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Dirac delta function in spherical cordinates

  1. Jun 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Calculate
    ##\int_{r=0}^\inf δ_r (r -r_0)\,dr##

    2. Relevant equations
    ##\int_V \delta^3(\vec{r} - \vec{r}') d\tau = 1##

    3. The attempt at a solution
    $$\int_V \delta^3(\vec{r} - \vec{r}') d\tau =
    \int_V \frac {1}{r^2 sinθ}\delta_r(r-r_0) \delta_θ (θ-θ_0) \delta_Φ (Φ-Φ_0) r^2 sinθ dr dθ dΦ = 1
    $$
    $$\int_{r=0}^ \inf \delta_r(r-r_0) dr \int_{θ=0}^{π/2}\delta_θ (θ-θ_0) dθ\int_{Φ=0}^{2π}\delta_Φ (Φ-Φ_0) dΦ = 1$$
    $$\int_{r=0}^ \inf \delta_r(r-r_0) dr = \int_{θ=0}^{π/2}\delta_θ (θ-θ_0) dθ = \int_{Φ=0}^{2π}\delta_Φ (Φ-Φ_0) dΦ = 1$$

    Is this correct?
     
  2. jcsd
  3. Jun 8, 2017 #1

    marcusl

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    I don't think it is, because you have assumed specific values [itex]\theta_0[/itex] and [itex]\Phi_0[/itex] but your problem is one-dimensional and independent of angle. I would approach it as a 1D problem and use the definition of a delta as the limit of a sequence of functions of unit area. Have you covered that in your class?
     
  4. jcsd
  5. Jun 8, 2017 #2
    No

    Are the above two steps correct even if they are not related to the current problem?
     
  6. Jun 8, 2017 #3

    marcusl

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    Only if [itex]\theta_0[/itex] and [itex]\Phi_0[/itex] lie within the limits specified in the integrals. In fact, I've not previously seen the form you used for [itex]\delta^3[/itex] in the first line under #3 of your original post.

    There is lots of information on the web about the delta as a sequence of functions. Page 4 of this link, for instance,
    https://redirect.viglink.com/?forma...p://links.uwaterloo.ca/amath731docs/delta.pdf
    contains a concise derivation of the 1D delta function that you may find helpful.
     
  7. Jun 10, 2017 #4

    rude man

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    Not sure what there is to compute. The definition of the dirac distribution (I still call it function but we must humor the pure mathematicians :rolleyes:) is ∫ f(x) δ(x - x0) dx over all x space = f(x0) in cartesian coordinates.

    In the case of cartesian coordinates, x ranges over -∞ < x < ∞ whereas in spherical coordinates, r ranges over 0 < r < ∞. So ∫f(r - r0) dr over 0 to ∞ = f(r0) and since f(r) = 1 here we simply get 1 for the answer.

    Unless the sub-r in δr is supposed to connote differentiation, in which case I plead ignorance.
     
  8. Jun 10, 2017 #5

    marcusl

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    You have quoted the answer, which is already given in the question. I assume that he needs to show how to get the answer.
     
  9. Jun 10, 2017 #6

    rude man

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    ??
     
  10. Jun 11, 2017 #7

    haruspex

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    Not sure what the expression means.
    As written, this is purely a 1-dimensional integral. The answer will depend whether r0 is positive or negative.
    If you mean it as a volume integral, ##\int_{\infty}\delta(\vec r -\vec {r_0}).\vec {dr}##, then as @rude man says, it is by definition 1.
     
  11. Jun 11, 2017 #8

    rude man

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    Hm, I didn't necessarily mean it as a volume integral and I don't see that it matters whether r0 is + or -.
    So, confusion all around! :confused:
     
  12. Jun 11, 2017 #9

    haruspex

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    You were interpreting it as r in spherical coordinates. If not as a volume integral then it must be a linear integral in some arbitrary but unspecified direction, and δ must be defined in terms of line integrals in that direction. In that case, it is a semi-infinite line and r0 could be negative, making the result zero.
     
  13. Jun 11, 2017 #10

    rude man

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    If a semi-infinite line then a negative r0 is not defined.
     
  14. Jun 11, 2017 #11

    haruspex

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    The integral range is a semi-infinite line. Just like x from 0 to ∞ in Cartesian.
     
  15. Jun 11, 2017 #12

    rude man

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    Right. So there are no negative numbers, by definition.
     
  16. Jun 11, 2017 #13

    haruspex

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    Eh? There are no negative values of r in the range, but there is nothing to stop r0 being negative.
     
  17. Jun 11, 2017 #14

    rude man

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    r0 must be in the range of r which in turn is in the range 0 to ∞, hence r0 cannot be <0.
     
  18. Jun 11, 2017 #15

    haruspex

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    I guess it depends on your view of the conventions.
    By convention, we represent points in the plane or 3-space in polars using only non-negative values of r, but that is merely a canonical form. In the same way, we pick ranges for the angles. But in plane (r, θ) coordinates I would say the point (-1, 0) exists, and it is the same as the points (1, π) and (1, 3π).
    Anyway, the whole discussion is academic. We need to know what the question intends. It makes no sense to me as it stands.
     
  19. Jun 11, 2017 #16

    rude man

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    Agreed.
     
  20. Jun 11, 2017 #17

    rude man

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    Don't agree. Different rules in Australia? :smile:
     
  21. Jun 13, 2017 #18
    Sorry for being late in replying,
    Thank you all for the replies,
    I asked this question because I thought that the rules for calculating integration in spherical coordinates are different but I was wrong.
    The above steps are correct by the definition of δ- function (Only if ##\theta_0 and \Phi_0## lie within the limits specified in the integrals).
     
  22. Jun 13, 2017 #19

    rude man

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    There is actually a small issue here still:
    Consider: f(x)δ(x - a)dx = f(a)/2 if a is either the upper or lower limit of integration. I.e. not f(a).
    So δ(r - r0)dr with limits of integration 0 and ∞ = 1/2 if r0 = 0.
     
  23. Jun 13, 2017 #20
    How do we get it?
    In Cartesian Coordinates system, if a is either lower or upper limit ,we can easily change the limit such that a is neither lower nor upper limit as the integrand is 0 for x>a or x<a because of δ - function.

    But in spherical coordinates , I can't do it for r=0 as a lower limit.
    So, is the result valid only for r=0 as a lower limit spherical or cylindrical systems?
     
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