Abel transform: Wikis

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Updated live from Wikipedia, last check: May 22, 2013 14:11 UTC (43 seconds ago)

In mathematics, the Abel transform, named for Niels Henrik Abel, is an integral transform often used in the analysis of spherically symmetric or axially symmetric functions. The Abel transform of a function f(r) is given by:

$F(y)=2\int_y^\infty \frac{f(r)r\,dr}{\sqrt{r^2-y^2}}.$

Assuming f(r) drops to zero more quickly than 1/r, the inverse Abel transform is given by

$f(r)=-\frac{1}{\pi}\int_r^\infty\frac{d F}{dy}\,\frac{dy}{\sqrt{y^2-r^2}}.$

In image analysis, the forward Abel transform is used to project an optically thin, axially symmetric emission function onto a plane, and the reverse Abel transform is used to calculate the emission function given a projection (i.e. a scan or a photograph) of that emission function.

In recent years, the inverse Abel transformation (and its variants) has become the cornerstone of data analysis in photofragment-ion imaging and photoelectron imaging. Among recent most notable extensions of inverse Abel transformation are the Onion Peeling and BAsis Set Expansion (BASEX) methods of photoelectron and photoion image analysis.

Geometrical interpretation

A geometrical interpretation of the Abel transform in two dimensions. An observer (I) looks along a line parallel to the x-axis a distance y above the origin. What the observer sees is the projection (i.e. the integral) of the circularly symmetric function f(r) along the line of sight. The function f(r) is represented in gray in this figure. The observer is assumed to be located infinitely far from the origin so that the limits of integration are ±∞

In two dimensions, the Abel transform F(y) can be interpreted as the projection of a circularly symmetric function f(r) along a set of parallel lines of sight which are a distance y from the origin. Referring to the figure on the right, the observer (I) will see

$F(y)=\int_{-\infty}^\infty f\!\left(\sqrt{x^2 + y^2}\right)\,dx$

where f(r) is the circularly symmetric function represented by the gray color in the figure. It is assumed that the observer is actually at x = ∞ so that the limits of integration are ±∞ and all lines of sight are parallel to the x-axis. Realizing that the radius r is related to x and y via r2 = x2 +  y2, it follows that

$dx=\frac{r\,dr}{\sqrt{r^2-y^2}}.$

The path of integration in r does not pass through zero, and since both f(r) and the above expression for dx are even functions, we may write:

$\int_{-\infty}^\infty f(r)\,dx=2\int_0^\infty f(r)\,dx, \;\; r= \sqrt{x^2+y^2}.$

Substituting the expression for dx in terms of r and rewriting the integration limits accordingly yields the Abel transform.

The Abel transform may be extended to higher dimensions. Of particular interest is the extension to three dimensions. If we have an axially symmetric function f,z) where ρ2 = x2 + y 2 is the cylindrical radius, then we may want to know the projection of that function onto a plane parallel to the z axis. Without loss of generality, we can take that plane to be the yz-plane so that

$F(y,z) =\int_{-\infty}^\infty f(\rho,z)\,dx =2\int_y^\infty \frac{f(\rho,z)\rho\,d\rho}{\sqrt{\rho^2-y^2}}$

which is just the Abel transform of f,z) in ρ and y.

A particular type of axial symmetry is spherical symmetry. In this case, we have a function f(r) where r2 = x2 +  y2 + z2. The projection onto, say, the yz-plane will then be circularly symmetric and expressible as F(s) where s2 = y2 +  z2. Carrying out the integration, we have:

$F(s) =\int_{-\infty}^\infty f(r)\,dx =2\int_s^\infty \frac{f(r)r\,dr}{\sqrt{r^2-s^2}}$

which is again, the Abel transform of f(r) in r and s.

Verification of the inverse Abel transform

Assuming f is continuously differentiable and f, f' drop to zero faster than 1/r, we can set u = f(r) and $v=\sqrt{r^2-y^2}$. Integration by parts then yields

$F(y) = -2 \int_y^\infty f'(r) \sqrt{r^2-y^2} \, dr.$

Differentiating formally,

$F'(y) = 2 y \int_y^\infty \frac{f'(r)}{\sqrt{r^2-y^2}} \, dr.$

Now plug this into the inverse Abel transform formula:

$-\frac{1}{\pi} \int_r^\infty \frac{F'(y)}{\sqrt{y^2-r^2}} \, dy = \int_r^\infty \int_y^\infty \frac{-2 y}{\pi \sqrt{(y^2-r^2) (s^2-y^2)}} f'(s) \, ds dy.$

By Fubini's theorem, the last integral equals

$\int_r^\infty \int_r^s \frac{-2 y}{\pi \sqrt{(y^2-r^2) (s^2-y^2)}} \, dy f'(s) \,ds = \int_r^\infty (-1) f'(s) \, ds = f(r).$

Relationship to other integral transforms

Relationship to the Fourier and Hankel transforms

The Abel transform is one member of the FHA cycle of integral operators. For example, in two dimensions, if we define A as the Abel transform operator, F as the Fourier transform operator and H as the zeroth-order Hankel transform operator, then the special case of the Projection-slice theorem for circularly symmetric functions states that:

$FA=H.\,$

In other words, applying the Abel transform to a 1-dimensional function and then applying the Fourier transform to that result is the same as applying the Hankel transform to that function. This concept can be extended to higher dimensions.