Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Fourier Integral help

Tags:
  1. Feb 15, 2017 #1
    upload_2017-2-15_14-5-3.png

    Where , rho 1 and rho 2 are two dimensional position vectors and K is a two dimensional vector in the Fourier domain. I encountered the above Eq. (27) in an article and the author claimed that after integration the right hand side gives the following result:
    upload_2017-2-15_14-7-50.png
    I tried to solve this integral but couldn't get it right. please can anyone here can guide me how to get this result from Eq. (27).

    This integral has been annoying me for a few weeks!
    Thank you so much
     
  2. jcsd
  3. Feb 15, 2017 #2

    jasonRF

    User Avatar
    Science Advisor
    Gold Member

    If it weren't for that factor ##Sign(i,j)## in the result I would have thought that we could treat ##\sigma_i##, ##\sigma_j## and ##\delta_{ij}## as arbitrary constants in (27) and that you were simply trying to find the Fourier transform of a multivariate Gaussian (which I would do by diagonalizing the quadratic form to get a bunch of 1-D Fourier transforms of Gaussians, which can be performed many ways). But it seems like something else is going on here.

    So I think we need more information. What do ##\sigma_i##, ##\sigma_j## and ##\delta_{ij}## stand for, and what is ##Sign(i,j)##?

    Jason
     
  4. Feb 15, 2017 #3
    Hi thanks for the reply. The parameters ##\sigma_i## and ##\sigma_j## corresponds to the spread of spectral density and ##\delta_{ij}## is the width of the correlation between the i and j components of a field. and sign (i, j) = 1 when i = j and sign (i,j) = -1 if i is not equal to j.
     

    Attached Files:

  5. Feb 15, 2017 #4

    jasonRF

    User Avatar
    Science Advisor
    Gold Member

    Okay. I think I get it now. We can treat the parameters as constants, but the final simplifications can take into account that the answer will depend on whether ##i=j## or ##i\neq j##.

    What part are you getting stuck on? Can you do the integrals and are just struggling with the algebra to simplify the expression, or are you having trouble doing the integrals? I've derived a simple expression for the general integral, but have not done the messy algebra to see if the final result agrees with the posted answer.

    jason
     
  6. Feb 15, 2017 #5
    Hi Jason,

    First, thank you so much for the reply.
    I am struggling with the doing the integrals. I tried and tried but couldn't get the result.
     
  7. Feb 15, 2017 #6

    jasonRF

    User Avatar
    Science Advisor
    Gold Member

    Okay. By the way, my "simple" formula I thought I had was wrong. The correct way is messy algebra, but conceptually quite simple.
    The basic form of your integral is:
    $$
    I = \int_{-\infty}^{\infty} d^2\mathbf{r}_1 \int_{-\infty}^{\infty} d^2\mathbf{r}_2\, e^{-i2\pi\mathbf{k \cdot }(\mathbf{r}_1-\mathbf{r}_2)} \, e^{-\mathbf{r}_1 \cdot \mathbf{r}_1 /a_1^2}\,e^{-\mathbf{r}_2 \cdot \mathbf{r}_2 /a_2^2}\, e^{-2 b \mathbf{r}_2 \cdot \mathbf{r}_1 }
    $$
    We can re-write this,
    $$
    I = \int_{-\infty}^{\infty} d^2\mathbf{r}_1\, e^{-i2\pi\mathbf{k \cdot } \mathbf{r}_1}\, e^{-\mathbf{r}_1 \cdot \mathbf{r}_1 /a_1^2} \, J(\mathbf{r}_1)
    $$
    where,
    $$
    \begin{eqnarray}
    J(\mathbf{r}_1) & = & \int_{-\infty}^{\infty} d^2\mathbf{r}_2\, e^{i2\pi\mathbf{k \cdot} \mathbf{r}_2} \,e^{-\mathbf{r}_2 \cdot \mathbf{r}_2 /a_2^2}\, e^{-2 b \mathbf{r}_2 \cdot \mathbf{r}_1 }
    \end{eqnarray}
    $$
    Now change variables using ##\mathbf{y} = \mathbf{r}_2/a_2## to get
    $$
    \begin{eqnarray}
    J(\mathbf{r}_1) & = & a_2^2 \int_{-\infty}^{\infty} d^2\mathbf{y} \,e^{i2\pi a_2 \mathbf{k \cdot} \mathbf{y}} \,e^{-\mathbf{y} \cdot \mathbf{y}}\, e^{-2 b a_2 \mathbf{r}_1 \cdot \mathbf{y} } \\
    & = & a_2^2\int_{-\infty}^{\infty} d^2\mathbf{y} \,e^f
    \end{eqnarray}
    $$
    Where all the mess is in ##f##. Re-writing,
    $$
    \begin{eqnarray}
    f & = & i2\pi a_2 \mathbf{k \cdot} \mathbf{y} -\mathbf{y} \cdot \mathbf{y} -2 b a_2 \mathbf{r}_1 \cdot \mathbf{y} \\
    & \equiv & \mathbf{s \cdot} \mathbf{y} -\mathbf{y} \cdot \mathbf{y} .
    \end{eqnarray}
    $$
    where the above defines ##\mathbf{s}=i2\pi a_2 \mathbf{k} -2 b a_2 \mathbf{r}_1##. Now we want to find a new integration variable ## \mathbf{z} = \mathbf{y} - \mathbf{w}## for a constant ##\mathbf{w}## that makes the linear term above go away. We look at ##f##,
    \begin{eqnarray}
    f & = & \mathbf{s \cdot} \mathbf{y} -\mathbf{y} \cdot \mathbf{y} \\
    & = & \mathbf{s \cdot} \mathbf{z} + \mathbf{s \cdot} \mathbf{w} - \mathbf{z} \cdot \mathbf{z} - \mathbf{w} \cdot \mathbf{w} - 2 \mathbf{w} \cdot \mathbf{z}
    \end{eqnarray}
    so if ##\mathbf{w} = \mathbf{s}/2## we get,
    \begin{eqnarray}
    f & = & \mathbf{s \cdot} \mathbf{s} /2- \mathbf{z} \cdot \mathbf{z} - \mathbf{s} \cdot \mathbf{s}/4 \\
    & = & \mathbf{s \cdot} \mathbf{s} /4- \mathbf{z} \cdot \mathbf{z}
    \end{eqnarray}
    and we have,
    \begin{eqnarray}
    J(\mathbf{r}_1) & = & a_2^2 e^{ \mathbf{s \cdot} \mathbf{s} /4} \int_{-\infty}^{\infty} d^2\mathbf{z} \,e^{-\mathbf{z} \cdot \mathbf{z}} \\
    & = & \pi \, a_2^2 e^{ \mathbf{s \cdot} \mathbf{s} /4}
    \end{eqnarray}
    Now, ##\mathbf{s \cdot} \mathbf{s} /4 = \zeta_1 \mathbf{k \cdot} \mathbf{k} + \zeta_2 \mathbf{r}_1 \cdot \mathbf{r}_1 + i \zeta_3 \mathbf{k \cdot} \mathbf{r}_1 ##; doing the algebra will give the the values of the real ##\zeta_1##,##\zeta_2##, and ##\zeta_3##.
    So our original integral becomes,
    $$
    I = \pi \, a_2^2 e^{\zeta_1 \mathbf{k \cdot} \mathbf{k}} \int_{-\infty}^{\infty} d^2\mathbf{r}_1 \, e^{-i(2\pi-\zeta_3)\mathbf{k \cdot } \mathbf{r}_1}\, e^{-\mathbf{r}_1 \cdot \mathbf{r}_1 (\zeta_2 + 1/a_1^2)}
    $$
    This can be integrated using the same trick above. The algebra will be really messy, but this is do-able. Hope this helps!

    Jason
     
    Last edited: Feb 15, 2017
  8. Feb 16, 2017 #7

    jasonRF

    User Avatar
    Science Advisor
    Gold Member

    I have another way that yields a simple result that may be easier to deal with. I will use matrix notation, so dot products will be ##\mathbf{k \cdot x} = \mathbf{k}^T\mathbf{x}## where the T superscript represents transpose.

    Let ##\mathbf{r}## be a 4 x 1 vector that contains ##\rho_1## and ##\rho_2## stacked on top of each other, and ##\mathbf{k}## be the 2x1 vector wave-vector. Then the integral is of the form
    \begin{eqnarray}
    I & = & \int_{-\infty}^{\infty} d^4 \mathbf{r} \, e^{i 2 \pi \mathbf{k}^T B \mathbf{r}}\, e^{-\mathbf{r}^T A \mathbf{r}}
    \end{eqnarray}
    Where ##B## is a 2x4 matrix and ##A## is a positive definite symmetric 4x4 matrix. For your problem I believe these have a very special form. If ##I_2## is a 2x2 identity matrix, then in block-matrix notation I think your problem can be represented
    \begin{eqnarray}
    B & = & \left( \begin{array} & I_2 & -I_2 \end{array} \right) \\
    A & = & \left( \begin{array} & a I_2 & b I_2 \\ b I_2 & c I_2 \end{array} \right) \\
    \end{eqnarray}
    The results below do not rely on these forms, but they are probably helpful in doing the algebra to get the final answer in a form you want.

    Anyway, now we use the fact that ##A## is positive definite to do an eigen-decomposition, ##A = P \Lambda P^T##, where ##\Lambda## is the diagonal matrix of eigenvalues and ##P## is an orthogonal matrix of eigenvectors. Then the integral can be written,
    \begin{eqnarray}
    I & = & \int_{-\infty}^{\infty} d^4 \mathbf{r} \, e^{i 2 \pi \mathbf{k}^T B \mathbf{r}}\, e^{-\mathbf{r}^T P \Lambda P^T \mathbf{r}}
    \end{eqnarray}
    Now we do a change of variable to ##\mathbf{y} = \Lambda^{1/2} P^T \mathbf{r}##. Note that the Jacobian is ##\left| \Lambda^{1/2} P^T\right|=\left| \Lambda\right|^{1/2} =\left| A \right|^{1/2}##, where ##| \cdot |## represents the determinant. So we have
    \begin{eqnarray}
    I & = & \left| A \right|^{-1/2}\int_{-\infty}^{\infty} d^4 \mathbf{y} \, e^{i 2 \pi \mathbf{k}^T B P \Lambda^{-1/2} \mathbf{y}}\, e^{-\mathbf{y}^T \mathbf{y}} \\
    & = & \left| A \right|^{-1/2}\int_{-\infty}^{\infty} d^4 \mathbf{y} \, e^{\mathbf{s}^T\mathbf{y} - \mathbf{y}^T \mathbf{y}}
    \end{eqnarray}
    where ##\mathbf{s} = i 2 \pi \Lambda^{-1/2} P^T B^T \mathbf{k}##. Now we do the same trick as before to get rid the of the term linear in ##\mathbf{y}## (by letting ##\mathbf{z}=\mathbf{y}-\mathbf{s}/2##) and end up with,
    \begin{eqnarray}
    I & = & \left| A \right|^{-1/2}e^{\mathbf{s}^T\mathbf{s}/4} \int_{-\infty}^{\infty} d^4 \mathbf{z} \, e^{ - \mathbf{z}^T \mathbf{z}} \\
    & = & \pi^2 \left| A \right|^{-1/2}e^{\mathbf{s}^T\mathbf{s}/4}
    \end{eqnarray},
    Now, ##\mathbf{s}^T\mathbf{s} = -4 \pi^2 \mathbf{k}^T B P \Lambda^{-1/2}\Lambda^{-1/2} P^T B^T \mathbf{k}=-4 \pi^2 \mathbf{k}^T B A^{-1} B^T \mathbf{k}##. So we finally have
    \begin{eqnarray}
    I = \pi^2 \left| A \right|^{-1/2}e^{-\pi^2 \mathbf{k}^T B A^{-1} B^T \mathbf{k}}.
    \end{eqnarray}
    The fact that ##A## and ##B## have such special forms will hopefully make this expression reducible. For example forumulas in
    https://en.wikipedia.org/wiki/Block_matrix
    can be used to find the inverse of ##A## quite easily, and I suspect the determinant of ##A## is also not so bad.

    Good luck,

    Jason
     
    Last edited: Feb 16, 2017
  9. Feb 16, 2017 #8

    jasonRF

    User Avatar
    Science Advisor
    Gold Member

    I found a nice formula for the determinant. Start from the equation between (1) and (2) of,
    https://en.wikipedia.org/wiki/Woodbury_matrix_identity
    \begin{eqnarray}
    \left(\begin{array} & A & U \\ V & C \end{array} \right) & = & \left(\begin{array} & I & UC^{-1} \\ 0 & I \end{array} \right) \left(\begin{array} & A-UC^{-1}V & 0 \\ 0 & C \end{array} \right) \left(\begin{array} & I & 0 \\ C^{-1}V & I \end{array} \right)
    \end{eqnarray}
    Taking the determinant yields
    \begin{eqnarray}
    \left| \left(\begin{array} & A & U \\ V & C \end{array} \right) \right| & = & \left|A-UC^{-1}V \right| \left| C \right|
    \end{eqnarray}
    In your case all of the blocks are proportional to ##I_2##, so this should be easy to compute.

    jason
     
    Last edited: Feb 16, 2017
  10. Feb 17, 2017 #9
    Dear Jason,

    Thank you so much for the great help and giving priceless time to guide me in solution of this integral. I am trying to solve it with the approaches you mentioned. I hope I can find a reasonable solution to this integral. I also tried to solve the integral with using the formula, integral (exp(-bx^2) exp(iax))dx = sqrt(pi/b)exp(-a^2/4b) but I couldn't get the required result.
     
  11. Feb 17, 2017 #10

    jasonRF

    User Avatar
    Science Advisor
    Gold Member

    Did you try the approach I gave in post 7? When i plug in the parameters of your problem into equation 20 I get a result that is almost the one you posted. Instead of ##sign(i,j)## I get ##\pi^2##.

    In retrospect I do not understand how the sign factor could possibly appear. For the special case ##\kappa=0## you have an inyegral of a Gaussian, which must be positive. Likewise, it seems inevitable that there should be a ##\pi^2##. So i do not see how your posted solution would be correct.
     
    Last edited: Feb 17, 2017
  12. Feb 28, 2017 #11
    Hi Jason,

    I tried your approaches but I find it a bit difficult to understand fully. Maybe I don't have enough skills and a more 'mechanical' or brute force approach may be easier for me to understand. Now I will try once again the approaches you mentioned if I can get answer with it.

    Thank you so much for the huge help.
    Regards
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Fourier Integral help
  1. Fourier Integral (Replies: 4)

Loading...