Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

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

In some of the articles which I read recently, I happen to see the following statement.

In Nonextensive statistical physics, it is inappropriate to use the original distribution $P=(p_i)$ instead of the escort distribution $P'=(p_i')$ where $p_i'=\frac{p_i^q}{\sum_j p_j^q}$ in calculating the expectation values of physical quantities.

Can someone clarify this point?

share|cite|improve this question
If you use the original distribution, you just won't get nonextensive stat phys. It's just a recipe to generate nonextensiveness. – Raskolnikov Sep 21 '11 at 11:27
Can you point to the articles? I am curious about this now. – Ron Maimon Sep 22 '11 at 6:21
For example and Chapter 9 of the book Thermodynamics of Chaotic Systems: An Introduction by C.Beck and F.Schlogl is on the escort distributions. – Ashok Sep 22 '11 at 10:29
up vote 1 down vote accepted

I don't know if this helps, but the escort distribution has a simple interpretation: you have q replicas of the system, and you are selecting to view only those instances when all q replicas have exactly the same state. This situation occurs with probability $p_i^q$, so the formula follows.

I can only imagine highly contrived situations where such a thing would be physically relevant. Perhaps you are looking at q atoms with k levels, which fluctuate statistically, but only when they are in the same level can you get electron flow through the system, because only then do you have resonant mixing of the states. Then if you restrict attention only to those cases where the electron manages to hop through the system, the average of the physical quantities of the atoms in those trials will be governed by the distribution you give.

Unfortunately, there seems to be a large literature on this, and I have no idea what physical systems it describes.

share|cite|improve this answer
As I am a mathematician, I do not understand fully. I understand your first paragraph to some extent, but I do not understand the second paragraph. Anyway, thanks a lot. – Ashok Sep 22 '11 at 10:31
@Ashok: sorry about that--- I meant only that I can't think of a real physical system where this sort of thing would be relevant. I made up a system where q atoms need to all be in the same state for something to happen, but this is contrived. – Ron Maimon Sep 22 '11 at 13:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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