Thermonuclear Fusion in the Sun and Other Stars — Question 3

Question 3

Solar neutrino number flux follows an inverse-square geometry

A star completes $9.20\times10^{37}$ helium-production cycles each second and emits two neutrinos per cycle isotropically. An observer is $1.50\times10^{11}$ m away. Ignore absorption between source and observer.

1. Find total neutrino number flux in both $\,\mathrm{m^{-2}s^{-1}}$ and $\,\mathrm{cm^{-2}s^{-1}}$.

2. If electron-neutrino survival probability is .550, find electron-flavor flux and explain why total flux is not reduced by flavor conversion.

Question 3 - Solution

Solar neutrino number flux follows an inverse-square geometry

(a) Method and calculation

$$\boxed{\dot N_\nu=2(9.20\times10^{37})=1.84\times10^{38}\,\mathrm{s^{-1}}}.$$$$\boxed{\Phi_{\mathrm{all}}=\frac{\dot N_\nu}{4\pi r^2}=6.50767\times10^{14}\,\mathrm{m^{-2}s^{-1}}=6.50767\times10^{10}\,\mathrm{cm^{-2}s^{-1}}}.$$

(b) Interpretation and further calculation

$$\boxed{\Phi_e=.550\Phi_{\mathrm{all}}=3.57922\times10^{10}\,\mathrm{cm^{-2}s^{-1}}}.$$$$\boxed{\Phi_e+\Phi_\mu+\Phi_\tau=\Phi_{\mathrm{all}}}\quad\text{for conversion among active flavors}.$$

Physical check

A flavor-selective detector does not directly measure total flux. Oscillation changes flavor populations without removing neutrinos in this model.