# How much energy, momentum, and/or angular momentum are lost to radiation in Rutherford scattering?

Inspired by this question:

Consider two charged particles, of masses $$m_1$$ & $$m_2$$ and charges $$q_1$$ and $$q_2$$. They approach each other from a great distance, interact via their electromagnetic fields, and end up going in different directions. (For simplicity, let's work in the CM frame.) Usually when we analyse this problem, we assume that the two particles interact via a potential energy $$V(\vec{r}_1, \vec{r}_2) = -\frac{q_1 q_2}{4 \pi \epsilon_0 r_{12}}.$$ Under these assumptions, via standard arguments, the total mechanical energy $$E_\text{mech} = \sum_i \frac{1}{2} m_i \vec{v}_i^2$$, mechanical momentum $$\vec{p}_\text{mech} = \sum_i m_i \vec{v}_i$$, and mechanical angular momentum $$\vec{L}_\text{mech} = \sum_i m_i \vec{r}_i \times \vec{v}_i$$ of the two particles are conserved in the scattering event.

However, we also know that these charges are accelerating during their collision, and accelerating charges (can) radiate. This radiation can in principle carry energy, linear momentum, and angular momentum away from the particles.

1. Do scattering charges radiate net energy? If so, how much?
2. Does the total mechanical momentum of the charges change in the scattering process? If so, what is $$\Delta \vec{p}_\text{mech}$$?
3. Does the total mechanical angular momentum of the charges change in the scattering process? If so, what is $$\Delta \vec{L}_\text{mech}$$?

It seems like this is a natural enough question that someone should have already addressed it, so pointers to the literature (in lieu of a full answer here) would still be helpful. I have a nagging suspicion that there is a nice symmetry argument to be made for question 2 (and possible question 3 as well), but I can't quite put my finger on it.

I'm most interested in an answer couched in the language of classical electrodynamics, though if insight can be gleaned from quantum probabilities I'd be happy to hear about it.

• My direct experience is with electron scattering. The energy loss (even for "elastic" kinematics) varies from event to event, and runs from less than the experimental resolution up to significant fractions (several percent or more) of the initial CoM energy. Your question assumes a classical continuous basis, but the radiation takes on the form of a population dominated by quantum field theory statistics. Alphas should be cleaner (lower accelerations, right?) but ... it's still going to be a statistical process with large variations from event to event. Aug 16, 2019 at 19:00
• For alpha particles at ~1 to 10MeV, radiation losses are not detectable and are not considered at all in the analysis of Rutherford Backscattering Spectrometry. The mass difference of >7300 is certainly one reason. Aug 16, 2019 at 19:36
• Isn't bremsstrahlung exactly the answer to (1)? Sep 24, 2020 at 20:36

1. Yes, scattering charges radiate energy. According to pages 717 and 718 of the aforementioned book, the total energy lost in radiation by a particle with charge $$ze$$ and mass $$M$$ traversing unit thickness of matter containing one fixed charge with a charge $$Ze$$ per unit volume is $$\frac{dE_{rad}}{dx}=\frac{16}{3}Z\left(\frac{Ze^2}{\hbar c}\right)\frac{z^4e^4}{Mc^2}\int_0^1\mathrm{d}x\ln\left(\frac{1+\sqrt{1-x}}{\sqrt{x}}\right)$$
2. I believe the change in momentum is $$(\Delta p)^2=2p^2(1-\cos\theta')$$ as given on page 714. $$\theta'$$ is the scattering angle.