Both the (current) answers to the question state that it is not possible for your idea to work. Aside from the fact that you are explicitly not asking for a feasibility study, the obligatory XKCD reference shows that it's all about the solid angle of the detector from the supernova and so, such a detector may indeed be feasible far into the future. In addition, this answer ignores a number of annoying details in the hope of getting a decent approximation to an answer relatively quickly and easily. It's not a hard-science question, after all!

It turns out that, after making these various estimates and approximations, the efficiency/average energy collected per antineutrino incident on the detector from a supernova would be around $5.3\times 10^{-26}\,\text{J}$.

Background

To start with, let's note the fundamental principles behind what's going on behind Cherenkov style neutrino detection:

1. Cherenkov radiation: Similar to a sonic boom, when a charged particle travels through a medium with refractive index $n$ and travels faster than the speed of light in that medium, $\frac{c}{n}$, (in the case of standard materials with $n>1$), it emits light, typically detected by lots of photomultiplier tubes. This is commonly observed in water, which has a refractive index of $n \approx \frac{4}{3}$ (for the purposes of this answer, this is a good enough estimate for wavelengths of light $\lesssim 17.5\,\mu m$).
2. Detecting neutrinos: Neutrinos are uncharged, so cannot cause Cherenkov radiation by themselves. As such, some sort of interaction needs to occur in order to get a charged particle that can in turn cause radiation. This is generally done by the weak interaction. The most common example in a Cherenkov detector is an $\bar\nu_e+ p \rightarrow n + e^+$. That is, an anti-neutrino converts a proton (Hydrogen) into a neutron and emits a positron in the process. The positron then emits Cherenkov radiation before getting absorbed by a similar process at some later point.
3. Detecting/using the emitted photons: Once generated, a photon has to be able to go from the point of generation to the point of detection. Of course, photons can also interact with and get lost in the surrounding environment. This is described by the transmittance, $T$, of light through the material, which is (in terms of the attenuation coefficient, $\alpha$, over a distance $d$) $T = e^{-\alpha d}$. For light with an initial intensity $I_0$, the final intensity is then $I_0T$, which can then be detected.

However, there are some Issues with the Standard Model of particle physics: (all of these will be ignored as the definite model for how these are actually fixed is obviously unknown, but are here for completion)

• It doesn't account for gravity (any effects of this on neutrinos are, to my knowledge, unknown). As this has effects on large energy collisions, which could occur in supernovae, this may have en effect on the production of neutrinos and antineutrinos.
• The Standard Model predicts that neutrinos are massless, yet the 2015 Nobel Prize was awarded for experimentally showing that neutrinos do in fact have mass This would have an effect on the scattering cross-section of neutrinos but as we don't yet know what the correct model is¹, this can't really be accounted for here.
• Recent particle physics experiments at e.g. CERN have begun to indicate physics beyond the standard model (note: this is not conclusive evidence by some margin, but is a good indication). It isn't impossible that this would have an effect on neutrinos.

Calculating the Efficiency

Cross section of the Interaction

the cross section for the relevant type of neutrino scattering is² $$\sigma = \frac{4G^2_FM_W^2E_{e^{+}}^2}{\pi\left(M_W^2 + 4E_{e^{+}}^2\right)},$$ where (in natural units) the Fermi constant, $G_F \approx 1.16\times 10^{-5}\,\text{GeV}^{-2}$ and the mass of the weak boson, $M_W \approx 80.4\, \text{GeV}$.

Generating Cherenkov Radiation

As mentioned above, any created positrons need to travel with a velocity $v \gtrsim\frac{3c}{4}$, or, have an energy $$E_{e^+} = \gamma_v m_{e^+}c^2 \gtrsim \frac{4m_{e^+}c^2}{\sqrt{7}}$$ to emit Cherenkov radiation. The total energy of radiation emitted will then be assumed to be $$E_{\text{emit}} = E_{e^+} - \frac{4m_{e^+}c^2}{\sqrt{7}}.$$ That is, the probability of any positron interacting with anything else in the time taken to emit the radiation is assumed to be negligible.

Boosting Frames

What is the energy of the created positrons? According to Hans-Thomas Janka, antineutrinos are created with an energy of around $14-16.5\,\text{MeV}$ for about half a second in a supernova. However, the energy used in the above cross section isn't this energy, but the energy in the zero momentum frame, which can be increased or decreased relative to the supernova by boosting the detector. The issue with this is that performing any boost on a detector is just going to cost energy and so, will defeat the purpose of using the detector to generate energy

Reaction Rate

For a flux of neutrinos $\Phi_0$ incident on the detector and number density of protons $N_p$, the rate of reactions can then be taken to be ³ $$R = N_p\sigma\Phi \implies \Phi = \Phi_0e^{-N_p\sigma z},$$ as in the case of transmitting photons. Each molecule then has 10 protons and using Very-high-density amorphous ice with a density of $1.25\,\text{g}\,\text{cm}^{-3}$ gives $N_p = 10\cdot \frac{1.25}{m_{\text{water}}}$ protons per cubic centimetre. One water molecule is approximately $2.99\times 10^{-23}\,\text{g}$, so the number of protons per cubic centimetre is approximately $4.18\times 10^{23}$.

Efficiency

Assuming a stationary detector, the positron energy is assumed to be around $0.015\,\text{GeV}$, giving a cross section of $\sigma \approx 1.518\times 10^6\,\text{J}^{-2} \approx 1.518\times 10^{-45}\,\text{m}^2$. This gives the number of neutrino interactions having occurred after a distance $z$ metres as approximately $\Phi_0\left(1 - e^{-6.35z\times 10^{-16}}\right)$, with each of these interactions emitting about $0.0142\,\text{GeV}$, $e^{\frac{-\alpha z}{2}}$ of which is assumed to be transmitted on average. Conveniently, the absorption coefficient of blue light is low, at around $0.02\,\text{m}^{-1}$. Or, per unit metre of area of detector, the energy collected is $$E \sim \Phi_0\left(1 - e^{-6.35z\times 10^{-16}}\right)\cdot e^{-0.01 z}\cdot 0.0142\,\text{GeV}.$$

Plot of efficiency of detector in stationary frame in GeV. Vertical axis shows average energy in GeV obtained per antineutrino. Horizontal axis shows depth ($z$) of detector

Plotting this gives a maximum of $$E_{\text{max}} \sim 3.3\Phi_0\times 10^{-16}\,\text{GeV} \approx 5.3\Phi_0\times10^{-26}\,\text{J}$$ with a detector with a depth of $100\,\text{m}$. If it were somehow possible to collect all $10^{58}$ neutrinos, this would indeed give a very impressive amount of energy of up to $\sim 10^{32}\,\text{J}$! It should be noted that this is due to the insane amount of energy produced by a supernova as the efficiency is... Somewhat pathetic.

^{1 It's generally thought that neutrinos are either Dirac neutrinos or Majorana neutrinos}

^{2 given in terms of the energy of the positron as that's both what's experimentally measured and calculating the energy of neutrinos generated by a supernova isn't exactly trivial. Also, this is just proton to neutron scattering, so ignores details such as water being a molecule, which might make a difference}

^{3 This is perhaps a bit of an assumption as perhaps there is a possibility that interactions with certain protons in the water molecule may cause e.g. an interaction with an emitted positron and a bound electron. On the other hand, this would just create more energy}

^{4 This is just a hack as the actual transmittance depends on (as well as the distance in the detector) the direction the positron is emitted, which is a random angle in the zero momentum frame, the details of which depend on the (unknown) neutrino mass}

posted over 6 years ago

Mithrandir24601‭ staff

99 reputation 3 6 16 1