Dirac-delta function in spherical polar coordinates

[amjad-sh]

< Mentor Note -- thread moved from the Homework physics forums to the technical math forums >[/color]

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.

 

[Orodruin]

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.

 

[amjad-sh]

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?

 

[Orodruin]

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.

 

[vanhees71]

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
$$\delta(\cos \vartheta-\cos \vartheta')=\frac{1}{\sin \vartheta} \delta(\vartheta-\vartheta').$$