Rutherford Scattering Derivation

In summary, the Rutherford scattering trajectory is derived by considering the conservation of linear momentum and the coulomb force between the incoming particle and the target nucleus. The acceleration is divided into two components - radial and centripetal - which are both dependent on the distance r. The second term, which represents the centripetal acceleration, is an artifact of the polar coordinate system and is independent of the force.
  • #1
Dalkiel
9
0

Homework Statement


I'm trying to work out the derivation of the the Rutherford scattering trajectory. I understand the conservation of linear momentum, and that the only force acting is the coulomb force between the incoming particle and the target nucleus. Early on in the derivation I'm told that essentially there are two components to the acceleration, a radial and centripetal. I understand each component separately, but I just don't understand why there are two parts to the acceleration.


Homework Equations


[itex]F = \frac{zZe^{2}}{4\pi\epsilon_{0}r^{2}}=M\left[\frac{d^{2}r}{dt^{2}}-r\left(\frac{d\varphi}{dt}\right)^{2}\right][/itex]

XLtHPl8.png


The Attempt at a Solution


The force is dependent on r, and thus so is the acceleration. The first term on the right (second derivative of position) I understand as the acceleration at a given distance r, and is the radial acceleration. The second term is where I'm running into a bit of trouble. It is the centripetal acceleration, and has a direction opposite that of the radial acceleration. I can also see that it is radial velocity squared times r. I just don't see why it's in the derivation. The acceleration is dependent on r, and we have the first term that takes that into consideration.
 
Physics news on Phys.org
Back
Top