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Mathematical physics

  1. Mar 12, 2009 #1
    What will be Kronecker Delta Function in Cylindrical co-ordinates as well as in spherical Polar coordinates?
     
  2. jcsd
  3. Mar 12, 2009 #2

    jambaugh

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    Do you mean Dirac delta function? The Kronecker delta function is for discrete variables.

    That having been said you can view the Dirac delta function as the derivative of a step function (in one variable). In many variables you take products of delta functions for each coordinate. You then must consider how the measure changes in different coordinate systems.

    Delta functions of general functions of the variable behave as follows.
    Let [itex]u = u(x)[/itex] be a continuous function of x with [itex] u(a)=b[/itex] and [itex] u'(a) \ne 0[/tex].

    Then:
    [tex] \int f(x) \delta(u(x)-b)dx = \int f(x)\delta(u-b)\frac{du}{u'} = \frac{f(x)}{u'}|_{u=b} [/tex]
    where we use variable substitution: [itex] u=u(x), du = u'dx, dx = du/u'[/itex].

    Since the equation:
    [tex]\int f(x)\delta(u(x)-b)dx = \frac{f(a)}{u'(a)}=\int f(x)\frac{\delta(x-a)}{u'(a)}dx[\tex]
    holds for arbitrary function [itex] f(x)[/tex] we have that:
    [tex] \delta(u(x)-b) = \frac{1}{u'(a)}\delta(x-a)[/tex]
    where
    [tex] u(a) = b[/tex]
    or equivalently:
    [tex]\delta(x-a) = u'(a)\delta(u(x)-u(a))[/tex]

    Similarly if you promote x and u to coordinate vectors and work out the same argument you get:
    [tex] \delta^n(\vec{x}-\vec{a}) = \left|\frac{\partial \vec{u}}{\partial \vec{x}}\right|_{\vec{x}=\vec{a}}\delta(\vec{u}(\vec{x})-\vec{u}(\vec{a})[/tex]
    where
    [tex] \left|\frac{\partial \vec{u}}{\partial \vec{x}}\right|_{\vec{x}=\vec{a}}[/tex]
    is the Jacobian determinant at [itex]\vec{a}[/tex] (and n is the dimension.)
     
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