Neutrinos and the Earth

Not all neutrinos that pass through the earth actually hit anything. What we wish to measure is the probability of the neutrino interacting with something in the earth assuming that it passes through the planet. One of the dominant modes of interaction with matter is neutrino-electron scattering. (Other modes will enhance our interaction rates by a factor of only 2 or 3, so we'll ignore them.) This interaction is a neutrino bouncing off an electron. We measure this interaction as a cross section, that is, what an electron looks like to an incoming neutrino.

To visualize this, imagine shooting an arrow at a distant apple. The chance of hitting the apple is proportional to the size of the apple's silloutte: it's cross section.

The cross section of an electron to neutrinos created in proton-proton reactions in the sun (the most common variety found near earth) is, on average, cm².¹

Next, we need to know how many of these electrons there are. The earth is electrically neutral, so there is 1 electron per 1 proton. Protons, on average, make up half the weight of the earth (the other half being neutrons). Thus, the number of electrons is

$$\begin{eqnarray*}\mathrm{Mass}\, \mathrm{of}\; \mathrm{the}\; \mathrm{Earth}\tim... ..., Electron}{2\, Nucleons} & & \\ =5.9\times 10^{50}electrons & \end{eqnarray*}$$

So, the total cross section of all the electrons in the earth is

$$11.6\times 10^{-46}cm^{2}/electron\times 5.9\times 10^{50}electrons=6.88\times 10^{5}cm^{2}$$

Let's convert this to meters squared for convience:

$$6.88\times 10^{5}cm^{2}\times \frac{1m}{100cm}\times \frac{1m}{100cm}=68.8m^{2}$$

So, to neutrinos, the earth looks about the size of a wall seven meters by ten meters! However, the 'bricks' in this wall are scattered over an area the size of the earth. If we look at the earth from space, we see a disk that has an area

$$\pi r^{2}=3.14\times \left( 6.3\times 10^{6}m\right) ^{2}=1.25\times 10^{14}m^{2}$$

If a neutrino goes wizzing through this area, it has a chance of hitting something that is

$$\frac{Area\; of\; Target}{Total\; Area}=\frac{68.8m^{2}}{1.25\times 10^{14}m^{2}}=5.5\times 10^{-13}$$

This is the probability of a given neutrino hitting something as it travels through ``all this howling emptiness''. The probability of an observable interaction is much less than this. As a brief side note, solar neutrino experiments are able to see actual results because the sun throws off truely incredible numbers of neutrinos. Even at our distance from the sun, there are ~10⁸neutrinos per second passing through an area of 1 meter square. Thus, the rate of neutrino hits occouring in the earth is roughly 7 billion interactions every second.