Angle between two momenta in particle physics (principal axis of a two-body decay vs. center-of-mass motion in the lab) Situation:
I have events with a W-Boson decaying into two leptons (e.g. electron and electron-neutrino). Now I want to see, whether there is an angle range into which the leptons are emitted favourably. I use the angle $\theta$ between the direction of the W-Boson momentum and the electron momentum, both transformed into the W-Boson restframe. 
The method is, to determine for each event $cos(\theta)$ and draw a histogram with the number of events over $cos(\theta)$.
Question(s) (about how to determine $cos(\theta)$):
I first straightforward boosted both momentum vectors into the W-restframe and calculated the angle between them.
Then somebody told me that I can use the formula $$ cos(\theta) = \frac{E_e-E_{\nu_e}}{E_e+E_{\nu_e}},$$ which I did then too, using the energy in the restframe of the experiment.
The question is, why those two methods should give the same result (I have the feeling that it is just something simple with vectors, what I don't see clearly right now).
And, do I first have to boost the Lorentz vectors of the leptons into the W-boson restframe before using the second method?
I am asking because at the moment the histograms look similar, but if I calculate the difference between the results for $cos(\theta)$ for each event I get something looking like a gauss distribution ranging from -1 to 1, which is quite much, I believe.
...sorry if the answer is obvious, I'm a newbie to the topic :)
 A: 
W-Boson events decaying into two leptons (e.g. electron and electron-neutrino)

The convention is of course to say that a leptonic decay of a $W^-$ boson produces a negatively charged lepton (such as an electron, $e^-$) together with an anti-neutrino (of the matching weak state, such as an anti-electron-neutrino, $\overline\nu_e$).

the angle $\theta$ between the direction of the W-Boson momentum and the electron momentum, both transformed into the W-Boson restframe.

Well, as far as I understand, the momentum of a particular W-Boson with respect to itself (and to any other participants at rest with respect to this particular W-Boson) is in all cases null. (Making it difficult to speak of an angle with respect to it.)
However, we may still recognize the momentum of a given W-Boson (of course with respect to the reference lab/detector, $\mathbf p_{\text{lab}}[~W~]$) as defining, decay by decay, the applicable "reference direction" (which is aligned with, but just opposite to, the direction of motion of the lab/detector with respect to a given W-Boson); and we can consider the angle $\theta$ between this "reference direction" and the direction of the motion of the electron, as determined by the members of the W-Boson restframe.

I first straightforward boosted both momentum vectors into the W-restframe and calculated the angle between them. 

I suppose that you've been using some ready-made software ("methods") for this purpose ... (which dealt with the difficulty mentioned above without you even noticing).

Then somebody told me that I can use the formula [...] why those two methods should give the same result (I have the feeling that it is just something simple with vectors, what I don't see clearly right now).

I can derive the suggested formula only as an approximation, provided that


*

*the masses of the two leptonic decay products of the W-Boson are considered negligible in comparison to (roughly half of) the W-Boson mass; and surely they are (for any "practical" purposes I can imagine); and

*the W-Boson mass is neglected, too; which may or may not be sensible, as may be parametrized by the momentum of the given W-Bosons with respect to the lab/detector.
Now, I'd set $\text{Sin}[~\theta~]$ := the ratio between the magnitude of the transverse momentum component (with respect to the "reference direction", decay by decay) of the electron, $p^{tr}_W[~e~] = \| \mathbf p^{tr}_W[~e~] \|$, and the entire electron momentum magnitude, $p_W[~e~] = \| \mathbf p_W[~e~] \|$; both as determined by the members of the applicable W-Boson restframe.
There are several useful relations to note, since we're dealing with a two-body decay of the given W-Bosons:      


*

*$\mathbf p_{\text{lab}}[~W~] = \mathbf p_{\text{lab}}[~e~] + \mathbf p_{\text{lab}}[~\overline\nu_e~]$, defining the "reference direction"; 

*$E_{\text{lab}}[~W~] = E_{\text{lab}}[~e~] + E_{\text{lab}}[~\overline\nu_e~]$; 

*$\mathbf p_W[~e~] = -\mathbf p_W[~\overline\nu_e~]$;

*$p_W[~e~] = \frac{c~m_W}{2} = p_W[~\overline\nu_e~]$;

*$p^{tr}_W[~e~] = p^{tr}_{\text{lab}}[~e~] = -p^{tr}_W[~\overline\nu_e~] = p^{tr}_{\text{lab}}[~\overline\nu_e~]$,

*$(p_{\text{lab}}[~W~])^2 := (\| \mathbf p_{\text{lab}}[~W~] \|)^2 := \mathbf p_{\text{lab}}[~W~] \cdot \mathbf p_{\text{lab}}[~W~] = (E_{\text{lab}}[~W~])^2 - (m[~W~])^2$, therefore

*$(\mathbf p_{\text{lab}}[~e~] + \mathbf p_{\text{lab}}[~\overline\nu_e~]) \cdot (\mathbf p_{\text{lab}}[~e~] + \mathbf p_{\text{lab}}[~\overline\nu_e~]) = $
$(p_{\text{lab}}[~e~])^2 + 2~\mathbf p_{\text{lab}}[~e~] \cdot \mathbf p_{\text{lab}}[~\overline\nu_e~] + p_{\text{lab}}[~\overline\nu_e~])^2 = $
$(E_{\text{lab}}[~e~] + E_{\text{lab}}[~\overline\nu_e~])^2 - (m[~W~])^2$,
and consequently

*$2~\mathbf p_{\text{lab}}[~e~] \cdot \mathbf p_{\text{lab}}[~\overline\nu_e~] = 2~E_{\text{lab}}[~e~] ~ E_{\text{lab}}[~\overline\nu_e~] + (m[~e~])^2 + (m[~\overline\nu_e~])^2 - (m[~W~])^2 \approx 2~E_{\text{lab}}[~e~] ~ E_{\text{lab}}[~\overline\nu_e~] - (m[~W~])^2.$
All this can be inserted into an expression of the transverse momentum magnitude of the electron:  
$$ p^{tr}_{\text{lab}}[~e~] := \| \mathbf p^{tr}_{\text{lab}}[~e~] \| = \| \mathbf p_{\text{lab}}[~e~] - \left( \frac{\mathbf p_{\text{lab}}[~e~] \cdot \mathbf p_{\text{lab}}[~W~]}{(p_{\text{lab}}[~W~])^2} \right) ~ \mathbf p_{\text{lab}}[~W~] \|.$$
Evaluating this seems rather involved to me ... but it gives a neat result:
$$ \text{Sin}[~\theta~] := \frac{p^{tr}_{W}[~e~]}{p_{W}[~e~]} = 2~\frac{p^{tr}_{\text{lab}}[~e~]}{m[~W~]} = \pm \sqrt{\frac{4~E_{\text{lab}}[~e~] ~ E_{\text{lab}}[~\overline\nu_e~] - (m[~W~])^2}{(E_{\text{lab}}[~e~] + E_{\text{lab}}[~\overline\nu_e~])^2 - (m[~W~])^2}}.$$
Consequently:
$$ \text{Cos}[~\theta~] := \sqrt{1 - (\text{Sin}[~\theta~])^2} = \sqrt{ 1 - \frac{4~E_{\text{lab}}[~e~] ~ E_{\text{lab}}[~\overline\nu_e~] - (m[~W~])^2}{(E_{\text{lab}}[~e~] + E_{\text{lab}}[~\overline\nu_e~])^2 - (m[~W~])^2}} = \frac{E_{\text{lab}}[~e~] - E_{\text{lab}}[~\overline\nu_e~]}{\sqrt{ (E_{\text{lab}}[~e~] + E_{\text{lab}}[~\overline\nu_e~])^2 - (m[~W~])^2}}.$$
Finally, neglecting the W-boson mass $m[~W~]$ leads to the formula shown in your question.
