Conservation of angular momentum in scattering processes

  • #1
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Greetings.

So... let us consider a particle moving in the yz plane, coming from the infinite towards a region were a gravitational potential is appreciable. The Lagrangian of the system is

[itex]\mathcal{L} = \frac{1}{2}\mu (\dot{r}^2+r^2{\dot \phi}^2) + \frac{G\,m\,M}{r}[/itex]

where [itex]\mu [/itex] is the reduced mass and [itex]r[/itex] is the relative distance between the scattered particle and the particle generating the gravitational potential. From this Lagrangian we take that the quantity

[itex]L = m\,r\,v [/itex]

is conserved, right? Now, from the angular momentum vector we have,

[itex] \vec{L} = m\,(y\,\dot{z} - z\,\dot{y}) [/itex]

and [itex]|\vec{L}| = m\,r\,v\,\sin\theta [/itex] where [itex]\theta[/itex] is the angle between [itex]r[/itex] and [itex]v [/itex].

So... my problem is, the angular momentum is then only conserved when [itex]r[/itex] and [itex]v [/itex] are orthogonal? So, it doesn't really apply to scattering, only for orbiting particles?

Thank you very much.
 

Answers and Replies

  • #2
sophiecentaur
Science Advisor
Gold Member
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So, it doesn't really apply to scattering, only for orbiting particles?
Scattering and orbiting involve an External Force. Momentum is transferred to the 'other' object which provides the Force. Just because that object is massive, (net) momentum conservation still applies.
The Energy lost to the massive object may be near-zero and ignored.
 

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