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Re: Re: Divergence and Dirac Delta Function
*To*: mathgroup at smc.vnet.net
*Subject*: [mg9186] Re: [mg9140] Re: Divergence and Dirac Delta Function
*From*: Elvis Dieguez <elvisum at ibm.net>
*Date*: Tue, 21 Oct 1997 02:02:59 -0400
*Sender*: owner-wri-mathgroup at wolfram.com
Paul Abbott wrote:
The package
> In[1]:= << Calculus`DiracDelta`
>
> adds rules so that you handle some derivatives and integrals involving
>
> distributions. For example,
>
> ?UnitStep
>
> UnitStep[x] is a function that is 1 for x > 0 and 0 for x <
> 0.
> UnitStep[x1, x2, ...] is 1 for (x1 > 0) && (x2 > 0) &&... and
> 0
> for (x1 < 0) || (x2 < 0) || ... .
>
> ?DiracDelta
>
> DiracDelta[x] is a distribution that is 0 for x != 0 and
> satisfies Integrate[DiracDelta[x], {x, -Infinity,Infinity}] =
> 1. DiracDelta[x1, x2, ...] is a distribution that is 0 for x1
> != 0 ||
> x2 != 0 || ... and satisfies Integrate[DiracDelta[x1, x2,
> ...],
> {x1, -Infinity, Infinity}, {x2, -Infinity, Infinity}, ...] =
> 1.
>
> To compute the divergence of a point charge with field proportional to
>
> 1/r^2 you effectively need to compute the radial derivative of a
> step-function, e.g.,
>
> In[2]:= D[UnitStep[r], r]
>
> Out[2]= DiracDelta[r]
>
> In other words, the DiracDelta arises because of a step discontinuity
> at
> r=0.
>
> You might want to also have a look at
>
>
> h
> tp://www.pd.uwa.edu.au/Physics/Courses/Second_Year/ElectroMagnetism.html
>
> In particular Chapter 3 and Appendix B.
>
> Cheers,
> Paul
I understand the theory behind the dirac delta function... however, I am
not clear as to how Mathematica treats the following divergence:
Div[1/r^2]. (Assuming of course that I am using spherical
coordinates--and that 1/r^2 is in the unit r direction). When I
compute that divergence it returns the value -> 0 but I know the
correct value is 4*Pi*DeltaFunction (because of the surface integral!).
Is there anyway of getting this result by computing the above
divergence (and not using the Unit Function). Thanks.
Elvis
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