How Does Non-Radial Symmetry Affect the Fourier Transform of 1/|r|?

In summary: 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}}Is this correct?Thanks!Yes, that is correct.
  • #1
Morberticus
85
0
Hi,

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?

Thanks
 
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  • #2
Hi!
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.
 
  • #3
mnb96 said:
Hi!
May I ask you a couple of questions before attempting to provide an answer?

Sure! Thanks for the help.

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.

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]

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)

Unfortunately no. It is a 6-dimensional integral.
 
  • #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
 
  • #5
aesir said:
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

[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.
 

FAQ: How Does Non-Radial Symmetry Affect the Fourier Transform of 1/|r|?

What is the Fourier transform of 1/|r|?

The Fourier transform of 1/|r| is a mathematical function that represents the transformation of a signal from its original domain (in this case, the spatial domain) to the frequency domain. It is commonly used in physics and engineering to analyze and interpret signals and images.

What is the physical significance of the Fourier transform of 1/|r|?

The Fourier transform of 1/|r| has various physical interpretations, depending on the context in which it is used. In general, it can be used to determine the frequency components of a signal or the spatial frequencies present in an image. It is also closely related to the concept of diffraction in physics and optics.

How is the Fourier transform of 1/|r| calculated?

The Fourier transform of 1/|r| is calculated using a mathematical formula that involves integration and complex numbers. It can be computed using various numerical methods and software programs, such as MATLAB or Mathematica.

What are the applications of the Fourier transform of 1/|r|?

The Fourier transform of 1/|r| has numerous applications in physics, engineering, and other fields. It is commonly used in signal processing, image analysis, and data compression. It also has applications in quantum mechanics, electromagnetism, and other areas of physics.

Are there any limitations or drawbacks to using the Fourier transform of 1/|r|?

While the Fourier transform of 1/|r| is a powerful tool for analyzing signals and images, it has some limitations and drawbacks. One limitation is that it assumes the signal or image is periodic, which may not always be the case. It also has some numerical limitations and can be computationally expensive for large datasets.

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