# fourier transforms

• Dec 31st 2007, 03:59 AM
ppyvabw
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.
• Dec 31st 2007, 12:55 PM
Opalg
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 $\widehat{B'}(p) = \widehat{B}(R(p))$.
• Jan 1st 2008, 03:01 PM
ppyvabw
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?
• Jan 2nd 2008, 10:17 AM
ppyvabw
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.
• Jan 2nd 2008, 11:21 AM
Opalg
Quote:

Originally Posted by ppyvabw
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→x–a. 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(x–a). This is done by making the substitution y=x–a 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:
. . . . . . $\begin{array}{rcl}\widehat{B'}(p) &=& \displaystyle \int_{\mathbb{R}^3}B(R(x))e^{-ix.p}dx \vspace{1ex} \\ &=& \displaystyle \int_{\mathbb{R}^3}B(y)e^{-iR^{\text{\textsf{T}}}(y).p}dy \vspace{1ex} \\ &=& \displaystyle \int_{\mathbb{R}^3}B(y)e^{-iy.R(p)}dy = \widehat{B}(R(p)). \end{array}$
• Jan 2nd 2008, 11:39 AM
ppyvabw
Yeah sorry, I meant R, not Lambda. Ok, thanks.