Particle Disintegration: Equation trouble

[Piano man]

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:

$$\tan\theta=\frac{v_0\sin\theta_0}{v_0\cos\theta_0+V}$$ where $$\theta$$ is the resultant angle in the Laboratory system and $$\theta_0$$ is the resultant angle in the Centre of Mass system.

Also given is $$v=V+v_0$$ 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 $$\cos\theta_0$$ one should obtain

$$\cos\theta_0=-\frac{V}{v_0}\sin^2\theta \pm \cos\theta\sqrt{1-\frac{V^2\sin^2\theta}{v_0^2}}$$

but I've gotten $$\cos\theta_0=\frac{V}{v_0}(\cos\theta-1)+\cos\theta$$

from the substitution $$\sin\theta_0=\sin\theta\left(\frac{V+v_0}{v_0}\right)$$ which seems geometrically sound.

Can anyone see where that other equation comes from for $$\cos\theta_0$$?

Thanks.

 

[vela]

Are you sure the equation you started with applies to this situation? It's valid for non-relativistic cases, but particle decay typically involves relativistic speeds.

 

[Piano man]

Yes, in the book, it follows on immediately, with no mention of relativistic effects.
I think it's just some algebraic reworking of the equation that I'm not seeing.

 

[vela]

From the equation

$$\tan\theta=\frac{v_0\sin\theta_0}{v_0\cos\theta_0+V}$$

I'm inferring that the angles are measured relative to the direction of V, the velocity of the primary particle in the lab frame. Your second equation from equating the y-components in the two frames is wrong. It should be

$$v_0\sin\theta_0=\sqrt{(v_0\cos\theta_0+V)^2+(v_0\sin\theta_0)^2}\sin\theta$$

The equation relating the velocities should be

$$\vec{v}=\vec{v}_0+\vec{V}$$

which is a vector equation, so you can't just add the magnitudes of v₀ and V to get the magnitude of v.

To derive the other equation, start with the tan θ equation, square it, and rewrite sin² θ₀ in terms of cos θ₀. You'll get a quadratic equation in cos θ₀.

 

[Piano man]

Excellent! I got it!

That was some marathon of a reworking - three pages and an hour later...

Thanks vela for your help. :)

 

[Piano man]

Excellent! I got it!

That was some marathon of a reworking - three pages and an hour later...

Thanks vela for your help.