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Fourier Transform Applied to Electrostatics

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data
    How would you solve the one-dimensional Poisson's equation:

    [tex]$\nabla ^2 \phi = \frac{\rho}{\epsilon_0}$[/tex]

    Using Fourier Transforms?

    [tex]$\phi (x) = \int ^{+\infty}_{-\infty} G(k) e^{-i k x} dk$

    $G(k) = \int^{+\infty}_{-\infty} \phi (x) dx$[/tex]

    I've been trying to understand Fourier transforms and I do not yet see how they help; I cannot solve this equation with Fourier. If someone could show me and explain that would be helpful, thanks.

    2. Relevant equations
    [tex]$\nabla ^2 \phi = \frac{\rho}{\epsilon_0}$

    $\phi (x) = \int ^{+\infty}_{-\infty} G(k) e^{-i k x} dk$

    $G(k) = \int^{+\infty}_{-\infty} \phi (x) dx$

    [/tex]
    3. The attempt at a solution[tex]
     
    Last edited: Oct 9, 2009
  2. jcsd
  3. Oct 9, 2009 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    Fourier transforms change linear differential equations into algebraic equations, which are usually easier to solve. What happens when you take your expression for phi, and differentiate it with respect to x?
     
  4. Oct 9, 2009 #3
    I get
    [tex]
    $\int_{-\infty}^{+\infty} (-k^2) G(k) e^{-ikx} dk = \frac{\rho}{\epsilon _0}$
    [/tex]

    If I fourier transform back the other side I just get:

    [tex]
    $\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} (- k^2) G(k) e^{ix(k'-k)} dk dx = \frac{\int_{-\infty}^{+\infty} \rho (x) e^{-ikx} dx}{\epsilon_0}
    [/tex]

    (Assuming one dimension to be simpler)

    I don't see how this turns into a linear equation. Forgive me if I am being naif, but I just can't see the step.

    Thanks for the reply
     
  5. Oct 18, 2009 #4
    I just realized I posted this a while ago, I must write a correction to it, in case anyone finds it.
    I cannot see what Ben was suggesting, it does not make sense, I was asking for some kind of proof, not some procedure that is done because everybody knows that's how it works

    But here is what I believe is why, if I am wrong, someone please correct it

    [itex]
    $\nabla ^2 \phi = \frac{\rho}{\epsilon _0}$


    Integrate with respect to $e^{i \vec{k} \cdot \vec{x}}d\vec{x}$ on both sides:

    $\int \nabla^2 \phi(\vec{x}) e^{i \vec{k} \cdot \vec{x}}d\vec{x} = \int e^{i \vec{k} \cdot \vec{x}} \frac{\rho}{\epsilon _0} d\vec{x}$


    Now, we use integration by parts, and assume that $\phi (\vec{x})$ goes off at $\pm \infty$ to get (you can do this yourselves). ($\phi$ not dying off at infinity is not something physical, or at least not so common in physics equations :-D)

    $-|k|^2 \int \phi(\vec{x}) e^{i \vec{k} \cdot \vec{x}} d\vec{x} = \int e^{i \vec{k} \cdot \vec{x}} \frac{\rho}{\epsilon _0} d\vec{x} $

    $-|k|^2 \hat{\phi}(\vec{k}) = \int e^{i \vec{k} \cdot \vec{x}} \frac{\rho}{\epsilon _0} d\vec{x} $

    I've used multiple dimensions here, you can just substitute $\vec{x} = x$ and $d\vec{x} = dx$, and $\phi(\vec{x}) = \phi(x_1, x_2, ..., x_N) \rightarrow \phi(x)$ for one dimension (same for k)

    If this is wrong, someone please correct me but I am pretty confident this is right. I wish someone would have explained this to me correctly...

    [\latex]
     
    Last edited: Oct 18, 2009
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