Calculation of transition radiation
intensity

$\quad$ Let us consider a neutrino with non-zero toroid dipole moment crossing the interface between two media, see Fig. 1, with refraction indices $n_1$ and $n_2$ ($n_1\neq n_2$). The electromagnetic interactions of neutrinos is described by the Hamiltonian:

$${\cal H}_{\rm int} = i e{\cal T}(q^2)\overline{\psi}(x)\varepsilon_{\mu \nu \lambda \rho} P^{\nu} q^{\lambda} \gamma^{\rho} \psi(x){\cal A}^\mu(x),$$

using the identity $\varepsilon_{\mu \nu \lambda \rho} \gamma_{\rho} = \frac{i}{2} \left( \gamma_{... ...gamma_{\lambda} - \gamma_{\lambda} \gamma_{\nu} \gamma_{\mu} \right) \gamma_{5}$, we we obtain

$${\cal H}_{\rm int} \Rightarrow e{\cal T}(q^2)\overline{\psi}(x)\gamma_\mu\gamma_5 \psi(x) \frac{\partial F^{\mu\nu}(x)}{\partial x^\nu}$$

$$= e{\cal T}(q^2)\overline{\psi}(x) \gamma_\mu \gamma_5\psi(x)J^\mu_{\rm ext}.$$ (6)

Here $\psi$, ${\cal T}(q^2)$, $J^\mu_{\rm ext}$, ${\cal A}^\mu$ and $F^{\mu\nu}$ are the neutrino wave function, neutrino toroid form factor, electromagnetic current, $4-$potential and tensor of the electromagnetic field, respectively (the Hamiltonian (6) was also obtained by Zel'dovich [10] using the anapole parametrization).

The transition $\nu(p_1)\rightarrow\nu(p_2)+\gamma(k)$ becomes possible due to the TDM of the neutrino . In a medium with refraction index $n$, the four-momentum vector of a photon is given by $k^\mu=(\omega,{\bf k}),\quad \vert{\bf k}\vert=n\omega$ ($\omega$ is the energy of a photon), and the transition probability reads $\Gamma=\vert{\cal S}_{fi}\vert^2\frac{Vd^3{\bf p}_2}{(2\pi)^3} \frac{Vd^3{\bf k}}{(2\pi)^3}$, where the transition matrix element is expressed as

$$\vert{\cal S}_{fi}\vert^2 = (2\pi)^3\ell^2t\frac{m_\nu}{E_1V}\frac{m_\nu}{E_2V} \frac{(1-n^2)^2\omega^4}{2\omega n^2 V}$$

$$\times \delta(p_{1x,y}-p_{2x,y}-k_{x,y}) \delta(E_1-E_2-\omega)$$

$$\times \left\vert\int^{\ell/2}_{-\ell/2}dz \exp[i(p_{1z}-p_{2z}-k_z)z]{\cal M}_{fi} \right\vert^2.$$ (7)

Here ${\cal M}_{fi}=e{\cal T}(0)\overline{u}_2 \widehat{\varepsilon}\gamma_5 u_1$ is the amplitude, and $t$, $\ell$ and $V=\ell^3$ denote time, length and volume of the transition region, respectively, and $\ell=\beta t$, where $\beta=p/E$ is the velocity of the neutrino. The phase of the integrand in (7) defines the formation-zone length of the medium as

$$Z(n)=(p_{1z}-p_{2z}-k_z)^{-1}= (p_{1z}-p_{2z}-n\omega\cos\theta)^{-1},$$

where $\theta$ is the angle between the photon and the direction of the incident neutrino. The details of further calculations are the same as in [12], and here we present only final results for the energy intensity $S$ per interface

$$\frac{d^2 S}{d\theta d\omega}= \frac{{\cal T}^2(0) \omega^6... ... +\frac{E_\nu E_2}{pp_{2z}}-1+\frac{m_\nu^2}{pp_{2z}}\right\},$$ (8)

where

$$R_i=\frac{1-n_i^2}{n_i}\frac{1}{p-p_{2z}-n_i\omega\cos\thet... ...omega\int_0^{\theta_{\rm max}} \frac{d^2 S}{d\theta d\omega},$$ (9)

and

$$p_1^\mu=(E_\nu,0,0,p),\quad p_{2z}=\sqrt{E_2^2-m_\nu^2-n^2\omega^2\sin^2\theta}, \quad E_2=E_\nu-\omega.$$

Using the numerical value (1), $\tau_{\nu_e}=e{\cal T}(0)$ with ${\cal T}(0)=\sqrt{2}G_F/\pi^2$, and assuming that the refractive index can be expressed as $n_i(\omega)=1-\omega_i^2/2\omega^2$ for $\omega \gg \omega _i$ ($\omega_i$ is the plasma frequency) for a medium-vacuum transition ($\omega_2=0$, $R_2=0$), we present the energy spectrum and angular distribution in Figs. 2 and 3. The total energy loss of the neutrino has been computed numerically and is shown as a function of the neutrino mass for $E_\nu =1$ MeV in Fig. 4. For $m_\nu<10$ eV the TR energy is approximately constant and equals $S\simeq 2\times10^{-40}$ keV. Because of the finite value of the TDM for massless neutrinos [4,5], the TR does not vanish in this limit and has the value $S\Bigr\vert _{m_\nu=0}=2.26\times10^{-40}$ keV.

In order to estimate the magnitude of this effect, let us consider a transition radiation detector (TRD) which can be used to measure experimentally such transition radiation of neutrinos. The TRD consists of sets ($N_1$) of ``radiator'' and xenon-gas chambers, where one radiator typically comprises of a few hundred layers ($N_2$) of a polypropylene film ($\omega_p=20$ eV) and a gas ($\omega_p\ll 1$ eV). The total energy deposition ($W$) in the TRD is given by

$$W=S\cdot F_\nu\cdot A^2\cdot T\cdot N_1\cdot N_2,$$

where $F_\nu$ is the neutrino flux, $A$ is the TRD area and $T$ is the time of measurement. For example, we have taken the neutrino flux coming from a nuclear reactor, $F_\nu\sim10^{13}\overline{\nu}_e/{\rm cm}^2\;{\rm sec}$, with the energy of $E_\nu =1$ MeV and TRD parameters as: $A=10$ m$^2$ $N_1=10$ sets and $N_2=10^4$ layers. The total energy deposition for the TDM $\tau_{\nu_e}=e\sqrt{2}G_F/\pi^2$ with TR energy $S=10^{-40}$ keV is

$$W=3\times10^{-10}\left(\frac{T}{1\; {\rm year}}\right).$$

Unfortunately, this value is extremely small and cannot be extracted from the background of an experimental setup. But this small radiation which always exist for both massive and massless neutrinos, may have interesting consequencies in astrophysics.