Morberticus
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Hi,
I have a question about the Fourier transform of \frac{1}{|\mathbf{r_1} - \mathbf{r_2}|} over a finite cube of unit volume. Where |\mathbf{r_1} - \mathbf{r_2}| is \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}
I know it looks like
\sum_\mathbf{k} f_k e^{-i\mathbf{k}\cdot (\mathbf{r_1}-\mathbf{r_2})}
where f_k is the Fourier coefficient
f_k = \frac{1}{V} \int_V \frac {e^{-i\mathbf{k} \cdot \mathbf{r} } } {|\mathbf{r}| } d\mathbf{r}
over the volume {-1,1}{-1,1}{-1,1}
My question is, what happens when \frac{1}{|\mathbf{r_1} - \mathbf{r_2}|} is not radially symmetric. Say |\mathbf{r_1} - \mathbf{r_2}| is
\sqrt{(x_1-x_2)^2 + a(y_1-y_2)^2 + b(z_1-z_2)^2}
would the expression then become
\sum_\mathbf{k} f_k e^{-i\mathbf{k}\cdot (x_1-x_2)}e^{-i\mathbf{k}\cdot a(y_1-y_2)}e^{-i\mathbf{k}\cdot b(z_1-z_2)}
and would the coefficient f_k be affected? My guess is yes it would be over the interval {-1,1},{-a,a},{-b,b}
Is this correct?
Thanks
I have a question about the Fourier transform of \frac{1}{|\mathbf{r_1} - \mathbf{r_2}|} over a finite cube of unit volume. Where |\mathbf{r_1} - \mathbf{r_2}| is \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}
I know it looks like
\sum_\mathbf{k} f_k e^{-i\mathbf{k}\cdot (\mathbf{r_1}-\mathbf{r_2})}
where f_k is the Fourier coefficient
f_k = \frac{1}{V} \int_V \frac {e^{-i\mathbf{k} \cdot \mathbf{r} } } {|\mathbf{r}| } d\mathbf{r}
over the volume {-1,1}{-1,1}{-1,1}
My question is, what happens when \frac{1}{|\mathbf{r_1} - \mathbf{r_2}|} is not radially symmetric. Say |\mathbf{r_1} - \mathbf{r_2}| is
\sqrt{(x_1-x_2)^2 + a(y_1-y_2)^2 + b(z_1-z_2)^2}
would the expression then become
\sum_\mathbf{k} f_k e^{-i\mathbf{k}\cdot (x_1-x_2)}e^{-i\mathbf{k}\cdot a(y_1-y_2)}e^{-i\mathbf{k}\cdot b(z_1-z_2)}
and would the coefficient f_k be affected? My guess is yes it would be over the interval {-1,1},{-a,a},{-b,b}
Is this correct?
Thanks