If I have a system consisting of 2 Einstein solids (A and B) is it equivalent to say that maximizing the multiplicity of the system is the same as setting the temperature to $\infty$? I have two reasons to support this Idea.

1) The total multiplicity of the system can be expressed as: $$\Omega_{tot} = \Omega_A(q_A,N_A) \Omega_B(q_B,N_B)$$ Which is the product of the individual multiplicities, and $q = q_A + q_B = $ Total quanta of energy shared between the system, and $N = N_A + N_B =$ total number of quantum harmonic oscillators.

I wish to maximize the multiplicity with respect to oscillator A's total internal energy ($U_A)$ $$\left( \frac{ \partial \Omega_{tot} }{ \partial U_A } \right) = 0 \implies \left( \frac{ \partial Ln( \Omega_{tot} )}{ \partial U_A } \right) = \frac{ \Omega_{tot}'}{ \Omega_{tot}} = 0$$

This is true because Ln(x) is a monotonic function. But it should be noted that the definition of temperature is such that:

$$\frac{1}{kT} = \left( \frac{ \partial Ln( \Omega_{tot} )}{ \partial U_A } \right) = 0 \implies T= \infty$$

k here is the boltzman constant.

2) This makes some intuitive sense. On the Gaussian distribution as a function of Internal energy, the maximum multiplicity lies on the peak. The change in temperature gives a quantification of the direction of heat flow. If the maximum multiplicity is attained, then heat cannot flow any higher, it can only flow downward. This corresponds to infinite temperature.

I hope my intuition here is correct.


1 Answer 1


My intuition was correct. For more information visit APS Website.

  • $\begingroup$ link rot. please delete answer or update $\endgroup$ Commented Feb 24 at 4:01

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