kinetic energies of all molecules of a body. Then, why do we consider it a different fundamental physical quantity altogether [K], and not an alternate form of energy, with a dimensional formula derived from three initial fundamental quantities, length [L], mass [M] and time [T]?

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See also my answer to a related question: physics.stackexchange.com/questions/17551/units-and-nature/… – Steve B Dec 3 '12 at 16:10
Possible duplicate: physics.stackexchange.com/questions/60830/… (Even if it's not a duplicate, you might find the answers helpful.) – Nathaniel May 5 at 2:23
Oh, sorry, I just realised this question is older than that one. – Nathaniel May 5 at 2:24

Temperature is nothing else than energy per degree of freedom. It is purely for historical reasons that energy per degree of freedom is measured in Kelvin, and not in, say, micro-eV. It is just that these systems of units got fixed and became widely used before the statistical meaning of temperature became clear.

For the same reason, mass measured in kg and not in, say, Tera-eV.

If you would correct all of this, and apply more rational choices of units, you would end up with a natural system of units. This is what many physicists do in their professional lives. In such a system constants like the speed of light and Boltzmann's constant end up as being defined equal to unity. This makes it clear that these are not constants of nature, but man-made artifacts caused by the use of clumsy systems of units. In that respect Boltzmann's constant k is no different than the constant measuring the number of cubic inches in a gallon.

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Particle physicists indeed do like to express temperature in the units of energy, usually in their most favorite unit of energy, a gigaelectronvolt.

Kelvins are used for historical reasons as well as for the sake of having reasonable numbers in everyday conditions. Before the temperature-energy relationship $$E \sim kT$$ was realized in the late 19th century ($k=1.38\times 10^{-23} {\rm J/K}$), people used various temperature scales such as the Celsius scale (they still do). The Celsius scale divides the interval between the freezing and boiling points of water to 100 parts (percent of the length of the interval). The linearity is given by the volume of an ideal gas at fixed pressure, $V\sim T$.

Later, it was appreciated that the freezing point isn't a terrible fundamental special temperature worth the label "zero", so a shifted Celsius scale, the Kelvin scale, was defined. It's still convenient to use the Celsius degrees because it's sensible to talk about temperatures near 0 and not 300 kelvins.

Today, we know it's sensible to express the temperature via the equivalent energy per degree of freedom (times two), $E=kT$, but it's still reasonable to use kelvins and Celsius degrees because the room temperature is of order $10^{-20}$ joules.

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From the phenomenological point of view of thermodynamics, the unit of temperature is somewhat arbitrary and has been fixed by historical 'accident' - as long as $$Q = T\cdot\Delta S$$ ends up being an energy, everything works out.

From the point of view of statistical mechanics, it makes sense to make the entropy $S$ unitless, same as the probabilities in terms of which it is defined.

This corresponds to setting Boltzmann's constant $k_B=1$ and makes temperature an energy. This as well as further unifications (like setting speed of light $c=1$ to unify the dimensions of space and time as well as mass and energy) are features of 'natural' systems of units.

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You actually explained why already by yourself and well done for noticing the problem. Dimensional analysis does show that temperature and energy are not the same.

Just to convince ourselves of this, lets look at the ideal gas equation for a single mole, (i.e. number moles n=1 below) of ideal gas:

$$pV=nRT=RT$$

With $nR=8.31\mathrm{m^2kg s^{-2}K^{-1}}$ looking like it has the units we use for energy over those for temperature (i.e. the units are $\mathrm{\text{ } m^2kg s^{-2}K^{-1}}=JK^{-1}$).

$$\frac{pV}{R}=T$$

What happens to the units now? The units in this expression do not amount tot units of energy. This means $\text{Energy}\neq\text{Temperature}$ and is what we expected.

As others have said, in various ways; temperature is a function of state. The sum of variables the function of state depends on tells us the number of degrees of freedom it has. These are represented by the number of invariables in the function of the state. i.e. general form:

$$g(x_1,x_2,..,x_n)$$

Where subscript n, is a positive integer which gives the degree of freedom. The Empirical temperature is a single valued result found at a particular time (i.e. so we have a result that is independent of time) when we examine our function of state. Look at the function of state variables volume, V and pressure, p which are represented as follows:

$$\Phi(p,V)=T_{empirical}$$

Proof:

Proof for existence of temperature (derivation is from C. J. Adkins equilibrium thermodynamics and paraphrased a bit here to illustrate the concept) added for anyone who wants to convince themselves of what I just said.

The condition for thermal equilibrium is described by the zeroth law of thermodynamics that is:

If two bodies A and B are in equilibrium with a third, C then they are also in equilibrium with each other.

Consider a simple example where we model a thermodynamic process involving fluid of fixed mass (i.e. independent of mass).

The condition for thermodynamic equilibrium for function, F of two arbitrary states,with body A at state 1 and and body C at state 3. (You may have plotted a p/V indicator diagram so it will help you to think of this now)

$$F_1=(p_1,V_1,p_3,V_3)=0$$

Next the condition for equilibrium for body B at state 2 and C at 3.

$$F_2=(p_2,V_2,p_3,V_3)=0$$

Solving both for $p_3$

$$p_3=f_1(p_1,V_1,V_3)=0$$

$$p_3=f_2(p_2,V_2,V_3)=0$$

and equating the two:

$$p_3=f_1(p_1,V_1,V_3)=f_2(p_2,V_2,V_3)$$

Solve for $p_1$. The reasoning for this change of variables did actually confuse me for a little while, but it is again just a matter of being comfortable manipulating expressions.

$$p_1=g(V_1,p_2,V_2,V_3)$$

$$F_3=(p_1,V_1,p_2,V_2)=0$$

$$p_1=f_3(V_1,p_2,V_2)$$

$$p_3=f_1(p_1,V_1,V_3)=f_2(p_2,V_2,V_3)$$

$$p_3=f_1(p_1,V_1)=f_2(p_2,V_2)$$

$$\Phi_1(p_1,V_1)=\Phi_2(p_2,V_2)$$

Which gives the conditions for thermal equilibrium the existence of temperature as a function of p and V.

Hence the result expressed earlier.

$$\Phi(p,V)=T_{empirical}$$

This should convince you that temperature and energy are not the same thing!

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Downvoter care to explain? – Magpie May 5 at 6:41
Concerning the first question, I think the first dimensional argument why temperature and energy are not the same is a good one, and indeed it could be that this is all what the OP wanted since the question is titled "Dimensional Analysis". – Dilaton May 8 at 11:05
the rest confuses me a little bit. For example as Johannes said in his answer, temperature is the kinetic energy of a degree of freedom, so I think it should not depend on the number of degrees of freedom as I see it. In a system with different kinds of degrees of freedom in extreme nonequilibrium each degree of freedom can have its own temperature. In theormodynamics and statistical mechanics, the temperature is the intensive quantitiy (it does not depend on the size of the system) which corresponds to the energy of the system, so they are related. – Dilaton May 8 at 11:10
When deriving the equilibrium state by maximazing the entropy, the temperature is the Lagrange multiplier corresponding to the energy. – Dilaton May 8 at 11:10
I saw ;-) I do also see what you mean now about the degrees of freedom bit now. I think I misused the word depend there. I'll change it. – Magpie May 9 at 20:57

Kinetic temperature is the average of the kinetic energies. The thermodynamic concept of temperature $T$ is more general.

Temperature measures the partial ratio of energy to entropy changes $T\equiv (\partial E / \partial S)$. Temperature cannot be considered an equivalent of energy, because the concept of entropy is needed for the definition as well. Notice from the definition that energy is an extensive quantity whereas temperature is an intensive quantity. It is possible to have a composite system with a given energy, but without a thermodynamic temperature (e.g. a composite system made of two thermally isolated solids: one hot and the other cold)

Using the fundamental equation of thermodynamics we find that $T=T(E,V,N,\dots)$, which implies that the concept of temperature surpasses that of energy. For instance, temperature can change whereas energy remain constant.

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