I dont understand the last step. I’m trying to understand how equation 10 follows, especially the last delta equation

Question

I dont understand following steps of a solution where I need to find the Normalization constant $A(E,P,N)$ . The normalization is given by: $$ \int \rho(\vec x)d\vec x = 1 $$ where $d \vec x = C_N d^Np d^Nq$ and $C_N = \frac{1}{N!} \frac{1}{(2 \pi \hbar)^N}$

Constant energy E and constant momentum is given by: $$ \sum_{i=1}^{N} \frac{p_i^2}{2m} = E \qquad \sum_{i=1}^{N} p_i = P $$ The microcanonical equilibrium density is: $$ \rho (\vec x) = A(E,P,N) \space \delta(\sum_{i=1}^{N} \frac{p_i^2}{2m}-E) \space \delta (\sum_{i=1}^{N} p_i - P) $$ With $d^Npd^Nq = \prod_{i=1}^{N}dp_i dq_i$ the normalization factor is given by:

$$ 1 = A C_N \int \prod_{i=1}^{N}dp_i dq_i \delta(\sum_{i=1}^{N} \frac{p_i^2}{2m}-E) \space \delta (\sum_{i=1}^{N} p_i - P) \\ = A C_N V^N \int \prod_{i=1}^{N}d\tilde p_i \delta(\sum_{i=1}^{N} \frac{\tilde p_i^2}{2m}-\tilde E) \space \delta (\sum_{i=1}^{N} \tilde p_i)\\ = A C_N V^N (2m)^\frac{N}{2} \tilde E^{\frac{N}{2}-1} \int \prod_{i=1}^{N}dz_i \delta(\sum_{i=1}^{N} z_i^2-1) \space \delta (\sum_{i=1}^{N} z_i)\\ $$

where

$$ \tilde p_i = p_i - \frac{P}{N}, \quad \tilde E = E - \frac{P}{2mN} > 0, \quad z_i = \frac{\tilde p_i}{\sqrt{2m \tilde E}} $$

I dont understant the last delta function $\delta (\sum_{i=1}^{N} z_i)$. It should be $\delta (\sum_{i=1}^{N} \sqrt{2m \tilde E} \space z_i)$ , no?

Answer 1

You’re right; this is wrong.

The substitution of the $z_i$ for the $\tilde p_i$ as integration variables yields a factor $(2m\tilde E)^{\frac N2}$. The first delta function is $\delta\left(\sum_{i=1}^N\frac{\tilde p_i^2}{2m}-\tilde E\right)=\delta\left(\tilde E\left(\sum_{i=1}^Nz_i^2-1\right)\right)$, so replacing it by $\delta\left(\sum_{i=1}^Nz_i^2-1\right)$ yields a factor $\tilde E^{-1}$. The second delta function is $\delta\left(\sum_{i=1}^N\tilde p_i\right)=\delta\left(\sqrt{2m\tilde E}\sum_{i=1}^Nz_i\right)$, so replacing it by $\delta\left(\sum_{i=1}^Nz_i\right)$ should have yielded another factor $\left(\sqrt{2m\tilde E}\right)^{-1}$, but the factors $(2m)^\frac N2\tilde E^{\frac N2-1}$ before the integral don’t reflect this last factor. Well spotted.