Deriving Covariant Form of $E_{1}E_{2}|\vec{v}|$

[bananabandana]

Given a two particle scattering problem with (initial) relative velocity $|\vec{v}|$, apparently the product $E_{1}$E_{2}|\mathb{v}|$ can be expressed in the covariant form:

$$ E_{1}E_{2}|\vec{v}| = \sqrt{ (p_{1}\cdot p_{2} - m_{1}^{2}m_{2}^{2}} $$

My textbook gives no further explanation - how has this been arrived at?
I tried expanding out:
$$ E_{1}^{2}E_{2}^{2}= (m_{1}^{2}- |\vec{p_{1}}|^{2})(m_{2}^{2}-|\vec{p_{2}}|^{2})$$
But this doesn't seem to go anywhere useful at all.

 

[Ibix]

The expression given doesn't look right on dimensional grounds. ##p_1\cdot p_2## has dimensions of ##[M^2]## in c=1 units, but ##m_1^2m_2^2## has dimension ##[M^4]##. I suspect a typo somewhere.

 

[Ibix]

More helpfully, I think that (whatever the correct answer is) this comes from the inner product of the two four-momenta. In the rest frame of one of the particles it's easy to work out its value, which is an invariant and includes your ##\vec v##. You've some work to do in a general frame in order to relate the particles' ##\gamma## factors to that relative velocity. Presumably it drops out somehow.

Note that I haven't worked through this myself - I'm just going on the combination of inner product of three momenta, product of energies and product of masses looking a bit like an inner product.

By the way, you have your signs wrong in your last expression. ##E^2=m^2+p^2##. The way you have it, energy decreases as momentum increases.

 

[sweet springs]

E1E2|→v|=√(p1⋅p2−m21m22

In the frame where 1 is staying still, p1=0, the right hand side becomes pure imaginary number. More information is needed to read this formula, if it is right, properly.