# Possible ambiguity in using the Dirac Delta function

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)?

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Have you ever tried to calculate an integral of $\delta(x^2)$? –  Vladimir Kalitvianski Feb 22 '13 at 18:56
I can calculate it by letting $t=x^2$, but maybe here the problem is that $\delta(x)$ (and its integration) is defined for $(-\infty,\infty)$, while $x^2$ is always $\ge0$. The point of my question is, what kind of guideline one should follow when translating a physical constraint into a $\delta\text{-function}$? –  DFJ Feb 22 '13 at 19:31
Well, then the question becomes, what do you exactly mean by "unnecessary complications" (this is actually what I meant in my original question)? What are the guidelines for avoiding "unnecessary complications"? Is $\delta(x)$ "just right" for the constraint $x=0$, while $\delta(2x)$ overly complicated? Sorry this might sound picky, but I am serious and this is really something I am curious about and uncomfortable with. –  DFJ Feb 22 '13 at 20:13
As far as I can see, they simply took $\delta(M\cos\theta)=\delta(\cos\theta)/M$. If $M=0$ is important for the integral, then this should be clearly stated from the beginning. I understand what you mean by "not getting additional factors", but isn't this a "tautology"? You see, I was asking "what should be put inside $\delta()$", and the answer is "not getting additional factors"? How do we know which are intrinsic and which are additional? –  DFJ Feb 22 '13 at 20:48
Yes, but the point is that $g$ is never unique. There is an ambiguity in its choice that's not removable by clever handling. Even your formulation is ambiguous: If $g(x,y,F(x,y))\equiv 0 \equiv g(G(y,z),y,z)$, then the integrals $\int f(x,y,F(x,y))dxdy$ and $\int f(G(y,z),y,z) dydz$ need not coincide, even though they are both "restrictions of $\int f(x,y,z)dxdydz$ to $g(x,y,z)=0$". Neither is a good candidate for that invariant. –  Emilio Pisanty Feb 22 '13 at 23:48

## 4 Answers

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.

1. 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 ${\bf p}=m{\bf v}$.

2. 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.

References:

1. Pitaevskii and Lifshitz, Physical Kinetics.
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What does $n$ mean in your $\delta^n$? And why determinant instead of just a function derivative? –  Vladimir Kalitvianski Feb 22 '13 at 21:39
@Qmechanic Thanks for answering. I have no problem understanding the mathematical manipulations here. What I am asking is not the mathematics, but the translation from "physics" to mathematics. Specifically for the footnote to equation (1.1) in "Physical Kinetics", when translating the sentence "$\mathbf{M}$ should be perpendicular to the molecular axis" into delta function, one could use $\delta(\mathbf{M}\cdot\mathbf{n})$, or $\delta(\mathbf{M}\cdot\mathbf{n}/|\mathbf{M}|)=\delta(\cos\theta)$. Which one is correct (or both?) and why? –  DFJ Feb 22 '13 at 22:41
@Qmechanic, the OP seems to mean delta functions of the type $\delta^1(g(x))$ for functions $g:\mathbb{R}^n\rightarrow\mathbb{R}$, so you the transformation you quote does not strictly apply. –  Emilio Pisanty Feb 23 '13 at 0:08
I updated the answer. –  Qmechanic Feb 23 '13 at 20:06
The second sentence of Section 4 is the key point: Rather than speculate what is the most natural convention, it is more important to be consistent. Different choice of volume element $d\tau$ by a (possibly $M$-dependent) factor leads to different choice of the distribution integrand $f$ by the corresponding inverse factor. The probability distribution should always integrate to one: $\int f~ d\tau =1$. –  Qmechanic Feb 25 '13 at 8:44

It depends what you want to calculate. As you rightly note, delta functions are not dimensionless, so that including one in your integral will change its dimensionality: you will be calculating something rather different! Most of the time this won't matter if you do it right, but you do need to think about what you want to calculate.

The integral $\int f(x,y,z)\delta(g(x,y,z))dxdydz$ may well be what you want, though of course you must note that its dimensionality will be $[f/g]\times\text{volume}$. In any application you need to look carefully at this expression and see if it matches, geometrically, what you want to calculate. This is made worse by the fact that the rate of change of $g$ across its zero may change on different parts of the surface. (This will be the case e.g. if you change $g$ for $g\times h$, where $h$ is never zero but is not a constant.) In that case the relative contribution of different parts of the surface to the integral may change, which is of course very bad!

If it is the pure surface integral, $$\int_{g=0}f\,\text d\mu,$$ that you want, then you really should "neutralize" the action of the delta function with the appropriate derivative of $g$: $$\int_{g=0}f\,\text d^2\mu=\int f(x,y,z)\,|\nabla g(x,y,z)|\delta(g(x,y,z))\,dxdydz.$$ The quantity $|\nabla g|\delta(g)$ is the geometric invariant I suspect you're really after, and the integral therefore does not depend on choice of $g$. For more information, the section Properties in $n$ dimensions if the Wikipedia article looks like a good place to start.

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Dear Emilio Pisanty. It seems to me that the integral in the last formula of your answer (v1) does depend on the choice of $g$ (unless further restrictions are put on $g$, such as, what the zero-locus set of $g$ is). For instance, if one takes $g(x,y,z)=x-a$, where $a$ is a parameter, then the integral becomes $\iint\! dy~dz~ f(a,y,z)$, which depends on $a$ for generic $f$. –  Qmechanic Feb 23 '13 at 17:02
Well, of course that will depend on $a$, since you want to restrict to the plane $x=a$. The point is that if you change $g$ to something like $g=2(x-a)$ or even $g=(x^2+x^4)(x-a)$ (for nonzero $a$), which doesn't change the surface $S:g=0$ you want to integrate over, the integral $\int_Sfd\mu$ shouldn't change. –  Emilio Pisanty Feb 23 '13 at 17:40

The Integral measure, after inserting the delta function, which is $\delta(g(x_i))\prod_i dx_i$, should have the correct dimension, I think. In this particular problem, each component of the angular momentum, if present in the integral measure, should contribute dimension [M]. Now one particular component is fixed, this component naturally appears in the delta function. As a consequence, the dimension associated to the component drops out of the integral. In the end, two degrees of freedom, two components of angular momentum, the dimension of the integral measure is [M]^2.

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If there is a constraint like $g(x,y,z)=0$, then variables $x,y,z$ are not all independent in the integral. You can express, for example, $x=F(y,z)$, so factually the integration is carried out over $y$ and $z$. The delta-function argument must be such that you obtain the same result as $$\int f(F(y,z),y,z)dydz$$.

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Yes, the result of the integral containing a delta function should equal the one with a constraint (though I have never seen anyone ever checked for this). But in some physics books the presence of delta function also amounts to change the dimension of the integral into the "correct" one (since the delta function is NOT dimensionless), for example, in the book page I mentioned above. –  DFJ Feb 22 '13 at 21:43