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Hi I had posted the same post 2 days ago but I am posting it again because of my bad handwriting. I apologize to the man who wanted to read my post.

I am not familiar with the tool which is used in this site. So I use another tool to write my solution.

I don't know whether my solution is right so I wanted verify it.

Moreover, I am not able to calculate the integral in the last line which is about $dx'dy'dz'$, please let me know how I can proceed to the next step.

here is my solution

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To evaluate the integral of interest, we have

$$\begin{align} \frac{1}{(2\pi)^3}\int_{R^3}\frac{e^{i\vec k\cdot (\vec r-\vec r')}}{k^2}d^3\vec k&=\frac{1}{(2\pi)^3}\int_0^{2\pi}\int_0^\infty \int_0^\pi \frac{e^{i\vec k\cdot (\vec r-\vec r')}}{k^2} \,k^2\,\sin (\theta)\,d\theta\,dk\,d\phi \tag 1\\\\ &=\frac{1}{(2\pi)^2}\int_0^\infty \int_0^\pi e^{i k|\vec r-\vec r'|\cos(\theta)}\,\sin (\theta)\,d\theta\,dk\tag 2\\\\ &=\frac{1}{(2\pi)^2}\int_0^\infty 2\frac{\sin(k|\vec r-\vec r'|)}{k|\vec r-\vec r'|}\,dk \tag 3\\\\ &=\frac{1}{4\pi |\vec r-\vec r'|}\tag 4 \end{align}$$

NOTES:

In arriving at $(1)$ we use a spherical coordinate system $(k,\theta,\phi)$ in $\vec k$-space. Note that $d^3\vec k=k^2\,\sin(\theta)\,d\theta\,dk\,d\phi$

In going from $(1)$ to $(2)$, we rotate our $\vec k$ coordinate system so that the polar axis aligns with $\vec r-\vec r'$. Then, $\vec k\cdot (\vec r-\vec r')=k|\vec r-\vec r'|\cos(\theta)$. Noting that the integrand is independent of $\phi$, we carry out the integration over $\phi$ to produce a factor of $2\pi$.

In going from $(2)$ to $(3)$, we carry out the integration over $\theta$.

In going from $(3)$ to $(4)$, we made use of the result for the Sine Integral$\int_0^\infty \frac{sin(kR)}{kR}\,dk=\frac{\pi}{2R}$.

Had we chosen to carry out the integral in $(2)$ in the reverse order we would write

$$\begin{align} \frac{1}{(2\pi)^2}\int_0^\infty \int_0^\pi e^{i k|\vec r-\vec r'|\cos(\theta)}\,\sin (\theta)\,d\theta\,dk&=\lim_{L\to \infty}\frac{1}{(2\pi)^2}\int_0^L \int_0^\pi e^{i k|\vec r-\vec r'|\cos(\theta)}\,\sin (\theta)\,d\theta\,dk\\\\ &=\lim_{L\to \infty}\frac{1}{(2\pi)^2}\int_0^\pi \int_0^L e^{i k|\vec r-\vec r'|\cos(\theta)}\,\sin (\theta)\,dk\,d\theta\\\\ &=\lim_{L\to \infty}\frac{1}{(2\pi)^2}\int_0^\pi \frac{e^{i|\vec r-\vec r'|L\cos(\theta)}-1}{i|\vec r-\vec r'|\cos(\theta)}\,\sin(\theta)\,d\theta\\\\ &=\lim_{L\to \infty}\frac{1}{(2\pi)^2}\int_{-1}^1 \frac{e^{i|\vec r-\vec r'|L\,x}-1}{i|\vec r-\vec r'|\,x}\,dx\\\\ &=\lim_{L\to \infty}\frac{1}{(2\pi)^2}\int_{-1}^1 \frac{\sin(|\vec r-\vec r'|L\,x)}{|\vec r-\vec r'|\,x}\,dx\\\\ &=\frac{1}{(2\pi)^2}\frac{1}{|\vec r-\vec r'|}\int_{-\infty}^\infty \frac{\sin(x)}{x}\,dx\\\\ &=\frac{1}{4\pi |\vec r-\vec r'|} \end{align}$$

as expected!

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  • $\begingroup$ @DKL Thank you for the best vote. Just FYI ... you can also give a vote up to questions you post. ;-)) - Mark $\endgroup$ – Mark Viola Mar 16 '16 at 23:13

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