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How would I go about calculating the integral $ \int d^3 \mathbf r {1\over 1+ \mathbf r \cdot \mathbf r} \delta(\mathbf r - \mathbf r_0) $ where $\mathbf r_0 = (2,-1,3)$

My attempt so far: I have been trying to do this in spherical polar coordinates. ${\rm d}^3 \mathbf r = r^{2}\sin\left(\theta\right)\,{\rm d}r\,{\rm d}\theta\,{\rm d}\phi $

$\mathbf r \cdot\mathbf r = r^2 $

I think $\delta(\mathbf r - \mathbf r_0)={1\over r^2\sin\left(\theta\right)}\, \delta(r-r_0)\delta(\theta-\theta_0)\delta(\phi-\phi_0) $ ?

Therefore the integral I am attempting is: $ \int {1\over r^2 }\delta(r-r_0)\delta(\theta-\theta_0)\delta(\phi-\phi_0) dr d\theta d\phi $ Presumably the $\phi$ integral will just give $2\pi$ , and the $ \theta $ integral will just give $\pi$ ? so then we have

$2\pi^2 \int {1\over r^2 }\delta(r-r_0)\ dr $

So where do I go from here? Thanks

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  • $\begingroup$ $\large{\bf r}\cdot{\bf r} = r^{2}$. Also, $\large\int{\rm f}\left({\bf r}\right)\delta\left({\bf r} - {\bf r_{0}}\right)\,{\rm d}^{3}{\bf r} = {\rm f}\left({\bf r_{0}}\right)$ $\endgroup$ Jan 12, 2014 at 17:38
  • $\begingroup$ Ok thanks. I've edited the question to include this. Obviously that is correct - I think I confused myself. $\endgroup$ Jan 12, 2014 at 17:38
  • $\begingroup$ Am I correct in my thoughts on the phi and theta integrals? $\endgroup$ Jan 12, 2014 at 17:46

1 Answer 1

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Observe that $$ \int \operatorname{d}^3\pmb r\; f(\pmb r)δ(\pmb r−\pmb r_0)=f(\pmb r_0) $$ so that $$ \int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)=\frac{1}{1+|\pmb r_0|^2} $$ In spherical coordinates $(r,\theta,\phi)$, where $r \in [0,\infty)$, $\theta\in [0,\pi]$, and $\phi\in [0,2\pi)$, $$ \delta(\pmb r -\pmb r_0)=\frac{\delta(r-r_0)}{r^2}\frac{\delta(\theta-\theta_0)}{\sin^2\theta}\delta(\phi-\phi_0)\quad ~\textrm{Eq. 1 (cf. [1])}$$ and $\operatorname{d}^3\pmb r=r^2\sin^2\theta \operatorname{d}r\operatorname{d}\theta\operatorname{d}\phi$ so that $$ \int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)= \int_0^{2\pi}\operatorname{d}\phi\,\delta(\phi-\phi_0) \int_0^{\pi}\operatorname{d}\theta \delta(\theta-\theta_0)\int_0^{\infty}\operatorname{d}r \frac{1}{1+r^2}\delta(r-r_0) $$ then $$ \int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)= \int_0^{\infty}\operatorname{d}r \frac{1}{1+r^2}\delta(r-r_0)=\frac{1}{1+r_0^2} $$

Alternatively, observe that if the problem involves spherical coordinates, but with no dependence on either $\theta$ or $\phi$, one has $$ \delta(\pmb r -\pmb r_0)=\frac{\delta(r-r_0)}{4\pi r^2} $$ and $$ \int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)= \int_0^{2\pi}\operatorname{d}\phi \int_0^{\pi}\operatorname{d}\theta \sin^2\theta \int_0^{\infty}r^2\operatorname{d}r \frac{1}{1+r^2}\frac{\delta(r-r_0)}{4\pi r^2}=\frac{1}{1+r_0^2} $$

Bibliography

[1] fen.bilkent.edu.tr/~ercelebi/mp03.pdf

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  • $\begingroup$ ok great I see now. So how would my values of r0 tie in? Do I just consider the magnitude of the vector to get a final numerical answer? $\endgroup$ Jan 13, 2014 at 18:48
  • $\begingroup$ @user1887919 yes! $r_0^2=2^2+(-1)^2+3^2=14$ so the integral is $f(\pmb r_0)=\frac{1}{1+14}=\frac{1}{15}$. $\endgroup$
    – alexjo
    Jan 13, 2014 at 20:51
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    $\begingroup$ See this fen.bilkent.edu.tr/~ercelebi/mp03.pdf $\endgroup$
    – alexjo
    Oct 3, 2019 at 5:41

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