Re: delta function
- From: "rancid moth" <rancidmoth@xxxxxxxxx>
- Date: Fri, 29 May 2009 11:56:36 +1000
rancid moth wrote:
hello,
you see in many physics books (particularly on electromagnetism) the
following statement
int(0,oo) k*J_m(k*r)*J_m(k*r') dk = 1/r * Delta(r-r')
where J_m is the m-th order bessel function of the first kind and
Delta is the diracdelta "function". I know why they write this -
because it's the orthogonality (although that's probably not the
right word) property of the Hankel transform. I know how Titchmarsh
derives the property (although he never uses delta function
representations - which i think is much better because the delta
function, to me, obscures a lot of things - but that might just be
me), and he uses subtle arguments about bounded variation, taking
estimates of integrals using mean value theorems. great.
However in many physics books you see this statement set as a problem
- i.e. "show that...." and yet no where in the text does it mention
or even assume you know anything about Hankel transforms, bounded
variation, function spaces, etc. They must obviously have a
different method in mind - most likley something to do with
eigenfunction expansions
So i'm wondering if anyone has another way of obtaining this result
that isnt the proof found in Titchmarsh's introduction to fourier
transforms and one that is possbily more aligned to what a book on EM
theory would have in mind.
I've been thinking about this a little more....could the following integral
(Weber's second integral) be used to prove this result
int(0,oo) exp(-p*t^2)*J_m(a*t)*J_m(b*t)*t dt = 1/(2*p)
*Exp[-(a^2+b^2)/(4*p) ] * I_m(a*b/(2*p))
where I_m is the modified bessel function, noting that I_m(z) ~
Exp[z]/Sqrt[2*pi*z] z->oo
in the limit p->0 could this be shown directly to represent a delta
function? for example
1/(2*p) *Exp[-(a^2+b^2)/(4*p) ] * I_m(a*b/(2*p)) ~ 1/(sqrt(a*b*2*p))
*Exp[-(a^2+b^2-2*a*b)/(4*p) ] = 1/(sqrt(a*b*2*p)) *Exp[-(a-b)^2/(4*p) ] =
1/a * delta(a-b)
that makes sense, because that form for the delta function is essentially
Wierstrass' singular integral...which i can definately show under the
integral sign, will give the equivilent of a delta function - not that i
think an EM book is going to expect that - i think they probably just want a
well known representation.
Not particularly satisfactory however.
.
- References:
- delta function
- From: rancid moth
- delta function
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