0

i have problems applying the fourier transformation. all these integrals confuse me. so here is my calculation:

$$ \Delta\phi(r)=-4\pi\delta(r) $$ $$ \text{Left hand side}:-\frac{1}{(2\pi)^{3/2}}\int_{}^{}e^{ikr}k^2\tilde{\phi}(k) dk $$ $$ \text{Right hand side}: =-\frac{4\pi}{(2\pi)^3}\int_{}^{}e^{ikr} dk=-\frac{1}{(2\pi)^{3/2}}\int_{}^{}e^{ikr}4\pi*(2\pi)^{-3/2} dk $$ $$ \Rightarrow\tilde{\phi}(k)=\frac{4\pi*(2\pi)^{-3/2}}{k^2} $$ $$ \phi(r)=\frac{1}{(2\pi)^{3/2}}\int_{}^{}e^{ikr}\tilde{\phi}(r) dk=\frac{4\pi}{(2\pi)^{3}}\int_{}^{}e^{ikr}\frac{1}{k^2} dk $$ and if i calculate this integral i dont get the expected result ($\phi(r)\sim\frac{1}{r}$)

Physiker
  • 1,469
peter mafai
  • 451

1 Answers1

3

First let me write the first equation correctly

$$\Delta\phi(\vec{r})=-4\pi\delta^{(3)}(\vec{r})\tag{1},$$

and note that $\delta^{(3)}(\vec{r})\neq\delta(r)$ (sorry I couldn't find it on the english Wikipedia). For the left hand side we have

$$\phi(\vec{r})=\frac{1}{(2\pi)^{3/2}}\int\tilde\phi(\vec{k})e^{i\vec{k}\cdot\vec{r}}d^3\vec{k}\tag{2}$$

$$\Delta\phi(\vec{r})=-\frac{1}{(2\pi)^{3/2}}\int k^2\tilde\phi(\vec{k})e^{i\vec{k}\cdot\vec{r}}d^3\vec{k},\tag{3}$$

where $k=|\vec{k}|$. On the other hand, the delta function is

$$\delta^{(3)}(\vec{r})=\frac{1}{(2\pi)^{3}}\int e^{i\vec{k}\cdot\vec{r}}d^3\vec{k}\tag{4},$$

so you get $$\tilde{\phi}(\vec{k})=\frac{4\pi(2\pi)^{3/2}}{(2\pi)^3}\frac{1}{k^2}\tag{5}.$$

Now, using $(2)$

$$\phi(\vec{r})=\frac{4\pi}{(2\pi)^3}\int\frac{1}{k^2} e^{i\vec{k}\cdot\vec{r}}d^3\vec{k}=\frac{4\pi·2\pi}{(2\pi)^3}\int_0^\pi \sin\theta ~d\theta\int_0^\infty k^2~dk \frac{1}{k^2} e^{ikr\cos\theta}=\\= \frac{1}{\pi}\int_{-1}^1 dx\int_0^\infty dk~e^{ikrx}=\frac{1}{\pi}\int_0^\infty\frac{e^{ikr}-e^{-ikr}}{ikr}~dk=\frac{2}{\pi r}\int_0^\infty\frac{\sin(kr)}{k} dk\tag{6}=\frac{1}{r},$$

where you can find the last integral here.

AFG
  • 2,363