I am reading this PDF talking about microcanonical ensemble : https://itp.uni-frankfurt.de/~valenti/WS13-14/all_1314_chap8.pdf

In this ensemble I know that the energy is fixed at a given $E$, we don't allow the energy to vary within an ensemble.

But, when he computes averages in this ensemble, the author allows the energy to be within $[E;E+\Delta]$. Is this $\Delta$ here for technical reasons ?

It is for example at the beginning of the paragraph 8.2 :

$$ <O>=\frac{1}{\Gamma(E,V,N)} \int \int_{E<H(p,q)<E+\Delta} d^{3N}q d^{3N}p ~ O(p,q) $$

With :

$$\Gamma(E,V,N)= \Delta \int \int d^{3N}q d^{3N}p \delta(E-H(p,q))$$

From my perspective, I would rather define :

$$<O>=\frac{1}{\Omega(E,V,N)} \int \int d^{3N}q d^{3N}p \delta(E-H(p,q)) O(p,q)$$

Where :

$$\Omega(E,V,N)=\int \int d^{3N}q d^{3N}p \delta(E-H(p,q)) $$

So for me I agree with their definition if we take $\Delta$ going to $0$, but for any non $0$ $\Delta$, the result is $\Delta$ dependant.

So why is there this $\Delta$ ? Is it only for technical purpose ? But why don't we define things like I proposed ? Am I wrong somewhere ?


When you decide to work in the microcanonical ensemble you consider the energy fixed to a value E. However, from a physical point of view, this is equivalent to assume to work in an absolutely isolated system, that is hard to find in the real world. In fact almost every system has some influences from its surroundings, however little the fluctuation of energy may be. As a consequence the energy of a known system cannot be defined sharply, and that's the reason for which the pdf considered $E \in [E, E + \Delta]$.

Of course the range over which the energy may vary is considered small in comparison with the mean value of the energy, so $\frac{\Delta}{E} = O(\tfrac{1}{\sqrt{N}}) $, where N is the number of the allowed microstates.

So, answering to your questions the $\Delta$ isn't only used for a technical purpose, but it has a physical meaning.


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.