Cross section of coulomb sacttering vs Rutherford

In summary, the differential cross section of Coulomb scattering can be expressed as a form factor times Rutherford's model. However, if the potential is non-spherical or the scattered particle penetrates too far into the charge distribution, Rutherford's model may not be adequate. In this case, one must go back to elementary principles and solve for the trajectory of the particle using conservation laws. The resulting equation can then be used to determine the distance of closest approach, scattering angle, and differential scattering cross section. In order to find the charge distribution, one can use Poisson's equation and solve some integral equations.
  • #1
oosgood
7
0
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?
 
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  • #2
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
 
Last edited:
  • #3
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.
 

1. What is the difference between Coulomb scattering and Rutherford scattering?

Coulomb scattering and Rutherford scattering are both types of elastic scattering, which means that the particles involved do not change their energy or identity during the interaction. The main difference between the two is the strength of the interaction between the particles. Coulomb scattering occurs when the particles interact through the electromagnetic force, while Rutherford scattering occurs when the particles interact through the strong nuclear force. In other words, Coulomb scattering is a result of the repulsion between like-charged particles, while Rutherford scattering is a result of the attraction between positively charged particles and negatively charged particles.

2. How does the cross section of Coulomb scattering compare to the cross section of Rutherford scattering?

The cross section of scattering is a measure of the probability of a particle being scattered by another particle. In Coulomb scattering, the cross section is typically larger than in Rutherford scattering because the electromagnetic force is much stronger than the strong nuclear force. This means that particles interacting through the electromagnetic force are more likely to be scattered than those interacting through the strong nuclear force.

3. What factors affect the cross section of Coulomb scattering and Rutherford scattering?

The cross section of both Coulomb scattering and Rutherford scattering is affected by several factors, including the charge and mass of the particles, the distance between the particles, and the energy of the particles. In general, the higher the energy of the particles, the larger the cross section will be. Additionally, the cross section for Coulomb scattering increases as the charge of the particles increases and as the distance between the particles decreases. For Rutherford scattering, the cross section increases as the mass of the particles increases and as the distance between the particles decreases.

4. What is the relationship between the cross section of Coulomb scattering and the impact parameter?

The impact parameter is the distance between the path of a particle and the center of the scattering particle. In Coulomb scattering, the cross section is directly related to the impact parameter. As the impact parameter decreases, the cross section increases. This is because when the particles are closer together, the electromagnetic force is stronger and the particles are more likely to be scattered. However, in Rutherford scattering, the cross section is not directly related to the impact parameter due to the strong nuclear force also playing a role in the interaction.

5. How does the cross section of Coulomb scattering vs Rutherford scattering change with increasing energy?

As the energy of the particles increases, the cross section of both Coulomb scattering and Rutherford scattering also increases. This is because higher energy particles have a greater chance of interacting and being scattered by other particles. However, the relationship between cross section and energy is different for Coulomb scattering and Rutherford scattering. In Coulomb scattering, the cross section increases more rapidly with increasing energy compared to Rutherford scattering, as the electromagnetic force is stronger at higher energies.

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