Take the Rutherford scattering, as for example in this picture:

enter image description here

What is the easiest way to show that the impact parameter "b" (see picture) is the same for the ingoing and outgoing trajectories? Are there different arguments to show this?



A summary of the answer by Mike (see below for more details):

Since the scattering is elastic, the magnitude of the momentum does not change: $$ |\vec p_1|=|\vec p_2|, $$ and because the deflecting force is a central force, the angular momentum is conserved: $$ L_1=L_2. $$ From $L_1=|\vec p_1|b_1$, and $L_2=|\vec p_2|b_2$ we see immediately that $b_1=b_2$.

  • $\begingroup$ Time reversal symmetry in the elastic scattering process... $\endgroup$ – Jon Custer Feb 10 '15 at 0:25
  • $\begingroup$ @Jon That could be expanded into an nice answer to complement Mike's effort below. $\endgroup$ – dmckee Feb 10 '15 at 1:17
  • $\begingroup$ Thanks Jon, this was also my first thought. But it seemed to me to far fetched, that's why I asked! Can we establish the time reversal symmetry directly from the fact that only central forces are involved? $\endgroup$ – Stan Feb 10 '15 at 1:18
  1. We know/assume that the mass of the particle is constant.

  2. Elastic scattering means no change in kinetic energy (in the center-of-mass frame). Since the mass of the particle is constant, this means there is no change in the speed (not velocity; speed) of the particle during an elastic collision.

  3. Torque is $\vec{\tau} = \vec{r} \times \vec{F}$. For a central-force law $\vec{F}$ is along $\vec{r}$, so the cross product is zero. There is therefore no torque, and hence no change in angular momentum from beginning to end.

Now, we just put these three facts together. Since the initial angular momentum (asymptotically far in the past) is $L_0 = m_0\, v_0\, b_0$, and the final angular momentum (asymptotically far in the future) is $L_1 = m_1\, v_1\, b_1$. But since $L$, $m$, and $v$ don't change, there can be no change in $b$.

As with most any use of conservation laws, you could -- in principle -- show the same thing using integrals of the dynamics. But such an approach would surely be painful and ugly. I can't think of any other approach.

  • $\begingroup$ Thanks Mike! That's a sort of answer I was looking for. I'll include a condensed summary in my original question. $\endgroup$ – Stan Feb 10 '15 at 1:19
  • $\begingroup$ Just for completeness regarding point 2.: usually we ignore the recoil of the gold nucleus, this is equivalent to saying $m_{Gold}\to\infty$. This, in turn, means that center-of-mass frame = rest frame of the gold atoms = lab frame. $\endgroup$ – Stan Feb 10 '15 at 1:36

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