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I have been looking everywhere but I am unable to prove $\delta(\vec{x}-\vec{a}) = \frac{1}{fgh}\delta(x_u-a_u)\delta(x_v-a_v) \delta(x_w-a_w)$

Where $f,g,h$ are scale factors for an orthogonal system $u,v,w$. If $\vec{a}$ lies on a degenerate coordinate then $\delta(\vec{x}-\vec{a}) = \frac{1}{fg\int hdw}\delta(x_u-a_u)\delta(x_v-a_v) \delta(x_w-a_w)$

I know that the delta function is a generalized function, and is generally used in the form $\int_{r_0\in V} f(\vec{r})\delta (\vec{r}-\vec{r_0})dV = f(\vec{r_0})$

But I am unsuccesful in using this to prove the above expressions.

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    @Willie For the latter part of your comment, could you post it as an answer.2011-04-30

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(Reposting as an answer by request of the OP.)

This is a fact about pullbacks of a distribution by functions. For a treatment, see section 7.2 of Friedlander and Joshi, Introduction to the theory of distributions.

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$F(\vec{a})=\int \delta(\vec{x}-\vec{a})\,F\,dx\,dy\,dz=\int\delta(\vec{x}-\vec{a})\,F\,|J|\,du\,dv\,dw$

where $J$ is the Jacobian, and

$\int\delta(x_u-a_u)\delta(x_v-a_v)\delta(x_w-a_w)\, F\,du\,dv\,dw=F(\vec{a}) \; .$

In other words,

$\delta(\vec{x}-\vec{a}) |J|=\delta(x_u-a_u)\delta(x_v-a_v)\delta(x_w-a_w) \; .$

For orthogonal coordinates, $J=fgh$.