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Delta function in spherical coords

  1. Dec 2, 2007 #1
    1. The problem statement, all variables and given/known data
    If we have a delta function in cartesian coords, how do we convert it into spherical.
    for example : delta (r) = delta(x-x0) delta(y-y0) delta(z-z0)

    2. Relevant equations

    3. The attempt at a solution
    I used
    delta (r) = delta(r-r0) delta(cos{theta}-cos{theta0}) delta (phi-phi0)/(r sin{theta})^2

    How do we find r0,cos{theta0} and phi0, if what im using is the right formula.

    do we use:

    z= r cos{theta}
    y= r sin{theta} cos {phi}
    x= r sin {theta} sin {phi}

    and then say the delta function is non-zero if x=x0 or x0=r sin {theta} sin {phi}
    and so on and then solve fir r, cos{theta} and phi?

  2. jcsd
  3. Dec 3, 2007 #2


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    Science Advisor

    I'm not sure what you mean by writing it in spherical coordinates. The Delta function is 0 if [itex]\rho[/itex] (your r) is non-zero.
  4. Dec 3, 2007 #3
    If you were working in polar coordinates for example, the 'obvious' thing to do would be to write [tex]\delta = \delta \left( r \right)[/tex] but this is incorrect since the delta function would not satisfy all of the required properties. I can't remember exactly off the top of my head but in cylindrical coordinates you define the delta function as something like [tex]\delta \left( {x,y} \right) = \frac{{\delta \left( r \right)}}{{2\pi r}}[/tex].

    I would imagine that the delta function would be defined in a similar way for spherical coordinates. My guess would be that the 2*pi*r would be replaced by 4*pi*r in spherical coordinates.
  5. Dec 3, 2007 #4
  6. Dec 3, 2007 #5


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    Science Advisor

    Lorna, your solution is correct, but you should really write the left-hand side as
    This is sometimes written as
    to emphasize that this is a 3-dimensional delta function
    Last edited: Dec 3, 2007
  7. Dec 3, 2007 #6
    thanks all
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