Reconcile different definitions of temperature

Question

In classical thermodynamics, one way to define the temperature is to measure the efficiency of the Carnot cycle, i.e.
$$w = 1-frac{T_c}{T_h}$$
And in some other texts (like in the statistical mechanics notes from David Tong), temperature is defined as
$$frac{1}{T} = frac{partial S}{partial E}$$
My question is how can one reconcile this two definitions? How do we know one is equivalent to the other?

jacob1729 · Accepted Answer

There are many definitions of temperature. Here are four:
Definitions 1,2 - Thermodynamics
The zeroth law implies the existance of a value called temperature, $T$, that is equal when bodies are in thermal equilibrium. The issue is that given one scale $T$, we can construct another scale $f(T)$ that also has this property for any monotonic function $f$. The second law lets us place restrictions on what the function is and single out a special one.
Definition 1 (Kelvin) - The temperature is the unique integrating factor of the heat. That is, it is the unique function such that:
$$frac{1}{T}(dU - delta W)$$
is an exact differential. This is shown in any thermodynamics text/course to be equivalent to:
Definition 2 (Carnot) - The temperature is the unique function such that the efficiency of a reversible engine operating between reservoirs $T_H,T_C$ (which is independent of the exact construction of the engine, so long as it is reversible) is:
$$ eta = 1- T_c / T_h $$
Definitions 3,4 - Statistical Physics
All very well but what is temperature? Well statistical physics has a few more ways to define things. Firstly, we have the definition in the OP:
Definition 3 (Boltzmann) - We assume a system at fixed energy $E$ can be in one of $Omega(E)$ many microstates and assign an entropy $S(E)=log Omega(E) $. Then define:
$$ T = left(frac{dS}{dE}right)^{-1} $$
this is shown in any statistical physics course to be equivalent to:
Definition 4 (Gibbs) - For a system of known mean energy $U$ connected to a heat bath, the temperature is the Lagrange multiplier $T=1/beta$ appearing in the canonical distribution:
$$p_alpha = frac{1}{Z} e^{-E_alpha beta}$$
Equivalence of Definitions
I personally think the easiest pairs to prove equivalence for are $1iff 2, 3iff 4, 1iff 4$. OP is asking for the equivalence $2 iff 3$ but perhaps they would be happy with the $1 iff 4$ argument (since, 2 is genuinely very unwieldy whilst 3 has the unfortunate side effect that the microcanonical ensemble doesn't really allow heat transfer since its not in contact with a heat bath... this can be avoided but it boils down to proving ensemble equivalence.)
If this does satisfy the OP, then the argument begins by noting the inserting the Gibbs distribution into the Shannon entropy $S=-sum p_alpha log p_alpha$  gives:
$$ S = sum p_alpha (beta E_alpha + log Z) = beta U + log Z$$
We then find (using $dZ = -sum(beta dE_alpha + dbeta E_alpha)e^{-beta E_alpha}$ and ommitting some of the algebra):
$$ dS = beta dU - sum p_alpha beta dE_alpha $$
$$ dS = beta (dU - delta W) $$
where we identify the term $sum_alpha p_alpha dE_alpha$ as being the work done on the system (the changes in energy levels are due to the volume changing etc). So it's clear that $beta$ is $1/T$ according to the Kelvin definition.
$1iff 2$ normally happens by showing that all reversible engines have the same efficiency, then additionally just calculating the efficiency of a specific reversible engine: the ideal gas Carnot engine.

Alexander · Answer

I recommend the following great article that reviews several independent definitions of temperature, how they coincide in equilibrium and depart out of equilibrium.

Casas-Vázquez, J., and D. Jou. "Temperature in non-equilibrium states: a review of open problems and current proposals." Reports on Progress in Physics 66.11 (2003): 1937.

occd2000 · Answer

Wikipedia derives this quite nicely. Recall that the definition of efficiency (I prefer $eta$ because $w$ usually denotes work) is the amount of energy being converted to work ($Q_H-Q_C$, the difference between heat entering the system and leaving it, since the Carnot engine is ideal) divided by the total heat entering the system from the hot bath ($Q_H$):
$$
eta = frac{Q_H-Q_c}{Q_H}
$$
so compared with your expression for the efficiency:
$$
1-frac{Q_C}{Q_H}=1-frac{T_C}{T_H}
frac{T_H}{Q_H}-frac{T_c}{Q_C}=0
$$
Let us now define a new state function, entropy $S$, so that for reversible processes (such as the Carnot engine),
$$
dS=frac{dQ}{T}
$$
Recall that in the Carnot engine, heat enters or leaves the system only after it is brought to the heat bath's temperature, so for a full cycle of the engine (the integral is zero since final and initial entropy are the same for a cycle)
$$
0 = oint dS = oint frac{dQ}{T} = underbrace{int frac{dQ}{T}}_{text{Isothermal expansion}} + underbrace{int frac{dQ}{T}}_{text{Adiabatic expansion}} + underbrace{int frac{dQ}{T}}_{text{Isothermal compression}} + underbrace{int frac{dQ}{T}}_{text{Adiabatic compression}}
= frac{1}{T_H}int dQ_{H} + 0 + frac{1}{T_C}int dQ_C + 0
0 = frac{T_H}{Q_H}-frac{T_c}{Q_C}
$$
in agreement with our previous derivation (note that the negative sign on $Q_C$ is due to heat leaving the system to the cold bath).
Finally, we note that heat only enters the system when no work is being performed, so from the first law
$$
dE=dQ-dW=dQ
$$
using this in the definition of entropy and rearranging gives
$$
frac{partial S}{partial Q} = frac{partial S}{partial E} = frac{1}{T}
$$

Reconcile different definitions of temperature

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