Relativistic Central Force Problem and Scattering

Question

I derived the relativistic Binet-equation for a particle of charge $q$ subject to electromagnetic interaction due to a fixd point charge $Q$ ($Qq>0$) held at the origin. Let $u=1/r$. The result I got is $$\frac{d^2u}{d\theta^2}+u=-\frac{m\gamma}{L^2}\frac{dV}{du}$$ where $V(u)=Qqu$ (I work in CGS units) and in my case of interest (that is, scattering of a particle with a given initial velocity $v_0$ and impact parameter $b$) $L=m\gamma_0bv_0$. Using conservation of energy $E=m\gamma_0c^2=m\gamma c^2+V(u)$ and rearranging, we can write the equation as $$\frac{d^2u}{d\theta^2}+u\left(1-\left(\frac{Qq}{Lc}\right)^2\right)=-\frac{QqE}{L^2c^2}$$ Now what I am struggling with: Let's say I want to get the non-relativistic limit from this equation, that is $\gamma \simeq 1$ at all times. If we write the velocity of the particle in polar coordinates and use conservation of angular momentum I get $$\gamma=\sqrt{1+\left(\frac{L}{mc}\right)^2\left(\left(\frac{du}{d\theta}\right)^2+u^2\right)}$$

Now, by dimensional analysis on the relevant quantities $u(\theta) \propto \frac{QqE}{L^2c^2}$ so I get the following condition for the non-relativistic limit to hold at all times: $Lc>>Qq$ ($E \simeq mc^2$ in the non relativistic limit). What I am struggling to understand is $1)$ where I could have got this same condition working in the Newtonian frame and $2)$ what happens if the angular momentum is really small due to a very low initial velocity. In particular, what I am struggling to understand is how the case in which the initial angular momentum is very small ($Qq/Lc>1$) is linked to the Newtonian limit of this equation. Surely if the angular momentum is small because the initial velocity is, then in the subsequent motion the velocity will be at most equal to the initial one so we are always in the non relativistic regime. So how can I deal with the fact that in the second equation, for very small angular momentum, I get a totally different solution from the one obtained in the non relativistic regime by putting $\gamma \simeq 1$?

For other comments, I've already looked up the link Mercury's Orbital Precession in Special Relativity and they assume implicitly the condition $Lc>Qq$.

Answer 1

To de-clutter the equations, I will set $m=Qq=c=1$. In general, it is often more convenient to use the integrated Binet formula rather than the differential one. Recall that in Newtonian mechanics, starting from the Hamiltonian: $$ H = \frac12p^2+\frac1r $$ you get the effective radial Hamiltonian: $$ H_r = \frac12p_r^2+\frac{L^2}{2r^2}+\frac1r $$ Using: $$ p_r = \dot r = -L\frac{du}{d\theta} $$ this gives the familiar Binet formula: $$ \frac{L^2}2\left[\left(\frac{du}{d\theta}\right)^2+u^2\right]+u = E $$ or in its derivative form: $$ L^2\left[\frac{d^2u}{d\theta^2}+u\right]+1 = 0 $$ Similarly in the relativistic case: $$ H = \sqrt{1+p^2}-1+\frac1r \\ H_r = \sqrt{1+p_r^2+\frac{L^2}{r^2}}-1+\frac1r $$ This time, you need to use: $$ \dot r = \frac{p_r}{\sqrt{1+p_r^2+\frac{L^2}{r^2}}} \\ p_r = \sqrt{\frac{1+\frac{L^2}{r^2}}{1-\dot r^2}}\dot r \\ \gamma = \sqrt{1+p_r^2+\frac{L^2}{r^2}} = \sqrt{\frac{1+\frac{L^2}{r^2}}{1-\dot r^2}} $$ so that: $$ \sqrt{\frac{1+L^2u^2}{1-L^2\left(\frac{du}{d\theta}\right)^2}}-1+u = E $$ If you insist in writing it in the differentiated version, you will get: $$ \frac{\gamma_r^2L^2}\gamma\left[\gamma^2\frac{d^2u}{d\theta^2}+u\right]+1 = 0 \\ \gamma_r^2 = \frac1{\sqrt{1-L^2\left(\frac{du}{d\theta}\right)^2}} $$ and in the non relativistic limit, both $\gamma_r\to1$ and $\gamma\to1$, so you do recover the previous Newtonian Binet formula.

Your non-relativistic limit criterion was not accurate. It is obtained when: $$ p_r^2+\frac{L^2}{r^2} \ll 1 $$ By energy conservation: $$ p_r^2+\frac{L^2}{r^2} = E-\frac1r \leq E $$ so the non relativistic limit is uniformly valid in the repulsive case as long as: $$ E\ll 1 $$ Notice that the bound is uniform in $L$, only $E$ matters.

Angular momentum is only relevant in the attractive case, where you need the minimal radius: $$ p_r^2+\frac{L^2}{r^2} = E+\frac1r \leq E+\frac{1+\sqrt{1+4EL^2}}{2L^2} $$ Therefore, the condition is non uniform in $L$: $$ E+\frac{1+\sqrt{1+4EL^2}}{2L^2}\ll 1 $$ This is to be expected since even at low energy, a low impact parameter can allow considerable speed up at the point of closest approach. This unlike the repulsive case where the particle necessarily slows down as it gets closer.

Back to the Binet formula, the limit is obtained when both: $$ \gamma_r\to1 \\ L^2\left[\left(\frac{du}{d\theta}\right)^2+u^2\right]\ll1 $$ The latter is precisely equal to quantity of the previous discussion, and the former condition is automatically satisfied if the latter is.

Hope this helps.