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Math Help - fourier transforms

  1. #1
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    fourier transforms

    if B(x) is a scalar field in (3d space) with fourier transform \widetilde{B}(p) and x\rightarrow x' is a finite isometry (i.e a rotation and translation x'=Rx-a what is the fourier transform of B(x')

    I keep wanting to write \widetilde{B'}(p)=e^{ip.x`}\widetilde{B}(p) (i.e, a change of phase) but I aint so sure and I'm too dumb to prove it. Maybe I should say that doing this works for what I intend it for.
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  2. #2
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    A translation gives a change of phase, so if B'(x) = B(x-a) then \widehat{B'}(p) = e^{-ia\mathbf{.}p}\widehat{B}(p) (using a hat to denote the Fourier transform).

    The Fourier transform commutes with rotations, so if B'(x) = B(R(x)) then \widehat{B'}(p) = \widehat{B}(R(p)).
    Last edited by Opalg; January 2nd 2008 at 11:46 AM. Reason: corrected formula
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  3. #3
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    Thanks very much for your reply. There is too much detail to go through the whole problem, but the thing about rotations leaves me with a problem in that it won't work for what i want. I'm not saying you are wrong, I'm just stupid.

    Can you prove it, or give me some pointers about how to. I would be gratefull. I have had a thought about it myself, and my argument is as follows.

    B(x')=\int d^3p e^{ipx'}\overline{B}(p)=\int d^3 p e^{ipx'-ipx+ipx}\overline{B}(p)=
    =\int d^3 p e^{ipx}e^{ip(x'-x)}\overline{B}(p)=\int d^3 p e^{ipx} \overline{B'}(p)

    with \overline{B'}(p)=e^{ip(x'-x)}\overline{B}(p)

    with x --> x' a rotation or a translation. This isn't quite the same as what i thought in my first post, but it works. Unless I have done something completely wrong, I don't see what is wrong with this?
    Last edited by ppyvabw; January 1st 2008 at 04:12 PM.
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  4. #4
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    ignore the above load of nonsense.

    The thing is, I need a general expression in terms of x',B' and p', and not explicitly involving \Lambda or a, if that makes sense.
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  5. #5
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    Quote Originally Posted by ppyvabw View Post
    I need a general expression in terms of x',B' and p', and not explicitly involving \Lambda or a, if that makes sense.
    I'm not sure what \Lambda is (it hasn't been mentioned before). In fact, I'm not too happy with any of this notation, so I'll use my own.

    I'll use x and p to denote points in 3-dimensional space: x will be the variable for B-space (the space on wich the scalar field B is defined) and p will be the variable for \widehat{B}-space (the space on which the Fourier trnsform is defined).

    Let T:x→x' be an isometry of B-space. Then T consists of a rotation R followed by a translation x→xa. If you are looking for a formula for the Fourier transform of B(Tx) that doesn't mention R and a explicitly, then I don't think you are going to find one, the reason being that the Fourier transform affects translations and rotations differently.

    The Fourier transform of B is given by \widehat{B}(p) = \int_{\mathbb{R}^3}B(x)e^{-ix.p}dx, where x.p denotes the inner product of x and p.

    To find the effect of a translation, we have to calculate the Fourier transform of B'(x)=B(xa). This is done by making the substitution y=xa in the integral:
    . . . . . . \begin{array}{rcl}\widehat{B'}(p) &=& \displaystyle \int_{\mathbb{R}^3}B(x-a)e^{-ix.p}dx \vspace{1ex} \\ &=& \displaystyle  \int_{\mathbb{R}^3}B(y)e^{-i(y+a).p}dy \\ &=& \displaystyle  e^{-ia.p}\int_{\mathbb{R}^3}B(y)e^{-iy.p}dy = e^{-ia.p}\widehat{B}(p). \end{array}

    To find the effect of a rotation R, we have to calculate the Fourier transform of B'(x)=B(Rx). This is again done by making a substitution, namely y=R(x). This time, we have to worry about how to express dy in terms of dx. For a general linear transformation R, the formula would be dy=Jdx, where J is the jacobian of the transformation, which is the determinant of R. Since R is a rotation, its determinant is 1, so in fact we just get dy=dx. We also need to use the fact that x.Rp = R^{\text{\textsf{T}}}x.p, where R^T is the transpose of R. This is also the inverse of R, so if y=R(x) then x = R^{\text{\textsf{T}}}(y). If B'(x)=B(R(x)), the calculation then goes:
    . . . . . . \begin{array}{rcl}\widehat{B'}(p) &=& \displaystyle \int_{\mathbb{R}^3}B(R(x))e^{-ix.p}dx \vspace{1ex} \\ &=& \displaystyle  \int_{\mathbb{R}^3}B(y)e^{-iR^{\text{\textsf{T}}}(y).p}dy \vspace{1ex} \\ &=& \displaystyle \int_{\mathbb{R}^3}B(y)e^{-iy.R(p)}dy = \widehat{B}(R(p)). \end{array}
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  6. #6
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    Yeah sorry, I meant R, not Lambda. Ok, thanks.
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