Need help with EM/Jackson problem

[pmb_phy]

In Classical Electrodynamics - 2nd Ed., J.D. Jackson on page 50 there is a problem I need help with. Its problem 1.2 which states

The Dirac delta function in three dimensions can be taken as the improper limit as $\alpha \rightarrow 0$ of the Gaussian function

$$D(\alpha;x,y,z) = (2\pi)^{-3/2} \alpha^{-3} exp[-\frac{1}{2\alpha^2}(x^2 + y^2 + z^2)]$$

Consider a general orthogonal coordinate system specified by the surfaces, u = constant, v = constant, w = constant, with length elements du/U, dv/V, dw/W in the three perpendicular directions. Show that

$\delta$(x - x') = $\delta$(u - u')$\delta$(v - v')$\delta$(w - w')UVW

by considering the limit of the above Gaussian. Note that as $\alpha \rightarrow 0$ only the infinitesimal length element need be used for the distance between the points in the exponent.

I can't seem to get started on this one. Note that $$D(\alpha;x,y,z)$$ is the product of three Gaussian functions i.e.

$$D(\alpha;x,y,z)$$ = G(x)G(y)G(z)

The product UVW reminds me of a Jacobian but I'm not quite sure how.

Any thoughts/solutions/answers? I want to know the answer more than I want to be walked through it with hints (I have hundreds of more problems to work through besides this one which I gave up on). Thanks.

Pete

 

[dextercioby]

HINT:( )
$$\iiint_{R^{3}} \lim_{\alpha\rightarrow 0} [D(\alpha;x,y,z)] \ dx \ dy \ dz$$

$$=\iiint_{R^{3}} \lim_{\alpha\rightarrow 0} [D(\alpha;x-x',y-y',z-z')] \ d(x-x') \ d(y-y') \ d(z-z') =1$$

Now make the change of coordinates...

Daniel.

 

[reilly]

Pete -- Think of a delta function as the inverse of a volume element, so if the volume at a point changes by dx dy dz -> dx dy dz/UVW, the delta function must transform as advertised.

More detail: as alpha-> zero, the Gaussian gets very narrow, and the transformation from (x,y,z) to (u,v,w) reduces to a constant one. So, the factor x*x + y*y + z*z in the original Gaussian, becomes, from dx=du/U, (u/U)**2 + (v/V)**2 + (w/W) **2. The rest is a bit of algebra.

All a delta function cares about is its own point -- the rest be damned.

Regards,
R