Categories: Physics, Plasma physics.

# Rutherford scattering

Rutherford scattering or Coulomb scattering is an elastic pseudo-collision of two electrically charged particles. It is not a true collision, and is caused by Coulomb repulsion.

The general idea is illustrated below. Consider two particles 1 and 2, with the same charge sign. Let 2 be initially at rest, and 1 approach it with velocity $\vb{v}_1$. Coulomb repulsion causes 1 to deflect by an angle $\theta$, and pushes 2 away in the process:

Here, $b$ is called the impact parameter. Intuitively, we expect $\theta$ to be larger for smaller $b$.

By combining Coulomb’s law with Newton’s laws, these particles’ equations of motion are found to be as follows, where $r = |\vb{r}_1 - \vb{r}_2|$ is the distance between 1 and 2:

\begin{aligned} m_1 \dv{\vb{v}_1}{t} = \vb{F}_1 = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}_1 - \vb{r}_2}{r^3} \qquad \quad m_2 \dv{\vb{v}_2}{t} = \vb{F}_2 = - \vb{F}_1 \end{aligned}

Using the reduced mass $\mu \equiv m_1 m_2 / (m_1 \!+\! m_2)$, we turn this into a one-body problem:

\begin{aligned} \mu \dv{\vb{v}}{t} = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}}{r^3} \end{aligned}

Where $\vb{v} \equiv \vb{v}_1 \!-\! \vb{v}_2$ is the relative velocity, and $\vb{r} \equiv \vb{r}_1 \!-\! \vb{r}_2$ is the relative position. The latter is as follows in cylindrical polar coordinates $(r, \varphi, z)$:

\begin{aligned} \vb{r} = r \cos{\varphi} \:\vu{e}_x + r \sin{\varphi} \:\vu{e}_y + z \:\vu{e}_z = r \:\vu{e}_r + z \:\vu{e}_z \end{aligned}

These new coordinates are sketched below, where the origin represents $\vb{r}_1 = \vb{r}_2$. Crucially, note the symmetry: if the “collision” occurs at $t = 0$, then by comparing $t > 0$ and $t < 0$ we can see that $v_x$ is unchanged for any given $\pm t$, while $v_y$ simply changes sign:

From our expression for $\vb{r}$, we can find $\vb{v}$ by differentiating with respect to time:

\begin{aligned} \vb{v} &= \big( r' \cos{\varphi} - r \varphi' \sin{\varphi} \big) \:\vu{e}_x + \big( r' \sin{\varphi} + r \varphi' \cos{\varphi} \big) \:\vu{e}_y + z' \:\vu{e}_z \\ &= r' \: \big( \cos{\varphi} \:\vu{e}_x + \sin{\varphi} \:\vu{e}_y \big) + r \varphi' \: \big( \!-\! \sin{\varphi} \:\vu{e}_x + \cos{\varphi} \:\vu{e}_y \big) + z' \:\vu{e}_z \\ &= r' \:\vu{e}_r + r \varphi' \:\vu{e}_\varphi + z' \:\vu{e}_z \end{aligned}

Where we have recognized the basis vectors $\vu{e}_r$ and $\vu{e}_\varphi$. If we choose the coordinate system such that all dynamics are in the $(x,y)$-plane, i.e. $z(t) = 0$, we have:

\begin{aligned} \vb{r} = r \: \vu{e}_r \qquad \qquad \vb{v} = r' \:\vu{e}_r + r \varphi' \:\vu{e}_\varphi \end{aligned}

Consequently, the angular momentum $\vb{L}$ is as follows, pointing purely in the $z$-direction:

\begin{aligned} \vb{L}(t) = \mu \vb{r} \cross \vb{v} = \mu \big( r \vu{e}_r \cross r \varphi' \vu{e}_\varphi \big) = \mu r^2 \varphi' \:\vu{e}_z \end{aligned}

Now, from the figure above, we can argue geometrically that at infinity $t = \pm \infty$, the ratio $b/r$ is related to the angle $\chi$ between $\vb{v}$ and $\vb{r}$ like so:

\begin{aligned} \frac{b}{r(\pm \infty)} = \sin{\chi(\pm \infty)} \qquad \quad \chi(t) \equiv \measuredangle(\vb{r}, \vb{v}) \end{aligned}

With this, we can rewrite the magnitude of the angular momentum $\vb{L}$ as follows, where the total velocity $|\vb{v}|$ is a constant, thanks to conservation of energy:

\begin{aligned} \big| \vb{L}(\pm \infty) \big| = \mu \big| \vb{r} \cross \vb{v} \big| = \mu r |\vb{v}| \sin{\chi} = \mu b |\vb{v}| \end{aligned}

However, conveniently, angular momentum is also conserved, i.e. $\vb{L}$ is constant in time:

\begin{aligned} \vb{L}'(t) &= \mu \big( \vb{r} \cross \vb{v}' + \vb{v} \cross \vb{v} \big) = \vb{r} \cross (\mu \vb{v}') = \vb{r} \cross \Big( \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}}{r^3} \Big) = 0 \end{aligned}

Where we have replaced $\mu \vb{v}'$ with the equation of motion. Thanks to this, we can equate the two preceding expressions for $\vb{L}$, leading to the relation below. Note the appearance of a new minus, because the sketch shows that $\varphi' < 0$, i.e. $\varphi$ decreases with increasing $t$:

\begin{aligned} - \mu r^2 \dv{\varphi}{t} = \mu b |\vb{v}| \quad \implies \quad \dd{t} = - \frac{r^2}{b |\vb{v}|} \dd{\varphi} \end{aligned}

Now, at last, we turn to the main equation of motion. Its $y$-component is given by:

\begin{aligned} \mu \dv{v_y}{t} = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{y}{r^3} \quad \implies \quad \mu \dd{v_y} = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{y}{r^3} \dd{t} \end{aligned}

We replace $\dd{t}$ with our earlier relation, and recognize geometrically that $y/r = \sin{\varphi}$:

\begin{aligned} \mu \dd{v_y} = - \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \frac{y}{r} \dd{\varphi} = - \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \sin{\varphi} \dd{\varphi} = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \dd{(\cos{\varphi})} \end{aligned}

Integrating this from the initial state $i$ at $t = -\infty$ to the final state $f$ at $t = \infty$ yields:

\begin{aligned} \Delta v_y \equiv \int_{i}^{f} \dd{v_y} = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos{\varphi_i} \big) \end{aligned}

From symmetry, we see that $\varphi_i = \pi \!-\! \varphi_f$, and that $\Delta v_y = v_{y,f} \!-\! v_{y,i} = 2 v_{y,f}$, such that:

\begin{aligned} 2 v_{y,f} = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos(\pi \!-\! \varphi_f) \big) = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( 2 \cos{\varphi_f} \big) \end{aligned}

Furthermore, geometrically, at $t = \infty$ we notice that $v_{y,f} = |\vb{v}| \sin{\varphi_f}$, leading to:

\begin{aligned} 2 |\vb{v}| \sin{\varphi_f} = \frac{q_1 q_2}{2 \pi \varepsilon_0 b |\vb{v}| \mu} \cos{\varphi_f} \end{aligned}

Rearranging this yields the following equation for the final polar angle $\varphi_f \equiv \varphi(\infty)$:

\begin{aligned} \tan{\varphi_f} = \frac{\sin{\varphi_f}}{\cos{\varphi_f}} = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu} \end{aligned}

However, we want $\theta$, not $\varphi_f$. One last use of symmetry and geometry tells us that $\theta = 2 \varphi_f$, and we thus arrive at the celebrated Rutherford scattering formula:

\begin{aligned} \boxed{ \tan\!\Big( \frac{\theta}{2} \Big) = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu} } \end{aligned}

In fact, this formula is also valid if $q_1$ and $q_2$ have opposite signs; in that case particle 2 is simply located on the other side of particle 1’s trajectory.

## References

1. P.M. Bellan, Fundamentals of plasma physics, 1st edition, Cambridge.
2. M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.