Finding the Energy of an Outgoing Particle in a Momentum Conservation Problem

[ChrisJ]

This problem assumes working in natural units where $c=1$, and using the Minkowski metric where the time component is positive and the space ones negative (as I know the opposite convention is just as commonly used).

EDIT: I had intended to display 4-vectors as bold and the 3-vectors with the arrow but forgot that it won't parse the Tex properly. So I left the 3-vectors with the arrows, and if it has no arrow and has no component subscript (i.e. x/y/z) assume its a 4-vector.

1. Homework Statement
Particle u of mass $m_u$ has total energy $E_u$ in the the frame of the lab. It decays into two new particles, particle v of mass $m_v$ and particle w of mass $m_w$.

Show that in particle u's rest frame, the energy of outgoing particle v is given by $E_v = \frac{m_u^2+m_v^2-m_w^2}{2m_u}$

Homework Equations

$p \cdot p = m^2$
$E^2 = m^2 + |\vec{p}|^2$

The Attempt at a Solution

Never done conservation of four momentum problem before, but I assume that it is the case that $p_u = p_v + p_w$ and with the energy components being conserved, and the 3-momentum components being conserved, so that $p_u = (E_v + E_w, \vec{p_v} + \vec{p_w})$ ?

So far I have done,

$$p_w = p_u - p_v \\ p_w \cdot p_w = (p_u - p_v) \cdot (p_u - p_v) \\ m_w^2 = m_u^2 + m_v^2 - 2 ( p_u \cdot p_v ) \\ p_u \cdot p_v = \frac{m_u^2+m_v^2-m_w^2}{2}$$

Which implies that $p_u \cdot p_v = E_v m_u$ but this is where I come unstuck and can't get that result.

I have that $p_u = (E_v + E_w, \vec{p_v} + \vec{p_w}) = (E_u , \vec{p_u})$
and that $p_v = (E_v , \vec{p_v})$

Doing the dot product I get,
$$p_u \cdot p_v = E_v(Ev + E_w) - \vec{p_v}(\vec{p_v}+\vec{p_w}) \\ p_u \cdot p_v = E_v^2+E_v E_w - p_{v_x}^2 - p_{w_x}^2 - p_{v_y}^2 - p_{w_y}^2 - p_{v_z}^2 - p_{w_z}^2$$

I am not sure if this is the correct way to go about it, but I can't seem to progress from here, not sure if it is because I have gone down the wrong path or if its just a algebra/connection issue. Any help/advice is much appreciated, thanks :)

 

[TSny]

$$p_w = p_u - p_v \\ p_w \cdot p_w = (p_u - p_v) \cdot (p_u - p_v) \\ m_w^2 = m_u^2 + m_v^2 - 2 ( p_u \cdot p_v ) \\ p_u \cdot p_v = \frac{m_u^2+m_v^2-m_w^2}{2}$$

Nice

Which implies that $p_u \cdot p_v = E_v m_u$ but this is where I come unstuck##

This follows easily if you keep in mind that you are in the rest frame of particle u.

 

[ChrisJ]

NiceThis follows easily if you keep in mind that you are in the rest frame of particle u.

Thanks for reminding about that, managed to get it now, since $p_u = (E_u, 0)=(m_u,0)$, Cheers :)