I have a question.
I was browsing different scripts on statistical mechanics and saw a definition of the "area of accessible phase space at energy $E$ " for a single harmonic oscillator. The author quickly corrected himself, stating, that this is more of a contour in phase space. So I calculated both, got different results and need someone explaining, where my mistake is or what is causing this discrepancy.
For simplicity let's define the Hamiltonian as: $$ H(q,p) = p^2 + q^2 $$
And the curve $\mathcal{C}$ in phase space $\Gamma=\{(q,p)\in\mathbb{R}^2\}$ is the circle, where $H=E$ with the radius $R=\sqrt{E}$.
Now the odd defintion for the mentioned area / contour given by the mentioned textbook author, where $\delta$ is the Dirac Distribution: $$ N(E) =\int\mathrm{d}\Gamma~\delta(H(q,p)-E) $$
If I remember correctly there is also a tool given by the Dirac $\delta$ to evaluate compositions with a function $f$ which will come in handy:
$$ \delta(f(x)) =\sum\limits_{i=1}^N\frac{\delta(x-x_i)}{|{f'(x_i)}|} $$
Here $x_i$ denote the roots of $f(x)$.
My calculation would now read as follows:
\begin{align*} N(E) &= \int\limits_{-\infty}^\infty \mathrm{d}q \int\limits_{-\infty}^\infty \mathrm{d}p ~ \delta(p^2+q^2-R^2) = \int\limits_{0}^\infty \mathrm{d}\rho ~ \rho ~ \delta(\rho^2-R^2) \int\limits_{0}^{2\pi} \mathrm{d}\varphi \\ &= 2\pi \frac{1}{2R} \int\limits_{0}^\infty \mathrm{d}\rho ~ \rho ~ \delta(\rho-R) =\pi \end{align*}
Here I used the know parameterization $(q,p)^T=(\rho\cos\varphi,\rho\sin\varphi)^T$. It's significant, that I get (if I'm not wrong) the area of a unit circle independent of the energy $E$.
That's completely against my intuition so I try it again with an line integral over $\mathcal{C}=\{(q,p)\in\mathbb{R}^2|q^2+p^2=R^2\}$.
$$ N(E) = \int\limits_\mathcal{C} \mathrm{d}s = \int\limits_0^{2\pi} \mathrm{d}\varphi \left| \frac{\partial \vec{s}}{\partial \varphi} \right| = \int\limits_0^{2\pi} \mathrm{d}\varphi \left| \begin{pmatrix} -R\sin\varphi \\ R\cos\varphi \end{pmatrix} \right| = 2\pi R =2\pi \sqrt{E} $$ This is of course a very simple calculation in this case of a circle.
So now my questions: These two results are different right? If yes, why thinks the author of this textbook, he could define a contour in phase space with the help of this Dirac Distribution and if he didn't intend to do that, what is the use of an "area" defined like this (with a DD)?
As I know little about the physical context that leads to the quantity $N(E)$, let me focus on the mathematical aspect. Note that
$$ \int_{\mathbb{R}^2} \mathrm{d}q\mathrm{d}p \, \delta(q^2+p^2-R^2) = \left[ \frac{\partial}{\partial \epsilon} \int_{\mathbb{R}^2} \mathrm{d}q\mathrm{d}p \, \mathbf{1}_{\{ q^2 + p^2 - R^2 \leq \epsilon \}} \right]_{\epsilon = 0}, $$
where $\mathbf{1}_{\{\cdots\}}$ is the indicator function notation, which takes value $1$ if $\cdots$ is true and value $0$ otherwise. On the other hand, the line integral in your second formulation corresponds to
$$ \int_{\mathcal{C}} \mathrm{d}s = \int_{\mathbb{R}^2} \mathrm{d}q\mathrm{d}p \, \delta(\sqrt{\smash[b]{q^2+p^2}}-R) = \left[ \frac{\partial}{\partial \epsilon} \int_{\mathbb{R}^2} \mathrm{d}q\mathrm{d}p \, \mathbf{1}_{\{ \sqrt{\smash[b]{q^2 + p^2}} - R \leq \epsilon \}} \right]_{\epsilon = 0}. $$
In both cases, the integral is supported on the contour $H(q, p) = E$, but the surface measure is weighted in a different way, i.e., $ \mathrm{d}q\mathrm{d}p \, \delta(q^2+p^2-R^2) = \mathrm{d}q\mathrm{d}p \, \frac{1}{2R} \delta(\sqrt{\smash[b]{q^2 + p^2}}-R) $. Since this "energy shell" as a hypersurface in the phase space does not alone encode the information about the surface measure, it should be specified by other means. I guess that the definition
$$ N(E) = \int \mathrm{d}\Gamma \, \delta(H(\Gamma) - E) = \left[ \frac{\partial}{\partial \epsilon} \int \mathrm{d}\Gamma \, \mathbf{1}_{\{ H(\Gamma) \leq E + \epsilon \}} \right]_{\epsilon = 0} $$
exactly serves this purpose, and perhaps the author's notion of "area" conforms to this choice.