cross section of coulomb sacttering vs Rutherfordby oosgood Tags: coulomb, cross, rutherford, sacttering 

#1
Feb1812, 04:13 AM

P: 7

differential cross section of coulomb scattering can be expressed as a form factor times Rutherford one, we model the charge we used spherically symmetric charge density, if we replace this point charge with other distribution like Gaussian , what is the difference between these two models, is there physical explain?




#2
Mar312, 07:41 AM

P: 84

If your potential is firstly spherically symmetric and secondly your scattered particle does not penetrate too far into the charge distribution then Rutherford's model is adequate. The phrase I used of 'too far' is a matter of accuracy. A Gaussian potential clearly is non zero at large distances and will affect the result.
If you want to consider non spherical symmetrical potentials or when the kinetic energy of the scattered particle is significant with respect to the electrostatic energy of the scattering particle ie it goes 'too far' then you will need to go back to elementary principles  see Goldstein's Classical Mechanics  but be warned there are very few 'closed' ie analytic solutions for such problems. A help maybe in a totally different area of scattering theory look up the extensions made to Mie's theory of scattering from ellipsoids  the math will help Hope this helps Regards and good luck Sam 



#3
Mar312, 08:22 AM

P: 3,015

Let us suppose the impact factor is [itex]b[/itex], and the initial velocity is [itex]v_\infty[/itex]. If the potential energy is spherically symmetric [itex]V = V(r)[/itex] and vanishes at infinity, then the law of conservation of angular momentum and energy give:
[tex] m \, v_\infty \, b = m \, r^2 \, \dot{\phi} [/tex] [tex] E = \frac{m \, v^2_\infty}{2} = \frac{m}{2} \, \left(\dot{r}^2 + r^2 \, \dot{\phi}^2 \right) + V(r) [/tex] One can then eliminate [itex]\dot{\phi}[/itex] from the first equation and use it to go over from time dependence to angular dependance: [tex] \dot{r} = \frac{d r}{d \phi} \, \dot{\phi} = \frac{v_\infty \, b}{r^2} \, \frac{d r}{d \phi} = v_\infty \, \frac{d}{d \phi} \left( \frac{b}{r} \right), \ z \equiv \frac{b}{r} [/tex] to get a 1st order differential equation for the trajectory of the particle: [tex] E = \frac{m \, v^2_\infty}{2} = \frac{m \, v^2_\infty}{2} \, \left[ \left( \frac{d z}{d \phi} \right)^2 + z^2 \right] + V \left( \frac{b}{z} \right) [/tex] [tex] \left( \frac{d z}{d \phi} \right)^2 = 1  \frac{V(b/z)}{E}  z^2 [/tex] The value for which the r.h.s. becomes zero determines the distance of closest approach [itex]r_0(b, E) = b/z_0(b, E)[/itex]. If we measure the angle [itex]\phi[/itex] from that point, then r increases and z decreases with increasing angle. This means we need to take the negative root of the above equation and the variables separate: [tex] \phi = \int_{b/r}^{z_0(b, E)}{\frac{d z}{\sqrt{1  \frac{V(b/z)}{E}  z^2}}} [/tex] As we take [itex]r \rightarrow \infty[/itex], the angle approaches the "scattering angle". Due to symmetry, we have: [tex] \theta(b, E) = \pi  2 \, \int_{0}^{z_0(b, E)}{\frac{d z}{\sqrt{1  \frac{V(b/z)}{E}  z^2}}} [/tex] where, again, [itex]z_0[/itex] is determined from: [tex] 1  \frac{V(b/z_0)}{E}  z^2_0 = 0 [/tex] Once you know [itex]\theta = \theta(b; E) \Rightarrow b = b(\theta; E)[/itex], you can find the differential scattering cross section from: [tex] \sigma_d(\theta; E) = \frac{d\sigma}{d \Omega} = \frac{b(\theta; E)}{\sin \theta} \, \frac{d b}{d \theta} [/tex] More interesting for you is the inverse problem: Given [itex]\sigma_d(\theta; E)[/itex], can you find [itex]V = V(r)[/itex]? This amounts to solving some intergral equations, and is rather involved, so I cannot present it right now. Once you have that, you can find the charge distribution from Poisson's equation: [tex] \nabla^2 V = \frac{1}{r^2} \, \frac{d}{d r} \left( r^2 \, \frac{d V}{d r} \right) = \frac{q\, \rho(r)}{\epsilon_0} [/tex] where q is the charge of the projectile. 


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