Invariant mass spectrum to a transverse momentum distribution For the decay of the Higgs boson in 2 photons having the following invariant mass formula:
$$M = 2E_{1}E_{2}(1 - \cos \theta)$$
How can I go from an invariant mass spectrum distribution to a transverse momentum distribution?
 A: Setting 
$$\theta := \theta_1 + \theta_2$$,     
the momentum of the Higgs boson (candidate) with respect to the lab
$$ \| \textbf p_{lab}[~H~] \| = \| \textbf p_{lab}[~\gamma_1~] \| ~\text{Cos}[~\theta_1~] + \| \textbf p_{lab}[~\gamma_2~] \| ~\text{Cos}[~\theta_2~] = (E_{lab}[~\gamma_1~] ~ \text{Cos}[~\theta_1~] + E_{lab}[~\gamma_2~] ~ \text{Cos}[~\theta_2~]) / 2,  $$ 
the transverse momentum magnitude of both photons equally as
$$ \| \textbf p_{lab}^{trans~H}[~\gamma_1~] \| = \| \textbf p_{lab}[~\gamma_1~] \| ~\text{Sin}[~\theta_1~] = \| \textbf p_{lab}^{trans~H}[~\gamma_2~] \| = \| \textbf p_{lab}[~\gamma_2~] \| ~ \text{Sin}[~\theta_2~] = E_{lab}[~\gamma_1~] ~ \text{Sin}[~\theta_1~] / c = E_{lab}[~\gamma_2~] ~ \text{Sin}[~\theta_2~] / c $$ 
and with
$$\sqrt{ (M~c^2)^2 + (\| \textbf p_{lab}[~H~] \| c)^2 } = E_{lab}[~\gamma_1~] + E_{lab}[~\gamma_2~]$$ 
I get, first of all, the invariant mass (of the Higgs boson (candidate)) as
$$M^2 = 2~E_{lab}[~\gamma_1~]~E_{lab}[~\gamma_2~]~(1 - \text{Cos}[~\theta~]) / c^4,$$
in other words: in the formula as presently stated in the question the $M$ seems to be missing a square;
and I get the "transverse momentum" (magnitude) in terms of $M$ and the two photon energies as
$$\| \textbf p_{lab}^{trans~H}[~\gamma_1~] \| = \| \textbf p_{lab}^{trans~H}[~\gamma_2~] \| = \frac{M~c}{2}~\sqrt{ \frac{4~E_{lab}[~\gamma_1~]~E_{lab}[~\gamma_2~] - M^2~c^4}{(E_{lab}[~\gamma_1~] + E_{lab}[~\gamma_2~])^2 - M^2~c^4}}.$$
