When doing integration over several variables with a constraint on the variables, one may (at least in some physics books) insert a $\delta\text{-function}$ term in the integral to account for this constraint.
For example, when calculating
$$\displaystyle\int f(x,y,z)\mathrm{d}x\mathrm{d}y\mathrm{d}z$$
subject to the constraint $$g(x,y,z)=0,$$ one may calculate instead
$$\displaystyle\int f(x,y,z)\delta\left(g(x,y,z)\right)\mathrm{d}x\mathrm{d}y\mathrm{d}z.$$
The ambiguity here is that, instead of $\delta\left(g(x,y,z)\right)$, one may as well use $\delta\left(g^2(x,y,z)\right)$, $\delta\left(k g(x,y,z)\right)$ where $k$ is a constant, or anything like these to account for the constraint $g(x,y,z)=0$.
But this leads to problems, since the result of the integration will surely be changed by using different arguments in the $\delta\text{-function}$. And we all know that the $\delta\text{-function}$ is not dimensionless.
My impression is that many physical books use the $\delta\text{-function}$ in a way similar to the above example. The most recent example I came across is in "Physical Kinetics" by Pitaevskii and Lifshitz, the last volume of the Landau series.
In their footnote to Equation (1.1) on page 3, there is a term $\delta(M\cos\theta)$ to account for the fact that the angular momentum $\mathbf{M}$ is perpendicular to the molecular axis. But then, why not simply $\delta(\cos\theta)$ instead of $\delta(M\cos\theta)$?
One may say that when using $\delta(\cos\theta)$, the dimension of result is incorrect. Though such an argument may be useful in other contexts, here for this specific example the problem is that it is not clear what dimension should the result have (the very reason for me to have this question is because I don't quite understand their Equation (1.1), but I am afraid that not many people read this book).
To be clear: I am not saying the calculations in this or other books using the $\delta\text{-function}$ in a way similar to what I show above are wrong. I am just puzzled by the ambiguity when invoking the $\delta\text{-function}$. What kind of "guideline" one should follow when translating a physical constraint into a $\delta$-function? Note that I am not asking about the properties (or rules of transformation) of the $\delta\text{-function}$.
Update: Since this is a stackexchange for physics, let me first forget about the $\delta\text{-function}$ math and ask, how would you understand and derive equation (1.1) in the book "Physical Kinetics" (please follow this link, which should be viewable by everyone)?
Best Answer
1) As OP basically notes, an $n$-dimensional delta function transforms under change of variables $f:\mathbb{R}^n \to \mathbb{R}^n$ with (the absolute value of) an inverse Jacobian
$$ \tag{1} \delta^n(f(x))~=~ \sum_{x_{(0)},f(x_{(0)})=0 }\frac{1}{|\det(\partial f(x_{(0)}))|} \delta^n(x-x_{(0)}), $$
where the sum $\sum$ is over all zeroes $x_{(0)}$ of $f$, i.e. points $x_{(0)}$ where the function $f(x_{(0)})=0$ vanishes.
2) A typical situation arises when one reparametrizes $m$ constraints
$$ \tag{2} \chi^{a}(x) ~\longrightarrow~ \chi^{\prime a}(x)~=~ \sum_{b=1}^{m} R^{a}{}_{b}(x) \chi^{b}(x) $$
in an $n$-dimensional space (say $\mathbb{R^n}$ for simplicity) with $m\leq n$. Here both sets of constraints
$$\tag{3} \chi^{1}(x)~=~0, \ldots, \chi^{m}(x)~=~0, \qquad \text{and} \qquad \chi^{\prime 1}(x)~=~0, \ldots, \chi^{\prime m}(x)~=~0,$$
are supposed to describe the same $n-m$ dimensional zero-locus set $\Sigma\subseteq\mathbb{R^n}$. Technically, let us also impose that the quadratic matrix $R^{a}{}_{b}$, and both the rectangular matrices $\partial_i \chi^{a}$ and $\partial_i \chi^{\prime a}$, have maximal rank $m$ on the zero-locus set $\Sigma$. Then the delta function of constraints transforms with an inverse Jacobian
$$ \tag{4} \delta^m(\chi^{\prime }(x)) ~=~ \frac{1}{|\det(R(x))|}\delta^m(\chi(x)). $$
3) In all branches of physics it is important to keep track of such Jacobian factors.
E.g. in the path integral formulation of gauge theories, where one imposes gauge-fixing conditions $\chi$ (also known as choice of gauge) via a delta function $\delta(\chi)$, it is important to insert a compensating Faddeev-Popov determinant (that transforms with the opposite Jacobian factor under reparametrizations of the gauge-fixing conditions), to keep the path integral independent of $\chi$.
4) As far as we can tell Ref. 1 is careful to keep the same convention throughout the book. Rather than speculate what is the most natural convention, it is more important to be consistent. Concretely, Ref. 1 is discussing volume elements for distributions of the form $$\tag{5} d\tau~=~dV d\Gamma, \qquad dV~:=~dx~dy~dz,$$ where $d\Gamma$ denote additional variables.
For a monoatomic gas, one may choose $d\Gamma=d^3p$, but one could also choose $d\Gamma=d^3v=m^{-3} d^3p$, where ${\bf p}=m{\bf v}$.
In the approximation that Ref. 1 works in, for a diatomic molecule, there is no moment of inertia (and hence no angular momentum component ${\bf M}\cdot {\bf n}=0$) in the direction ${\bf n}$ of the (symmetry) axis of the diatomic molecule. Here $|{\bf n}|=1$. Hence it is natural to choose $$\tag{6} d\Gamma~=~d^3p~d^3M \iint_{{\bf n}}\! d^2n~\delta({\bf M}\cdot {\bf n}). $$ With spherical decomposition of ${\bf M}=M{\bf m}$, where ${\bf m}\cdot {\bf n}=\cos\theta$, this becomes $$ d\Gamma~=~d^3p~M^2dM~d^2m \iint_{{\bf n}}\! d^2n ~\delta(M\cos\theta)$$ $$~=~d^3p~MdM~d^2m \int_{0}^{2\pi}\! d\varphi \int_{-1}^{1}\!d\cos\theta ~\delta(\cos\theta)$$ $$\tag{7}~=~2\pi ~d^3p~MdM~d^2m, $$ which is eq. (1.1) in Ref. 1.
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