Special relativity - Transformation of angles

[Aleolomorfo]

Homework Statement

Let's have a three-particle decay of equal mass $m$; in the CM frame the three particles have equal energy $E$ and they form angles of $\frac{2\pi}{3}$ between each other. Which is the angle between two of the three particles in the rest frame of the other one. (The z-axis is along the direction of particle 1)

Homework Equations

The Attempt at a Solution

I need a confirmation about my reasoning. I have written the 4-momentum in both frames;
$$p_1=(E,0,0,\sqrt{E^2-m^2})$$
$$p_2=(E,0,\frac{\sqrt{3}}{2}\sqrt{E^2-m^2},-\frac{1}{2}\sqrt{E^2-m^2})$$
$$p_3=(E,0,-\frac{\sqrt{3}}{2}\sqrt{E^2-m^2},-\frac{1}{2}\sqrt{E^2-m^2})$$
$$k_1=(m,0,0,0)$$
$$k_2=(E',\vec{k_2})$$
$$k_3=(E',\vec{k_3})$$
With $|\vec{k_2}|=\sqrt{E'^2-m^2}$ and $|\vec{k_3}|=\sqrt{E'^2-m^2}$.
I have used the invariance of $p_2^\mu p_{3\mu}=k_2^\mu k_{3\mu}$ with:
$$k_2^\mu k_{3\mu}=E'^2-|\vec{k_1}||\vec{k_2}|\cos\theta=E'^2-(E'^2-m^2)cos\theta$$
$$p_2^\mu p_{3\mu}=\frac{1}{2}(3E^2-m^2)$$
I have to find the value of $E'$ with a lorentz boost of $v=\frac{\sqrt{E^2-m^2}}{E}$: $E'=\gamma(E-vp_z)$. With a bit of calculus I have found:
$$cos\theta=\frac{3E^4-4E^2m^2+m^4}{3E^4-2E^2m^2-m^4}$$
The result seems a bit strange and so I have the doubt that my solution is wrong, maybe I have made a calculus mistake and the reasoning is right.

 

[TSny]

I think your result is correct. If you factor the numerator and denominator in your final result, it will simplify.

A nice way to get an expression for $E'$ is to set up $p_1^{\mu} p_{2 \mu} = k_1^{\mu} k_{2 \mu}$

 

[Aleolomorfo]

I think your result is correct. If you factor the numerator and denominator in your final result, it will simplify.

A nice way to get an expression for $E'$ is to set up $p_1^{\mu} p_{2 \mu} = k_1^{\mu} k_{2 \mu}$

Thank you very much