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Fourier representation of fields

  1. Jul 12, 2006 #1
    Hello,

    From Jean Zinn-Justin:

    "The conducting plates impose to the electric field to be perpendicular to the plates. It is easy to verify that this condition is satisfied if the vector field [itex]A_\mu[/itex] itself vanishes on the plates. Calling L the distance between the plates, z=0 and z=L the plate positions, we thus integrate over fields which have the fourier representation,

    [tex]A_\mu (\mathbf{x}_{\perp},z)=\int d^{d-1}p_{\perp}\sum_{n\ge1} e^{ip_{\perp} \cdot \mathbf{x}_{\perp}} \sin(\pi z/L) \tilde{A}(\mathbf{p}_{\perp},n)[/tex],

    where [itex]\mathbf{x_{\perp}}[/itex] are the space coordinates in the remaining directions."

    OK so the bit that I dont understand is the fourier representation buisness. Am I right in thinking that the exponential factor is the imaginary part of the fields?

    Also what is meant by [itex]\tilde{A}()[/itex]?

    Answers to these questions or comments that will help to make this equation clear to me would be very much appreciated.

    Thank you for your time.
     
    Last edited: Jul 12, 2006
  2. jcsd
  3. Jul 13, 2006 #2
    Please? *insert post filler here*
     
  4. Jul 13, 2006 #3
    The Fourier representation is a representation of the function in the sinusoidal basis. The quantity [tex]\tilde{A}(\mathbf{p},n)[/tex] is the (Fourier) amplitude at coordinate (p, n). Because the vector potential [tex] A[/tex] is real, the Fourier amplitudes satisfy certain symmetry conditions.

    BTW, I think you are missing a factor of n inside you sine function.
     
  5. Jul 17, 2006 #4
    Consider a function f(x) of one real variable x. We define the "Fourier transform of f(x)" as the function F(p), of the single real variable p, which is given by

    F(p) ≡ (2π)-1 ∫dx e-ipx f(x) .

    It then follows by theorem that

    f(x) = ∫dp eipx F(p) ,

    which is called the "inverse Fourier transform of F(p)".
    ____________________

    This pattern can be applied to any number of dimensions m. In that case,

    F(p) ≡ (2π)-m ∫dmx e-ipx f(x) ,

    and again, by theorem, it follows that

    f(x) = ∫dmp eipx F(p) .
    ____________________

    For the case at hand, we are dealing with a function

    a(x,y,z) = ∑n≥1 sin(nπz/L) a(n,x,y) ... [1] ,

    and with Fourier transforms A(n,p1,p2), in m=2 dimensions, of the functions
    a(n,x,y) such that

    A(n,p1,p2) ≡ (2π)-2 ∫dx dy exp[-ip1∙x] exp[-ip2∙y] a(n,x,y) ... [2] ,

    so that, by theorem,

    a(n,x,y) = ∫ dp1 dp2 exp[ip1∙x] exp[ip2∙y] A(n,p1,p2) ... [3] .

    Substituting [3] into [1] gives the original expression posted at the beginning of this thread, which in the current notation is just

    a(x,y,z) = ∫ dp1 dp2n≥1 exp[ip1∙x] exp[ip2∙y] sin(nπz/L) A(n,p1,p2) ... [4] .
    _____________________________

    IN ANSWER TO THE QUESTIONS:

    The exponential factors

    exp[ip1∙x] exp[ip2∙y]

    that appear in [4] are not the imaginary parts of the field. Those exponentials are just parts of the "inverse Fourier transform" which takes

    A(n,p1,p2) → a(n,x,y) ,

    for each value of n, according to equation [3].

    Also, what is meant by A(n,p1,p2) is just, quite simply, the Fourier transform of
    a(n,x,y), according to equation [2], for each value of n.
     
    Last edited: Jul 17, 2006
  6. Jul 17, 2006 #5
    MadMax,

    What is your end goal by applying this fourier method to a parallel plate geometry.

    I am interested in calculating capacitance of parallel plates, taking into account fringing fields etc... all the 'non-ideal' properties of real life capacitances.

    My methods so far have only included a very basic Boundary element method, but I am beginning to look into conformal mapping and I am always looking for new modelling techniques.

    The BEM method allows me to create a surf / mesh plot of the variation in charge across the plates using MATLAB - my aim is to include field lines and the such like to these models.

    Regards

    Tom
     
  7. Jul 18, 2006 #6
    Wow, thank you very much Eye_in_the_ sky.

    I should have said earlier when beautiful1 mentioned it but there wasn't an n missing in the sin.. at least the book didn't have an n in there.. :S do you think there is an erratum?

    Also you say that the electric field before fourier transformation was given by a(x,y,z) = ∑n≥1 sin(nπz/L) a(n,x,y) . Is that simply a sum over all possible standing waves? If not where does that come from?

    tommyers: I'm looking to learn and understand calculation of the cassimir force for various geommetries and materials. I thought I'd start with parallel conducting plates since it was the simplest example in the text book recommended to me =)
     
  8. Jul 18, 2006 #7
    Tom,

    MadMax is considering here something very different from electrostatic fields.

    He considers time-dependent fields inside an extremely narrow gap in between two plates, in the absence of 'externally' applied fields. In quantum mechanics, there is a so-called "vacuum field", extremely weak that pervades everywhere. This vacuum field cannot be used as a source of energy (although some people pretend the contrary) but it is necessary for the consistency of quantum mechanics as well as to explain some experiments. One of these experiments is the Casimir force.

    The Casimir force, extremely small, occurs because in between the parallel plates some "vacuum photons" cannot propagate, they cannot exist, they are short-circuited by the plates (so to speak). At the same time, these same photons can exist outside of the plates. Therefore, the (radiation) pressure inside the plates is smaller than outside and this produces a small attraction of the two plates, the so-called Casimir force.

    Michel
     
    Last edited: Jul 18, 2006
  9. Jul 18, 2006 #8
    I think I once read an article on this - it discussed the collision of two closely moored ships!

    Tom
     
  10. Jul 19, 2006 #9
    There is a typo in that book.

    We have boundary conditions

    a(x,y,0) = a(x,y,L) = 0 .

    This is just like saying for a function f(z)

    f(0) = f(L) = 0 ,

    in which case we would want to write (presuming that doing so would simplify the problem for us)

    f(z) = ∑n≥1 cnsin(nπz/L) .

    Those functions sin(nπz/L), for n=1,2,... , are "complete" on the interval [0,L]. That is, any (reasonably well-behaved) function f(z) can be written in terms of them for appropriate choices of the coefficients cn.

    So, similarly in our case we write

    a(x,y,z) = ∑n≥1 sin(nπz/L) a(n,x,y) .
     
  11. Jul 19, 2006 #10
    wow, thats awesome Eye_in_the_Sky. :O :O

    You have explained everything beautifully =)

    Thank you very very much =)
     
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