In a class discussion last week, someone pointed out that a typical core collapse supernova releases $\sim10^{46}\text{ J}$ of energy in the form of (anti-)neutrinos while only radiating $\sim10^{44}\text{ J}$ in the form of photons. This then means that the energy flux from neutrinos is about two orders of magnitude higher than the energy flux from photons. If you could find a way to capture that energy, you'd have . . . quite lot of energy. I'm considering this as a possible power generation mechanism in a world of mine.

The problem is, you'd need to capture a lot of neutrinos, and that's not easy. From SN 1987A, in the Large Magellanic Cloud, 25 neutrinos were captured by several detectors out of a total of $\sim10^{58}$. Granted, we're about $1.5\times10^{5}$ light-years away, and so, with the same efficiency, we could capture perhaps $\sim5\times10^9$ neutrinos at a distance of $10$ light-years. But that wouldn't even get us $1\text{ J}$ of energy, assuming we captured all of the energy from each neutrino!

Neutrino detection might be an implausible energy source, even given the high energies of supernovae. I currently don't think it's possible to wrest anything useful from it. However, I'd at least like to give it a shot before dismissing it. Therefore, I have one question: How efficient can a neutrino detector be? In other words, given a certain flux of neutrinos, what is the upper bound to the fraction that the detector could absorb?

Some clarifications:

  • I'm not asking for a general plausibility study, nor am I asking how close a detector could be to such a supernova. I just want to know about efficiency.
  • I'd like to use a Cherenkov-style detector, along the lines of Super-Kamiokande. However, if there's a much more efficient design, I'm open to hearing it.
  • I'll assume that the builders of such a detector are from several Type II civilizations, on the Kardashev scale.
  • Let's keep things pretty firmly grounded in science, please. I'd like to really, really minimize hand-waving.
  • $\begingroup$ Before I put some real thought into this I'd say that your main probleem is going to be the scale of the thing. Neutrinos are so weakly interacting that you would have to have a contraption so enormous that it might take more energy to build than you could potentially get out $\endgroup$
    – bendl
    Dec 14, 2017 at 15:52
  • $\begingroup$ @bendl I'm assuming that this thing is going to at least have to be on the order of kilometers, which is why a Type II civilization (at the least) is needed. I don't know how this type of detector would scale with size, which is hopefully something people will address. $\endgroup$
    – HDE 226868
    Dec 14, 2017 at 15:54
  • 1
    $\begingroup$ Well, more in the order of the light-year. It is commonly stated that a light-year of lead would only stop half of the neutrinos flying through it. $\endgroup$
    – Keelhaul
    Dec 14, 2017 at 16:04
  • $\begingroup$ @Keelhaul Quite true, but I don't need 50% of the neutrinos; many orders of magnitude below that would be fine. Even only one one-thousandth of 1% would yield a lot of energy. $\endgroup$
    – HDE 226868
    Dec 14, 2017 at 16:24
  • $\begingroup$ You need cosmic events where neutrino/photon ratios are skewed much heavier to neutrino side. At any distance, it would be easier to capture 1 photon than 1 out of 100 neutrinos. $\endgroup$
    – Alexander
    Dec 14, 2017 at 17:47

3 Answers 3


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 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 $5.3\times 10^{-26}\,\text{J}$.


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)

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 $4.18\times 10^{23}$.


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 detectorPlot 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

  • 1
    $\begingroup$ This is the best answer I've seen in a long time! It's what I would write if I knew more than half of what's in it. :-) I do have one (minor) question - is the quantity $N_p\sigma z$ essentially analogous to the optical depth used to determine the intensity of light as it travels through a medium? It looks identical; I just haven't seen the expression used for particles besides photons. $\endgroup$
    – HDE 226868
    Dec 28, 2017 at 3:37
  • $\begingroup$ @HDE226868 Thanks! Yes, it's the same principle, with the obvious difference that it's 'neutrino' instead of optical and caused by different interactions $\endgroup$ Dec 28, 2017 at 19:26

A useful fairly easy to understand reference on neutrino interactions is found here: http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/neutrino3.html

From the above source:

enter image description here

This gives us a means of estimating possible neutrino absorptions and identifies some areas we can tweak to improve performance.


We can't really control the flux of neutrinos, other than placing our detectors inside operating fission reactors which produce anti-neutrinos, but potentially advanced civilizations could place their detectors closer to the sources (adjacent to supernova) or potentially use gravity lensing to increase the number of neutrinos passing through their detectors, but it seems likely these extreme means could be used more wisely for other power generation methods.

Neutrino Cross Section

This is a function of the energy level of the neutrino not much we can change here and sadly this is where we get really screwed, the values are on the order of 10^-45 an exceedingly small number reflecting the very unlikely possibility of a neutrino interacting with anything.

Density of Nucleons

This we can effect, making your target more dense will increase the likelihood of interaction, unfortunately it is not going to have a huge effect.

For water we have 1000kg/m^3, that 1000 kg is going to be 1x10^6 / 18 (the molar mass of water) = ~55,555 moles of water or 55,555 x 6.023*10^23 (that big number helps a little) = 3.3 x 10^28 molecules x 18 nucleons per molecule gives us 6 x 10^29 nucleons / m^3 for water, that's a big number it should help right, but no we are still 16 orders of magnitude short compared to our 10^-45 (for fun compare 10^-16 to 10^-9 or 1 in a trillion, really bad odds)

For fun let's express that as a percentage 1 m^3 of water will absorb ~0.000000000000001% of the neutrinos passing through it!

We can do better than water, lead is about 11x denser than water, but that one order of magnitude is still up against that 10^-45 number not going to affect it much (are you sensing a pattern yet).

We could theoretically push it up a few more notches as a Type II civilization and go for degenerate neutron star matter or otherwise insanely dense material, estimates for neutron star dense matter are on the order of 10,000 kg/cc or ~10^7 x greater than water, which does push us up to the 1 in a trillion chance.

Volume of Detector

Again here we can make improvements, make the detector bigger! Unfortunately for comparison sake a light year is ~10^15 m (remember 10^-45!!) so even light year sized detectors are not likely to absorb much more than a fraction of any neutrinos passing through it. The anecdote of a light year of lead stopping half the neutrinos is actually being generous, it would really only stop about 10-15%.


So you could absorb most neutrinos with an absorber of neutron star material density a million km thick, which is probably a nontrivial enterprise for even a type II or III civilization. For anything less you are not going to absorb very many neutrinos, and definitely not enough to make any usable power (the power density numbers are astoundingly worse than terrible). When compared to other sources, you'd be better of extracting power with the photons of distant stars.

  • $\begingroup$ In other words you agree, the maximum possible efficiency of a detector is currently in determinable $\endgroup$
    – anon
    Dec 14, 2017 at 20:44
  • $\begingroup$ Yes, indeterminable unless you have the exact size and density of the neutrino detector, which aren't given in the question, but given those it is fairly easy to calculate the percentage of the neutrino flux passing through it that would be absorbed. $\endgroup$
    – Josh King
    Dec 14, 2017 at 20:55
  • $\begingroup$ as well as chemical composition, fidelity and dispersion of sensors $\endgroup$
    – anon
    Dec 14, 2017 at 20:59
  • $\begingroup$ Yeah, I just focused on neutrinos absorbed, it only gets worse if you start talking about how to actually detect them. $\endgroup$
    – Josh King
    Dec 14, 2017 at 21:00
  • 1
    $\begingroup$ I'm fine with people trying to figure out the parameters of a possible detector and what the resulting efficiency would be. $\endgroup$
    – HDE 226868
    Dec 14, 2017 at 21:08

How efficient can a neutrino detector be? In other words, given a certain flux of neutrinos, what is the upper bound to the fraction that the detector could absorb?

Oh boy......

So to start on this it should first be made clear what the largest factors are in determining this fraction.

The first statistic:

∼10 46 J ∼1046 J of energy in the form of (anti-)neutrinos

does not represent the amount of neutrinos received by Earth but is the amount of Neutrinos emitted by the source spherically outwards. Thus the first largest factor is the distance between the source and the detector. As this limits the total possible neutrinos a detector can receive.

The next obvious factor is the cross section surface area of the detector with respect to the emission source. As this also impacts how many neutrinos can possibly be collected.

Now the way most these detectors work is by the simple fact that IF a neutrino collides with a hydrogen atom it releases an electron which can be sensed with a sensor. This is why most neutrino arrays are surrounded by water containing electron sensing sensors. Even then this is dependent on the energy of the neutrino with low energy neutrinos being harder to detect with plenty passing through the Earth every day.

So the question of:

How efficient can a neutrino detector be?

Is essentially impossible to determine as it is dependent on the source, energy of the neutrino, and scale and sophistication of the detector.

As for the root objective of this being about energy production that is even more absurd as the electric output from these interactions is so rare and miniscule that often the hardest part in building these arrays is finding big enough and isolated locations that the external environment isn't releasing electrons that would contaminate the data. This is why they cant be built in the ocean with moving sea water against metal sources or lightning strikes.

  • $\begingroup$ PS: IceCube: masterclass.icecube.wisc.edu/en/analyses/cosmic-neutrinos is the array im more familiar with, regardless, as I understand it, its the same methodology across most neutrino arrays: hope and pray a neutrino collides and releases a detectable electron emission. $\endgroup$
    – anon
    Dec 14, 2017 at 17:32
  • 1
    $\begingroup$ I only quoted the $10^{46}$ figure to note that relative fluxes through any given area would be much higher for neutrinos than photons. Regardless of the area of the detector, the difference will be a factor of 100. Also, you don't really tell me anything about efficiency; you just say what the relevant factors are - which I already am aware of. I also mentioned in the question that I don't care about distance at this point in time; it's not important to efficiency. The distance determines the flux, but not what fraction of that flux will be absorbed - which is what I'm interested in. $\endgroup$
    – HDE 226868
    Dec 14, 2017 at 18:16
  • $\begingroup$ I did however say that that does come down to the specific energy of the neutrino, scale and sophistication of the array. Different sources release different varieties of neutrinos with lower energy neutrinos being harder to detect. Thus efficiency is indeterminable unless you relate it a specific energy spectrum. If you simply say all energy spectrum the fraction becomes ambiguously low $\endgroup$
    – anon
    Dec 14, 2017 at 18:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .