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 v1\vb{v}_1. Coulomb repulsion causes 1 to deflect by an angle θ\theta, and pushes 2 away in the process:

Two-body repulsive 'collision'

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

By combining Coulomb’s law with Newton’s laws, these particles’ equations of motion are found to be as follows, where r≡∣r1 ⁣− ⁣r2∣r \equiv |\vb{r}_1 \!-\! \vb{r}_2| is the distance between 1 and 2:

m1dv1dt=F1=q1q24πε0r1−r2r3m2dv2dt=F2=−F1\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 \qquad m_2 \dv{\vb{v}_2}{t} = \vb{F}_2 = - \vb{F}_1 \end{aligned}

Using the reduced mass μ≡m1m2/(m1 ⁣+ ⁣m2)\mu \equiv m_1 m_2 / (m_1 \!+\! m_2), we turn this into a one-body problem:

μdvdt=q1q24πε0rr3\begin{aligned} \mu \dv{\vb{v}}{t} = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}}{r^3} \end{aligned}

Where v≡v1 ⁣− ⁣v2\vb{v} \equiv \vb{v}_1 \!-\! \vb{v}_2 is the relative velocity, and r≡r1 ⁣− ⁣r2\vb{r} \equiv \vb{r}_1 \!-\! \vb{r}_2 is the relative position. The latter is as follows in cylindrical polar coordinates (r,φ,z)(r, \varphi, z):

r=rcos⁡φ e^x+rsin⁡φ e^y+z e^z=r e^r+z e^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 r1=r2\vb{r}_1 = \vb{r}_2. Crucially, note the symmetry: if the “collision” occurs at t=0t = 0, then by comparing t>0t > 0 and t<0t < 0 we can see that vxv_x is unchanged for any given ±t\pm t, while vyv_y simply changes sign:

Equivalent one-body deflection

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

v=r′=(r′cos⁡φ−rφ′sin⁡φ) e^x+(r′sin⁡φ+rφ′cos⁡φ) e^y+z′ e^z=r′ (cos⁡φ e^x+sin⁡φ e^y)+rφ′ ( ⁣− ⁣sin⁡φ e^x+cos⁡φ e^y)+z′ e^z=r′ e^r+rφ′ e^φ+z′ e^z\begin{aligned} \vb{v} = \vb{r}' &= \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 e^r\vu{e}_r and e^φ\vu{e}_\varphi. If we choose the coordinate system such that all dynamics are in the (x,y)(x,y)-plane, i.e. z(t)=0z(t) = 0, we have:

r=r e^rv=r′ e^r+rφ′ e^φ\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 L\vb{L} is as follows, pointing purely in the zz-direction:

L(t)=μr×v=μ(re^r×rφ′e^φ)=μr2φ′ e^z\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, in the figure above, imagine a right-angled triangle with hypotenuse r\vb{r} and short side bb. When t→+∞t \to +\infty, trigonometry tells us the following, where χ\chi is the final angle between v\vb{v} and r\vb{r}:

lim⁡t→+∞br(t)=sin⁡χχ≡lim⁡t→+∞∡(r(t),v(t))\begin{aligned} \lim_{t \to +\infty} \frac{b}{r(t)} = \sin{\chi} \qquad \qquad \chi \equiv \lim_{t \to +\infty} \measuredangle(\vb{r}(t), \vb{v}(t)) \end{aligned}

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

lim⁡t→+∞∣L(t)∣=μr∣v∣sin⁡χ=μb∣v∣\begin{aligned} \lim_{t \to +\infty} \big| \vb{L}(t) \big| = \mu r |\vb{v}| \sin{\chi} = \mu b |\vb{v}| \end{aligned}

This is useful, because angular momentum is conserved, i.e. L\vb{L} is constant in time. We prove this by using the product rule of differentiation, and replacing μv′\mu \vb{v}' with the reduced equation of motion:

L′(t)=μ(r×v′+v×v)=r×(μv′)=r×(q1q24πε0rr3)=0\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}

Thanks to this, we can equate the two preceding expressions for the magnitude ∣L∣|\vb{L}|, leading to the relation below. Note the appearance of a new minus, because the sketch shows that φ′<0\varphi' < 0, i.e. φ\varphi decreases with increasing tt:

−μr2dφdt=μb∣v∣  ⟹  dt=−r2b∣v∣dφ\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 yy-component is given by:

μdvydt=q1q24πε0yr3  ⟹  μdvy=q1q24πε0yr3dt\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 dt\dd{t} with our earlier relation, and recognize geometrically that y/r=sin⁡φy/r = \sin{\varphi}:

μdvy=−q1q24πε0b∣v∣yrdφ=−q1q24πε0b∣v∣sin⁡φdφ=q1q24πε0b∣v∣d(cos⁡φ)\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 ii at t→−∞t \to -\infty to the final state ff at t→+∞t \to +\infty yields:

Δvy≡∫ifdvy=q1q24πε0b∣v∣μ(cos⁡φf−cos⁡φi)\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 Δvy=vy,f ⁣− ⁣vy,i=2vy,f\Delta v_y = v_{y,f} \!-\! v_{y,i} = 2 v_{y,f}, and that φi=π ⁣− ⁣φf\varphi_i = \pi \!-\! \varphi_f, such that:

Δvy=2vy,f=q1q24πε0b∣v∣μ(cos⁡φf−cos⁡(π ⁣− ⁣φf))=q1q24πε0b∣v∣μ(2cos⁡φf)\begin{aligned} \Delta v_y = 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 for t→+∞t \to +\infty we notice that vy,f=∣v∣sin⁡φfv_{y,f} = |\vb{v}| \sin{\varphi_f}, leading to:

2∣v∣sin⁡φf=q1q22πε0b∣v∣μcos⁡φf\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 φf\varphi_f:

tan⁡φf=sin⁡φfcos⁡φf=q1q24πε0b∣v∣2μ\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 the deflection angle θ\theta, not φf\varphi_f. One last use of symmetry and geometry tells us that θ=2φf\theta = 2 \varphi_f, and we thus arrive at the celebrated Rutherford scattering formula:

tan⁡ ⁣(θ2)=q1q24πε0b∣v∣2μ\begin{aligned} \boxed{ \tan\!\bigg( \frac{\theta}{2} \bigg) = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu} } \end{aligned}

In fact, this formula is also valid if q1q_1 and q2q_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.