2
$\begingroup$

According to wikipedia, the sun emits enough neutrinos that the number passing through a square meter of area oriented perpendicular to the sun at Earth distance is around $6.5 \times 10^{14}$ per second.

What is the momentum flux of these neutrinos? If you counted up the momentum of all the solar neutrinos passing through that square meter of area at $1 \cdot au$ from the sun, what's the order of magnitude? For example, if you made an impossible sci-fi material that was opaque to neutrinos, could the "lift" generated by the neutrino "wind" through a $1\cdot m^2$ "sail" overcome Earth gravity?

$\endgroup$
2
  • 1
    $\begingroup$ "if you made an impossible sci-fi material that was opaque to neutrinos, could the "lift" generated by the neutrino "wind" through a 1 m^2 "sail" overcome Earth gravity?"" this belongs in worldbuilding $\endgroup$ Apr 8 '17 at 19:48
  • $\begingroup$ Please note that you also have to mind the typical cross-section of neutrinos: $\sigma_{\nu} \sim 10^{-45}$cm$^2$. $\endgroup$
    – Stefano
    Jul 3 '18 at 15:07
6
$\begingroup$

You really only need one more piece of data to finish the problem: the typical solar neutrino has a momentum of a few $\mathrm{MeV}/c \approx 1.5 \times 10^{-21} \,\mathrm{kg \cdot m / s}$.

Multiplying that by the flux you list above gives about $P_\nu \approx 1 \times 10^{-6} \,\mathrm{Pa}$ for the pressure on a neutrino absorber at about 1 AU.


For reference this should be compared to the effectiveness of a solar sail which is around \begin{align*} P_\gamma &= 2 \left(\frac{\text{solar constant}}{c}\right) \\ &= 2 \left(\frac{1400\,\mathrm{W/m^2}}{3 \times 10^8\,\mathrm{m/s}}\right) \\ &\approx 1 \times 10^{-5} \,\mathrm{Pa} \end{align*} or an order of magnitude larger.

$\endgroup$
1
  • $\begingroup$ It should be 2 orders of magnitude larger, since the neutrino flux is 2% of the photon flux. $\endgroup$
    – ProfRob
    Jul 5 '20 at 13:57
5
$\begingroup$

There are two approaches to the calculation.

One is to know that the vast majority of solar neutrinos come from a p+p reaction that yields neutrinos with a maximum of 0.42 MeV. The spectrum yields an average neutrino energy of around 0.3 MeV.

As the neutrinos are highly relativistic, then $p = E/c$. The momentum flux is therefore $1.04\times 10^{-7}$ N/m$^2$.

The other approach is to know that 2.3% of the Sun's luminosity emerges in the form of neutrinos. So the neutrino "solar constant" at the Earth is $0.023 \times 1.37\times 10^3$ W/m$^2$. To get the momentum flux, one divides by $c$ to get $1.05\times 10^{-7}$ N/m$^2$.

Can this overcome gravity? Your perfectly absorbing neutrino sail (which we assume is transparent to electromagnetic radiation, which has 1/0.023 times more momentum flux?) would have to have a mass of less than $10^{-8}$ kg/m$^2$.

$\endgroup$
2
$\begingroup$

Here's a relevant reference:

"On the feasibility of neutrino sails" by R. H. Peck, H. W. Buttery, C. D.Y. Moore, C. J. Middleton (2016)

In this paper we consider whether a neutrino sail, a sheet of material absorbing neutrinos and gaining thrust from their momentum change, is a viable method of spacecraft propulsion. We calculate the thrust that could be achieved per unit area and compare this to that possible using a photon solar sail. We also calculate the thickness of sail necessary assuming that there are no special conditions under which the cross section for neutrino interactions with nuclei can be increased. We find that a thickness of 34000 light years would be necessary if a sheet of osmium were used, whereas neutron star matter could achieve this at 189 km thickness. We conclude that a neutrino sail is not a practical method of propulsion.

The calculations agree with the other answers here.

We have shown that a photon sail gives 70 times the thrust per unit area than the upper bound of what a neutrino sail could reach

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.