Calculation of the decay rate of tachyonic neutrinos against charged-lepton-pair and neutrino-pair Cerenkov radiation

We consider the process shown in figure 1(a), which is LPCR. The threshold condition reads as ${q}^{2}={({E}_{3}-{E}_{1})}^{2}-{({\vec{k}}_{3}-{\vec{k}}_{1})}^{2}\geqslant 4{m}_{e}^{2}$, where q is the four-momentum of the virtual Z⁰ boson, while the incoming and outgoing neutrino momenta are ${p}_{1}^{\mu }=({E}_{1},{\vec{k}}_{1})$ and ${p}_{3}^{\mu }=({E}_{3},{\vec{k}}_{3})$. Threshold is reached when, depending on the geometry, the energy transfer from initial to final state is maximum, while the spatial momentum transfer is minimum. This implies that a larger spatial momentum transfer actually is disfavored from a point of view of pair production, because it leads to lesser values of q². In other words, the greater the spatial momentum transfer, the smaller is the four-momentum transfer. Geometrically, we want the outgoing spatial momentum to be as close to the incoming spatial momentum as possible. At threshold, we can thus safely assume that the final neutrino state actually propagates into the same direction as the initial state.

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Threshold is reached for a collinear geometry of maximum symmetry. The incoming and outgoing tachyonic particles are on the mass shell, i.e., ${E}_{1}=\sqrt{{k}_{1}^{2}-{m}_{\nu }^{2}}$ and ${E}_{3}=\sqrt{{k}_{3}^{2}-{m}_{\nu }^{2}}$. The four-vector notation can thus be reduced to just two components, $q=({E}_{3},{k}_{3})-({E}_{1},{k}_{1})$, and the momentum transfer q² carried by the Z⁰ boson therefore reads as follows,

$$\begin{eqnarray}&&{q}^{2}={({E}_{1}-{E}_{3})}^{2}-{({k}_{1}-{k}_{3})}^{2}.\end{eqnarray} \tag{ 2.1 }$$

Electron–positron pair production threshold is reached at

$$\begin{eqnarray}&&{q}^{2}={(\sqrt{{k}_{1}^{2}-{m}_{\nu }^{2}}-\sqrt{{k}_{3}^{2}-{m}_{\nu }^{2}})}^{2}-{({k}_{1}-{k}_{3})}^{2}=4{m}_{e}^{2}.\end{eqnarray} \tag{ 2.2 }$$

For minimum energy and momentum of the final neutrino state, we have ${E}_{3}=0$ and ${k}_{3}={m}_{\nu }$. Then, the condition (2.2) transforms into

$$\begin{eqnarray}&&{({k}_{1})}_{\mathrm{th}}=2\displaystyle \frac{{m}_{e}^{2}}{{m}_{\nu }}+{m}_{\nu }.\end{eqnarray} \tag{ 2.3 }$$

The energy of the tachyonic neutrino at threshold is given as

$$\begin{eqnarray}&&{({E}_{1})}_{\mathrm{th}}=\sqrt{{({k}_{1})}_{\mathrm{th}}^{2}-{m}_{\nu }^{2}}=2\,\displaystyle \frac{{m}_{e}}{{m}_{\nu }}\,\sqrt{{m}_{e}^{2}+{m}_{\nu }^{2}}\approx 2\,\displaystyle \frac{{m}_{e}^{2}}{{m}_{\nu }}+{m}_{\nu }+{ \mathcal O }\left(\displaystyle \frac{{m}_{\nu }^{3}}{{m}_{e}^{2}}\right),\end{eqnarray} \tag{ 2.4 }$$

where the latter approximation is valid for $\delta \ll 1$. One can rewrite this result, based on the tachyonic dispersion relation ${m}_{\nu }={E}_{1}\,\sqrt{{v}_{\nu }^{2}-1}={E}_{1}\,\sqrt{{\delta }_{\nu }}$,

$$\begin{eqnarray}&&{({E}_{1})}_{\mathrm{th}}\approx {({k}_{1})}_{\mathrm{th}}=2\displaystyle \frac{{m}_{e}^{2}}{{m}_{\nu }}+{m}_{\nu }\approx 2\displaystyle \frac{{m}_{e}^{2}}{{m}_{\nu }}=2\displaystyle \frac{{m}_{e}^{2}}{{({E}_{1})}_{\mathrm{th}}\,\sqrt{{\delta }_{\nu }}}\,\Rightarrow \,{({E}_{1})}_{\mathrm{th}}\approx \sqrt{2}\,\displaystyle \frac{{m}_{e}}{{\delta }_{\nu }^{1/4}}.\end{eqnarray} \tag{ 2.5 }$$

We note that the tachyonic threshold is a definite function of the mass parameters m_ν and m_e of the tachyonic neutrino and of the electron (positron), respectively.

In the previous section, we found that for electron–positron (charged lepton) pair production, threshold is reached for a collinear geometry. In order to investigate the presence or absence of a threshold for tachyonic pair production (i.e, with two outgoing tachyons), it is instructive to have a look at various geometries. Let us consider a pair of tachyons, both of them on the mass shell, ${E}^{2}-{\vec{k}}^{2}=-{m}_{\nu }^{2}$. The outgoing particles of the pair are labeled with the indices 2 and 4, as in figure 1. If we assume that the tachyons are emitted collinearly and with the same energy, then ${p}^{\mu }=(E,\vec{k})={p}_{2}^{\mu }=({E}_{2},{\vec{k}}_{2})={p}_{4}^{\mu }=({E}_{4},{\vec{k}}_{4})$, and

$$\begin{eqnarray}&&{q}^{2}={({p}_{2}+{p}_{4})}^{2}=4{p}^{\mu }{p}_{\mu }=4\,({\vec{k}}^{2}-{m}_{\mu }^{2})-4\,{\vec{k}}^{2}=-4\,{m}_{\mu }^{2},\end{eqnarray} \tag{ 2.6 }$$

which is negative. For two neutrinos of different energy, emitted collinearly. i.e., with ${\vec{k}}_{2}={k}_{2}\,{\hat{{\rm{e}}}}_{z}$ and ${\vec{k}}_{4}={k}_{4}\,{\hat{{\rm{e}}}}_{z}$), one has

$$\begin{eqnarray}&&{E}_{2}=\,\sqrt{{k}_{2}^{2}-{m}_{\mu }^{2}},\qquad {E}_{4}=\sqrt{{k}_{4}^{2}-{m}_{\mu }^{2}},\end{eqnarray} \tag{ 2.7a }$$

$$\begin{eqnarray}&&{q}^{2}=\,{(\sqrt{{k}_{2}^{2}-{m}_{\mu }^{2}}+\sqrt{{k}_{4}^{2}-{m}_{\mu }^{2}})}^{2}-{({k}_{2}+{k}_{4})}^{2}.\end{eqnarray} \tag{ 2.7b }$$

For small tachyonic mass parameter m_ν, a Taylor expansion yields

$$\begin{eqnarray}&&{q}^{2}=-\left(2+\displaystyle \frac{{k}_{2}}{{k}_{4}}+\displaystyle \frac{{k}_{4}}{{k}_{2}}\right)\,{m}_{\nu }^{2}+{ \mathcal O }({m}_{\nu }^{4}).\end{eqnarray} \tag{ 2.8 }$$

In the limits ${k}_{2}\to 0,{k}_{4}\to \infty$ (or vice versa), q² assumes very large negative numerical values, demonstrating the absence of a lower threshold.

One might ask, however, if there is perhaps a higher cutoff for the allowed q² in relativistic tachyonic pair production kinematics. For the production of an anti-collinear pair, one has

$$\begin{eqnarray}&&{\vec{k}}_{2}=\,{k}_{2}\,{\hat{{\rm{e}}}}_{z},\qquad {\vec{k}}_{4}=-{k}_{4}\,{\hat{{\rm{e}}}}_{z},\end{eqnarray} \tag{ 2.9a }$$

$$\begin{eqnarray}&&{E}_{2}=\,\sqrt{{k}_{2}^{2}-{m}_{\mu }^{2}},\qquad {E}_{4}=\sqrt{{k}_{4}^{2}-{m}_{\mu }^{2}},\end{eqnarray} \tag{ 2.9b }$$

$$\begin{eqnarray}{{q}}^{2} & = & {(\sqrt{{{k}}_{2}^{2}-{{m}}_{{\mu }}^{2}}+\sqrt{{{k}}_{4}^{2}-{{m}}_{{\mu }}^{2}})}^{2}-{({{k}}_{2}-{{k}}_{4})}^{2}\\ & = & 4{{k}}_{2}{{k}}_{4}+{O}({{m}}_{{\nu }}^{2}).\end{eqnarray} \tag{ 2.9c }$$

For large k₁ and k₂, this expression assumes arbitrarily large positive numerical values. The only condition relevant to the allowed range of q² for tachyonic pair production thus is

$$\begin{eqnarray}&&-\infty \lt {q}^{2}\lt \infty ,\qquad {q}^{0}\gt 0.\end{eqnarray} \tag{ 2.10 }$$

This result has important consequences for the calculation of neutrino-pair Cerenkov radiation (see figure 1(b)).

One crucial question one might ask concerns the applicability of Fermi theory for the decay processes shown in figures 1(a) and (b), in the high-energy regime. The question is whether the condition ${q}^{2}\ll {M}_{Z}^{2}$, which ensures the applicability of Fermi theory, remains valid for a highly energetic, oncoming neutrino. Concerning this question, we first recall that, as already shown, threshold for pair production is reached for collinear geometry, i.e., when the final neutrino momentum k₃ is along the same direction as the initial state momentum k₁. This implies that the maximum momentum transfer, for given energy E₁ of the incoming particle, also is reached for collinear geometry. Reducing space to one dimension, we find for the square q² of the momentum transfer,

$$\begin{eqnarray}{q}^{2} & = & {(\sqrt{{k}_{1}^{2}-{m}_{\nu }^{2}}-\sqrt{{k}_{3}^{2}-{m}_{\nu }^{2}})}^{2}-{({k}_{1}-{k}_{3})}^{2}\\ & = & 2({k}_{1}{k}_{3}-\sqrt{{k}_{1}^{2}-{m}_{\nu }^{2}}\,\sqrt{{k}_{3}^{2}-{m}_{\nu }^{2}}-{m}_{\nu }^{2}).\end{eqnarray} \tag{ 2.11 }$$

For given k₁, maximum four-momentum transfer is reached when the momentum of the outgoing particle is equal to ${k}_{3}={m}_{\nu }$, and thus ${E}_{3}=0$. This implies that

$$\begin{eqnarray}&&{q}_{\max }^{2}=2\,{m}_{\nu }({k}_{1}-{m}_{\nu })\approx 2{k}_{1}{m}_{\nu }.\end{eqnarray} \tag{ 2.12 }$$

The condition for using the effective Fermi theory for the virtual Z⁰ boson exchange in figures 1(a) and (b) is ${q}^{2}\ll {M}_{Z}^{2}$, which in the high-energy limit can be reformulated as

$$\begin{eqnarray}&&{q}^{2}\approx 2{k}_{1}{m}_{\nu }\approx 2{E}_{1}{m}_{\nu }\ll {M}_{Z}^{2},\qquad {E}_{1}\ll \displaystyle \frac{{M}_{Z}^{2}}{{m}_{\nu }}\sim \displaystyle \frac{{({10}^{11}\mathrm{eV})}^{2}}{1\,\mathrm{eV}}\sim {10}^{22}\,\mathrm{eV},\end{eqnarray} \tag{ 2.13 }$$

where we have conservatively estimated the neutrino mass parameter to be on the order of $1\,\mathrm{eV}$. (In general, one estimates the neutrino masses to be of order $(0.01\div0.05)\,\mathrm{eV}$, see section 1 of [48] and the discussion around equation (14.21) of [49].) The condition ${E}_{1}\ll {10}^{22}\,\mathrm{eV}={10}^{7}\,\mathrm{PeV}$ is easily fulfilled by the most energetic neutrinos seen by the IceCube collaboration [22, 23], which do not exceed $\sim 2\,\mathrm{PeV}$ in energy. Hence, we can safely assume Fermi theory to be valid in the entire range of incoming neutrino energies relevant for the current investigation.

Let us briefly analyze the role of the 'rest frame' of the faster-than-light, incoming neutrino in the context of the tachyonic dispersion relation $E=\sqrt{{k}^{2}-{m}_{\nu }^{2}}$. As is evident from a Minkowski diagram (see figure 2), the rest frame of the tachyonic 'space-like' neutrino cannot be reached via a Lorentz transformation. By contrast, for a tachyon, it is possible to transform into a frame where the time interval (not the space interval!) swept on the tachyonic trajectory is zero, i.e., the tachyonic particle assumes an infinite velocity. This frame of infinite velocity, in some sense, constitutes the equivalent of the rest frame; namely, the incoming particle has zero time evolution (as opposed to zero space evolution), and thus infinite velocity. According to tachyonic kinematics, it then has zero energy. For illustration, we consider the boost into a frame with energy $0\lt u\lt 1$ (see figure 3),

$$\begin{eqnarray}&&E^{\prime} =\gamma \,(E-u\,k),\qquad k^{\prime} =\gamma \,(k-u\,E),\qquad E=\sqrt{{k}^{2}-{m}_{\nu }^{2}}.\end{eqnarray} \tag{ 2.14 }$$

For a boost velocity $u=E/k=\sqrt{{k}^{2}-{m}_{\nu }^{2}}/k\lt 1$, we have $E^{\prime} =0,k^{\prime} =m$.

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One might be tempted to suggest that the decay rate calculation could be simplified in the tachyonic 'rest' frame (with respect to the time, not space, i.e., $E^{\prime} =0$). However, in this frame, one cannot calculate the decay rate. This is most easily seen from an energy conservation condition. The oncoming neutrino energy vanishes for infinite velocity ($E^{\prime} =0$). Hence, the oncoming particle cannot provide the energy necessary to produce an electron–positron pair.

By contrast and for comparison, for a tardyonic (subluminal) particle, the Dirac 'gap' between positive-energy and negative-energy states ensures that the energy of an oncoming, say, muon, is always bound by its rest mass m_μ from below, even under a Lorentz transformation. Hence, the muon decay from rest, with $E={m}_{\mu }\gg 2{m}_{e}$, is kinematically possible. Because the oncoming muon is timelike, the emitted virtual W boson can still carry enough momentum transfer ${q}^{2}\gt 0$ in order to produce an electron, and an electron antineutrino. This is not the case for an oncoming tachyonic neutrino, whose energy is not bound from below, and can in fact vanish. When the oncoming neutrino energy vanishes, so does the decay rate.

Alternatively, one can observe that in its own rest frame (the 'real rest frame' where the tachyon has a vanishing spatial momentum), the neutrino has the following properties,

$$\begin{eqnarray}&&{v}_{\nu }=0,\qquad \qquad {k}_{\nu }=\displaystyle \frac{{m}_{\nu }{v}_{\nu }}{\sqrt{{v}_{\nu }^{2}-1}}=0,\end{eqnarray} \tag{ 2.15 }$$

but

$$\begin{eqnarray}&&{E}_{\nu }={m}_{\nu }/\sqrt{{0}^{2}-1}=\sqrt{{k}_{\nu }^{2}-{m}_{\nu }^{2}}=\sqrt{{0}^{2}-{m}_{\nu }^{2}}={\rm{i}}\,{m}_{\nu }.\end{eqnarray} \tag{ 2.16 }$$

The energy becomes imaginary in its own rest frame. According to figure 3, the rest frame of a space-like, tachyonic neutrino cannot be reached via a Lorentz transformation, consistent with the purely real (rather than complex) quantities which enter equation (2.14). A further kinematic consideration is illustrative. Namely, according to figure 3, the energy of the tachyonic particle decreases as one 'chases' it, then approaches zero and eventually flips sign at boost velocity u. For boost velocities beyond this point, the energy becomes negative, or alternatively, the time ordering of the start and end point of the trajectory of the tachyon reverses. A left-handed tachyonic neutrino, for boost velocities beyond u, would be seen as a right-handed antitachyon moving in the opposite spatial direction, for the moving observer. The spatial momentum $k^{\prime}$, as seen from figure 3, always remains in the region $k^{\prime} \geqslant {m}_{\nu }$. The region with imaginary energies ${E}_{\nu }=\pm {\rm{i}}\,\sqrt{{m}_{\nu }^{2}-{\vec{k}}_{\nu }^{2}}$ with ${k}_{\nu }\lt {m}_{\nu }$, never can be reached for an initial plane-wave tachyonic state with ${k}_{\nu }\gt {m}_{\nu }$, via a Lorentz transformation.

These considerations, together, afford an immediate explanation for the observation that the final result for the decay rate must necessarily vanish with the energy of the oncoming neutrino, and in fact, shall later be seen to contain the neutrino energy as a linear term. The calculation of the decay rate of the tachyonic neutrino needs to be done in the laboratory frame (lab frame).

A few final considerations regarding the time ordering of tachyonic world lines and the calculation of the decay rate are in order. As already emphasized, decay rates are normally calculated most easily in the rest frame of the decaying particle. For tachyons, we cannot go into the true rest frame of the decaying particle, because the frame with $k^{\prime} =0$ cannot be reached for a tachyon, as already described. In the case of a tardyonic particle, there is an energy gap of twice the rest mass between the spectrum of positive-energy (particle) versus negative-energy (anti-particle) states. This energy gap vanishes for tachyons. A tardyonic oncoming particle state cannot transform into an incoming anti-particle state, irrespective of the Lorentz frame in which the process is observed. This implies that, e.g., for the decay of an oncoming muon into a muon neutrino, electron and electron antineutrino, there is no Lorentz frame in which the same process would be observed as a time-reversed process, i.e., the annihilation of an incoming muon antineutrino and an incoming muon, into a W boson, and the eventual production of an electron and an electron antineutrino.

Furthermore, it is known that the energy of a tachyonic particle may change sign upon a Lorentz transformation (see figure 3), so that particle trajectories may become anti-particle trajectories (with the time ordering of start and end points reversed). Indeed, the fact that some particle creation and annihilation operators transform into anti-particle operator upon a Lorentz transformation, has been mentioned as an important problematic aspect of tachyonic field theories [13–21], while possible re-interpretations have recently been discussed in [6].

For a tachyonic decay of an oncoming initial tachyonic neutrino into an electron–positron pair, and an energetically lower neutrino, this means the following. The interpretation of the process may depend on the Lorentz frame in which it is observed; tachyonic trajectories have no definite time ordering. (The only ordering in the tachyonic case concerns the helicity: a left-handed particle state will transform into a right-handed anti-particle state, and vice versa.) The decay of a highly energetic oncoming neutrino ('Big Bird', see [22, 23]) into a energetically lesser one ('Tweety') via electron–positron pair production is interpreted equivalently as the annihilation of an incoming tachyonic neutrino, and an incoming tachyonic antineutrino, in specific Lorentz frames (see figure 2(b)). In other Lorentz frames, it is even reinterpreted as the decay of an incoming highly energetic antineutrino, into a less energetic antineutrino and an electron–positron pair (see figure 2(c)).

We now consider the kinematics of the tachyonic decays displayed in figure 2 in detail. In figure 2(a), the world-line trajectories of the oncoming neutrino (joining space-time points labeled 1 and 2) and of the final zero-energy neutrino (joining space-time points labeled 2 and 3) are displayed. When 'chasing' the decaying neutrino with a Lorentz boost, transforming the x and t axes into $x^{\prime}$ and $t^{\prime}$, respectively, then from visual inspection, it is evident that the time ordering on the final decay product trajectory has reversed (${t}_{2}={t}_{3}$ but ${t}_{3}^{{\prime} }\lt {t}_{2}^{{\prime} }$). The decay product has turned into an incoming antineutrino, and the Lorentz-transformed process describes neutrino–antineutrino annihilation (into an electron–positron pair, but the world lines of the decay products are not displayed in figure 2). Physical reality has to be ascribed to both interpretations [13–16]. The observation of the moving ('primed') observer is equally valid. For the lab frame, this means that unless we have a counter-propagating beam of antineutrinos, the neutrino–antineutrino annihilation process does not contribute to the discussion of the 'decay', which only converts the oncoming highly energetic neutrino into one with lesser energy.

Let us now consider figure 2(b). The incoming neutrino is chased by a 'faster' Lorentz boost. The transformed axes become $x^{\prime\prime}$ and $t^{\prime\prime}$, and the first, the decaying neutrino, now constitutes a zero-energy decay product, for the decay of an incoming antineutrino (time-ordered trajectory $3\mapsto 2$). As the boost velocity crosses the $x^{\prime\prime}$ and $t^{\prime\prime}$ axes, the decay 'product' (from the point of view of the lab frame) has turned into an incoming, highly energetic, antineutrino, which in the triple-primed frame in figure 2(c), decays into an energetically lower antineutrino (from the point of view of the boosted frame). At the point where the $x^{\prime\prime}$ and $t^{\prime\prime}$ axes are crossed, the initial, incoming neutrino has transformed into an outgoing zero-energy neutrino or antineutrino state (the interpretation changes exactly at the point where the energy changes sign).

What do these considerations imply for the description of the tachyonic decay of a neutrino? We are working in the lab frame, and we need to calculate the process in the lab frame. Processes with incoming antineutrinos must be excluded from the integration, because they cannot contribute to the decay of an incoming neutrino. The interpretation of a process involving tachyons may depend on the Lorentz frame; for the calculation of the decay rate, only processes with incoming and outgoing positive-energy neutrino states may be considered, even if these states may transform into anti-particle states upon a Lorentz transformation. The final results are still Lorentz-invariant, as discussed below in section 5.3.

In order to proceed to the calculation of the decay rate of the tachyonic, incoming neutrino, we briefly compile known Lagrangians from standard electroweak theory (see also appendix A). We denote the weak coupling constant as g_w. According to chap 12 of [50], quantum electrodynamics is described by the coupling of the electron to the photon,

$$\begin{eqnarray}&&{{ \mathcal L }}_{1}=-{g}_{w}\,\sin {\theta }_{{\rm{W}}}(\overline{e}\,{\gamma }^{\mu }e)\,{A}_{\mu },\end{eqnarray} \tag{ 3.1 }$$

where ${\theta }_{{\rm{W}}}$ is the Weinberg angle and e and $\overline{e}$ describe the electron–positron field operators, while A_μ is the electromagnetic field operator. Furthermore, the charged vector boson ${W}^{\pm }$ interacts with a neutrino-electron current,

$$\begin{eqnarray}{{ \mathcal L }}_{2} & = & \left\{-\displaystyle \frac{{g}_{w}}{2\sqrt{2}}[\overline{e}\,{\gamma }^{\rho }(1-{\gamma }^{5})\,{\nu }_{e}]\,{W}_{\rho }^{+}+{\rm{h}}.{\rm{c}}.\right\}\\ & & +\left\{-\displaystyle \frac{{g}_{w}}{2\sqrt{2}}[{\overline{\nu }}_{\mu }\,{\gamma }^{\rho }(1-{\gamma }^{5})\,\mu ]\,{W}_{\rho }^{-}+{\rm{h}}.{\rm{c}}.\right\},\end{eqnarray} \tag{ 3.2 }$$

where the addition of the Hermitian adjoint is necessary in order to include the ${W}^{+}$ boson. For the calculation of the muon decay, one needs the full Lagrangian given in equation (3.2), even twice, namely, once for the muon–muon-neutrino current, and a second time for the decay of the W into the electron and electron antineutrino, i.e., the same current is used in the electron and in the neutrino sector.

For the decay process of the tachyonic neutrino, one needs the coupling of the neutrino to the Z⁰ boson,

$$\begin{eqnarray}&&{{ \mathcal L }}_{3}=-\displaystyle \frac{{g}_{w}}{4\,\cos {\theta }_{{\rm{W}}}}[\overline{\nu }\,{\gamma }^{\mu }(1-{\gamma }^{5})\,\nu ]\,{Z}_{\mu },\end{eqnarray} \tag{ 3.3 }$$

as well as the coupling of the left- and right-handed electron to the Z⁰ boson,

$$\begin{eqnarray}{{ \mathcal L }}_{4} & = & \,\displaystyle \frac{{g}_{w}}{4\,{\rm{\cos }}\,{\theta }_{{\rm{W}}}}\,\overline{e}[{\gamma }^{\mu }(1-{\gamma }^{5})-4\,{{\rm{\sin }}}^{2}({\theta }_{{\rm{W}}})\,{\gamma }^{\mu }]\,e\,{Z}_{\mu }\\ & = & -\displaystyle \frac{{g}_{w}}{2\,{\rm{\cos }}\,{\theta }_{{\rm{W}}}}\,\overline{e}[{c}_{{\rm{V}}}\,{\gamma }^{\mu }-{c}_{{\rm{A}}}\,{\gamma }^{\mu }\,{\gamma }^{5}]\,e\,{Z}_{\mu },\\ {c}_{{\rm{V}}} & = & \,-\displaystyle \frac{1}{2}\,+\,2\,{{\rm{\sin }}}^{2}({\theta }_{{\rm{W}}}),\qquad {c}_{{\rm{A}}}=-\displaystyle \frac{1}{2}.\end{eqnarray} \tag{ 3.4 }$$

The latter form allows us to identify the vector coupling and axial-vector coupling coefficient c_V and c_A. According to equation (12.237) of [50], the vacuum-expectation value v of the Higgs, the weak coupling constants g_w and ${g}_{w}^{{\prime} }$, and the masses of the vector gauge bosons ${W}^{\pm },{Z}^{0}$ and A, are related by

$$\begin{eqnarray}{M}_{W} & = & \displaystyle \frac{1}{2}\,v\,{g}_{w},\ {M}_{Z}=\displaystyle \frac{1}{2}\,v\,{({g}_{w}^{2}+{g}_{w}^{{\prime} 2})}^{1/2},\\ {M}_{A} & = & 0,\ \displaystyle \frac{{M}_{W}}{{M}_{Z}}=\displaystyle \frac{{g}_{w}}{{({g}_{w}^{2}+{g}_{w}^{{\prime} 2})}^{1/2}}=\cos {\theta }_{W}\approx \displaystyle \frac{\sqrt{3}}{2}\approx 0.877.\end{eqnarray} \tag{ 3.5 }$$

These values match the experimental observations of ${M}_{W}=80.385(15)\,\mathrm{GeV}/{c}^{2}$ and ${M}_{Z}=91.1876(21)\,\mathrm{GeV}/{c}^{2}$. The matching with Fermi's effective coupling constant is given as

$$\begin{eqnarray}&&\displaystyle \frac{{G}_{{\rm{F}}}}{\sqrt{2}}=\displaystyle \frac{{g}_{w}^{2}}{8\,{M}_{W}^{2}}.\end{eqnarray} \tag{ 3.6 }$$

Let us anticipate a certain consideration regarding the prefactors encountered in the calculation of invariant matrix elements, in the weak decay of the muon, and in the weak decay of a tachyonic neutrino. For the weak decay of the muon, one uses the Lagrangian (3.2), whose prefactors give a numerical factor $1/{(2\sqrt{2})}^{2}=1/8$. For the weak decay by LPCR, we need to use the Lagrangians (3.3) and (3.4), whose combination results in a prefactor

$$\begin{eqnarray}&&\left(-\displaystyle \frac{1}{4\,{\rm{\cos }}\,{\theta }_{{\rm{W}}}}\right)\times \left(-\displaystyle \frac{1}{2\,{\rm{\cos }}\,{\theta }_{{\rm{W}}}}\right)=\displaystyle \frac{1}{8\,{{\rm{\cos }}}^{2}{\theta }_{{\rm{W}}}}.\end{eqnarray} \tag{ 3.7 }$$

However, the weak decay of the tachyonic neutrino is mediated by a Z boson as opposed to a W boson, which results in a factor

$$\begin{eqnarray}&&\displaystyle \frac{1}{8\,{{\rm{\cos }}}^{2}{\theta }_{{\rm{W}}}}\,\displaystyle \frac{{g}_{w}^{2}}{{M}_{Z}^{2}}=\displaystyle \frac{{g}_{w}^{2}}{8\,{({\rm{\cos }}{\theta }_{{\rm{W}}}{M}_{Z})}^{2}}=\displaystyle \frac{{G}_{{\rm{F}}}}{\sqrt{2}},\end{eqnarray} \tag{ 3.8 }$$

which is the same prefactor that we encounter in the invariant matrix element for the weak decay of the muon.

Let us analyze the degrees of freedom in the phase-space of the final state, in three-body decay of a tachyonic neutrino into a less energetic neutrino, and a light fermion–antifermion pair. The momentum transfer is ${q}^{2}\gt {(2{m}_{e})}^{2}$ from the first fermion line. The decay rate is then obtained as an integral over the differential decay rate,

$$\begin{eqnarray}&&{\rm{d}}{\rm{\Gamma }}={{\rm{d}}}^{3}{k}_{1}\,{{\rm{d}}}^{3}{k}_{2}\,{{\rm{d}}}^{3}{k}_{3}\,{\delta }^{(4)}({p}_{1}+{p}_{2}+{p}_{3}).\end{eqnarray} \tag{ 3.9 }$$

This decay rate is 9 times differential, with 4 conservation conditions. We thus have 5 effective free variables.

These can be assigned as follows: for the decay of a tachyonic neutrino via pair production, we may fix the three momentum components of the outgoing neutrino. Because both the incoming as well as the outgoing neutrino have to be on the mass shell, this fixes the four-vector ${q}^{\mu }={p}_{1}^{\mu }-{p}_{3}^{\mu }=({q}^{0},\vec{q})$ completely. We can then go into the rest frame of the virtual Z⁰ boson and argue that the decay must be completely symmetric there; i.e., the electron and positron should come out in directions exactly opposite of each other. This gives us two more degrees of freedom, namely, the polar and azimuthal angles of one of the outgoing fermions.

The three momenta of the outgoing neutrino and the two light fermion angles add up to the five effective degrees of freedom. So, once we have ${q}^{\mu }=({q}^{0},\vec{q})$, we have only two degrees of freedom left for the electron–positron pair.

The rationale behind the calculations reported below can be summarized as follows. We shall approach the eventual calculation of the decay rate of a tachyonic neutrino due to electron–positron pair production in two steps.

• Step 1 (Complexities in the lab frame): as already emphasized, the tachyonic calculation, in which we are eventually interested, requires us to consider amplitudes in the lab frame, as opposed to the rest frame of the decaying particle. We thus need experience with calculations in the lab frame. The calculation of the muon decay rate is in principle very well-known for the rest frame of the decaying particle. Here, we generalize the calculation to a muon decay rate calculation in the lab frame, where as we shall see, the allowed ${\vec{k}}_{3}$ momenta (in the conventions used for figure 1) are inside an ellipsoid. Lorentz invariance of the integral over the allowed outgoing momenta is explicitly shown.
• Step 2 (Decay of tachyonic, space-like particles): in the calculation of the decay rate of the tachyonic neutrino, we assume (in the spirit of figure 1) that both the incoming as well as the outgoing neutrinos are on the tachyonic mass shell, ${E}_{1}={({\vec{k}}_{1}^{2}-{m}_{\nu }^{2})}^{1/2}$ and ${E}_{3}={({\vec{k}}_{3}^{2}-{m}_{\nu }^{2})}^{1/2}$. Under these circumstances, tachyonic decay is made possible exclusively due to the mass terms in the dispersion relations; hence we cannot ignore these terms Furthermore, as we have already discussed, we need to remember that the region with $| {\vec{k}}_{3}| \lt {m}_{\nu }$ actually is excluded from the region of allowed tachyonic momenta. We find that the physically allowed outgoing momenta are located inside the rotationally symmetric, shallow hull of a cupola-like structure, centered about the axis of the oncoming (decaying) neutrino (which we choose to be the positive z axis). As already anticipated in section 2.5, we shall need to explicitly exclude from the calculation any processes related to neutrino–antineutrino annihilation. This necessity, in turn, makes the use of the explicit spinor solutions of the tachyonic Dirac equation [6, 9] necessary.

In the calculation, we also need to overcome the pitfall connected with the time ordering of the tachyonic trajectories, anticipated in section 2.5.

We shall consider the weak decay of the muon, in the conventions of figure 4. The interaction terms from the electroweak standard model is used according to equation (3.2). For momentum transfers ${q}^{2}\ll {M}_{W}^{2}$, the effective four-fermion Lagrangian thus is

$$\begin{eqnarray}{ \mathcal L } & = & \displaystyle \frac{{g}_{w}^{2}}{8\,{M}_{W}^{2}}({\overline{\nu }}_{\mu }\,{\gamma }_{\lambda }\,(1-{\gamma }^{5})\mu )(\overline{e}\,{\gamma }^{\lambda }\,(1-{\gamma }^{5}){\nu }_{e})\\ & = & \displaystyle \frac{{G}_{{\rm{F}}}}{\sqrt{2}}({\overline{\nu }}_{\mu }\,{\gamma }_{\lambda }\,(1-{\gamma }^{5})\mu )(\overline{e}\,{\gamma }^{\lambda }\,(1-{\gamma }^{5}){\nu }_{e}),\end{eqnarray} \tag{ 3.10 }$$

where we use the matching (3.6). The Lorentz-invariant matrix element thus is (within the conventions of figure 4)

$$\begin{eqnarray}&&{ \mathcal M }=\displaystyle \frac{{G}_{{\rm{F}}}}{\sqrt{2}}(\overline{u}({p}_{3})\,{\gamma }_{\lambda }\,(1-{\gamma }^{5})u({p}_{1}))(\overline{u}({p}_{4})\,{\gamma }^{\lambda }\,(1-{\gamma }^{5})v({p}_{2})).\end{eqnarray} \tag{ 3.11 }$$

Summing over the final spin states and averaging over the spin projections of the initial state leads to

$$\begin{eqnarray}\displaystyle \frac{1}{2}\,\displaystyle \sum _{\mathrm{spins}}| { \mathcal M }{| }^{2} & = & \,\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{4}\,{\rm{Tr}}[{\not\hspace{-2pt}{p}}_{3}\,{\gamma }_{\lambda }\,(1-{\gamma }^{5})({\not\hspace{-2pt}{p}}_{1}+{m}_{\mu }){\gamma }_{\nu }\,(1-{\gamma }^{5})]\\ & & \times {\rm{Tr}}[{\not\hspace{-2pt}{p}}_{2}\,{\gamma }^{\lambda }\,(1-{\gamma }^{5})({\not\hspace{-2pt}{p}}_{4}+{m}_{e}){\gamma }^{\nu }\,(1-{\gamma }^{5})]\\ & = & \,\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{4}[256\,({p}_{1}\cdot {p}_{2})\,({p}_{3}\cdot {p}_{4})]=64\,{G}_{{\rm{F}}}^{2}\,({p}_{1}\cdot {p}_{2})\,({p}_{3}\cdot {p}_{4}).\end{eqnarray} \tag{ 3.12 }$$

The procedure of integration in the rest frame of the decaying particle is discussed in equation (10.16) ff. of [12]. Furthermore, a mixed approach, where certain intermediate integrals are carried out covariantly, and only the final stages of the calculation are carried out in the rest frame of the decaying particle, is outlined in chap 7.2.2 of [51]. In the actual evaluation, in the conventions of figure 4, we keep the outgoing neutrino momentum as our final integration variable and write the decay rate in the lab frame as follows (for the expression in the lab frame, see [52]),

$$\begin{eqnarray}&&{\rm{\Gamma }}=\displaystyle \frac{1}{2{E}_{1}}\,\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{3}}{{(2\pi )}^{3}\,2{E}_{3}}\left(\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{2}}{{(2\pi )}^{3}\,2{E}_{2}}\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{4}}{{(2\pi )}^{3}\,2{E}_{4}}{(2\pi )}^{4}\,{\delta }^{(4)}({p}_{1}-{p}_{3}-{p}_{2}-{p}_{4})\right.\\ &&\quad \times \,\left.\left[\displaystyle \frac{1}{2}\,\displaystyle \sum _{\mathrm{spins}}| { \mathcal M }{| }^{2}\right]\right)\\ &&=\,\displaystyle \frac{1}{2{E}_{1}}\,\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{3}}{{(2\pi )}^{3}\,2{E}_{3}}\left(\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{2}}{{(2\pi )}^{3}\,2{E}_{2}}\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{4}}{{(2\pi )}^{3}\,2{E}_{4}}\right.\\ &&\quad \times {(2\pi )}^{4}\,{\delta }^{(4)}({p}_{1}-{p}_{3}-{p}_{2}-{p}_{4})[64{G}_{{\rm{F}}}^{2}({p}_{1}\cdot {p}_{2})\,({p}_{3}\cdot {p}_{4})]\phantom{\rule[-0000]{}{3.0ex}})\\ &&=\,\displaystyle \frac{2\,{G}_{{\rm{F}}}^{2}}{{\pi }^{5}\,(2{E}_{1})}\,\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{3}}{2{E}_{3}}\left(\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{2}}{2{E}_{2}}\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{4}}{2{E}_{4}}{(2\pi )}^{4}\,{\delta }^{(4)}({p}_{1}-{p}_{3}-{p}_{2}-{p}_{4})\right.\\ &&\quad \times ({p}_{1}^{\lambda }\cdot {p}_{2\lambda })\,({p}_{3\rho }\cdot {p}_{4}^{\rho })\phantom{\rule[-0000]{}{3.3ex}})\\ &&=\,\displaystyle \frac{2\,{G}_{{\rm{F}}}^{2}}{{\pi }^{5}\,(2{E}_{1})}\,\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{3}}{2{E}_{3}}({p}_{1}^{\lambda }{p}_{3}^{\rho }\,{J}_{\lambda \rho }({p}_{1}-{p}_{3}))=\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{12\,{\pi }^{4}\,(2{E}_{1})}\\ &&\quad \times \displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{3}}{2{E}_{3}}\,{p}_{1}^{\lambda }{p}_{3}^{\rho }({g}_{\lambda \rho }\,{q}^{2}+2\,{q}_{\lambda }\,{q}_{\rho })\\ &&=\,\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{12\,{\pi }^{4}\,(2{E}_{1})}\,\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{3}}{2{E}_{3}}({p}_{1}\cdot {p}_{3}\,{q}^{2}+2\,({p}_{1}\cdot q)\,({p}_{3}\cdot q)).\end{eqnarray} \tag{ 3.13 }$$

We have used the following result, derived in equation (B.1), which is obtained for two outgoing particles with labels 2 and 4 which are on the electronic mass shell ${p}_{2}^{2}={p}_{4}^{2}={m}_{e}^{2}$,

$$\begin{eqnarray}{J}_{\lambda \rho }(q) & = & \displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{2}}{2{E}_{2}}\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{4}}{2{E}_{4}}{\delta }^{(4)}(q-{p}_{2}-{p}_{4})({p}_{2\lambda }\,{p}_{4\rho })\\ & = & \sqrt{1-\displaystyle \frac{4\,{m}_{e}^{2}}{{q}^{2}}}\left[{g}_{\lambda \rho }\,\displaystyle \frac{\pi }{24}({q}^{2}-4{m}_{e}^{2})+{q}_{\lambda }\,{q}_{\rho }\,\displaystyle \frac{\pi }{12}\left(1+\displaystyle \frac{2{m}_{e}^{2}}{{q}^{2}}\right)\right]\\ & \approx & \,\displaystyle \frac{\pi }{24}\,{g}_{\lambda \rho }\,{q}^{2}+\displaystyle \frac{\pi }{12}\,{q}_{\lambda }\,{q}_{\rho },\qquad {m}_{e}\to 0.\end{eqnarray} \tag{ 3.14 }$$

By symmetry, we have ${J}_{\lambda \rho }(q)={J}_{\lambda \rho }(q)$. We have carried out the p₂ and p₄ integrals covariantly. Then, for the remaining integral over p₃, we need the appropriate integration limits. We thus need to integrate

$$\begin{eqnarray}&&{\rm{\Gamma }}=\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{12\,{\pi }^{4}\,(2{E}_{1})}\,\int \displaystyle \frac{{{\rm{d}}}^{3}{k}_{3}}{2{E}_{3}}({p}_{1}\cdot {p}_{3}\,{q}^{2}+2\,({p}_{1}\cdot q)\,({p}_{3}\cdot q)),\end{eqnarray} \tag{ 3.15 }$$

assuming an incoming muon with energy E₁ in the positive z direction, with the incoming p₁ on the muon mass shell, ${p}_{1}^{2}={m}_{\mu }^{2}$. The final p₃ describes the muon neutrino, so that within our approximations ${({p}_{3})}^{2}\approx 0$. The domain of the ${{\rm{d}}}^{3}{p}_{3}$ integration in equation (3.15) contains all four-momenta p₃ for which ${q}^{2}={({p}_{1}-{p}_{2})}^{2}\gt 0$.

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In the rest frame of the decaying muon, the integration domain would consist of a sphere composed of vectors ${p}_{3}=(| {\vec{k}}_{3}| ,{\vec{k}}_{3})$, with ${p}_{1}=({E}_{1},\vec{0})$ and $| {\vec{k}}_{3}| \leqslant {m}_{\nu }/2$. By contrast, in the lab fame, we consider a muon moving up the z axis, with energy E₁ and wave vector ${\vec{k}}_{1}$, and an outgoing muon neutrino with energy E₃ and wave vector ${\vec{k}}_{3}$,

$$\begin{eqnarray}&&{\vec{k}}_{1}={k}_{1}\,{\hat{e}}_{z},\ {E}_{1}=\sqrt{{k}_{1}^{2}+{m}_{\mu }^{2}},\ {\vec{k}}_{3}={k}_{\rho }\,{\hat{e}}_{\rho }+{k}_{z}{\hat{e}}_{z},\ {E}_{3}=| {\vec{k}}_{3}| =\sqrt{{k}_{\rho }^{2}+{k}_{z}^{2}}.\end{eqnarray} \tag{ 3.16 }$$

The momentum transfer reads as follows,

$$\begin{eqnarray}&&{q}^{2}=2{k}_{1}\,{k}_{z}+{m}_{\mu }^{2}-2\sqrt{{k}_{\rho }^{2}+{k}_{z}^{2}}\,\sqrt{{k}_{1}^{2}+{m}_{\mu }^{2}}.\end{eqnarray} \tag{ 3.17 }$$

The allowed vectors ${\vec{k}}_{3}$ are located inside a rotationally symmetric ellipsoid (see figure 5), which is centered at the point $(0,0,{k}_{z0})$ on the z axis. Let us denote by a the half axis of the ellipsoid in the radial direction ('away' from the z axis) and by b the half axis of the ellipsoid in the z direction (see figure 5). These half axes are given as follows,

$$\begin{eqnarray}&&a=\displaystyle \frac{{m}_{\mu }}{2},\qquad b=\displaystyle \frac{1}{2}\,\sqrt{{k}_{1}^{2}+{m}_{\mu }^{2}},\qquad {k}_{z0}=\displaystyle \frac{{k}_{1}}{2}.\end{eqnarray} \tag{ 3.18 }$$

For the integration of the final phase-space in the expression (3.15), we need to calculate

$$\begin{eqnarray}&&{\rm{\Gamma }}=\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{12\,{\pi }^{4}\,(2{E}_{1})}\,{\int }_{{q}^{2}={({p}_{1}-{p}_{3})}^{2}\gt 0}\displaystyle \frac{{{\rm{d}}}^{3}{p}_{3}}{2{E}_{3}}({p}_{1}\cdot {p}_{3}\,{q}^{2}+2\,({p}_{1}\cdot q)\,({p}_{3}\cdot q)).\end{eqnarray} \tag{ 3.19 }$$

One uses the following parameterization (${k}_{x}\equiv {k}_{3x},{k}_{y}\equiv {k}_{3y}$, and ${k}_{z}\equiv {k}_{3z}$),

$$\begin{eqnarray}&&{k}_{x}=a\,\xi \,{\rm{\sin }}\,\theta \,{\rm{\cos }}\,\varphi ,\ {k}_{y}=a\,\xi \,{\rm{\sin }}\,\theta \,{\rm{\sin }}\,\varphi ,\ {k}_{z}={k}_{z0}+b\,\xi \,{\rm{\cos }}\,\theta .\end{eqnarray} \tag{ 3.20 }$$

The Jacobian is

$$\begin{eqnarray}{{\rm{d}}}^{3}{k}_{3}=\left|\det \left(\begin{array}{ccc}\tfrac{\partial {k}_{x}}{\partial \xi } & \tfrac{\partial {k}_{x}}{\partial \theta } & \tfrac{\partial {k}_{x}}{\partial \varphi }\\ \tfrac{\partial {k}_{y}}{\partial \xi } & \tfrac{\partial {k}_{y}}{\partial \theta } & \tfrac{\partial {k}_{y}}{\partial \varphi }\\ \tfrac{\partial {k}_{z}}{\partial \xi } & \tfrac{\partial {k}_{z}}{\partial \theta } & \tfrac{\partial {k}_{z}}{\partial \varphi }\end{array}\right)\right|={a}^{2}\,b\,\xi \,\sin \theta .\end{eqnarray} \tag{ 3.21 }$$

Then, with $u=\cos \theta$ and ${k}_{1}=\chi {m}_{\mu }$, one has after the trivial integration over φ,

$$\begin{eqnarray}&&{\rm{\Gamma }}=\displaystyle \frac{{G}_{{\rm{F}}}^{2}{m}_{\mu }^{6}}{12\,{\pi }^{3}\,(2{E}_{1})}\,{\displaystyle \int }_{0}^{1}{\rm{d}}\xi \,{\displaystyle \int }_{-1}^{1}{\rm{d}}u\,\displaystyle \frac{{\xi }^{2}\,\sqrt{1+{\chi }^{2}}}{8\,\sqrt{\xi (\xi (1+{u}^{2}\,{\chi }^{2})+2u\sqrt{1+{\chi }^{2}}\,\chi )+{\chi }^{2}}}\\ &&\ \times \,[(4\xi \,u\chi ({\chi }^{2}+1)+(4{\chi }^{2}+3)\,\sqrt{{\chi }^{2}+1})\,\sqrt{\xi (\xi (1+{u}^{2}\,{\chi }^{2})+2u\sqrt{1+{\chi }^{2}}\,\chi )+{\chi }^{2}}\\ &&\ -\,\xi \,u\,\sqrt{{\chi }^{2}\,+\,1}\,(8{\chi }^{2}+7)\,\chi -2{\xi }^{2}({\chi }^{2}+1)\,(2{u}^{2}{\chi }^{2}+1)-4{\chi }^{2}-5{\chi }^{2}].\end{eqnarray} \tag{ 3.22 }$$

After a somewhat tedious u integration, the result can be written in terms of the variable

$$\begin{eqnarray*}| \xi \,\sqrt{1+{\chi }^{2}}-\chi | =\left\{\begin{array}{cc}\chi -\xi \,\sqrt{1+{\chi }^{2}} & 0\lt \xi \lt \tfrac{\chi }{\sqrt{1+{\chi }^{2}}}\\ \xi \,\sqrt{1+{\chi }^{2}}-\chi & \tfrac{\chi }{\sqrt{1+{\chi }^{2}}}\lt \xi \lt 1\end{array}\right..\end{eqnarray*}$$

The decay rate is naturally written as ${\rm{\Gamma }}={{\rm{\Gamma }}}_{1}+{{\rm{\Gamma }}}_{2}$, where the integration domains are such that $| \xi \,\sqrt{1+{\chi }^{2}}-\chi |$ assumes either of the values given in equation (3.23). Here,

$$\begin{eqnarray}{{\rm{\Gamma }}}_{1} & = & \displaystyle \frac{{G}_{{\rm{F}}}^{2}{m}_{\mu }^{6}}{12\,{\pi }^{3}\,(2{E}_{1})}\,{\displaystyle \int }_{0}^{\chi /\sqrt{1+{\chi }^{2}}}\displaystyle \frac{{\rm{d}}\xi }{8\chi }\xi \left(\sqrt{1+{\chi }^{2}}\,\mathrm{ln}\left(\displaystyle \frac{1+\xi }{1-\xi }\right)\right.\\ & & -2\xi (1+{\chi }^{2})(\chi (4\chi \,(\sqrt{1+{\chi }^{2}}-\chi )-3)+\sqrt{1+{\chi }^{2}})),\end{eqnarray} \tag{ 3.23a }$$

$$\begin{eqnarray}{{\rm{\Gamma }}}_{2} & = & \displaystyle \frac{{G}_{{\rm{F}}}^{2}{m}_{\mu }^{6}}{12\,{\pi }^{3}\,(2{E}_{1})}\,{\displaystyle \int }_{\chi /\sqrt{1+{\chi }^{2}}}^{1}\displaystyle \frac{{\rm{d}}\xi }{8\chi }\xi \left(\sqrt{1+{\chi }^{2}}\,\mathrm{ln}\left(\displaystyle \frac{1+\sqrt{1+{\chi }^{2}}}{1-\sqrt{1+{\chi }^{2}}}\right)\right.\\ & & +2\,\chi (2(1-{\xi }^{2})\,{\chi }^{2}-2{\xi }^{2}+3\xi -1)(1+{\chi }^{2})).\end{eqnarray} \tag{ 3.23b }$$

The two contributions evaluate to the expressions,

$$\begin{eqnarray}&&{{\rm{\Gamma }}}_{1}=\,\displaystyle \frac{{G}_{{\rm{F}}}^{2}{m}_{\mu }^{6}}{12\,{\pi }^{3}\,(2{E}_{1})}\\ &&\quad \times \,\displaystyle \frac{2\chi (3+4{\chi }^{2})\left(\sqrt{1+{\chi }^{2}}+2{\chi }^{2}(\chi -\sqrt{1+{\chi }^{2}})-3\,\mathrm{ln}\left(\tfrac{\sqrt{1+{\chi }^{2}}+\chi }{\sqrt{1+{\chi }^{2}}-\chi }\right)\right)}{48\,\chi \,\sqrt{1+{\chi }^{2}}},\end{eqnarray} \tag{ 3.24a }$$

$$\begin{eqnarray}&&{{\rm{\Gamma }}}_{2}=\displaystyle \frac{{G}_{{\rm{F}}}^{2}{m}_{\mu }^{6}}{12\,{\pi }^{3}\,(2{E}_{1})}\\ &&\quad \times \,\left[\displaystyle \frac{1}{8}-\displaystyle \frac{2\chi (3+4{\chi }^{2})\left(\sqrt{1+{\chi }^{2}}+2{\chi }^{2}(\chi -\sqrt{1+{\chi }^{2}})-3\,\mathrm{ln}\left(\tfrac{\sqrt{1+{\chi }^{2}}+\chi }{\sqrt{1+{\chi }^{2}}-\chi }\right)\right)}{48\,\chi \,\sqrt{1+{\chi }^{2}}}\right],\end{eqnarray} \tag{ 3.24b }$$

so that

$$\begin{eqnarray}&&{\rm{\Gamma }}={{\rm{\Gamma }}}_{1}+{{\rm{\Gamma }}}_{2}=\displaystyle \frac{{G}_{{\rm{F}}}^{2}{m}_{\mu }^{6}}{12\,{\pi }^{3}\,(2{E}_{1})}\times \displaystyle \frac{1}{8}=\displaystyle \frac{{G}_{{\rm{F}}}^{2}{m}_{\mu }^{5}}{192\,{\pi }^{3}}\,\displaystyle \frac{{m}_{\mu }}{{E}_{1}}.\end{eqnarray} \tag{ 3.25 }$$

This is the expected result for the muon decay width, with the $1/{E}_{1}$ prefactor already included, which is here obtained directly by an explicit integration in the lab frame.

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We have also verified [25] the results of Cohen and Glashow for superluminal neutrino decay, based on the noncovariant dispersion relation used in [31], via an independent calculation in the lab frame, as envisaged in [33]. The treatment described in [31] is based on a Lorentz-violating dispersion relation $E=| \vec{k}| {v}_{\nu }$ with ${v}_{\nu }\,\gt \,1$, which constitutes a fundamentally different theoretical model as compared to the Lorentz-invariant tachyonic treatment presented here. In addition to the decay rate, we shall also consider the energy loss rate of an incoming neutrino beam in the lab frame, due to LPCR and NPCR. A remark is in order: the calculation of the energy loss per time of an incoming muon beam, to complement a corresponding calculation for the tachyonic neutrino beam, is not applicable, because the end product of the decay is not a less energetic muon, but the muon disappears from the beam altogether (see figure 4).

We calculate the decay width of the incoming tachyonic neutrino, in the lab frame, employing a relativistically covariant (tachyonic) dispersion relation, with both incoming as well as outgoing neutrinos on the tachyonic mass shell (${E}_{1}={({\vec{k}}_{1}^{2}-{m}_{\nu }^{2})}^{1/2}$, and ${E}_{3}={({\vec{k}}_{1}^{2}-{m}_{\nu }^{2})}^{1/2}$, in the conventions of figure 1). In the lab frame, in full accordance with [52], the decay rate simply is

$$\begin{eqnarray}{\rm{\Gamma }} & = & \displaystyle \frac{1}{2{E}_{1}}\,\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{3}}{{(2\pi )}^{3}\,2{E}_{3}}\left(\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{2}}{{(2\pi )}^{3}\,2{E}_{2}}\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{4}}{{(2\pi )}^{3}\,2{E}_{4}}\right.\\ & & \times \left.{(2\pi )}^{4}\,{\delta }^{(4)}({p}_{1}-{p}_{3}-{p}_{2}-{p}_{4})\left[{\widetilde{\displaystyle \sum }}_{\mathrm{spins}}| { \mathcal M }{| }^{2}\right]\right).\end{eqnarray} \tag{ 3.26 }$$

Here, ${\widetilde{\displaystyle \sum }}_{\mathrm{spins}}$ refers to the specific way in which the average over the oncoming helicity states, and the outgoing helicities, needs to be carried out for tachyons. As explained in section 2.4, we cannot reach the rest frame of the decaying particle by a Lorentz transformation, in the case of a tachyonic neutrino. Furthermore, as outlined in section 2.3, a calculation with just the Fermi effective coupling constant actually is sufficient for the tachyonic case. For the relevant interaction terms, we use the same expression as in section 3.5. We recall the coupling of the decaying neutrino to the Z⁰ boson according to equation (3.3),

$$\begin{eqnarray}&&{{ \mathcal L }}_{3}=-\displaystyle \frac{g}{4\cos {\theta }_{{\rm{W}}}}\,\bar{\nu }\,{\gamma }^{\lambda }\,(1-{\gamma }^{5})\,\nu \,{Z}_{\lambda }.\end{eqnarray} \tag{ 3.27 }$$

For the coupling of the electron to the Z boson, we have according to equation (3.4),

$$\begin{eqnarray}&&{{ \mathcal L }}_{4}=-\displaystyle \frac{{g}_{w}}{2\,\cos {\theta }_{{\rm{W}}}}\,\overline{e}[{c}_{{\rm{V}}}\,{\gamma }^{\lambda }-{c}_{{\rm{A}}}\,{\gamma }^{\lambda }\,{\gamma }^{5}]\,e\,{Z}_{\lambda },\qquad {c}_{{\rm{V}}}\approx 0,\qquad {c}_{{\rm{A}}}=-\displaystyle \frac{1}{2}.\end{eqnarray} \tag{ 3.28 }$$

In view of the compensation mechanism given by equation (3.8), the effective four-fermion Lagrangian thus is given by

$$\begin{eqnarray}&&{ \mathcal L }=\displaystyle \frac{{G}_{{\rm{F}}}}{\sqrt{2}}\{\bar{\nu }\,{\gamma }^{\lambda }\,(1-{\gamma }^{5})\,\nu \}\{\overline{e}[{c}_{{\rm{V}}}\,{\gamma }^{\rho }-{c}_{{\rm{A}}}\,{\gamma }^{\lambda }\,{\gamma }^{5}]\,e\}.\end{eqnarray} \tag{ 3.29 }$$

The matrix element of the fundamental spinor solutions reads as follows,

$$\begin{eqnarray}&&{ \mathcal M }=\displaystyle \frac{{G}_{{\rm{F}}}}{\sqrt{2}}[{\overline{u}}^{{ \mathcal T }}({p}_{3})\,{\gamma }_{\lambda }\,(1-{\gamma }^{5})\,{u}^{{ \mathcal T }}({p}_{1})][\overline{u}({p}_{4})({c}_{{\rm{V}}}{\gamma }^{\lambda }-{c}_{{\rm{A}}}\,{\gamma }^{\lambda }\,{\gamma }^{5})\,v({p}_{2})].\end{eqnarray} \tag{ 3.30 }$$

Here, the ${u}^{{ \mathcal T }}({p}_{1})$, and ${u}^{{ \mathcal T }}({p}_{3})$, constitute Dirac spinor solutions of the tachyonic Dirac equation. In the helicity basis [9, 53], denoted by a subscript $\sigma =\pm$, the tachyonic particle and antiparticle spinors are

$$\begin{eqnarray}&&{u}_{+}^{{ \mathcal T }}(\vec{k})=\,\left(\begin{array}{c}\sqrt{| \vec{k}| +m}\,{a}_{+}(\vec{k})\\ \sqrt{| \vec{k}| -m}\,{a}_{+}(\vec{k})\end{array}\right),\qquad {u}_{-}^{{ \mathcal T }}(\vec{k})=\left(\begin{array}{c}\sqrt{| \vec{k}| -m}\,{a}_{-}(\vec{k})\\ -\sqrt{| \vec{k}| +m}\,{a}_{-}(\vec{k})\end{array}\right),\end{eqnarray} \tag{ 3.31a }$$

$$\begin{eqnarray}&&{v}_{+}^{{ \mathcal T }}(\vec{k})=\left(\begin{array}{c}-\sqrt{| \vec{k}| -m}\,{a}_{+}(\vec{k})\\ -\sqrt{| \vec{k}| +m}\,{a}_{+}(\vec{k})\end{array}\right),\qquad {v}_{-}^{{ \mathcal T }}(\vec{k})=\left(\begin{array}{c}-\sqrt{| \vec{k}| +m}\,{a}_{-}(\vec{k})\\ \sqrt{| \vec{k}| -m}\,{a}_{-}(\vec{k})\end{array}\right),\end{eqnarray} \tag{ 3.31b }$$

where the ${a}_{\sigma }(\vec{k})$ are the fundamental helicity spinors (see p 87 of [50]).

The properties of the tachyonic bispinor solutions differ somewhat from those of the 'normal' tardyonic bispinors. The well-known sum formula for the tardyonic states is (σ denotes the spin orientation)

$$\begin{eqnarray}&&\displaystyle \sum _{\sigma }{u}_{\sigma }(p)\otimes {\overline{u}}_{\sigma }(p)={\not\hspace{-2pt}{p}}+{m}_{e}.\end{eqnarray} \tag{ 3.32 }$$

For the tachyonic spin sums, one has the following sum rule for the positive-energy spinors,

$$\begin{eqnarray}&&\displaystyle \sum _{\sigma }(-\sigma )\,{u}_{\sigma }^{{ \mathcal T }}(p)\otimes {\overline{u}}_{\sigma }^{{ \mathcal T }}(p)\,{\gamma }^{5}=\displaystyle \sum _{\sigma }(-\vec{{\rm{\Sigma }}}\cdot \hat{p})\,{u}_{\sigma }^{{ \mathcal T }}(p)\otimes {\overline{u}}^{{ \mathcal T }}(p)\,{\gamma }^{5}={\not\hspace{-2pt}{p}}-{\gamma }^{5}\,m,\end{eqnarray} \tag{ 3.33 }$$

where we use $p=(E,\vec{p})$ as the convention for the four-momentum and $\hat{p}=\vec{p}/| \vec{p}|$ is the unit vector in the $\vec{p}$ direction. Upon promotion to a four-vector, we have ${\hat{p}}^{\mu }=(0,\hat{p})$. The sum rule can thus be reformulated as

$$\begin{eqnarray}\displaystyle \sum _{\sigma }{u}_{\sigma }^{{ \mathcal T }}(p)\otimes {\overline{u}}_{\sigma }^{{ \mathcal T }}(p) & = & \,(-\vec{{\rm{\Sigma }}}\cdot \hat{p})\,({\not\hspace{-2pt}{p}}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}=-{\gamma }^{5}\,{\gamma }^{0}\,{\gamma }^{i}\,{\hat{p}}^{i}\,({\not\hspace{-2pt}{p}}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}\\ & = & \,-{\not\hspace{-2pt}{\tau}}\,{\gamma }^{5}\,{\gamma }^{i}\,{\hat{p}}_{i}\,({\not\hspace{-2pt}{p}}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}=-{\not\hspace{-2pt}{\tau}}\,{\gamma }^{5}\,{\not\hspace{-2pt}{\hat{p}}}\,({\not\hspace{-2pt}{p}}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5},\end{eqnarray} \tag{ 3.34 }$$

where $\tau =(1,0,0,0)$ is a time-like unit vector. In [6, 9], it has been established that a consistent formulation of the tachyonic propagator is achieved when we postulate that the right-handed neutrino states, and the left-handed antineutrino states, acquire a negative Fock-space norm after quantization of the tachyonic spin-1/2 field. Hence, in order to consider the decay process of an oncoming, left-handed, positive-energy neutrino, we should consider the projection onto negative-helicity states,

$$\begin{eqnarray}&&\displaystyle \frac{1}{2}(1-\vec{{\rm{\Sigma }}}\cdot \hat{p})\,\displaystyle \sum _{\sigma }{u}_{\sigma }^{{ \mathcal T }}(p)\otimes {\overline{u}}_{\sigma }^{{ \mathcal T }}(p)={u}_{\sigma =-1}(p)\otimes {\overline{u}}_{\sigma =-1}(p)\\ &&\qquad =\,\displaystyle \frac{1}{2}(1-{\not\hspace{-2pt}{\tau}}\,{\gamma }^{5}\,{\not\hspace{-2pt}{\hat{p}}})\,({\not\hspace{-2pt}{p}}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}.\end{eqnarray} \tag{ 3.35 }$$

The squared and spin-summed matrix element is

$$\begin{eqnarray}\mathop{\widetilde{\displaystyle \sum }}\limits_{\mathrm{spins}}| { \mathcal M }{| }^{2} & = & \displaystyle \frac{{G}_{{\rm{F}}}^{2}}{2}\,{\rm{Tr}}\left[\displaystyle \frac{1}{2}(1-{\not\hspace{-2pt}{\tau}}\,{\gamma }^{5}{{\not\hspace{-2pt}{\hat{p}}}}_{3})\,({\not\hspace{-2pt}{p}}_{3}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}\,{\gamma }_{\lambda }\,(1-{\gamma }^{5})\right.\\ & & \times \left.\displaystyle \frac{1}{2}(1-{\not\hspace{-2pt}{\tau}}\,{\gamma }^{5}\,{{\not\hspace{-2pt}{\hat{p}}}}_{1})\,({\not\hspace{-2pt}{p}}_{1}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}\,{\gamma }_{\nu }\,(1-{\gamma }^{5})\right]\\ & & \times {\rm{Tr}}[({\not\hspace{-2pt}{p}}_{4}+{m}_{e})({c}_{{\rm{V}}}{\gamma }^{\lambda }-{c}_{{\rm{A}}}\,{\gamma }^{\lambda }\,{\gamma }^{5})\,({\not\hspace{-2pt}{p}}_{2}+{m}_{e})({c}_{{\rm{V}}}{\gamma }^{\rho }-{c}_{{\rm{A}}}\,{\gamma }^{\rho }\,{\gamma }^{5})]\\ & = & \displaystyle \frac{{G}_{{\rm{F}}}^{2}}{2}\,{ \mathcal S }({p}_{1},{p}_{2},{p}_{3},{p}_{4}),\end{eqnarray} \tag{ 3.36 }$$

with ${c}_{{\rm{V}}}\approx 0,{c}_{{\rm{A}}}\approx -1/2$ (the last step implicitly defines the expression ${ \mathcal S }$). Here, the meaning of the notation ${\widetilde{\displaystyle \sum }}_{\mathrm{spins}}$ becomes clear: we have summed over the spins of the outgoing electron–positron pair, while only one specific helicity is taken into account for the oncoming (decaying) neutrino.

The Dirac γ traces in equation (3.36) give rise to a rather lengthy expression, which can be simplified somewhat because incoming and outgoing particles are on their respective mass shells, ${p}_{2}^{2}={p}_{4}^{2}={m}_{e}^{2}$, while on the tachyonic mass shell, we have ${p}_{1}^{2}={p}_{3}^{2}=-{m}_{\nu }^{2}$. Some other scalar products vanish, e.g., the scalar product of the time-like unit vector τ and the space-like unit vector, which is $\tau \cdot \hat{p}=0$.

The result of the Dirac γ traces from equation (3.36) can then be inserted into equation (3.26), and the ${{\rm{d}}}^{3}{p}_{2}$ and ${{\rm{d}}}^{3}{p}_{4}$ integrals can be carried out using the following formulas, which we recall from appendix B (see equation (B.1)),

$$\begin{eqnarray}&&I(q)=\,\int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{2}}{2{E}_{2}}\int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{4}}{2{E}_{4}}{\delta }^{(4)}(q-{p}_{2}-{p}_{4})=\displaystyle \frac{\pi }{2}\,\sqrt{1-\displaystyle \frac{4\,{m}_{e}^{2}}{{q}^{2}}},\end{eqnarray} \tag{ 3.37 }$$

$$\begin{eqnarray}{J}_{\lambda \rho }(q) & = & \displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{2}}{2{E}_{2}}\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{4}}{2{E}_{4}}{\delta }^{(4)}(q-{p}_{2}-{p}_{4})({p}_{2\lambda }\,{p}_{4\rho })\\ & = & \sqrt{1-\displaystyle \frac{4\,{m}_{e}^{2}}{{q}^{2}}}\left[{g}_{\lambda \rho }\,\displaystyle \frac{\pi }{24}({q}^{2}-4{m}_{e}^{2})+{q}_{\lambda }\,{q}_{\rho }\,\displaystyle \frac{\pi }{12}\left(1+\displaystyle \frac{2{m}_{e}^{2}}{{q}^{2}}\right)\right],\end{eqnarray} \tag{ 3.38 }$$

$$\begin{eqnarray}&&K(q)=\,\int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{2}}{2{E}_{2}}\int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{4}}{2{E}_{4}}{\delta }^{(4)}(q-{p}_{2}-{p}_{4})({p}_{2}\cdot {p}_{4})=\displaystyle \frac{\pi }{4}\,\sqrt{1-\displaystyle \frac{4\,{m}_{e}^{2}}{{q}^{2}}}({q}^{2}-2{m}_{e}^{2}).\end{eqnarray} \tag{ 3.39 }$$

After the ${{\rm{d}}}^{3}{p}_{2}$ and ${{\rm{d}}}^{3}{p}_{4}$ integrations, we are left with an expression of the form

$$\begin{eqnarray}&&{\rm{\Gamma }}\,=\,\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{2}\displaystyle \frac{1}{{(2\pi )}^{5}}\,{\int }_{{q}^{2}\gt 4{m}_{e}^{2}}\displaystyle \frac{{{\rm{d}}}^{3}{p}_{3}}{2{E}_{3}}\,{ \mathcal F }({p}_{1},{p}_{3}),\end{eqnarray} \tag{ 3.40 }$$

where

$$\begin{eqnarray}&&{ \mathcal F }({p}_{1},{p}_{3})=\int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{2}}{2{E}_{2}}\int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{4}}{2{E}_{4}}{\delta }^{(4)}({p}_{1}-{p}_{2}-{p}_{3}-{p}_{4})\,{ \mathcal S }({p}_{1},{p}_{2},{p}_{3},{p}_{4}).\end{eqnarray} \tag{ 3.41 }$$

Both the expressions for ${ \mathcal S }({p}_{1},{p}_{2},{p}_{3},{p}_{4})$ as well as ${ \mathcal F }({p}_{1},{p}_{3})$ are too lengthy to be displayed in the context of the current paper. However, approximate formulas can be given, e.g., when the incoming energy E₁ is near threshold.

In order to obtain a better intuitive picture for the domain of allowed p₃ four-vectors, we have to analyze the tachyonic kinematics in some more detail. We calculate in the lab frame and assume that the oncoming neutrino has the energy-momentum four-vector

$$\begin{eqnarray}{p}_{1}^{\mu } & = & ({E}_{1},0,0,{k}_{1}),\ {E}_{1}\geqslant {({E}_{1})}_{{\rm{th}}}=2\displaystyle \frac{{m}_{e}}{{m}_{\nu }}\,\sqrt{{m}_{e}^{2}+{m}_{\nu }^{2}},\\ {k}_{1} & \geqslant & {({k}_{1})}_{{\rm{th}}}=2\displaystyle \frac{{m}_{e}^{2}}{{m}_{\nu }}+{m}_{\nu },\ {E}_{1}^{2}-{k}_{1}^{2}=-{m}_{\nu }^{2}\end{eqnarray} \tag{ 3.42 }$$

(see equations (2.3) and (2.4)). The final-state four-vector is conveniently parameterized as

$$\begin{eqnarray}&&{p}_{3}^{\mu }=({E}_{3},{k}_{3}\,{\rm{\sin }}\,\theta \,{\rm{\cos }}\,\varphi ,{k}_{3}\,{\rm{\sin }}\,\theta \,{\rm{\sin }}\,\varphi ,{k}_{3}\,{\rm{\cos }}\,\theta ),\qquad {k}_{3}\gt {m}_{\nu }.\qquad {E}_{3}^{2}-{k}_{3}^{2}=-{m}_{\nu }^{2}.\end{eqnarray} \tag{ 3.43 }$$

The condition ${k}_{3}\gt {m}_{\nu }$ is naturally imposed for tachyonic kinematics (see figure 3). The squared four-momentum transfer then reads as

$$\begin{eqnarray}{q}^{2} & = & \,2(\sqrt{{E}_{1}^{2}+{m}_{\nu }^{2}}\sqrt{{E}_{3}^{2}+{m}_{\nu }^{2}}\,{\rm{\cos }}\,\theta -{E}_{1}{E}_{3}-{m}_{\nu }^{2})\\ & = & 2({k}_{1}\,{k}_{3}\,{\rm{\cos }}\,\theta -\sqrt{{k}_{1}^{2}-{m}_{\nu }^{2}}\,\sqrt{{k}_{3}^{2}-{m}_{\nu }^{2}}-{m}_{\nu }^{2}),\end{eqnarray} \tag{ 3.44 }$$

where it is convenient to define $u=\cos \theta$. One may solve for the threshold angle $\cos {\theta }_{\mathrm{th}}$,

$$\begin{eqnarray}&&{q}^{2}=4{m}_{e}^{2}\quad \Rightarrow \quad u={\rm{\cos }}\,\theta ={u}_{{\rm{th}}}={\rm{\cos }}\,{\theta }_{{\rm{th}}}=\displaystyle \frac{{E}_{1}\,{E}_{3}+2{m}_{e}^{2}+{m}_{\nu }^{2}}{\sqrt{{E}_{1}^{2}+{m}_{\nu }^{2}}\sqrt{{E}_{2}^{2}+{m}_{\nu }^{2}}}.\end{eqnarray} \tag{ 3.45 }$$

For given E₁ and E₃, all angles θ with ${\rm{\cos }}\,\theta \gt {\rm{\cos }}\,{\theta }_{\mathrm{th}}$, i.e., for $\theta \lt {\theta }_{\mathrm{th}}$, are permissible. Conversely, setting $\cos {\theta }_{\mathrm{th}}=1$ and ${E}_{3}=0$, one may solve for E₁ and rederive the threshold condition (2.3). All of this implies that the domain of permissible ${\vec{k}}_{3}$ vectors, near threshold, is centered about the z axis and forms a 'cupola' of inner radius m_ν (see figure 6). Within the kinematically allowed region, the tachyonic momentum transfer q² is plotted in figure 7.

Zoom In Zoom Out Reset image size

Zoom In Zoom Out Reset image size

For given E₁, the widest opening angle $\theta ={\theta }_{\mathrm{th}}$ is reached when E₃ becomes zero. One finds

$$\begin{eqnarray}&&{\cos {\theta }_{{\rm{th}}}| }_{{E}_{3}=0}=\displaystyle \frac{2{m}_{e}^{2}+{m}_{\nu }^{2}}{{m}_{\nu }\,\sqrt{{E}_{1}^{2}+{m}_{\nu }^{2}}}\approx \left(\displaystyle \frac{2{m}_{e}^{2}}{{m}_{\nu }}+{m}_{\nu }\right)\,{E}_{1}^{-1}=\displaystyle \frac{{k}_{{\rm{th}}}}{{E}_{1}},\end{eqnarray} \tag{ 3.46 }$$

where the latter form is valid in the high-energy limit. Here, ${k}_{\mathrm{th}}=\tfrac{2{m}_{e}^{2}}{{m}_{\nu }}+{m}_{\nu }$ is the threshold momentum. It means that the produced pairs will be emitted in a very narrow cone, centered in the forward direction with respect to the decaying neutrino.

Maximum squared momentum transfer is reached at ${E}_{3}=0$ and $\theta =0$,

$$\begin{eqnarray}&&{q}^{2}={q}_{\max }^{2}=2\,{m}_{\nu }(\sqrt{{E}_{1}^{2}+{m}_{\nu }^{2}}-{m}_{\nu })=2\,{m}_{\nu }({k}_{1}-{m}_{\nu }),\end{eqnarray} \tag{ 3.47 }$$

confirming equation (2.12). Maximum outgoing energy E₃ is reached for minimum momentum transfer ${q}^{2}={q}_{\min }^{2}=4{m}_{e}^{2}$, with the final-state neutrino propagating into the positive z direction. Its energy is

$$\begin{eqnarray}&&{({E}_{3})}_{\max }={E}_{1}\left(2\,\displaystyle \frac{{m}_{e}^{2}}{{m}_{\nu }^{2}}+1\right)-\displaystyle \frac{2{m}_{e}\,\sqrt{{E}_{1}^{2}+{m}_{\nu }^{2}}\,\sqrt{{m}_{e}^{2}+{m}_{\nu }^{2}}}{{m}_{\nu }^{2}}\approx \displaystyle \frac{{E}_{1}\,{m}_{\nu }^{2}}{4{m}_{e}^{2}},\end{eqnarray} \tag{ 3.48 }$$

where the latter form is valid in the limit of large E₁. This corresponds to the maximum allowed k₃,

$$\begin{eqnarray}&&{({k}_{3})}_{\max }=\displaystyle \frac{{k}_{1}(2{m}_{e}^{2}+{m}_{\nu }^{2})-2{m}_{e}\sqrt{{k}_{1}^{2}-{m}_{\nu }^{2}}\,\sqrt{{m}_{e}^{2}+{m}_{\nu }^{2}}}{{m}_{\nu }^{2}}\approx \displaystyle \frac{{k}_{1}\,{m}_{\nu }^{2}}{4{m}_{e}^{2}},\end{eqnarray} \tag{ 3.49 }$$

where, again, the latter form is valid in high-energy limit ${k}_{1}\to \infty$. A plot of the physically relevant range for q² is given in figure 7.

Because of azimuthal symmetry, it is easily possible to find a convenient parameterization of the regime of allowed ${\vec{k}}_{3}$ in spherical coordinates (using the parameterization given in equation (3.43)). We may finally express the integrand ${ \mathcal F }$ from equation (3.41) in terms of the initial and final energies E₁ and E₃ of the decay process, and of the scattering angle θ (with $u=\cos \theta$). The decay rate given by equation (3.40) can thus be written as follows,

$$\begin{eqnarray}{\rm{\Gamma }} & = & \,\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{2}\displaystyle \frac{1}{{(2\pi )}^{5}}\,{\displaystyle \int }_{0}^{2\pi }{\rm{d}}\varphi \,{\displaystyle \int }_{{({k}_{3})}_{{\rm{\min }}}}^{{({k}_{3})}_{{\rm{\max }}}}\displaystyle \frac{{\rm{d}}{k}_{3}\,{k}_{3}^{2}}{2{E}_{3}}\,{\displaystyle \int }_{{u}_{{\rm{th}}}}^{1}{\rm{d}}u\,{ \mathcal F }({E}_{1},{E}_{3},u)\\ & = & \,\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{4}\displaystyle \frac{1}{{(2\pi )}^{4}}\,{\displaystyle \int }_{{({E}_{3})}_{{\rm{\min }}}}^{{({E}_{3})}_{{\rm{\max }}}}{\rm{d}}{E}_{3}\,\sqrt{{E}_{3}^{2}+{m}_{\nu }^{2}}{\displaystyle \int }_{{u}_{{\rm{th}}}}^{1}{\rm{d}}u\,{ \mathcal F }({E}_{1},{E}_{3},u).\end{eqnarray} \tag{ 3.50 }$$

Here, ${({k}_{3})}_{\min }={m}_{\nu }$, while ${({k}_{3})}_{\max }$ is given by equation (3.49). Furthermore, we have ${({E}_{3})}_{\min }=0$, while ${({E}_{3})}_{\max }$ is given by equation (3.48), and we have used the identity

$$\begin{eqnarray}&&{\rm{d}}{k}_{3}\,{k}_{3}={\rm{d}}{E}_{3}\,{E}_{3},\qquad {k}_{3}=\sqrt{{E}_{3}^{2}+{m}_{\nu }^{2}}.\end{eqnarray} \tag{ 3.51 }$$

It is instructive to consider the double-differential energy loss ${{\rm{d}}}^{2}{E}_{1}$, for a particle traveling at velocity ${v}_{\nu }\approx c$ (we restore factors of c for the moment), as it undergoes a decay with energy loss ${E}_{1}-{E}_{3}$, due to the energy-resolved decay rate $({\rm{d}}{\rm{\Gamma }}/{\rm{d}}E)\,{\rm{d}}E$, in time ${\rm{d}}t={\rm{d}}x/c$. It reads as follows,

$$\begin{eqnarray}&&{{\rm{d}}}^{2}{E}_{1}=-({E}_{1}-{E}_{3})\,\displaystyle \frac{{\rm{d}}{\rm{\Gamma }}}{{\rm{d}}{E}_{3}}{\rm{d}}{E}_{3}\,\displaystyle \frac{{\rm{d}}x}{c}.\end{eqnarray} \tag{ 3.52 }$$

Now we revert to natural units with c = 1, divide both sides of the equation by ${\rm{d}}x$ and integrate over the energy loss. One obtains

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{d}}{E}_{1}}{{\rm{d}}x}=-\int {\rm{d}}{E}_{3}\,({E}_{1}-{E}_{3})\,\displaystyle \frac{{\rm{d}}{\rm{\Gamma }}}{{\rm{d}}{E}_{3}}.\end{eqnarray} \tag{ 3.53 }$$

Hence, the energy loss rate is obtained as

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{d}}E}{{\rm{d}}x}=-\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{4}\displaystyle \frac{1}{{(2\pi )}^{4}}\,{\int }_{{({E}_{3})}_{\min }}^{{({E}_{3})}_{\max }}{\rm{d}}{E}_{3}\sqrt{{E}_{3}^{2}+{m}_{\nu }^{2}}\,({E}_{1}-{E}_{3})\,{\int }_{{u}_{\mathrm{th}}}^{1}{\rm{d}}u\,{ \mathcal F }({E}_{1},{E}_{3},u).\end{eqnarray} \tag{ 3.54 }$$

After a long, and somewhat tedious integration one finds the following expressions, which have been briefly indicated in [25],

$$\begin{eqnarray}{\rm{\Gamma }}=\,\left\{\begin{array}{cc}\displaystyle \frac{3}{2}\,\displaystyle \frac{{G}_{{\rm{F}}}^{2}\,{m}_{\nu }^{5}}{192\,{\pi }^{3}}\,\displaystyle \frac{{m}_{\nu }\,{({E}_{1}-{E}_{{\rm{th}}})}^{2}}{{m}_{e}^{2}\,{E}_{{\rm{th}}}} & {E}_{1}\gtrapprox {E}_{{\rm{th}}}\\ \displaystyle \frac{2}{3}\,\displaystyle \frac{{G}_{{\rm{F}}}^{2}\,{m}_{\nu }^{5}}{192{\pi }^{3}}\,\displaystyle \frac{{E}_{1}}{{m}_{\nu }} & {E}_{1}\gg {E}_{{\rm{th}}}\end{array}\right.,\end{eqnarray} \tag{ 3.55a }$$

$$\begin{eqnarray}\displaystyle \frac{{\rm{d}}E}{{\rm{d}}x}=\,\left\{\begin{array}{cc}3\,\displaystyle \frac{{G}_{{\rm{F}}}^{2}\,{m}_{\nu }^{5}}{192\,{\pi }^{3}}\,\displaystyle \frac{{({E}_{1}-{E}_{{\rm{th}}})}^{2}}{{E}_{{\rm{th}}}} & {E}_{1}\gtrapprox {E}_{{\rm{th}}}\\ \displaystyle \frac{4}{3}\,\displaystyle \frac{{G}_{{\rm{F}}}^{2}\,{m}_{\nu }^{5}}{192{\pi }^{3}}\,\displaystyle \frac{{E}_{1}^{2}}{{m}_{\nu }} & {E}_{1}\gg {E}_{{\rm{th}}}\end{array}\right..\end{eqnarray} \tag{ 3.55b }$$

In the high-energy limit, one may rewrite the expressions as follows,

$$\begin{eqnarray}&&{\rm{\Gamma }}=\displaystyle \frac{{G}_{{\rm{F}}}^{2}\,{E}_{1}^{5}\,{\delta }^{2}}{288\,{\pi }^{3}},\qquad \displaystyle \frac{{\rm{d}}E}{{\rm{d}}x}=\displaystyle \frac{{G}_{{\rm{F}}}^{2}\,{E}_{1}^{6}\,{\delta }^{2}}{144\,{\pi }^{3}},\qquad {E}_{1}\gg {E}_{\mathrm{th}}.\end{eqnarray} \tag{ 3.56 }$$

The ratio of the energy loss rate to the decay rate is given as

$$\begin{eqnarray}\displaystyle \frac{1}{{\rm{\Gamma }}}\,\displaystyle \frac{{\rm{d}}E}{{\rm{d}}x}=\left\{\begin{array}{cc}2\,\displaystyle \frac{{m}_{e}^{2}}{{m}_{\nu }}\approx {E}_{{\rm{th}}} & {E}_{1}\geqslant {E}_{{\rm{th}}}\\ 2{E}_{1} & {E}_{1}\gg {E}_{{\rm{th}}}\end{array}\right..\end{eqnarray} \tag{ 3.57 }$$

Here, according to equation (2.4), the threshold energy is

$$\begin{eqnarray}&&{E}_{\mathrm{th}}={({E}_{1})}_{\mathrm{th}}=\sqrt{{({k}_{1})}_{\mathrm{th}}^{2}-{m}_{\nu }^{2}}=2\,\displaystyle \frac{{m}_{e}}{{m}_{\nu }}\,\sqrt{{m}_{e}^{2}+{m}_{\nu }^{2}}\approx 2\,\displaystyle \frac{{m}_{e}^{2}}{{m}_{\nu }}.\end{eqnarray} \tag{ 3.58 }$$

For all results given in equations (3.55)–(3.57), we have assumed that ${m}_{e}\gg {m}_{\nu }$. Interpolating formulas are given as

$$\begin{eqnarray}{\rm{\Gamma }} & \approx & \displaystyle \frac{{G}_{{\rm{F}}}^{2}\,{m}_{\nu }^{6}}{128\,{\pi }^{3}\,{m}_{e}^{2}}\displaystyle \frac{{({E}_{1}-{E}_{{\rm{th}}})}^{2}}{{E}_{{\rm{th}}}}{\left(1+\displaystyle \frac{9{m}_{\nu }^{2}}{4{m}_{e}^{2}}\displaystyle \frac{{E}_{1}-{E}_{{\rm{th}}}}{{E}_{{\rm{th}}}}\right)}^{-1},\\ \displaystyle \frac{{\rm{d}}E}{{\rm{d}}x} & \approx & \displaystyle \frac{{G}_{{\rm{F}}}^{2}{m}_{\nu }^{5}}{64{\pi }^{3}}\displaystyle \frac{{({E}_{1}-{E}_{{\rm{th}}})}^{2}}{{E}_{{\rm{th}}}}\left(\displaystyle \frac{4{E}_{1}{E}_{{\rm{th}}}}{4{E}_{{\rm{th}}}^{2}+9{m}_{\nu }({E}_{1}-{E}_{{\rm{th}}})}\right).\end{eqnarray} \tag{ 3.59 }$$

These formulas interpolate between the regimes ${E}_{1}\gtrapprox {E}_{\mathrm{th}}$ and ${E}_{1}\gg {E}_{\mathrm{th}}$ given in equation (3.55).

Having laid out the formalism in section 3, we can be brief in the current section. In the lab frame, again, the decay rate evaluates to

$$\begin{eqnarray}{\rm{\Gamma }} & = & \displaystyle \frac{1}{2{E}_{1}}\,\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{3}}{{(2\pi )}^{3}\,2{E}_{3}}\left(\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{2}}{{(2\pi )}^{3}\,2{E}_{2}}\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{4}}{{(2\pi )}^{3}\,2{E}_{4}}\right.\\ & & \left.\times {(2\pi )}^{4}\,{\delta }^{(4)}({p}_{1}-{p}_{3}-{p}_{2}-{p}_{4})\left[{\widetilde{\displaystyle \sum }}_{\mathrm{spins}}| { \mathcal M }{| }^{2}\right]\right).\end{eqnarray} \tag{ 4.1 }$$

Here, ${\widetilde{\displaystyle \sum }}_{\mathrm{spins}}$ refers to the specific way in which the average over the oncoming helicity states, and the outgoing helicities, needs to be carried out for tachyons [25]. We use the Lagrangian (3.3). The effective four-fermion interaction thus is

$$\begin{eqnarray}&&{ \mathcal L }=\displaystyle \frac{{G}_{{\rm{F}}}}{2\sqrt{2}}[\overline{\nu }\,{\gamma }^{\mu }(1-{\gamma }^{5})\,\nu ][\overline{\nu }\,{\gamma }^{\mu }(1-{\gamma }^{5})\,\nu ].\end{eqnarray} \tag{ 4.2 }$$

The matrix element ${ \mathcal M }$ evaluates to

$$\begin{eqnarray}&&{ \mathcal M }=\displaystyle \frac{{G}_{{\rm{F}}}}{2\sqrt{2}}[{\overline{u}}^{{ \mathcal T }}({p}_{3})\,{\gamma }_{\lambda }\,(1-{\gamma }^{5})\,{u}^{{ \mathcal T }}({p}_{1})][{\overline{u}}^{{ \mathcal T }}({p}_{4})\,{\gamma }^{\lambda }\,(1-{\gamma }^{5})\,{v}^{{ \mathcal T }}({p}_{2})],\end{eqnarray} \tag{ 4.3 }$$

in the notation for the tachyonic bispinors adopted previously. We use, again, the helicity-projected sum rule

$$\begin{eqnarray}\displaystyle \sum _{\sigma }{u}_{\sigma }^{{ \mathcal T }}(p)\otimes {\overline{u}}_{\sigma }^{{ \mathcal T }}(p) & = & (-\vec{{\rm{\Sigma }}}\cdot \hat{k})\,({\not\hspace{-2pt}{p}}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}=-{\gamma }^{5}\,{\gamma }^{0}\,{\gamma }^{i}\,{\hat{k}}^{i}\,({\not\hspace{-2pt}{p}}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}\\ & = & -{\not\hspace{-2pt}{\tau}}\,{\gamma }^{5}\,{\not\hspace{-2pt}{\hat{k}}}\,({\not\hspace{-2pt}{p}}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5},\end{eqnarray} \tag{ 4.4 }$$

where $\tau =(1,0,0,0)$ is a time-like unit vector. This leads to

$$\begin{eqnarray}&&\displaystyle \frac{1}{2}(1-\vec{{\rm{\Sigma }}}\cdot \hat{k})\,\displaystyle \sum _{\sigma }{u}_{\sigma }^{{ \mathcal T }}(\vec{k})\otimes {\overline{u}}_{\sigma }^{{ \mathcal T }}(\vec{k})={u}_{\sigma =-1}(p)\otimes {\overline{u}}_{\sigma =-1}(p)\\ &&\quad =\displaystyle \frac{1}{2}(1-{\not\hspace{-2pt}{\tau}}\,{\gamma }^{5}\,{\not\hspace{-2pt}{\hat{k}}})\,({\not\hspace{-2pt}{p}}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}.\end{eqnarray} \tag{ 4.5 }$$

The squared and spin-summed matrix element for the tachyonic decay process thus is

$$\begin{eqnarray}&&\mathop{\widetilde{\displaystyle \sum }}\limits_{\mathrm{spins}}| { \mathcal M }{| }^{2}=\,\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{8}\,{\rm{Tr}}\left[\displaystyle \frac{1}{2}(1-{\not\hspace{-2pt}{\tau}}\,{\gamma }^{5}{{\not\hspace{-2pt}{\hat{k}}}}_{3})\,({\not\hspace{-2pt}{p}}_{3}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}\,{\gamma }_{\lambda }\,(1-{\gamma }^{5})\right.\\ &&\quad \times \left.\displaystyle \frac{1}{2}(1-{\not\hspace{-2pt}{\tau}}\,{\gamma }^{5}\,{{\not\hspace{-2pt}{\hat{k}}}}_{1})\,({\not\hspace{-2pt}{p}}_{1}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}\,{\gamma }_{\nu }\,(1-{\gamma }^{5})\right]\\ &&\quad \times {\rm{Tr}}\left[\displaystyle \frac{1}{2}(1-{\not\hspace{-2pt}{\tau}}\,{\gamma }^{5}{{\not\hspace{-2pt}{\hat{k}}}}_{4})\,({\not\hspace{-2pt}{p}}_{4}-{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}\,{\gamma }^{\lambda }(1-{\gamma }^{5})\right.\\ &&\quad \left.\times \displaystyle \frac{1}{2}(1-{\not\hspace{-2pt}{\tau}}\,{\gamma }^{5}{{\not\hspace{-2pt}{\hat{k}}}}_{2})\,({\not\hspace{-2pt}{p}}_{2}+{\gamma }^{5}\,{m}_{\nu })\,{\gamma }^{5}\,{\gamma }_{\lambda }(1-{\gamma }^{5})\right].\end{eqnarray} \tag{ 4.6 }$$

Again, we have chosen the convention to denote the by p₂ the momentum of the outgoing antiparticle.

We now turn to the integration over the four-momenta of the outgoing particles. In the calculation, one may use the fact that the helicity projector is well approximated equal to the chirality projector for tachyonic particle in the high-energy limit (with the energy being significantly larger than the tachyonic mass). On the tachyonic mass shell, one has ${p}_{1}^{2}={p}_{2}^{2}={p}_{3}^{2}={p}_{4}^{2}=-{m}_{\nu }^{2}$. The trace over the Dirac γ matrices can be evaluated with standard computer algebra [54, 55] and is inserted into equation (3.26). The ${{\rm{d}}}^{3}{p}_{2}$ and ${{\rm{d}}}^{3}{p}_{4}$ integrals are carried out with the help of the formulas (B.1a)–(B.1c), under the appropriate replacement ${m}_{e}^{2}\to -{m}_{\nu }^{2}$. After the ${{\rm{d}}}^{3}{p}_{2}$ and ${{\rm{d}}}^{3}{p}_{4}$ integrations, we are left with an expression of the form

$$\begin{eqnarray}&&{\rm{\Gamma }}=\,\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{8}\displaystyle \frac{1}{{(2\pi )}^{5}}\,{\int }_{{q}^{2}\gt 4{m}_{e}^{2}}\displaystyle \frac{{{\rm{d}}}^{3}{p}_{3}}{2{E}_{3}}\,\overline{{ \mathcal F }}({p}_{1},{p}_{3}),\end{eqnarray} \tag{ 4.7 }$$

where

$$\begin{eqnarray}&&\overline{{ \mathcal F }}({p}_{1},{p}_{3})=\int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{2}}{2{E}_{2}}\int \displaystyle \frac{{{\rm{d}}}^{3}{p}_{4}}{2{E}_{4}}{\delta }^{(4)}({p}_{1}-{p}_{2}-{p}_{3}-{p}_{4})\,\overline{{ \mathcal S }}({p}_{1},{p}_{2},{p}_{3},{p}_{4}).\end{eqnarray} \tag{ 4.8 }$$

The expressions for $\overline{{ \mathcal S }}({p}_{1},{p}_{2},{p}_{3},{p}_{4})$ as well as $\overline{{ \mathcal F }}({p}_{1},{p}_{3})$ are too lengthy to be displayed in the context of the current paper. We assume the same kinematics as in equations (3.42) and (3.43). The integrations are done with under the conditions that all $0\lt {E}_{3}\lt {E}_{1}$, and all ${q}^{2}={({p}_{2}+{p}_{4})}^{2}$ for the pair are allowed (see equation (2.10)), leading to

$$\begin{eqnarray}{\rm{\Gamma }} & = & \displaystyle \frac{{G}_{{\rm{F}}}^{2}}{8}\displaystyle \frac{1}{{(2\pi )}^{5}}\,{\displaystyle \int }_{0}^{2\pi }{\rm{d}}\varphi \,{\displaystyle \int }_{{k}_{3}={m}_{\nu }}^{{k}_{\max }}\displaystyle \frac{{\rm{d}}{k}_{3}\,{k}_{3}^{2}}{2{E}_{3}}\,{\displaystyle \int }_{-1}^{1}{\rm{d}}u\,\overline{{ \mathcal F }}({E}_{1},{E}_{3},u)\\ & = & \displaystyle \frac{{G}_{{\rm{F}}}^{2}}{16}\displaystyle \frac{1}{{(2\pi )}^{4}}\,{\displaystyle \int }_{0}^{{E}_{1}}{\rm{d}}{E}_{3}\,\sqrt{{E}_{3}^{2}+{m}_{\nu }^{2}}{\displaystyle \int }_{-1}^{1}{\rm{d}}u\,\overline{{ \mathcal F }}({E}_{1},{E}_{3},u),\end{eqnarray} \tag{ 4.9 }$$

in full analogy with equation (2.10). Finally, the energy loss rate is obtained in full analogy with equation (3.54),

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{d}}E}{{\rm{d}}x}=-\displaystyle \frac{{G}_{{\rm{F}}}^{2}}{4}\displaystyle \frac{1}{{(2\pi )}^{4}}\,{\int }_{0}^{{E}_{1}}{\rm{d}}{E}_{3}\sqrt{{E}_{3}^{2}+{m}_{\nu }^{2}}\,({E}_{1}-{E}_{3})\,{\int }_{-1}^{1}{\rm{d}}u\,{ \mathcal F }({E}_{1},{E}_{3},u).\end{eqnarray} \tag{ 4.10 }$$

After rather tedious integration one finds the following expressions,

$$\begin{eqnarray}&&{\rm{\Gamma }}=\,\displaystyle \frac{1}{3}\,\displaystyle \frac{{G}_{{\rm{F}}}^{2}\,{m}_{\nu }^{4}}{192{\pi }^{3}}\,{E}_{1},\end{eqnarray} \tag{ 4.11a }$$

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{d}}{E}_{1}}{{\rm{d}}x}=\,\displaystyle \frac{1}{3}\,\displaystyle \frac{{G}_{{\rm{F}}}^{2}\,{m}_{\nu }^{4}}{192{\pi }^{3}}\,{E}_{1}^{2}.\end{eqnarray} \tag{ 4.11b }$$

Strictly speaking, these formulas are valid only for ${E}_{1}\gg {m}_{\nu }$, but this condition is easily fulfilled for all phenomenologically relevant neutrino energy, assuming that the modulus of neutrino masses $| {m}_{\nu }|$ does not exceed $1\,\mathrm{eV}$. We reemphasize that, unlike in equation (3.55), there is no further threshold condition. Parametrically, the results in equations (3.55) and (4.11) are of the same order-of-magnitude. Hence, neutrino pair emission is the dominant decay channel in the medium-energy domain, for an oncoming tachyonic neutrino flavor eigenstate.

If we assume that the tachyonic neutrino hypothesis is real, then a natural question to ask concerns the phenomenological consequences of the calculations outlined above. Parametrically, the decays by LPCR and NPCR described by equations (3.55) and (4.11) might set important limits on the observability of tachyonic neutrinos, provided the absolute magnitude of the decay energy loss rates are sufficiently large in order to induce a significant decay probability for neutrinos traveling across the Universe. This is because neutrinos registered by IceCube have to 'survive' the possibility of energy loss by decay, and if they are tachyonic, then lepton and NPCR processes become kinematically allowed.

Indeed, it is known that even very small Lorentz-violating parameters in a Lorentz-violating extension of the standard model may induce very significant energy loss processes at high energies [39, 40]. This is because at high energies, small violations of the Lorentz symmetry correspond to very high virtualities of the particles (kinematic deviations from the mass shell), and therefore, the magnitude of the Lorentz-violating parameters is in fact severely constrained by the 37 neutrinos with $E\gt 60\,\mathrm{TeV}$ which are believed to be of cosmological origin and which have been registered by the IceCube collaboration [22, 23]. Meanwhile, preliminary evidence for a through-going muon depositing an energy of $\geqslant (2.6\pm 0.3)\mathrm{PeV}$ has been presented by some members of the IceCube collaboration [56]. The event could be interpreted in terms of a decay product of a neutrino of even higher energy [56]. (The difference of the energy deposited inside the detector and the neutrino energy, according to figure 4 of [23], is small.) If confirmed, this event would lead to even more restrictive bounds on the Lorentz-violating parameters.

The results given in equation (4.11) for the decay rate and energy loss rate due to NPCR are not subject to a threshold energy. Parametrically, they are of the same order-of-magnitude as those given for LPCR in equation (3.55), but the threshold energy is zero. Let us take as a typical cosmological distance 15 billion light years,

$$\begin{eqnarray}&&L=15\times {10}^{9}\,\mathrm{ly}=1.42\times {10}^{26}\,{\rm{m}}\end{eqnarray} \tag{ 5.1 }$$

and assume a (relative large) neutrino mass parameter of ${m}_{0}={10}^{-2}\,\mathrm{eV}$. One obtains for the relative energy loss according to equation (4.11a),

$$\begin{eqnarray}&&\displaystyle \frac{L}{{E}_{1}}\,\displaystyle \frac{{\rm{d}}{E}_{1}}{{\rm{d}}x}=\displaystyle \frac{1}{3}\,\displaystyle \frac{{{G}}_{{\rm{F}}}^{2}\,{m}_{0}^{4}}{192{\pi }^{3}}\,{E}_{1}\,L=5.02\times {10}^{-20}\,\displaystyle \frac{{E}_{1}}{\mathrm{MeV}}.\end{eqnarray} \tag{ 5.2 }$$

Even at the large 'Big Bird' energy of ${E}_{\nu }\approx 2\,\mathrm{PeV}$, the relative energy loss over 15 billion light years does not exceed $5\times {10}^{-20}$.

Again, assuming that ${m}_{\nu }={10}^{-2}\,\mathrm{eV}$, one obtains for the decay rate the result

$$\begin{eqnarray}&&{\rm{\Gamma }}=\displaystyle \frac{1}{3}\,\displaystyle \frac{{G}_{{\rm{F}}}^{2}\,{m}_{0}^{4}}{192{\pi }^{3}}\,{E}_{1}=1.06\times {10}^{-37}\left(\displaystyle \frac{{E}_{1}}{\mathrm{MeV}}\right)\left(\displaystyle \frac{\mathrm{rad}}{{\rm{s}}}\right).\end{eqnarray} \tag{ 5.3 }$$

Even for ${E}_{\nu }\approx 2\,\mathrm{PeV}$, this means that the decay rate is only of order ${10}^{-28}\,\tfrac{\mathrm{rad}}{{\rm{s}}}$, corresponding to a lifetime of $\sim {10}^{20}$ years, far exceeding the age of the Universe. Within the tachyonic model, quite surprisingly, both LPCR as well as NPCR are phenomenologically irrelevant, even for the highest-energy neutrinos registered by IceCube.

The tachyonic Dirac equation [1, 6] reads as

$$\begin{eqnarray}&&({\rm{i}}{\gamma }^{\mu }{\partial }_{\mu }-{\gamma }^{5}{m}_{\nu })\psi (x)=0,\qquad ({\rm{i}}{\gamma }^{\mu }{p}_{\mu }-{\gamma }^{5}{m}_{\nu }){u}^{{ \mathcal T }}(p)=0,\end{eqnarray} \tag{ 5.4 }$$

where the latter form holds for the plane-wave ansatz $\psi (x)={u}^{{ \mathcal T }}(p)\,\exp (-{\rm{i}}p\cdot x)$. The bispinor solutions ${u}^{{ \mathcal T }}(p)$ have been discussed at length in [6, 9] and are used here in equations (3.30) and (4.3). They apply, first and foremost, to a mass eigenstate, with a definite tachyonic mass parameter. The Fermi couplings are universal among all neutrino flavors, and hence, the interaction Lagrangians used in our paper share this property. Our results (3.55) and (4.11) for the decay and energy loss rates thus apply, at face value, to an incoming neutrino mass eigenstate. The results are thus relevant to the non-sterile neutrino flavor if at least one of the three observed non-sterile neutrino mass eigenstates is tachyonic.

We recall that the flavor eigenstates ${\nu }_{f}$ are connected to the mass eigenstates m_i by the Pontecorvo–Maki–Nakagawa–Sakata matrix,

$$\begin{eqnarray}&&{m}^{2}({\nu }_{f})=\displaystyle \sum _{i=1}^{3}| {U}_{{fi}}{| }^{2}\,{m}_{i}^{2},\end{eqnarray} \tag{ 5.5 }$$

where U_fi denote the elements of the flavor-mass mixing matrix. The decay and energy loss rates of the flavor eigenstates are given as

$$\begin{eqnarray}&&{\rm{\Gamma }}({\nu }_{f})=\,\displaystyle \sum _{i=1}^{3}| {U}_{{fi}}{| }^{2}\,{\rm{\Gamma }}({m}_{i}),\end{eqnarray} \tag{ 5.6 }$$

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{d}}E}{{\rm{d}}x}({\nu }_{f})=\displaystyle \sum _{i=1}^{3}| {U}_{{fi}}{| }^{2}\,\,\displaystyle \frac{{\rm{d}}E}{{\rm{d}}x}({m}_{i}).\end{eqnarray} \tag{ 5.7 }$$

For a slower-than light mass eigenstate i, one sets ${\rm{\Gamma }}({m}_{i})$ and $\tfrac{{\rm{d}}E}{{\rm{d}}x}({m}_{i})$ to zero. Here, just to be pedantic, we should point out that the calculation of NPCR in this case has to be modified to include all tachyonic mass eigenstates in the exit channel, conceivably modifying the overall results as much as by adding a multiplicative factor three (if all mass eigenstates are available in the exit channel of the tachyonic NPCR decay, see figure 1(b)).

The tachyonic dispersion relation ${E}^{2}-{\vec{k}}^{2}=-{m}_{\nu }^{2}$ conserves Lorentz invariance. Hence, one might ask about the Lorentz invariance of our results, and in particular, about the Lorentz invariance of the threshold condition (2.5); finally, one might 'chase' the high-energy neutrino, lowering its energy in the Lorentz-transformed, moving frame to a value below threshold. This question finds an answer in the subtleties of the tachyonic theory; we follow the discussion in [17]. Namely, upon a Lorentz transformation of the vacuum state, because there is no 'energy mass gap' between the positive- and negative-energy states, some of the annihilation operators of quantized fields will turn into creation operators, and vice versa. This point is explained in detail around equations (4.7)–(4.9) of [17]. (Incidentally, it is observed at the same place that the fundamental creation and annihilation operators of tachyonic fields have to be quantized according to fermionic statistics, which is another argument in favor of spin-1/2 rather than spinless tachyonic theories.) Furthermore, around equation (5.7) of [17], it is argued that the vacuum state in a tachyonic theory cannot be Lorentz-invariant, but is filled with those (real) anti-fermions whose energies are 'pushed down' to energies below zero, from initially positive-energy states, under the Lorentz transformation. Our figure 2 illustrates how the decay and energy loss rates, under a Lorentz transformation, turn into neutrino–antineutrino collision rates (leading to decay and energy loss) with the 'downshifted' real antiparticle states which are the result of the Lorentz transformation, finally restoring the Lorentz invariance of the results for the decay and energy loss rates, given in equations (3.55) and (4.11).

A very important question regarding the conceivable existence of tachyonic neutrinos concerns the possibility of superluminal signal propagation. We thus follow appendix A of [47] and ask how difficult it is to reliably 'stamp' any information onto the superluminal neutrinos. When assuming the dispersion relation $E={({\vec{k}}^{2}-{m}_{\nu }^{2})}^{1/2}$ with its classical equivalent $E={m}_{\nu }/\sqrt{{v}_{\nu }^{2}-1}=0$, the dilemma is that high-energy tachyonic neutrinos approach the light cone and travel only infinitesimally faster than light itself. In the high-energy limit, their interaction cross sections may be sufficiently large to allow for good detection efficiency but this is achieved at the cost of sacrificing the 'speed advantage' in comparison to the speed of light. Low-energy tachyonic neutrinos may a substantially faster than light, but their interaction cross sections are small and the information sent via them may be lost. The smallness of the cross sections sets important boundaries for the possibility to transmit information, as follows. In Appendix A of [47], it has been shown that, by postulating that superluminal particles should not have the capacity to transport any 'imprinted' information into the past, one is naturally led to the assumption that any conceivable superluminal particles have to be very light, and weakly interacting.

Following figure 1 of [57] (see also [58]), we now supplement these considerations with a numerical estimate. Neutrino-electron cross sections for $1\,\mathrm{GeV}\lt {E}_{\nu }\lt 1\,\mathrm{PeV}$ can be estimated to good accuracy using the formula

$$\begin{eqnarray}&&\sigma ={A}_{0}\,\displaystyle \frac{{E}_{\nu }}{{E}_{0}},\end{eqnarray} \tag{ 5.8 }$$

with ${A}_{0}\sim 0.0095$ fb and ${E}_{0}=1\,\mathrm{GeV}$. By order-of-magnitude, equation (5.8) remains valid for neutrino scattering off electrons, for all three neutrino flavors, even if additional charged-current interactions exist for electron neutrinos, due to exchange graphs with virtual W bosons (for muon and tau neutrinos, only the Z boson contributes at tree level). A particle typically cannot be localized to better than an area equal to the square of its (reduced) Compton wavelength (we temporarily restore factors of ℏ and c),

$$\begin{eqnarray}&&{A}_{\min }={\lambda\!\!\!{-}}^{2}={\left(\displaystyle \frac{{\hslash }}{{m}_{\nu }c}\right)}^{2}.\end{eqnarray} \tag{ 5.9 }$$

The detection probability P for a perfectly focused particle therefore cannot exceed

$$\begin{eqnarray}&&P=\displaystyle \frac{\sigma }{{A}_{\min }^{2}}=\displaystyle \frac{{A}_{0}\,{c}^{4}\,{m}_{\nu }^{3}}{{E}_{0}\,{{\hslash }}^{2}}\,\sqrt{\displaystyle \frac{1}{{\delta }_{\nu }}}.\end{eqnarray} \tag{ 5.10 }$$

If we are to send information reliably, then the detection probability should be of order unity. Setting P = 1 leads to

$$\begin{eqnarray}&&{\delta }_{\nu }=\displaystyle \frac{{A}_{0}^{2}\,{c}^{8}\,{m}_{\nu }^{6}}{{E}_{0}^{2}{{\hslash }}^{4}}.\end{eqnarray} \tag{ 5.11 }$$

When traveling at a speed $c+\delta c$ for a path length s, the neutrino acquires a path length difference of $\delta s$, which compares to its Compton wavelength ${\lambda\!\!\!{-}}={\hslash }/({m}_{\nu }\,c)$ as follows,

$$\begin{eqnarray}&&\delta s=s\,\displaystyle \frac{\delta c}{c}=s\,\displaystyle \frac{{\delta }_{\nu }}{2}.\end{eqnarray} \tag{ 5.12 }$$

The distance traveled by the superluminal neutrino exceeds the distance traveled by a light beam by an amount $\delta s={\lambda\!\!\!{-}}$ when $s={s}_{0}$ where

$$\begin{eqnarray}&&{s}_{0}=\displaystyle \frac{2\,{E}_{0}^{2}\,{{\hslash }}^{5}}{{A}_{0}^{2}\,{c}^{9}\,{m}_{\nu }^{7}}=6.63\times {10}^{74}\,{\rm{m}}{\left(\displaystyle \frac{{m}_{\nu }}{\mathrm{eV}/{c}^{2}}\right)}^{-7}.\end{eqnarray} \tag{ 5.13 }$$

Even at a (larger-than-realistic) mass square ${m}_{\nu }^{2}=1\,{\mathrm{eV}}^{2}$ for the tachyonic neutrino flavor eigenstate, the value of ${s}_{0}\sim {10}^{74}\,{\rm{m}}$ far exceeds the commonly assumed size of the Universe of ${10}^{26}\,{\rm{m}}$ by many orders of magnitude. The permissibility of slightly superluminal propagation on small length and distance scales has been discussed in the literature previously (see, e.g., [59]). Furthermore, we refer to the experiments in the group of Nimtz [60–62], which also use a compact apparatus and rely on the quantum mechanical tunneling effect, which lies outside the regime of classical mechanics. Thus, a very slightly superluminal neutrino flavor eigenstate with a light mass does not necessarily lead to a detectable violation of causality.