Dirac痴 Delta in Polar Coordinates

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    We define the Dirac Delta, δ(r-r0), through the statement that

(Eq地 1)

in which D=f(r0) when r0 lies inside the region of integration and D=0 when r0 lies outside the region of integration. We have already seen how the Dirac Delta works when r=(x,y,z), the classical rectilinear Cartesian coordinates. But now we want to see what it does when we use orthogonal curvilinear coordinates.

    Whatever we come up with, it痴 only correct if the Dirac Delta has a form that satisfies the normalization condition,

(Eq地 2)

In Cartesian coordinates in three dimensions we have

(Eq地 3)

so that we have the normalization condition as

(Eq地 4)

    Because for any generalized coordinate q we must have

(Eq地 5)

the integration over the volume element necessitates that we incorporate the appropriate scale factors into our calculation. For convenience in keeping the subscripts clear I have set q0=0. We need the scale factors to convert the units of the coordinates into units of distance and we work them out by way of the Pythagorean theorem in the form (in three dimensions)

(Eq地 6)

Dividing dqn out of that equation gives us the formula for calculating the scale factor hn. To calculate the partial derivatives in that formula we use the transformation equations that convert the q-coordinates into the equivalent Cartesian coordinates.

    For spherical polar coordinates r=(r,θ,ϕ) we have those equations as

(Eq地s 7)

We thus get our scale factors as

(Eq地s 8)

That result gives us our differential vector distance as

(Eq地 9)

and the associated differential volume element as

(Eq地 10)

    We see by way of that result that the Dirac Delta in spherical polar coordinates must have the form

(Eq地 11)

That statement must stand true to mathematics so that the integral

(Eq地 12)

as required by Equation 5. Note that Equation 5 tells us to integrate the Dirac Delta with respect to the coordinates and not necessarily with respect to spatial distance: that痴 why we have to use the scale factors.

    Interpreting the Dirac Delta seems straightforward, certainly in the case of the Cartesian coordinates. In three-dimensional space the function δ(x) has infinite value on the y-z plane (x=0) and zero value elsewhere. The function δ(y) has infinite value on the x-z plane (y=0) and δ(z) has infinite value on the x-y plane (z=0). Taken together, as in Equation 3, the combined deltas describe a function that has infinite value at a single point, r=(0,0,0), and zero value everywhere else.

    The individual Dirac Deltas that go into Equation 11 have a similar interpretation. The azimuthal (longitudinal) function δ(ϕ) has infinite value on the plane defined by the statement ϕ=0 and has zero value elsewhere. That plane originates on the straight line passing from the south pole (θ=0) to the north pole (θ=π) of an imaginary sphere centered on the origin of the coordinate grid, the line being extended infinitely in both directions. The plane ϕ=0 thus spans half of infinite space. The colatitudinal function δ(θ) has infinite value on the south polar line (θ=0), extending from the origin of the coordinate grid out toward infinity. If we had δ(θ-θ0),then the delta would have its infinite value on the conical sheet that has its vertex on the origin of the coordinate grid and all of its generators on lines for which θ=θ0. And the Dirac Delta δ(r) has its infinite value at a single point, r=0, which coincides with the origin of the coordinate grid.

    If the function f(r) in Equation 1 has no dependence on the azimuthal coordinate ϕ, then we must modify the Dirac Delta to eliminate δ(ϕ). We have the integration as

(Eq地 13)

That calculation necessitates that

(Eq地 14)

The same reasoning tells us that when f(r) has no angular dependence at all, we have the associated Dirac Delta as

(Eq地 15)

    As a simple example of the application of the Dirac Delta, consider the situation in which we have an electric charge Q spread uniformly and with infinitesimal thickness over a spherical shell of radius r0 centered on the origin of our spherical polar coordinate grid. We want to calculate the electrostatic potential emanating from that charge at any point in space, so we know that we want to calculate

(Eq地 16)

The symmetry that we have put into the situation makes the calculation dependent only upon radial distance. For the function describing the electric charge density we have

(Eq地 17)

Our integral thus takes the form (with dV=4πr2dr)

(Eq地 18)

For all r<r0, which excludes the region manifesting the non-zero value of the Dirac Delta, the integral equals zero plus the constant of integration, which constant must satisfy the boundary conditions of the problem (in this case equaling the potential on the outside of the shell in order to avoid having an infinite electric field on the shell). That fact tells us that we have a zero electric field inside the shell. For all r>r0, which includes the region manifesting the Dirac Delta, the integral gives us

(Eq地 19)

    How did we get

(Eq地 20)

It痴 certainly correct: in any situation in which the electric charge has a spherically symmetric distribution, we must calculate the potential as if all of the charge occupied a single point at the center of the sphere of symmetry, as if r0=0. But if we apply Equation 1 directly, we should, we think on first impression, get an infinity.

    Or should we? Equation 1 doesn稚 actually apply to the integral in Equation 18. In this case we have, in essence, f(x)δ(x)dx: we don稚 have a difference in the argument of the Dirac Delta as Equation 1 requires. Unable to use Equation 1, we can try integrating by parts, ∫fdg=fg-∫gdf. For this calculation we have f(x)=1/x and g(x)= δ(x)dx. We thus get

(Eq地 21)

which must be wrong, because f(x) goes to infinity where δ(x) has its non-zero value.

    On page 61 (in Section 15) of The Principles of Quantum Mechanics (4th Edition) Paul Dirac wrote that any number divided by a function that can go to zero must include some constant multiple of the Dirac Delta in order to avoid having a discontinuity at that zero point. In the present case we must have

(Eq地 22)

So how do we know that our constant must equal 1/r0?

    We want to avoid having a discontinuity on our sphere, so the potential on the sphere must equal the constant, non-zero potential inside the sphere. We can figure out what that is by imagining a point-like charge e occupying the point at the center of the sphere. Relative to that charge the charge Q has potential energy

(Eq地 23)

In concept we can harness that energy through either charge, so, in accordance with the conservation of energy theorem, the point charge must have the same potential energy as does Q. That means that inside the sphere the potential conforms to

(Eq地 24)

To avoid a discontinuity the potential must have the same value on the sphere itself, so we get the constant shown in Equation 22.

    Finally we invoke Equation 1 in the form

(Eq地 25)

In the present case we have f(a)=1/r0, so we have f(x)=1/r, as shown in Equation 19.

    Thus we have the Dirac Delta in polar coordinates and some of the stranger aspects of its behavior.

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