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I'm trying to express the relation of the scattering angle $\theta$ as a function of the impact parameter $b$ in the case of the scattering through a potential of the form $V(r) = \frac{k}{r^2}$. I'm following the steps of Griffiths(Introduction to Elementary Particles).

The angles involved in the deduction are these:

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We can see that $$\theta = \pi - 2 \phi_m$$

So if we know the form of $\phi_m$ as a function of $b$, the problem is solved.

I followed the steps of Griffiths and proved that in the general case(for any $V(r)$) the relation is $$\phi_m = b \int_{0}^{u_m} \frac{du}{\sqrt{1-b^2 u^2 - \frac{V}{E}}}$$

where the $u \equiv \frac{1}{r}$, and $u_m$ is associated with the closest approach.

My question: from this point it's just a matter of integrate this expression in the case where $V(r) = \frac{k}{r^2}$ but Griffiths say that $$u_m^2 = \frac{1}{b^2 + k/E}$$

Can anyone understand this last step? If I use this result from $u_m$ I obtain the correct answer: $$\theta = \pi - 2\phi_m = \pi - 2\left(\frac{\pi}{2}\frac{1}{\sqrt{1+\frac{k}{b^2E}}}\right) = \pi \left(1 - \frac{1}{\sqrt{1+\frac{k}{b^2E}}}\right)$$

I really appreciate any insight on this topic.

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  • $\begingroup$ You should double-check your equation for $\phi_m$. In particular, the $b^2$ term is not quite right. As a hint for your main question, the distance of closest approach is the position where the particle's radial velocity becomes zero. If you write the equation for $u_m^2$ in terms of $r_m$, you might see the connection. $\endgroup$
    – Endulum
    Mar 28, 2022 at 21:01
  • $\begingroup$ @Endulum thank you, I forgot the $u^2$ multiplying the $b^2$. I'm not sure if I got what you suggested, but I come up with a different reasoning (I don't know if it is correct, probably not): in the closest approach we'll have that the P.E=K.E (which is just E) so we can obtain a distance $r^2 = \frac{K}{E}$. We can add this distance to $b^2$ to obtain $r^2_m$ and as $u^2_m=\frac{1}{r2_m}$ we have $u^2_m = \frac{1}{b^2+\frac{K}{E}}$ $\endgroup$ Mar 28, 2022 at 22:06

1 Answer 1

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To see connection to the distance of closest approach, it helps to go back a few steps in the derivation. I don't have Griffiths' book, but hopefully my notation is not too different. Since you have already derived the formula for the scattering angle, I will not repeat that and will only present parts relevant to determining the distance of closest approach.

For motion in a central force, we can start from the Lagrangian in polar coordinates $$ L = \frac{1}{2} m \dot r^2 + \frac{1}{2} m r^2 \dot\theta^2 - V(r) $$ The Euler-Lagrange equations give us two equations of motion. $$ \frac{d}{dt}\frac{\partial L}{\partial \dot r} = \frac{\partial L}{\partial r} \implies \frac{d}{dt}(m\dot r) = -\frac{\partial}{\partial r}\left( \frac{1}{2}mr^2\dot\theta^2 + V(r) \right) $$ $$ \frac{d}{dt}\frac{\partial L}{\partial\dot\theta} = \frac{\partial L}{\partial \theta} \implies \frac{d}{dt}\left(mr^2\dot\theta\right) = 0 $$ The second of these implies that the angular momentum $\ell = mr^2\dot\theta$ is constant. Or, rearranging, $\dot\theta = \ell/mr^2$. We use this to eliminate $\dot\theta$ from the radial equation of motion, $$ m\ddot r =-\frac{\partial}{\partial r}\left(\frac{\ell^2}{2mr^2} + V(r)\right) $$ This now looks just like Newton's equation of motion for 1-d motion in the "effective potential" $U(r) = \frac{\ell^2}{2mr^2} + V(r)$. In this 1-d picture, the particle's kinetic energy is $\frac{1}{2}m\dot r^2$, so the conservation of energy during the scattering can be expressed by the relationship $$ E = \frac{1}{2}m\dot r^2 + U(r)$$ Now, the particle's distance of closest approach, $r_m$, is the distance where $\dot r=0$ (where it stops getting closer to the center). This implies that $U(r_m) = E$, which you can use to solve for $r_m$.

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