Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A particle has two energy states having energies $E_0$ and $E_1$ with degeneracies $g_0$ and $g_1$. The respective probabilities are $p_1$ and $p_2$. What is the entropy in terms of $p_1$, $p_2$, $g_1$ and $g_2$ ?

share|cite|improve this question

closed as off-topic by Nathaniel, Emilio Pisanty, Qmechanic Sep 13 '13 at 14:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Nathaniel, Emilio Pisanty, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

See statistical entropy – Trimok Sep 12 '13 at 17:56

In most problems like this, the data $g_1$, $g_2$, $E_0$ and $E_1$ would let you calculate the state occupancy probabilities $p_1$ and $p_2$, through the Boltzmann Distribution. To do this calculation, you would also need the thermodynamic temperature (see the first equations on the Wiki page).

But here $p_1$ and $p_2$ are given to you, as though someone has already done this calculation for you.

So, ask yourself do you really need to know the $E_0$ and $E_1$?

share|cite|improve this answer
The OP does need to know the degeneracies, as there are $g_1$ states with probability $p_1$ and $g_2$ states with probability $p_2$, so the entropy has $g_1 + g_2$ terms. – Nathaniel Sep 12 '13 at 14:50
@Nathaniel OOPS! Changed the answer. Thanks – WetSavannaAnimal aka Rod Vance Sep 12 '13 at 15:16

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