Question

In class we derived the Rutherford cross section employing the fact of cancellation of the r in the integral we had. There is another method that works in principle for any central force. Use $\psi$, the projectile's polar angle, measured from the direction of $\vec{u}$, then $\psi \to \pm \psi_0$ as $t \to \pm \infty$. Angle $\psi_0$ is equal $\int \dot{\psi} dt$ taken from the time of the closest approach to $\infty$. Rewrite this integral as $\int \frac{\dot{\psi}}{\dot{r}} dr$. Next write $\dot{\psi}$ in terms of angular momentum l and r, rewrite $\dot{r}$ in terms of energy E and the effective potential $U_{eff}$. You should be able by doing this prove that: $\theta = \pi - 2 \int_{r_{min}}^{\infty} \frac{(\frac{b}{r^2}) dr}{\sqrt{1 - (\frac{b}{r})^2 - U(r)/E}}$

          In class we derived the Rutherford cross section employing the fact of cancellation of the r in the integral we had. There is another method that works in principle for any central force. Use $\psi$, the projectile's polar angle, measured from the direction of $\vec{u}$, then $\psi \to \pm \psi_0$ as $t \to \pm \infty$. Angle $\psi_0$ is equal $\int \dot{\psi} dt$ taken from the time of the closest approach to $\infty$. Rewrite this integral as $\int \frac{\dot{\psi}}{\dot{r}} dr$. Next write $\dot{\psi}$ in terms of angular momentum l and r, rewrite $\dot{r}$ in terms of energy E and the effective potential $U_{eff}$. You should be able by doing this prove that:
$\theta = \pi - 2 \int_{r_{min}}^{\infty} \frac{(\frac{b}{r^2}) dr}{\sqrt{1 - (\frac{b}{r})^2 - U(r)/E}}$

In class we derived the Rutherford cross section employing the fact of cancellation of the r in the integral we had. There is another method that works in principle for any central force. Use ψ, the projectile's polar angle, measured from the direction of u⃗, then ψ→±ψ0 as t →±∞. Angle ψ0 is equal ∫ψ̇ dt taken from the time of the closest approach to ∞. Rewrite this integral as ∫(ψ̇)/(ṙ) dr. Next write ψ̇ in terms of angular momentum l and r, rewrite ṙ in terms of energy E and the effective potential Ueff. You should be able by doing this prove that:
θ = π - 2 ∫rmin^∞(((b)/(r^2)) dr)/(√(1 - ((b)/(r))^2 - U(r)/E))

Added by John H.

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