- #1
Piano man
- 75
- 0
Hi, I'm reading about particle disintegration at the moment and there's a step I don't follow.
I've got the following equation:
[tex]\tan\theta=\frac{v_0\sin\theta_0}{v_0\cos\theta_0+V} [/tex] where [tex] \theta [/tex] is the resultant angle in the Laboratory system and [tex] \theta_0 [/tex] is the resultant angle in the Centre of Mass system.
Also given is [tex]v=V+v_0[/tex] which are respectively the velocity of a resulting particle in the L system, the velocity of the primary particle in the L system, and the velocity of the resulting particle in the C system.
Solving for [tex]\cos\theta_0[/tex] one should obtain
[tex]\cos\theta_0=-\frac{V}{v_0}\sin^2\theta \pm \cos\theta\sqrt{1-\frac{V^2\sin^2\theta}{v_0^2}}[/tex]
but I've gotten [tex]\cos\theta_0=\frac{V}{v_0}(\cos\theta-1)+\cos\theta[/tex]
from the substitution [tex]\sin\theta_0=\sin\theta\left(\frac{V+v_0}{v_0}\right)[/tex] which seems geometrically sound.
Can anyone see where that other equation comes from for [tex]\cos\theta_0[/tex]?
Thanks.
I've got the following equation:
[tex]\tan\theta=\frac{v_0\sin\theta_0}{v_0\cos\theta_0+V} [/tex] where [tex] \theta [/tex] is the resultant angle in the Laboratory system and [tex] \theta_0 [/tex] is the resultant angle in the Centre of Mass system.
Also given is [tex]v=V+v_0[/tex] which are respectively the velocity of a resulting particle in the L system, the velocity of the primary particle in the L system, and the velocity of the resulting particle in the C system.
Solving for [tex]\cos\theta_0[/tex] one should obtain
[tex]\cos\theta_0=-\frac{V}{v_0}\sin^2\theta \pm \cos\theta\sqrt{1-\frac{V^2\sin^2\theta}{v_0^2}}[/tex]
but I've gotten [tex]\cos\theta_0=\frac{V}{v_0}(\cos\theta-1)+\cos\theta[/tex]
from the substitution [tex]\sin\theta_0=\sin\theta\left(\frac{V+v_0}{v_0}\right)[/tex] which seems geometrically sound.
Can anyone see where that other equation comes from for [tex]\cos\theta_0[/tex]?
Thanks.