# Form factor in scattering

• A

## 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?

Related High Energy, Nuclear, Particle Physics News on Phys.org
Homework Helper
Gold Member
2018 Award
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:
• vanhees71
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 $f(\vec r)$.

Last edited:
• 