Derivation of the atomic form factor

  • Thread starter mjordan2nd
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  • #1
My book defines the atomic form factor as

[tex]\int dV n_{j}(\vec{r})e^{-i \vec{G} \bullet \vec{r}}[/tex].

where n(r) is the electron density of the jth atom at r, and G is the reciprocal lattice vector. It says the above expression is equal to

[tex]2 \pi \int dr r^{2} d(cos \alpha) n_{j}(r) e^{-iGr cos \alpha}[/tex]

in the case that the electron distribution is spherically symmetric where alpha is the angle between the reciprocal lattice vector and r. I don't understand how the author goes from the first equation to the second. In fact, I'm not sure I'm comfortable with this d(cos) notation at all. If someone could point me to a site where I could read up about that, I would appreciate it. The author then proceeds to integrate d(cos) from -1 to 1. Why he does this is unclear to me as well. Any help is appreciated.
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Answers and Replies

  • #2
The integral is done in spherical coordinates.
The z axis is taken along the direction of the G vector so the angle between r and G si theta (the polar angle).
The volum element in spherical coordinates is
dV=r^2 sin(theta) dr d(theta) d(phi)
The author writes sin(theta)d(theta) as d(cos theta).
n(r) is asumed to have spherical symmetry here so it depends on the r (magnitude, not vector) only.

For more on sperical coordinates, you can see this, for example: