Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

what is the relationship between the total power of a function given in spherical coordinates in the Fourier domain:

$E_k=\int_{\mathbb{R}^3}|F(k,\Theta,\Phi)|^2k^2 \sin(\Theta)\,dk\,d\Theta\, d\Phi$

and the total power of its Fourier pair in the ordinary domain:

$E_r=\int_{\mathbb{R}^3}|f(r,\theta,\phi)|^2r^2 \sin(\theta)\,dr\,d\theta \,d\phi?$

Thanks in advance for any answers!

share|improve this question
    
What is the relation between $F$ and $f$ (with all $2\pi$'s)? –  Fabian Apr 27 '11 at 12:37
    
$f(r, \theta, \phi) = \int_{R^3}F(k, \Theta,\Phi)e^{i\vec{k}\vec{r}}k^2sin(\Theta)dkd\Theta d\Phi$ –  Andy Apr 27 '11 at 12:55
add comment

1 Answer

up vote 2 down vote accepted

We can plug the definition $$f(r, \theta, \phi) = \int_{\mathbb{R}^3}F(k, \Theta,\Phi)e^{i\mathbf{k}\cdot\mathbf{r}}k^2 \sin(\Theta)dk\,d\Theta\, d\Phi$$ into the expression for $E_r$ and obtain $$\begin{multline} E_r=\int_{\mathbb{R}^3}dr\,d\theta \,d\phi \left[ \int_{\mathbb{R}^3}F(k, \Theta,\Phi)e^{i\mathbf{k}\cdot\mathbf{r}}k^2 \sin(\Theta)dk\,d\Theta\, d\Phi\right]\\ \times \left[\int_{\mathbb{R}^3}F(k', \Theta',\Phi')^* e^{-i\mathbf{k'}\cdot\mathbf{r}}k'^2 \sin(\Theta')dk'\,d\Theta'\, d\Phi' \right]r^2 \sin(\theta) \end{multline}.$$

Exchanging the order of integration and using the completeness relation $$\int_{\mathbb{R}^3} e^{i \mathbf{k} \cdot \mathbf{r}} d^3 r = (2\pi)^3\delta^3(\mathbf{k}) = \frac{\delta(k) \delta(\Theta) \delta(\Phi)}{k^2 \sin(\Theta)}$$ yields $$E_r= (2\pi)^3 \int_{\mathbb{R}^3} |F(k,\Theta,\Phi)|^2 k^2 \sin(\Theta)\,dk\,d\Theta\,d\Phi =(2\pi)^3 E_k.$$

share|improve this answer
    
Thank you Fabian! –  Andy Apr 27 '11 at 19:38
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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