A partition function problem under conserved total momentum, involving Gaussian integration under a delta constraint

Question

Recently I learned about the following expression showing the kinetic part of the partition function in an N-atom ideal gas under conserved total momentum:

\begin{aligned} Q_{\mathrm{Kin}}^{\mathrm{CM}} & =\int \mathrm{d} \mathbf{p}^{d N} \prod_{i=1}^N \exp \left[-\left(\beta / 2 m_i\right) p_i^2\right] \delta\left(\sum_{i=1}^N \mathbf{p}_i\right) \\ & =\left(\frac{\beta}{2 \pi M}\right)^{d / 2} \prod_{i=1}^N\left(\frac{2 \pi m_i}{\beta}\right)^{d / 2} \\ \end{aligned} (1)

where $M=\sum_{i=1}^N m_i$

Although in physics, I was able to rationalize this result by showing that the system's degrees of freedom is decreased by one because of a conserved total momentum, and therefore the partition function should be divided by $(2\pi M/\beta)^{d/2}$, I had a hard time doing this integration mathematically.

Transforming all $\mathbf{p}_i$ into the relative momentums $\mathbf{p}_i'=\mathbf{p}_i-(m_i/M)\mathbf{P}$, where $\mathbf{P}=\sum_{i=1}^N \mathbf{p}_i$ seems to be the first idea, but the Jacobian for such a transformation does not seem to be 1, neither will the sum in the $\delta$ function be trivially expressed.

Another idea is to transform all $\mathbf{p}_i'=\mathbf{p}_i$ when $i=1\to N-1$ while $\mathbf{p}_N'=\sum_{i=1}^N \mathbf{p}_i$. This transformation is area-conserving, and after which the integrand becomes:

\begin{aligned} Q_{\mathrm{Kin}}^{\mathrm{CM}} & =\int \mathrm{d} \mathbf{p}'^{d N} \exp \left\{-\beta / 2 \left[\sum_{i=1}^{N-1} p_i^{'2}/m_i + \left(\mathbf{p}'_N - \sum_{i=1}^{N-1} \mathbf{p}'_i\right)^2/m_N \right] \right\} \delta\left(\mathbf{p}'_N\right)\\ &=\int \mathrm{d} \mathbf{p}'^{d N -3} \exp \left\{-\beta / 2 \left[\sum_{i=1}^{N-1} p_i^{'2}/m_i + \left( \sum_{i=1}^{N-1} \mathbf{p}'_i\right)^2/m_N \right] \right\} \end{aligned} (2)

The quadratic form within the exponent is not trivial to diagonalize neither.

My question is: Would there be a neat and rigorous way to derive the result in equation (1)? Your ideas will be highly appreciated!

Answer 1

Here, the "trick" consists in replacing the Dirac delta by its integral representation, namely $$ \delta(\mathbf{P}) = \int_{\mathbb{R}^d} \frac{\mathrm{d}^d\mathbf{s}}{(2\pi)^d}\, \exp(i\mathbf{P}\cdot\mathbf{s}) $$ in the present case, so that $$ \begin{align} Q_{kin}^{CM} &= \int_{\mathbb{R}^d}\frac{\mathrm{d}^d\mathbf{s}}{(2\pi)^d} \prod_{i=1}^N \int_{\mathbb{R}^d}\mathrm{d}^d\mathbf{p}_i \exp\left(-\frac{\beta}{2m_i}\mathbf{p}_i^2 + i\mathbf{p}_i\cdot\mathbf{s}\right) \\ &= \int_{\mathbb{R}^d}\frac{\mathrm{d}^d\mathbf{s}}{(2\pi)^d} \prod_{i=1}^N \left(\frac{2\pi m_i}{\beta}\right)^{d/2} \exp\left(-\frac{m_i}{2\beta}\mathbf{s}^2\right) \\ &= \int_{\mathbb{R}^d}\frac{\mathrm{d}^d\mathbf{s}}{(2\pi)^d}\, \exp\left(-\frac{M}{2\beta}\mathbf{s}^2\right) \prod_{i=1}^N \left(\frac{2\pi m_i}{\beta}\right)^{d/2} \\ &= \left(\frac{\beta}{2\pi M}\right)^{d/2} \prod_{i=1}^N \left(\frac{2\pi m_i}{\beta}\right)^{d/2} \end{align} $$

Answer 2

Many thanks to @Abezhiko! Besides, I have found a less elegant solution if you want to reference.

We know that for any symmetric PSD matrix $\mathbf{A}$, we have:

\begin{equation} \int \mathrm{d}^n \mathbf{x}\ \mathrm{exp}(-\mathbf{x}^T \mathbf{A} \mathbf{x}) = \sqrt{\frac{(2\pi)^n}{\mathrm{det(\mathbf{A})}}} \end{equation}

The quadratic form in my equation (2) corresponds to the following $(N-1) \times (N-1)$ matrix:

\begin{equation} \mathbf{A} = \left( \begin{matrix} m_1^{-1} + m_N^{-1} & m_N^{-1} & m_N^{-1} & \cdots & m_N^{-1}\\ m_N^{-1} & m_N^{-1} + m_N^{-1} & m_N^{-1} & \cdots & m_N^{-1}\\ m_N^{-1} & \cdots & \cdots & \cdots & m_N^{-1} \\ m_N^{-1} & m_N^{-1} & m_N^{-1} & \cdots & m_{N-1}^{-1} + m_N^{-1}\\ \end{matrix} \right) \end{equation}

Thus the problem resides in computing $\mathrm{det}(\mathbf{A})$. Doing so is slightly inconvenient but should not be too hard. One subtract the 1st row from the 2nd to the (N-1)'th rows, then multiply the 2nd to the (N-1)'th columns by the corresponding factor $m_i/m_1$ respectively, and add each of the multiplied columns to the first column. The process above gives an upper-trangle matrix whose determinant should equals to

$(m_1^{-1}+m_N^{-1}+m_N^{-1}m_2/m_1+\cdots+m_N^{-1}m_{N-1}/m_1)*m_2^{-1}*\cdots*m_{N-1}^{-1} = M/(m_1 m_2 \cdots m_N)$

where $M=\sum_{i=1}^N m_i$ is the total mass.

This should give exactly the same result as equation (1).