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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?

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?