# Dirac-delta function in spherical polar coordinates

In summary: This is most easily seen by the substitution ##\cos \vartheta = \cos \vartheta'## that has the only solution ##\vartheta = \vartheta'##.We can then use the fact that$$\delta(\vartheta - \vartheta') = \delta(\cos \vartheta - \cos \vartheta')\frac{d \cos \vartheta}{d \vartheta} = \frac{1}{\sin \vartheta} \delta(\cos \vartheta - \cos \vartheta').$$(Note: if you are not comfortable with the notation, ##\frac{d \cos \vartheta}{d \vartheta} = - \sin \var
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Hello.I was reading recently barton's book.I reached the part corresponding to dirac-delta functions in spherical polar coordinates.
he let :##(\theta,\phi)=\Omega## such that ##f(\mathbf {r})=f(r,\Omega)##

##\int d^3r...=\int_0^{\infty}drr^2\int_0^{2π }dφ\int_0^π dθsinθ...##

define
##\int_0^{2π}dφ\int_0^πdθsinθ=\int_0^{2π}dφ\int_{-1}^{1}dcosθ=\int d\Omega##
then## δ(\mathbf {r-r^{'}})=\frac {1}{r^2} δ(r-r')δ(\Omega-\Omega^{'})##>>>(1)
where ##δ(\Omega-\Omega^{'})=δ(φ-φ')δ(cosθ-cosθ')##>>>(2)

My problem is that I really didn't get how he switched from ##\int_0^{2π }dφ\int_0^π dθsinθ## into ##\int_{-1}^{1}dcosθ##

same thing corresponding to relations (1) and (2), I didn't get how he obtained them?
If somebody can give me a hint for obtaining them? thanks.

Relevant equations

##\int d^3rf(\mathbf{r})δ(\mathbf{r})=f(0)##
where##δ(r)=δ(x)δ(y)δ(z)##
such that ##δ(r)=1/(2π)^3\int d^3kexp(i\mathbf{k}.\mathbf{r})##

## The Attempt at a Solution

I need hints to know where yo start.

Last edited by a moderator:
My problem is that I really didn't get how he switched from ##\int_0^{2π }dφ\int_0^π dθsinθ## into ##\int_{-1}^{1}dcosθ##
That is just a regular variable substitution. Note that ##d\cos(\theta) = -\sin(\theta)d\theta##. Also, the ##\phi## integral is still there.

same thing corresponding to relations (1) and (2), I didn't get how he obtained them?
If somebody can give me a hint for obtaining them?
The three-dimensional delta function should have the property
$$\int f(\vec r) \delta(\vec r - \vec r_0) dV = f(\vec r_0).$$
You know that the delta function is zero everywhere except for at ##\vec r'## so make the ansatz ##\delta(\vec r - \vec r_0) = N \delta(r-r_0)\delta(\theta - \theta_0) \delta(\phi - \phi_0)## and start computing.

Orodruin said:
The three-dimensional delta function should have the property

∫f(⃗r)δ(⃗r−⃗r0)dV=f(⃗r0).∫f(r→)δ(r→−r→0)dV=f(r→0).​

\int f(\vec r) \delta(\vec r - \vec r_0) dV = f(\vec r_0).
You know that the delta function is zero everywhere except for at ⃗r′r→′\vec r' so make the ansatz δ(⃗r−⃗r0)=Nδ(r−r0)δ(θ−θ0)δ(ϕ−ϕ0)δ(r→−r→0)=Nδ(r−r0)δ(θ−θ0)δ(ϕ−ϕ0)\delta(\vec r - \vec r_0) = N \delta(r-r_0)\delta(\theta - \theta_0) \delta(\phi - \phi_0) and start computing.

I started by the relation
##\int f(\vec r)δ(\vec r- \vec r')dV=f(\vec r')##
we have## f(\vec r)=f(r,θ,φ)##
##dV=drr^2sinθdθdφ##
##\int f(\vec r)δ(\vec r -\vec r')drr^2sinθdθdφ=f(\vec r')##
Then I let## δ(\vec r -\vec r')=A_1δ(r-r')A_2δ(θ-θ')A_3δ(φ-φ')##
then I let ##f(r,θ,φ)=f(r)f(θ)f(φ)## ... I don't know if this is allowed.
Then
##\int_0^{\infty}f(r)r^2drA_1δ(r-r')\int_0^πA_2f(θ)sin(θ)dθδ(θ-θ')\int_0^{2π}A_3f(φ)δ(φ-φ')dφ=f(r',θ',φ')##
then by doing the change of variable ##u=sinθdθ## ##\int_0^πA_2f(θ)sinθδ(θ-θ')dθ=\int_{-1}^{1}f(θ)dcosθδ(cosθ-cosθ')A_2## (*)
This will lead to ##A_1=1/r^2 and A_2=1 and A_3=1##
and then ##δ(\vec r-\vec r')=1/r^2δ(r-r')δ(φ-φ')δ(cosθ-cosθ')##
What I'm not convinced about is relation (*),how the change of variable let ##δ(θ-θ')=δ(cosθ-cosθ')## If you can clarify it to me?

No, it is not true that ##\delta(\theta - \theta') = \delta(\cos\theta - \cos\theta')## (there is a constant factor between the two). However, you can just as well assume a factor of ##\delta(\cos\theta -\cos\theta')## instead of a factor of ##\delta(\theta-\theta')##. This is in effect what you have done.

One should clarify a few further things. First of all you cannot express ##\delta^{(3)}(\vec{x})## in spherical coordinates since spherical coordinates are singular along the entire polar axis and particularly at the origin.

Second you can use the general formula
$$\delta[f(x)]=\sum_{j} \frac{1}{|f'(x_j)|} \delta(x-x_j),$$
where ##f## is a function that has only 1st-order roots ##x_j##.

Thus for ##\vartheta, \vartheta' \in ]0,\pi[## you have
$$\delta(\cos \vartheta-\cos \vartheta')=\frac{1}{\sin \vartheta} \delta(\vartheta-\vartheta').$$

## 1. What is the Dirac-delta function in spherical polar coordinates?

The Dirac-delta function in spherical polar coordinates is a mathematical function that is used to describe the distribution of mass or charge in a three-dimensional space. It is commonly denoted as δ(𝜃, 𝜑), where 𝜃 and 𝜑 are the polar and azimuthal angles, respectively.

## 2. How is the Dirac-delta function defined in spherical polar coordinates?

The Dirac-delta function in spherical polar coordinates is defined as a delta function that is multiplied by a spherical coordinate Jacobian, which takes into account the change in coordinates from Cartesian to spherical. It can be written as δ(𝜃, 𝜑) = δ(x)δ(y)δ(z)𝑟²sin𝜃, where r is the distance from the origin.

## 3. What is the physical interpretation of the Dirac-delta function in spherical polar coordinates?

The Dirac-delta function in spherical polar coordinates represents a point source of mass or charge at a specific location in a three-dimensional space. It is used to model situations where the mass or charge is concentrated at a single point rather than being distributed over a volume.

## 4. How is the Dirac-delta function used in solving problems in physics and engineering?

The Dirac-delta function in spherical polar coordinates is a powerful tool in solving problems in physics and engineering. It is commonly used in solving integral equations, differential equations, and boundary value problems. It is also used in the analysis of electric and magnetic fields, as well as in quantum mechanics and fluid dynamics.

## 5. What are some properties of the Dirac-delta function in spherical polar coordinates?

Some properties of the Dirac-delta function in spherical polar coordinates include: it is an even function, it is non-zero only at the origin, its integral over all space is equal to 1, and it satisfies the sifting property. It also has a scaling property, which allows it to be scaled by a constant factor without changing its properties.

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