Form factor in scattering

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Main Question or Discussion Point

I'm reading through Thomson's "Modern Particle Physics", and I've gotten stuck at a point in the derivation of the form factor for electron scattering in a static potential due to an extended charge distribution. It's just a mathematical "trick" i don't quite get.

He goes from

$$\int\int e^{i\vec{q}.(\vec{r}-\vec{r}')} e^{i\vec{q}.\vec{r}'} \frac{Q\rho(\vec{r}')}{4\pi|\vec{r}-\vec{r}'|} d^3\vec{r}' d^3\vec{r}$$

Then asks us to "fix" ##\vec{r}'## and integrate over ##d^3\vec{r}## with substitution ##\vec R = \vec r - \vec r'##, which separates the integral into two parts:

$$ \int e^{i\vec{q}.\vec R} \frac{Q}{4\pi|\vec R|} d^3\vec R \int \rho(\vec r') e^{i\vec{q}.\vec{r}'} d^3 \vec r'$$

Where the latter integral is the form factor.

How does this make sense? ##\vec R ## is a function of ##\vec r'## and thus needs to be included in the integral over ##d^3\vec r'##. If they really want to fix ##\vec r'## then ##d^3 \vec r' = 0## and the whole expression should be zero, no?
 

Answers and Replies

Charles Link
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Suggestion: Do the integral over ## d^3 r ## first, (leaving the ## r' ## in there), (and pulling the terms with just ## r' ## out front as a constant), and I think it is apparent that the integral over ## d^3 r ## that involves terms with ## r-r' ## will be independent of the value of the fixed ##r' ##, (also, since the limits on the ## d^3 r ## integral are essentially ## -\infty ## and ## +\infty ##, and will not be changed by the ## r' ##). Thereby, it is permissible to drop ## r' ## in those terms and replace it with an integration over ## R ##.
 
Last edited:
13
2
Suggestion: Do the integral over ## d^3 r ## first, (leaving the ## r' ## in there), (and pulling the terms with just ## r' ## out front as a constant), and I think it is apparent that the integral over ## d^3 r ## that involves terms with ## r-r' ## will be independent of the value of the fixed ##r' ##, (also, since the limits on the ## d^3 r ## integral are essentially ## -\infty ## and ## +\infty ##, and will not be changed by the ## r' ##). Thereby, it is permissible to drop ## r' ## in those terms and replace it with an integration over ## R ##.
I see. The ##\vec r'## will simply shift the origin of ## f(\vec r) = e^{i\vec q \cdot (\vec r-\vec r')} \frac{Q}{4\pi |\vec r-\vec r'|} ## and because the integral limits are ##-\infty## and ##+\infty##, the integral will be the same for every value of ##\vec r'##. We can therefore set ## \vec r' = \vec {r_0} ## under this integral, where ##\vec r_0## is any constant vector. We can now pull this integral out, since it is independent of ##\vec r'##
$$\int_{-\infty}^{\infty} e^{\vec q \cdot (\vec r - \vec r_0)} \frac{Q}{4\pi |\vec r - \vec r_0|} d^3 \vec r \int_{-\infty}^{\infty} \rho(\vec r') e^{i\vec q \cdot \vec r'} d^3\vec r' $$ and if we now define ## \vec R = \vec r - \vec r_0 ##, we get ##d^3\vec R = d^3\vec r## (Since ##\vec {r_0} ## is constant), and thus

$$M_{fi} = \int_{-\infty}^{\infty} e^{\vec q \cdot \vec R} \frac{Q}{4\pi |\vec R|} d^3 \vec R \int_{-\infty}^{\infty} \rho(\vec r') e^{i\vec q \cdot \vec r'} d^3\vec r' $$

Had the integral limits been finite, this would in general not have worked since shifting the origin of ##f(\vec r)## would mean integrating over a different part of ##f(\vec r)##.
 
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