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Dirac-delta function in spherical polar coordinates

  1. Oct 7, 2017 #1
    < Mentor Note -- thread moved from the Homework physics forums to the technical math forums >

    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})##


    3. The attempt at a solution
    I need hints to know where yo start.
     
    Last edited by a moderator: Oct 8, 2017
  2. jcsd
  3. Oct 7, 2017 #2

    Orodruin

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    That is just a regular variable substitution. Note that ##d\cos(\theta) = -\sin(\theta)d\theta##. Also, the ##\phi## integral is still there.

    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.
     
  4. Oct 11, 2017 #3
    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 dont 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?
     
  5. Oct 11, 2017 #4

    Orodruin

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    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.
     
  6. Oct 30, 2017 #5

    vanhees71

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