Timeline for Spherical wave as sum of plane waves
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 17, 2022 at 22:34 | comment | added | rtmd | The final answer is: $\mathcal F[f](k\hat{\mathbf z})=\sqrt{\frac{2}{\pi}}\frac{1}{k^2-k'^2}+\frac{i}{k}\sqrt{\frac{\pi}{2}}\left(\delta(k-k')-\delta(k+k')\right)$. | |
| Feb 17, 2022 at 22:32 | comment | added | rtmd | $\frac{1}{2i}\int_0^{+\infty}e^{i(k'-k)r}dr=\frac{1}{2i}\left(\frac{1}{-i(k'-k)}+\pi\delta(k'-k)\right)=\frac{1}{2(k'-k)}-i\frac{\pi}{2}\delta(k'-k)$. $$ \int_0^{+\infty} e^{ik'r}\sin{kr} dr=\frac{1}{2(k'+k)}-\frac{1}{2(k'-k)}-i\frac{\pi}{2}\delta(k'+k)+i\frac{\pi}{2}\delta(k'-k)=\\ =\frac{k}{k^2-k'^2}+i\frac{\pi}{2}\left(\delta(k-k')-\delta(k+k')\right) $$ | |
| Feb 17, 2022 at 22:31 | comment | added | rtmd | $\int_0^{+\infty} e^{ik'r}\sin{kr} dr=\frac{1}{2i}\int_0^{+\infty}e^{i(k'+k)r}dr-\frac{1}{2i}\int_0^{+\infty}e^{i(k'-k)r}dr$. $\int_0^{+\infty}e^{i\omega t}dt=\int_{-\infty}^{+\infty}\theta(t)e^{i\omega t}dt=\frac{1}{-i\omega}+\pi\delta(\omega)$. See en.wikipedia.org/wiki/… (item 313, third version of the Fourier transform), here $\theta(t)$ is the Heaviside function. $\frac{1}{2i}\int_0^{+\infty}e^{i(k'+k)r}dr=\frac{1}{2i}\left(\frac{1}{-i(k'+k)}+\pi\delta(k'+k)\right)=\frac{1}{2(k'+k)}-i\frac{\pi}{2}\delta(k'+k)$. | |
| Sep 4, 2021 at 16:18 | comment | added | Ruslan | This answer misses the imaginary part. See my answer for the fix. | |
| May 29, 2016 at 21:14 | comment | added | user118698 | When doing the last integral one has to ensure the convergence of the integral by shifting $k'$ slightly in to the complex: $(k'+i\epsilon), \epsilon > 0$. For an incoming wave $e^{-ik'r}/r$ one has to choose $(k'-i\epsilon), \epsilon > 0$. Otherwise incoming and outgoing spherical waves would have the same Fourier transform. | |
| May 29, 2016 at 21:06 | review | Suggested edits | |||
| May 30, 2016 at 0:56 | |||||
| Dec 1, 2010 at 20:53 | comment | added | Boy S | PS, Yeha, I think that it's the same: it's similar to the antitrasform in the solution of the Poisson Equation, when you find the Green Function...! :-) | |
| Dec 1, 2010 at 20:47 | vote | accept | Boy S | ||
| Sep 8, 2021 at 15:40 | |||||
| Dec 1, 2010 at 20:34 | comment | added | Boy S | Yes, then I arrived at that consideration too. My doubt is just it, a k' purely real means no absorption, but I think that you can't explain a purely spherical wave with plane waves is very strange, because the inverse is possible...Maybe there's the way to do convergence of that integral in distrubution sense...like $\int e^{ikx}=2\pi \delta(k)$...I post the question on the mathematical section too, because that's really purely mathematical...Thank you :-) | |
| Dec 1, 2010 at 20:06 | comment | added | kennytm | @Boy: Change $\sin kr$ to $(e^{ikr}-e^{-ikr})/2i$. Then notice that $\int_0^\infty e^{iKr} dr = i/K$ (if we ignore convergence stuff. Actually the answer isn't right when $k'$ doesn't have a positive imaginary part since the integral diverges.) | |
| Dec 1, 2010 at 20:06 | comment | added | Lagerbaer | I think he wants to know how your last step of the calculation is done, i.e. the integral over $e^{ik' r}\sin kr$. | |
| Dec 1, 2010 at 19:56 | comment | added | kennytm | @Boy: What do you mean? | |
| Dec 1, 2010 at 19:53 | comment | added | Boy S | thank you! And how have you do the last passage? $1/k \int_0^{+infty} e^{ik'r}sinkrdr= \frac{1}{k^2+k'^2}$ | |
| Dec 1, 2010 at 19:31 | history | answered | kennytm | CC BY-SA 2.5 |