Graduate What is the Vector Manipulation Formula for Elastic Scattering Angle?

[JD_PM]

I was reading *Introduction to Nuclear Physics* by Krane and stumbled on the following (page 47):

In Elastic scattering, the initial electron wave function is of the form $e^{i k_i r}$ (free particle of momentum $p_i = \hbar k_i$). The scattered electron can also be regarded as a free particle of momentum $p_f = \hbar k_f$ and wave function $e^{i k_f r}$.

The interaction $V(r)$ converts the initial wave into the scattered wave; the probability for the transition will be proportional to the square of the following quantity:

$$F(q) = \int V(r) e^{iqr}dv$$

Plugging both Coulomb's potential and charge-per-unit-volume into $F(q)$:

$$F(q) = \int e^{iqr'} \rho(r') dv'$$

Normalizing and knowing that $\rho(r')$ just depends on $r'$ (and not on $\theta'$ nor $\phi'$) we get:

$$F(q) = \frac{4\pi}{q}\int r' sin (qr') \rho(r') dr'$$

Where $q = k_i - k_f$. The scattering is elastic, so momentum is conserved ($p_i = p_f$) and $q$ is merely a function of the scattering angle $\alpha$ between $p_i$ and $p_f$.

**Now a bit of vector manipulation shows:**

$$q = \frac{2p}{\hbar}sin(\frac{\alpha}{2})$$

**I do not know how to get the last expression**

 

[DrClaude]

Using $| \mathbf{k}_i | = |\mathbf{k}_f| = k$, calculate $q^2$:
$$
\begin{align*}
q^2 &= (\mathbf{k}_i - \mathbf{k}_f)^2 \\
&= k^2 - 2 \mathbf{k}_i \cdot \mathbf{k}_f - k^2 \\
&= 2 k^2 (1 - \cos \alpha) \\
&= 4 k^2 \sin^2 (\alpha /2 )
\end{align*}
$$
Taking the square root and using $p = \hbar k$ will give you your equation.

 

[JD_PM]

Using $| \mathbf{k}_i | = |\mathbf{k}_f| = k$, calculate $q^2$:
$$
\begin{align*}
q^2 &= (\mathbf{k}_i - \mathbf{k}_f)^2 \\
&= k^2 - 2 \mathbf{k}_i \cdot \mathbf{k}_f - k^2 \\
&= 2 k^2 (1 - \cos \alpha) \\
&= 4 k^2 \sin^2 (\alpha /2 )
\end{align*}
$$
Taking the square root and using $p = \hbar k$ will give you your equation.

In your third line; I do not see how you get the $-2k^2 cos \alpha$ term

 

[DrClaude]

In your third line; I do not see how you get the $-2k^2 cos \alpha$ term

$$
\begin{align*}
\mathbf{k}_i \cdot \mathbf{k}_f &= | \mathbf{k}_i | |\mathbf{k}_f| \cos \alpha \\
&= k^2 \cos \alpha
\end{align*}
$$
where $\alpha$ is the angle between $\mathbf{k}_i$ and $\mathbf{k}_f$ (or $\mathbf{p}_i$ and $\mathbf{p}_f$).

 

[JD_PM]

$$
\begin{align*}
\mathbf{k}_i \cdot \mathbf{k}_f &= | \mathbf{k}_i | |\mathbf{k}_f| \cos \alpha \\
&= k^2 \cos \alpha
\end{align*}
$$
where $\alpha$ is the angle between $\mathbf{k}_i$ and $\mathbf{k}_f$ (or $\mathbf{p}_i$ and $\mathbf{p}_f$).

It is the dot product definition! My bad, thanks DrClaude.