= Transverse mass
{title2=$m_T^2=2p_{T\ell}p_T^{\rm miss}(1-\cos\Delta\phi)$}
{wiki}
For a two-body <W boson> decay into a <charged lepton> $\ell$ and a <neutrino>, neglect both daughter <masses> and use <natural units> $c=1$. Let $\Delta\phi$ be the angle between the daughter <transverse momenta>. Identifying the <missing transverse momentum> with the <neutrino>'s gives
$$
m_T^2=2p_{T\ell}p_T^{\rm miss}(1-\cos\Delta\phi).
$$
This observable uses no $p_{z\nu}$. The 1983 <UA1 experiment> paper already used <transverse mass> as a lower bound on parent <mass>, before making additional decay-model assumptions for a mass estimate.<Experimental Observation of Isolated Large Transverse Energy Electrons with Associated Missing Energy at sqrt(s) = 540 GeV by G. Arnison>[Section 10, pp.113–114, Figs.8–10: bound versus model-dependent mass fit]{ref}
The bound follows directly from the <energy–momentum relation>. For each massless daughter write its <four-momentum>, using <rapidity> $y$ and azimuth $\phi$, as
$$
p^\mu=(p_T\cosh y,\ p_T\cos\phi,\ p_T\sin\phi,\ p_T\sinh y).
$$
Using the <Minkowski inner product> with positive time sign, $p^2=E^2-|\mathbf p|^2$, the two daughter <four-momenta> give the true parent <invariant mass> $M$:
$$
M^2=2p_{T\ell}p_{T\nu}\bigl[\cosh(y_\ell-y_\nu)-\cos\Delta\phi\bigr],\qquad
M^2-m_T^2=2p_{T\ell}p_{T\nu}\bigl[\cosh(y_\ell-y_\nu)-1\bigr]\geq0.
$$
Since $\cosh u\geq1$, \b[$m_T\leq M$], with equality at equal daughter <rapidities> when both <transverse momenta> are nonzero. Thus the unmeasured longitudinal information can enlarge the true <invariant mass>, but cannot lower it below this ideal <transverse mass>.
For an original example, stipulate $p_{T\ell}=p_{T\nu}=40\,\mathrm{GeV}$ and $\Delta\phi=\pi$. Then $m_T=80\,\mathrm{GeV}$. Equal <rapidities> give $M=80\,\mathrm{GeV}$; choosing $y_\ell-y_\nu=1$ with exactly the same transverse readout gives $M=80\cosh(1/2)\simeq90.210\,\mathrm{GeV}$. One measured transverse configuration therefore permits different true parent <masses>.
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