4
$\begingroup$

I am evaluating the integral over all space

$$\int \delta \left(r^2 - R^2\right) d \vec r$$

At first, I did this:

$$\int \delta \left(r^2 - R^2\right) d \vec r = 4 \pi \int_0^\infty \delta \left(r^2 - R^2\right) r^2 dr = 4 \pi R^2 $$

But then someone made me notice that we can use the property

$$\delta[f(r)] = \sum_i \frac{\delta (r-r_i)}{\mid f'(r_i) \mid} $$

where $r_i$ are the roots of $f$. The only root we have to consider, since $r\geq0$, is $+R$; I thus obtained

$$\delta \left(r^2 - R^2\right) = \frac{\delta(r-R)}{2R}$$

which holds the result

$$ \int \delta \left(r^2 - R^2\right) d \vec r = \frac{4 \pi}{2R} \int \delta(r-R) r^2 dr = 2 \pi R$$

Which result is the right one?

Update

I am starting to think that the first result is wrong because I am basically assuming that

$$\int_0^\infty \delta[f(x)] g(x) dx = g(x_i)$$

where $x_i$ is the only root of $f(x)$ contained in the interval $[0,\infty)$. But this is wrong because the normalization ($\mid f'(x_i) \mid^{-1}$) is missing...

$\endgroup$
  • $\begingroup$ @5xum You are right. There was also -I think- an error in my question. I have updated it. $\endgroup$ – valerio Jul 18 '16 at 9:02
  • $\begingroup$ @5xum But isn't that also true for, say, $\int_{-\infty}^{\infty} \delta(x) dx = 1$? $\endgroup$ – valerio Jul 18 '16 at 9:10
  • $\begingroup$ @5xum $\int_{-\infty}^{\infty} \delta(x) dx$ has no bounds and nothing I am integrating over, but it is not ill-defined. $\endgroup$ – valerio Jul 18 '16 at 9:12
  • $\begingroup$ @5xum Integral over all space means over $\mathbb{R}^3$, if I'm not mistaken. $\endgroup$ – pregunton Jul 18 '16 at 9:13
  • $\begingroup$ Ohhh, I missed the part "integral over all space". Dang. Sorry about that... $\endgroup$ – 5xum Jul 18 '16 at 9:13
3
$\begingroup$

Suppose you make the substitutions $u=r^2$, $U=R^2$. Then $du=2r\,dr$ and so $$4\pi \int_0^\infty \delta(r^2-R^2)\,r^2\,dr =4\pi \int_0^\infty \delta(u-U)\,\frac{\sqrt{u}}{2}\,du=2\pi \sqrt{U}=2\pi R$$ in accordance with the normalized formula. So a substitution which linearizes the argument of the delta reduces the integral to the standard case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.