- #1
gabu
- 5
- 0
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.
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.