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Fourier transform of 1/|r|

  1. Dec 18, 2011 #1

    I have a question about the fourier transform of [itex]\frac{1}{|\mathbf{r_1} - \mathbf{r_2}|}[/itex] over a finite cube of unit volume. Where [itex]|\mathbf{r_1} - \mathbf{r_2}|[/itex] is [itex]\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}[/itex]

    I know it looks like

    [itex]\sum_\mathbf{k} f_k e^{-i\mathbf{k}\cdot (\mathbf{r_1}-\mathbf{r_2})}[/itex]

    where f_k is the fourier coefficient

    [itex]f_k = \frac{1}{V} \int_V \frac {e^{-i\mathbf{k} \cdot \mathbf{r} } } {|\mathbf{r}| } d\mathbf{r}[/itex]

    over the volume {-1,1}{-1,1}{-1,1}

    My question is, what happens when [itex]\frac{1}{|\mathbf{r_1} - \mathbf{r_2}|}[/itex] is not radially symmetric. Say [itex]|\mathbf{r_1} - \mathbf{r_2}|[/itex] is

    [itex]\sqrt{(x_1-x_2)^2 + a(y_1-y_2)^2 + b(z_1-z_2)^2}[/itex]

    would the expression then become

    [itex]\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)}[/itex]

    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?

  2. jcsd
  3. Dec 20, 2011 #2
    May I ask you a couple of questions before attempting to provide an answer?

    1) First of all, from the second formula in your post, the one where you have a summation, it seems that you are asking for the Fourier series expansion, not the Fourier transform, as you wrote in the title. Which one are you interested in?

    2) Can you write explicitly the function (with all its variables) you want to find the Fourier transform of?

    3) you have not defined r inside the integral.

    4) Is r1 your variable and r2 some constant?
    (note: if the answer is yes, you can first of all apply the shift-theorem to get rid of r2 inside the integral)

    I personally feel that your question is formulated in a very confusing way, or maybe it's just me. Please clarify those points I asked, and then maybe we can provide some help to answer your last question on symmetry.
  4. Dec 25, 2011 #3
    Sure! Thanks for the help.

    Sorry I will try and be more clear. Ultimately I am looking for the Fourier expansion, but the Fourier transform is giving me the most trouble. The function is

    [itex]F(\mathbf{r_1},{r_2}) = \frac{1}{|\mathbf{r_1-r_2}|}[/itex]

    where r_1 and r_2 are 3 dimensional vectors. More explicitly

    [itex]F(x_1,y_1,z_1,x_2,y_2,z_2) = \frac{1}{\sqrt{(x_1 - x_2)^2 +(y_1 - y_2)^2 + (z_1-z_2)^2}}[/itex]

    Normally, I just employ the Fourier coefficient

    [itex]f_k = \frac{4 \pi}{k^2}[/itex]

    to get the expansion. But this time I am dealing with an extra complication. If any of the variables [itex]x_1,y_1,z_1,x_2,y_2,z_2[/itex] are greater than 1 or less than 0, the value of the function is 0. I.e. Instead of vanishing when the separation between r_1 and r_2 approaches infinity, the function abruptly vanishes if r_1 or r_2 is outside the unit cube. Effectively, the function outside this cuboidal boundaries is "chopped off". The form of f_k becomes:

    [itex]f_k = \frac{1}{V} \int_V \frac {e^{-i\mathbf{k} \cdot \mathbf{r} } } {|\mathbf{r}| } d\mathbf{r}[/itex]

    This is a three-dimensional integral (r is a three-dimensional vector) and is integrated over {-1,1} in each dimension. I.e.

    [itex]f_k = \frac{1}{8} \int^1_{-1} \int^1_{-1} \int^1_{-1} \cdots dxdydz[/itex]

    Unfortunately no. It is a 6-dimensional integral.
  5. Dec 26, 2011 #4
    You can extend the domain to infinity writing the function as the product of f and the characteristic function of the cube.
    Then by the convolution theorem the fourier transform is the convolution of [itex]f_k[/itex] and something like sinc(k).

    Don't know if you can go any further
  6. Dec 26, 2011 #5
    [itex]f_k[/itex] itself isn't too hard to evaluate numerically. I don't expect to find a simple analytical form. What I am wondering is how things change when I instead deal with the function

    [itex]F(x_1,y_1,z_1,x_2,y_2,z_2) = \frac{1}{\sqrt{(x_1 - x_2)^2 +a(y_1 - y_2)^2 + b(z_1-z_2)^2}}[/itex]

    where a and b are arbitrary constants.
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