IOM 2017, Теория — Задача 3. Neutrino

PROPOSED SOLUTION AND ANSWERS

Part I.

1. The particle mass in STR (rest mass) is determined from the relation between the particle energy and momentum. For ultrarelativistic particle

Therefore,

and the «mean» mass of electron neutrino in vacuum

The numerical value $\langle m_e\rangle\approx2{,}64\ \mathrm{meV}/c^2\approx4{,}69\cdot10^{-39}\ \mathrm{kg}$.

Similarly, for muon neutrino, $\langle m_\mu\rangle=\sqrt{m^2+(\Delta m)^2/4+m\Delta m\cos2\vartheta}$ and $\langle m_\mu\rangle\approx5{,}43\ \mathrm{meV}/c^2\approx9{,}66\cdot10^{-39}\ \mathrm{kg}$.

2. An electron neutrino created at $t=0$ and $x=0$ is a superposition of states of certain masses: $\vec u_e=\vec u_1\cos\vartheta+\vec u_2\sin\vartheta$. Each state propagates at a distance $x$ as a wave of certain frequency and wavevector, i.e.

Substituting expressions for $\vec u_{1,2}$ in this formula one finds that at any point the wave contains two «flavor polarizations»:

The wave of muon neutrinos (the second term) equals to the expression

corresponding to a propagating wave of the «mean» frequency $(\omega_2+\omega_1)/2$ and «amplitude» modulated by the harmonic factor of the half-difference frequency. Neutrinos travel a distance $x$ from the point of emission for a time $t=x/c$ and the half-difference frequency is $(\omega_2-\omega_1)/2=(E_2-E_1)/(2\hbar)=\Delta E/(2\hbar)$. It is important to note that small corrections due to neutrino masses can be neglected when calculating the mean energy ($E\equiv(E_1+E_2)/2\approx c|\vec p|$) and must be taken into account when calculating the energy difference:

hence $\Delta E=E_2-E_1\approx(m_2^2-m_1^2)c^4/(2E)=m\cdot\Delta m\cdot c^4/E=6\cdot10^{-12}\ \mathrm{eV}$. Obviously, $\Delta E\ll E$, so the «amplitude» factor can be regarded as almost constant during the oscillation period corresponding to the «mean» frequency. Therefore, the flux of muon neutrinos as a function of $x$ obtained by averaging $\vec u^{\,2}$ over the period corresponding to the «mean» frequency is:

Here $I_0$ is the initial flux of neutrinos. Therefore, the oscillation length corresponding to the spatial period of the second factor in $I_\mu$ equals $L=2\pi E\hbar/(m\cdot\Delta m\cdot c^3)\approx2\cdot10^5\ \mathrm m=200\ \mathrm{km}$.

ANSWERS-1: the «mean» mass of $\nu_e$ equals $\langle m_e\rangle\approx2{,}64\ \mathrm{meV}/c^2\approx4{,}69\cdot10^{-39}\ \mathrm{kg}$, the «mean» mass of $\nu_\mu$ equals $\langle m_\mu\rangle\approx5{,}43\ \mathrm{meV}/c^2\approx9{,}66\cdot10^{-39}\ \mathrm{kg}$, and the oscillation length in vacuum is $L=2\pi E\hbar/(m\cdot\Delta m\cdot c^3)\approx2\cdot10^5\ \mathrm m$.

Note: the contesters could estimate the «mean» masses somewhat differently, e.g. by following this line of argument: «The particle mass is proportional to its energy at zero momentum, therefore, the «mean» mass of electron neutrino in vacuum is $\langle m_e\rangle=m_1\cos^2\vartheta+m_2\sin^2\vartheta=m-\frac{\Delta m}{2}\cos2\vartheta$. Then $\langle m_e\rangle\approx2{,}59\ \mathrm{meV}/c^2\approx4{,}61\cdot10^{-39}\ \mathrm{kg}$. Similarly, for muon neutrino

and $\langle m_\mu\rangle\approx5{,}41\ \mathrm{meV}/c^2\approx9{,}62\cdot10^{-39}\ \mathrm{kg}$».

This approach is not logical from the STR perspective since these «masses» are not invariant under Lorentz transformations, the correct mass must be defined via the invariant $E^2-c^2|\vec p|^2$. However, the numerical values turn out to be quite reasonable. That is why such an approach should be given a partial credit.

Part II.

3. To estimate the impact of vacuum oscillations on the flux of electron neutrinos detected on the Earth, notice, that neutrinos entering the detector traveled various distances: they have been created at different points of active solar core of size $r_A\gg L$. Besides, during the data accumulation the Earth changes its position with respect to the Sun (the distance between the Sun and the Earth varies by $\varepsilon a\approx2{,}6\cdot10^9\ \mathrm m\gg L$). Therefore, the flux must be averaged over variations of $x$ which significantly exceed the oscillation length. Since $\langle\sin^2(\pi x/L)\rangle=1/2$, the flux of muon neutrinos corresponding to the decrement of the flux of electron neutrinos $\langle I_\mu(x)\rangle\approx\frac12I_0\sin^22\vartheta\approx0{,}058\cdot I_0$ which is less than 6% of the initial value! Thus, vacuum oscillations cannot explain the significant decrease of the $\nu_e$ flux.

4. Proceeding to description of the oscillations in matter, notice, that although the energy of neutrino interaction with solar electrons is very small compared to the neutrino energy (even for the maximum electron density in the core $\delta E\approx7{,}5\cdot10^{-12}\ \mathrm{eV}$), it is of the same order as $\Delta E=6\cdot10^{-12}\ \mathrm{eV}$! Therefore, this interaction cannot have a significant impact on the neutrino propagation and/or the total flux but it can noticeably affect the oscillations.

Thus, taking into account the interaction of neutrinos with matter, the mean energy of electron neutrino becomes

where the «tilde» over a quantity corresponds to that one in matter. The mean energy of muon neutrinos and the energy of interaction of electron and muon neutrinos remain unaltered:

Using the first two equations one finds that $\widetilde E=E+\delta E/2$ and $\Delta\widetilde E\cos2\widetilde\vartheta=\Delta E\cos2\vartheta-\delta E$. Combing this and the third equation yields the remaining parameters:

(the minus sign in the expression for $\Delta\widetilde E$ corresponds to $\delta E>\Delta E\cos2\vartheta$).

5. First of all, let us look at the new value of the mixing angle: if $\delta E=\delta E_{\mathrm{rez}}\equiv\Delta E\cos2\vartheta$, the angle turns out to be $\vartheta_{\mathrm{rez}}=\pi/4=45^\circ$! This situation is called «maximal mixing» because the fractions of muon and electron neutrinos in the flow change periodically in antiphase between 0 и 100%. The phenomenon of sharp amplification of the oscillations in matter is called «Mikheyev–Smirnov–Wolfenstein effect» or MSW-effect. The condition of MSW-effect for neutrino with our parameters reads $1{,}27\cdot10^{-12}\frac{n_e}{10^{31}\ \mathrm{m^{-3}}}\approx6\cdot10^{-12}\cdot\cos20^\circ$, i.e. $n_e=n_{\mathrm{rez}}\approx4{,}44\cdot10^{31}\ \mathrm{m^{-3}}$. Notice, that the mixing angle away from the resonance remains small (the oscillations in these regions do not significantly affect the flux). One can also see that there is a layer outside the active solar core where electron density takes the «resonance» value. The width of the region in which the density differs from the resonant one by less than 1% is

which exceeds the oscillation length by more than order of magnitude. Therefore, a neutrino passing through this layer undergoes many oscillation cycles in which the fraction of muon neutrinos changes harmonically from 0% to 100%. The initial phase of the waves from different points in the solar core varies in a wide range. When evaluating the flux of $\nu_\mu$ at the exit of resonance region one should average over these variations as well. Hence, the flux of muon neutrinos can be as large as 50% of the total flux (compare to 6% in vacuum). Thus, the solar matter increases the flux of $\nu_\mu$ 8.5 times, or approximately by 750%!

6. Another interesting observation concerns neutrino masses. It follows from the equations for $\Delta\widetilde E$ and $\widetilde\vartheta$ that the difference of energies of muon and electron neutrinos with the same momentum $\langle E_\mu\rangle-\langle E_e\rangle=\Delta\widetilde E\cdot\cos\widetilde\vartheta$ becomes negative for $\delta E>\Delta E\cos2\vartheta$ (i.e. when the electron density is higher than the resonant value)! Thus, in the solar core below the resonant layer the mass of $\nu_\mu$ is less than the mass of $\nu_e$. Outside the resonant layer $\delta E<\Delta E\cos2\vartheta$ and the mass of $\nu_\mu$ exceeds the mass of $\nu_e$. Obviously, the mass difference tends to its «vacuum» value when the density is small. Therefore, the plot looks approximately as:

Note: An accurate calculation yields the same result. Using the expression for mean energy (see 1)) $E\equiv(E_1+E_2)/2\approx c|\vec p|+(m_1^2+m_2^2)c^4/(4E)$, one obtains for parameters $m,\Delta m$ (and similarly for $\widetilde m,\Delta\widetilde m$):

(the quantity $M_1$ is introduced to simplify the calculation). Similarly, using the expression for $\Delta E\equiv E_2-E_1$ (see 1)) and the formula for $\Delta\widetilde E$ derived in 4), one obtains:

(here $M_2$ is another shorthand notation).

Solving the set of these two equations allows one to determine $\widetilde m$ and $\Delta\widetilde m$. Mass is positive, therefore, $2\widetilde m>\Delta\widetilde m$ and

The quantity $M_2$ reaches its minimum at the resonant density (the second term under the radical in the expression for $M_2$ vanishes) while $M_1$ changes monotonically (slowly decreases with density since $\delta E$ is proportional to the electron density and other terms in the expression for $M_1$ remain constant). One can see that $\Delta\widetilde m$ decreases when approaching the resonant density. Besides, $\cos2\widetilde\vartheta=0$ at the resonance. Therefore, the difference of masses of $\nu_\mu$ and $\nu_e$ vanishes. It is obvious that $\cos2\widetilde\vartheta$ switches sign when crossing the resonant layer. When the density exceeds the resonant value the sign is negative, i.e. electron neutrino «on average» is heavier than the muon neutrino. When the density is below the resonant value, $\nu_e$ is lighter than $\nu_\mu$ and the masses tend to their vacuum values when the density decreases. Clearly, the accurate analysis leads to the same conclusion as the proposed solution although the information obtained and the amount of calculations have significantly grown. Since the difference of mean masses of $\nu_\mu$ and $\nu_e$ in matter is expressed via $\widetilde m$, $\Delta\widetilde m$, and $\widetilde\vartheta$ one could write down the explicit formula for the mass difference as a function of electron density (it turns out to be quite cumbersome).

ANSWERS-2: $\operatorname{tg}2\widetilde\vartheta=\frac{\operatorname{tg}2\vartheta}{1-\delta E/(\Delta E\cos2\vartheta)}$, solar matter enhances the flux of $\nu_\mu$ approximately by 750%, the mass difference of $\nu_\mu$ and $\nu_e$ at small electron density corresponds to the vacuum value, decreases when approaching the resonant density, goes through zero and changes sign when passing through the resonance layer, and then increases as the density grows.

Part III

7. Firstly, it is necessary to estimate the energy scale of the phenomenon, i.e. to determine the energy released during the collapse. The «binding energy» of a gravitating sphere of mass $M$ and radius $r$ equals $W_g=-\frac35GM^2/r$. To estimate the radius of neutron star let us equate its density to the density of nuclear matter equal to the ratio of proton mass to its volume. The mass of $N_A$ (1 mole) of protons approximately equals one half of molar mass of molecular hydrogen ($\mu=2\ \text{г/моль}$), hence $M/r^3\approx\mu/(2N_Ar_p^3)$ and $r\approx r_p(2N_AM/\mu)^{1/3}$. For a neutron star of mass $1{,}5M_C$ one obtains $r\approx12\ \mathrm{km}$. The initial radius of the stellar core is greater by three orders of magnitude, so the released energy is $W^{(0)}\approx\frac35GM^2/r\approx3\cdot10^{46}\ \mathrm J$.

Actually, this number is an overestimation because it has been assumed that the core mass does not change during the collapse. However, according to STR a decrease of energy is equivalent to the decrease of mass, i.e. one should actually write $W=\frac35G(M-W/c^2)^2/r$. Since $Mc^2\approx2{,}7\cdot10^{47}\ \mathrm J\approx9W^{(0)}$, this effect is not too small. Solving the quadratic equation yields:

So, the better estimate of the released energy gives $W\approx0{,}825W^{(0)}\approx2{,}475\cdot10^{46}\ \mathrm J$. In any case, $W\sim10^{46}\ \mathrm J$ by the order of magnitude.

8. According to the problem statement the total energy carried away by neutrinos and antineutrinos $E_\nu\approx0{,}8W\approx2\cdot10^{46}\ \mathrm J$. Since the average energy of emitted neutrinos and antineutrinos equals $\langle E_1\rangle\approx10\ \mathrm{MeV}$, their total number is $N\approx E_\nu/\langle E_1\rangle\approx1{,}25\cdot10^{58}$. The number of proton-electron pairs in the initial stellar core is $N_{pe}\approx M/(2m_p)\approx N_AM/\mu\approx0{,}09\cdot10^{58}$, while in the final state almost only neutrons are left. Therefore, the number of neutrinos emitted during neutronization is $0{,}09\cdot10^{58}$, and the rest are emitted during the cooling. Thus, the total number of emitted antineutrinos and neutrinos is:

9. The number of neutrinos and antineutrinos from SN1987A passing through the detector on Earth is proportional to the ratio of the detector cross-section to the area of a sphere of radius $R\approx50\ \mathrm{kpc}$: $N^{(\mathrm{det})}\approx S/(4\pi R^2)N\approx3{,}5\cdot10^{-42}N$. Respectively, $N_{\bar\nu}^{(\mathrm{det})}\approx2\cdot10^{16}$ and $N_\nu^{(\mathrm{det})}\approx2{,}35\cdot10^{16}$. Since only the fraction $\alpha=3\cdot10^{-14}\%$ is detected, the expected number of detected antineutrinos and neutrinos is estimated to be $N_{\bar\nu}^{(\mathrm{reg})}=\alpha N_{\bar\nu}^{(\mathrm{det})}\approx6$ and $N_\nu^{(\mathrm{reg})}=\alpha N_\nu^{(\mathrm{det})}\approx7$.

10. Since all emitted neutrinos and antineutrinos are ultrarelativistic, the momentum carried by them per small solid angle is $dI/c$. Therefore, the component of the momentum carried by the radiation projected on the magnetic axis $z$ of the star is:

Since $\int I\,dt=E_\nu$ and $\int\cos^2\theta\,d\Omega=2\pi\int_0^\pi\sin\theta\cos^2\theta\,d\theta=4\pi/3$, one obtains $p_z=\kappa E_\nu/(3c)$. The recoil velocity is opposite to the magnetic axis and its value $V=p_z/\widetilde M$, where $\widetilde M$ is the mass of the created neutron star. Taking into account the mass defect $\widetilde M\approx M-W/c^2$, one finally obtains $V=\frac{\kappa E_\nu}{3(Mc^2-W)}c\approx80\ \mathrm{km/s}$. This speed is already noticeable against the background motion of stars in galaxies (a typical velocity of this motion is several hundred km/s).

ANSWERS-3: the total number of neutrinos and antineutrinos emitted by supernova is $N_\nu\approx6{,}7\cdot10^{57}$ and $N_{\bar\nu}\approx5{,}8\cdot10^{57}$; the number of neutrinos and antineutrinos passing through the detector is $N_\nu^{(\mathrm{det})}\approx2{,}35\cdot10^{16}$ and $N_{\bar\nu}^{(\mathrm{det})}\approx2\cdot10^{16}$; and the expected number of detected neutrinos and antineutrinos is $N_\nu^{(\mathrm{reg})}\approx7$ and $N_{\bar\nu}^{(\mathrm{reg})}\approx6$. The recoil speed of neutron star is $V\approx80\ \mathrm{km/s}$.

TABLE OF ANSWERS

Archive source notes: The final grading table prints $10^{-36}\ \mathrm{kg}$ for the mass ranges, whereas the worked solution and ANSWERS-1 print $10^{-39}\ \mathrm{kg}$. Its muon kilogram range prints a second $\ge$; both native defects remain literal above. The grading row2 has the native “2+3 (2 for the equation +1 for the value)=5” wording. Worked item6 prints $\cos\widetilde\vartheta$, although item4's energy difference has $\cos2\widetilde\vartheta$; no factor 2 is silently inserted. The accurate-calculation note says $M_1$ “slowly decreases with density” although its displayed defining expression increases with positive $\delta E$. The question uses solar radius $r_C$ and active-core radius $r_a$, while the key also uses $R_C$ and $r_A$; the graph labels are retained. The key explains the stated 2 g/mole hydrogen value as molecular hydrogen. The grading table's item10 momentum numerator uses Latin $k$, while the worked solution and question use $\kappa$. The two key graphs retain their actual native bytes. Blank native PDF pages16 and27 remain in the retained original.