In an answer to the post about Microscopic Definition of Heat and Work, Ronan says,

$$<dE> = \sum \epsilon_idp_i + p_id\epsilon_i$$

We can see that the change in average energy is partly due to a change in the distribution of probability of occurrence of microscopic state $\epsilon_i$ and partly due to a change in the eigen values $\epsilon_i$ of the N-particles microscopic eigen states.

and the first term corresponds to heat: $$TdS = \sum \epsilon_idp_i.$$ I am having a hard time imagining what it means. If the following is true:

  1. probabilities $p_i$ can not change the shape of Gibbs distribution if we assume that the new state is in equilibrium
  2. $\sum p_i = 1$

then the temperature must have risen (without any work done)! Is that correct, or can anything else happen?


1 Answer 1


When you heat a single-phase system without performing any work, then yes, the temperature must increase, which is interpreted as a change in the dispersion of energy levels.

Some alternatives are:

  • A two-phase system, described by expanding the fundamental relation $dU=T\,dS-P\,dV+\Sigma_i\mu_i\,dN_i$ to include the final term, where $\mu$ is the chemical potential and $N$ is the amount of stuff and where the temperature stays constant (in a first-order transition) even though the amount of each phase changes.

  • Doing work on a system, in which the energy levels might vary even though their dispersion is unchanged. An example is accelerating a mass, in which the speed of each constitute particle changes by a constant amount even though its speed relative to the center of mass remains the same, as does the temperature.


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