The Dirac delta function

Some remarks on delta functions as related to electricity and magnetism.

First, what is the Dirac delta function?

Recall that in elementary E&M, some problems were about point charges, and some problems were about charge distributions. Those distributions could be spread out over lines, surfaces, or volumes, depending on the facts. In first-year problems the distributions were almost always given as uniform, but the discussion would indicate that the density could vary from place to place, and that the same physics concepts would apply but the math would be more challenging.

Of course the idea of a smooth distribution is clearly an approximation, given that we know about atomic structure and the graininess of matter. But given quantum mechanical uncertainties, it might be less of an approximation that it seems. In any case, we know that around the house the graininess of atomic structure will almost never be relevant, whereas trying to deal with electrical matters in terms of lots of point charges is going to be unrealistically complicated; dealing with smooth, or "averaged", distributions will clearly be much simpler. Further, for most purposes any observation will determine only an average field, and we show in this course that the average of the exact field is quite accurately given by using only an averaged source.

For these various reasons, then, we generally write the physics relations in terms of the charge densities.

This then raises the question, given that the equations use only charge densities, how do we use them to deal with a true point charge? - with an object containing a nonzero specific amount of charge in a volume which is experimentally indistinguishable from zero?

The answer is, we use the Dirac delta function. Specifically, the number density (number per volume) for a point particle at the origin is a physical model for the Dirac delta function
delta(r),
sometimes less ambiguously written as
delta³(r).
For a point particle elsewhere than the origin, say at a, we write
delta(r - a).
In either case, we postulate a "function" which has a value of zero everywhere else, but a value of infinity at some particular point.

This delta function is a "three-dimensional" delta function. Easier to indicate graphically is a "one-dimensional" Dirac delta function, usually written as
delta(x) or delta(x-a).
I believe the one-dimensional version is actually the function Dirac introduced, since he introduced it in discussing the theory of quantum mechanics. I am confident he was discussing quantum mechanics in one dimension, since the expressions being discussed would be more elaborate without being more informative if he used three-dimensional illustrations.
The three-dimensional delta function can be constructed from the one-dimensional function by writing
delta(r) = delta(x) delta(y) delta(z),
or in general
delta(V) = delta(V_x) delta(V_y) delta(V_z).
Conversely, the one-dimensional function is a slice through the three-dimensional function. (Note that a "graph" of the three-dimensional function needs four-dimensional "graph paper": three for x,y,z and one for the function value, call that axis f; the "slice" that gives the one-dimensional function is a slice in the plane defined by the f axis and one of the others, say the x-f plane as a slice in x,y,z,f space.)

The question arises, however, how drastically infinite is this function? To show how this can be relevant, consider comparing densities for a point charge of 1 unit and for a point charge of 2 units. Both are zero almost everywhere, and 2*0 is 0, which is sensible. But both have infinity as the only nonzero value of charge density, but isn't infinity the same as infinity? We need a way of saying that one delta-function density is twice the other. (And, in fact, infinity = 2*infinity is an accepted truth in most branches of math that accept infinity as a value. But 3*infinity and infinity*infinity are also infinity; how do I tell whether my infinity is twice yours, or infinity times, or only 1/infinity times, yours?)

In the case of the delta function, we answer questions like these by postulating a result for the integral of the function. Specifically,
integral,all space, delta(r) = 1,
or in one dimension
integral,-infinity to infinity, delta(x) = 1.
This is consistent with the model I introduced; the integral over all space of a number density is the number of particles, which in the case of a single point particle is the number 1.

Note that since the value of the delta function is zero almost everywhere, we don't have to integrate over all space to get this definite answer. In fact, the integral for any region that excludes getting argument zero, will give zero; the integral for any region that includes the point where the argument is zero, will give 1. (However, if the zero-argument point is on the edge of the integration region, the result is ambiguous. This is associated with the fact that the Dirac delta is not a function in the usual technical sense, since infinity is not a number in the usual sense. The technical term is that it is a "distribution"; distributions act like functions in most ways, but not in all ways. It is also true that if integrating a delta function arises in a physics problem and the zero-argument point is on the region boundary, then the physical problem probably was not well-posed, and the question should be restated in some way. Often that will mean with the region slightly shifted, or the 'point' contribution spread out, or both. Occasionally just changing the mathematical description of the region can make the delta function's treatment clear.)

However, this also means that the delta function by itself can't represent a point charge's charge density function; the integral over all space of a charge density is the total charge, not a number. So we multiply the delta function by a factor of the charge value; now the integral over all space gives the charge value times 1, or the charge, as it should.

All of this discussion is based on a discussion of delta(r) or delta(r - a), or delta(x) or delta(x-a). A delta function with some other form of argument still satisfies the description that it is zero unless the argument is zero (whether the zero vector or the number zero), and that it is infinite when the argument is zero. But how strong the infinity is, which usually relates to what value arises from an integral with this delta function in it, has to be addressed by using rules for changing variables in integrals.

To start off, consider
integral,-3 to 4, of f(x) delta(x - 2) dx.
Unless x=2, it doesn't matter what f(x) gives, since whatever-it-is will multiply zero. Hence we get the same result if we replace x with 2 in f(x). This gives
integral,-3 to 4, of f(2) delta(x - 2) dx.
(Notice that we replace x with 2 in what multiplies the delta; we do not replace x in the delta function itself. That would change the integrand to something*delta(0)=something*infinity=infinity at every point, whereas the original integrand was something*delta(x-1) giving something*zero=0 almost everywhere.) Now f(2) is a constant; it can factor out of the integral, giving
f(2) integral,-3 to 4, of delta(x - 2) dx.
Now the integral gives 1, and the answer is f(2).
This generalizes to the rule
integral,a to b, f(x) delta(x - c) dx = f(c) if a < c < b; = 0 if c < a < b or a < b < c.
If a > b we use the rule integral,a to b = - integral,b to a.
(If c = a or c = b, we go back and look at whether we are asking a realistic, well-posed problem.)

Now consider
integral,-5 to 5, cos x delta(x² - 9) dx.
Here we will have two contributions, since x² - 9 is zero for each of two values of x, and both are in the integration region. Introduce
I₁ = integral,-5 to 0, cos x delta(x² - 9) dx
and
I₂ = integral,0 to 5, cos x delta(x² - 9) dx.
Then as a first step,
I₁ = cos(-3) integral,-5 to 0, delta(x² - 9) dx.
Next we can change variables, u = x² - 9, giving du = 2 x dx and
I₁ = cos(-3) integral,16 to -9, delta(u) du/(-2 sqrt(u + 9)).
Notice that, besides having delta(u), both dx and the integration limits have changed. (The square root that replaces x must be the negative square root, because for the original integral x was negative or zero.) Notice that the order of the limits on u is backward, too; we started a sum on x at -5 and progressed to 0; the corresponding terms for u start at +16 and progress to -9. This change in order of limits introduces a sign change, so we get
I₁ = cos(-3) integral,-9 to 16, delta(u) du/(2 sqrt(u + 9))
= cos(-3)/(2 sqrt(9)) = cos(3)/6.
An additional sign check is provided by noting that
integral,-5 to 0, delta(x² - 9) dx
is the integral, in the normal sense, of a nonnegative function; it should come out nonnegative, and we obtained +1/6 for it.
Similar manipulations apply to I₂; this time there is no minus on the square root but also there is no reversal of integration order (check it out). The result is that I₂ also equals cos(3)/6 and the original integral equals cos(3)/3.

If we consider our final results as if we had obtained them from basic-form delta functions, we find
I₁ = cos(-3)/(2*3)
= integral,-5 to 0, cos x delta(x - (-3))/|2(-3)| dx
and
I₂ = integral,0 to 5, cos x delta(x - (3))/|2(3)| dx.
Generalizing leads to the relationship
delta( f(x) ) = sum,a, delta(x - a)/|f '(a)|,
where a runs through all the zeros of f and the denominator is the magnitude of the value at a of the derivative of f.

A special case is delta(m x) = 1/|m| delta(x) (if the delta is one-dimensional), and in terms of this result the general case can be obtained by another argument:
First, because we get a contribution wherever delta(0) arises, the result for delta( f(x) ) will consist of a sum over zeros of f. Second, near where any of these contributions arise, we can use approximations based on x being close to a. (If it isn't close, the contribution to the integral will be zero anyway.) Specifically, we can use a Taylor series for f(x). That gives
delta( f(x) ) = delta( [f(a) + f '(a)(x-a) + 1/2 f "(a) (x-a)² + ...] ).
But by assumption a is a zero so f(a) = 0, and (x-a)² will be negligible. Hence, near a,
delta( f(x) ) = delta ( [f '(a) (x-a)] ), which by the special case result
= delta(x-a)/|f '(a)|.

Some WARNINGS:

i. Frequently delta is written where a purist would put delta³. You can always be sure if you check out the argument; a vector argument requires that delta³ is meant. Note that this does not apply if merely there is a vector in the argument; delta(a^.x - 5) is a one-dimensional delta, the argument is a scalar even though it is in terms of vectors.

ii. If the formula for delta( f(x) ) gives you a problem such as division by zero, or if something using a delta function gives you delta(0), then you need to go back to basic relationships, and possibly to the idealizations that led to using the delta function. A physically well-posed problem should generally not have such difficulties, but taking that well-posed problem and using insufficiently-justified idealizations, or occasionally ill-advised variable choices, can bring in extraneous problems.