# How come that $\int \delta(H(p,q)-E)dpdq=\Omega(E)$ not infinity?

In microcanonical ensembles we have (for one particle in 1 dimension)

$$\int \delta(H(p,q)-E)dpdq=\Omega(E)$$

I am not convinced and believe that this integral diverges. Take for example a harmonic oscillator in one dimension, with $$H(p,q)=p^2/2+q^2/2=E$$ ($$m=k=1$$ for simplicity). One can write the above integral as

$$\int f(p,q)dpdq$$

The integrand is $$f(p,q)=\infty$$ for all points $$(p,q)$$ with $$H(p,q)=E$$. Thus, one can think of $$f(p,q)$$ as a distribution with infinitely many (uncountably infinite) Dirac functions sprinkled around every point $$(p,q)$$ on the circle $$H(p,q)=p^2/2+q^2/2=E$$ (see figure). Each Dirac function will contribute 1 to the integral, but there are infinitely many of them on the circle so the integral will diverge. This argument is generalizable to higher dimensions.

I need help in clearing up my confusion as to why 1) this integral does not diverge and 2) it is proportional to $$\Omega(E)$$, i.e., the circumference of this circle.

• Note that the Dirac delta function isn't equal to infinity when its argument is zero. That's a common kind of intuition people build about the delta function, but it isn't the definition. Rather, it's simply defined in such a way that integrating it picks out the value at which its argument is zero. See this answer physics.stackexchange.com/q/371700 for a lot of detail on integrating dirac deltas with non-trivial functions as arguments. – Metropolis May 10 at 13:44
• You should also have a look at the wikipedia page, in particular the section "Properties in n dimensions". – Yvan Velenik May 10 at 13:48
• @Metropolis That is not my problem. My main contention is that there are infinitely many Dirac delta functions (one for each point $(p,q)$) on the circle, where each Dirac function which will contribute one to the total integral: $1+1+1+...=\infty$ – Omar Nagib May 10 at 14:00
• @OmarNagib No, the $\delta$ "function" does not decompose like that. Read the references suggested. – Yvan Velenik May 10 at 14:01
• that's just not how the Dirac $\delta$ works. – fqq May 10 at 14:03

The integrand is $$f(p,q)=\infty$$ for all points $$(p,q)$$ with $$H(p,q)=E$$.

No, it's not.

Thus, one can think of $$f(p,q)$$ as a distribution with infinitely many (uncountably infinite) Dirac functions sprinkled around every point $$(p,q)$$ on the circle

No, you can't.

The hand-waving tale that the Dirac delta $$\delta(x)$$ equals "zero at $$x\neq 0$$ and infinity at $$x=0$$" is just that: a hand-waving tale. It is useful for building intuition in one dimension, but that's it.

What the Dirac delta actually represents is a distribution (also sometimes known as a "generalized function"), in the formal sense of distribution theory. A distribution is a function $$\varphi: \mathscr F\to \mathbb R$$ which acts on the space of well-behaved functions $$\mathscr F$$ and assigns a real number to each function. (For example, for any function $$g$$ you can create a distribution that takes $$f$$ to $$\int_{-\infty}^\infty f(x)g(x)dx$$.) The Dirac delta distribution is the function \begin{align} \delta_0 : \mathscr F &\to \mathbb R \\ f & \mapsto \delta_0(f) = f(0) \end{align} which takes an arbitrary well-behaved function $$f$$ and returns the value of $$f$$ at the origin, $$f(0)$$.

In your integral, $$\int \delta(H(p,q)-E)dpdq,$$ the Dirac delta is being taken over a one dimensional space, and evaluated at the variable $$\epsilon = H(p,q)-E$$. If your analysis does not account for that, then it is wrong.

This is easier to handle for the specific example of the harmonic oscillator, where you're trying to calculate $$\mathrm{int}(E) = \iint \delta(\tfrac12(p^2+q^2)-E)dpdq.$$ This now has a Dirac delta evaluated over a one-dimensional energy variable which itself depends on a function of two separate integration variables $$-$$ which should look like quite a nightmare! To handle this, the thing to do is to make a suitable change of variables (in this case, to radial coordinates) that will separate the two. So, if we define $$h=\tfrac12(p^2+q^2)$$ and $$\phi = \arctan(p/q)$$ (so that $$q=\sqrt{2h}\cos(\phi)$$ and $$p=\sqrt{2h}\sin(\phi)$$, and $$dp\,dq = dh\,d\phi$$), which gives us $$\mathrm{int}(E) = \int_0^{2\pi}\int_0^\infty \delta(h-E)dh\,d\phi= \int_0^{2\pi}1 d\phi \times \int_0^\infty \delta(h-E)dh.$$ This separation shows what's really happening: the Dirac delta is indeed singular, and it is being integrated over one dimension to give a finite result. The other integration variable then gives the circumference of the ring of equal-energy states, which is what we want to calculate with $$\Omega(E)$$.

• Thank you for your detailed answer. Quick follow up. It seems to me that your last integral $\mathrm{int}(E)=2 \pi$ (for all positive $E$). I thought that it should rather be equal to the circumference of a circle with a radius $\sqrt{E}$, i.e., $\mathrm{int}(E)= \Omega(E)= 2 \pi \sqrt{E}$? – Omar Nagib May 10 at 15:50
• @OmarNagib That's appealing, but it is wrong starting with the dimensional analysis. The Dirac delta has dimensions of $[\delta(x)]=[1/x]$, which in this case gives $[\delta(h)]=[1/h]=[1/p^2]=[1/q^2]=[1/pq]$, so $\mathrm{int}(E)$ must be dimensionless. – Emilio Pisanty May 10 at 15:58
• From the statistical-mechanics side, $\mathrm{int}(E)$ is counting the number of microstates between energies $E$ and $E+dE$ (heuristically speaking), which forms a circular strip. As $E$ increases, the radius of this circular strip increases, but its width decreases proportionally. For more, see this answer. – Emilio Pisanty May 10 at 15:59
• Thank you very much. You have been very helpful. – Omar Nagib May 10 at 20:15