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.


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 M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.