I am reading Blundell and it's said here that if there are two systems with energies $E_1$ and $E_2$ in contact with each other and if $\Omega(E)$ represents the number of microstates for given energy $E$ then the two systems will be at equilibrium when the product $\Omega_1(E_1)\Omega_2(E_2)$ is maximised. So differentiating this with respect to either $E_1$ or $E_2$ and setting it to zero we get: $$\Omega_2(E_2)\frac{d\Omega_1(E_1)}{dE_1} + \Omega_1(E_1)\frac{d\Omega_2(E_2)}{dE_1} = 0$$ Since $dE_1 = -dE_2$ after some rearranging we get: $$\frac{d \ln\Omega_1}{dE_1} = \frac{d \ln\Omega_2}{dE_2} $$

Now it's said that since the systems are at equilibrium we can say both are at the same temperature T and we define temperature T as $$\frac{1}{k_B T}= \frac{d \ln\Omega}{dE}$$ where $k_B = 1.3807 \times 10^{-23} J K^{-1}$

My question is why exactly was the definition of temperature chosen in this way? Does equilibrium guarantee that both systems will arrive at the same temperature?


The concept of temperature is defined by the zeroth law of thermodynamics, which says that no heat will spontaneously flow between two thermal reservoirs (large systems in good internal thermal contact) at the same temperature. So any quantity that can be defined in such a way that it is always the same for any two reservoirs that will experience no mutual heat transfer will be a satisfactory definition of temperature, so we choose one that is as simple as possible. The proof you gave shows that this definition suffices-- for a fixed total energy, the quantity you define will indeed be the same for two reservoirs whenever heat does not flow, because no flow of heat is the same thing as saying you have maximized the number of microstates accessible by that energy partition between the reservoirs. Had you not maximized the number of accessible microstates, energy would repartition until you do-- i.e., heat would flow.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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