# Fraction of solar neutrinos arriving at the Earth

## Homework Statement

Consider solar neutrinos of energy 1 MeV (EDIT: 10 MeV not 1 MeV) which are formed at the center of the sun in the ##\nu_2## eigenstate. What fraction of it do you expect to arrive at earth as ##\nu_\mu## and what fraction as ##\nu_\tau##? Assume that it evolves adiabaticaly inside the sun.

## Homework Equations

$$\left| \psi(x, t) \right> = \left| \nu_2 \right> e^{-i\phi_2}$$

## The Attempt at a Solution

Because the neutrino is formed in the ##\left| \nu_2 \right>## eigenstate,
$$\left| \psi(x, t) \right> = \left| \nu_2 \right> e^{-i\phi_2} \\ = \left( U_{e_2}^* \left| \nu_e \right> + U_{\mu_2}^* \left| \nu_\mu \right> + U_{\tau_2}^* \left| \nu_\tau \right>\right) e^{-i\phi_2} \\ = \left( c_e \left| \nu_e \right> + c_\mu \left| \nu_\mu \right> + c_\tau \left| \nu_\tau \right>\right) e^{-i\phi_2}$$

Then,
$$P(\nu_e \to \nu_\mu ) = \left| \left< \nu_\mu | \psi(x, t) \right> \right|^2 \\ = \left| c_\mu \right|^2 \\ = \left| U_{\mu_2}^* \right|^2$$

But this probability depends neither on the neutrino energy nor on the length it has traveled. What did I do wrong?

Last edited:

Orodruin
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The fact that it is a 1 MeV neutrino is important only to know that it is produced above resonance (so that it actually starts out in the second eigenstate).

Why do you expect the probability to depend on the distance travelled?

Because since it will oscillate between different flavor states, "how far along" the neutrino has traveled would affect which state it is on?

Orodruin
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Because since it will oscillate between different flavor states, "how far along" the neutrino has traveled would affect which state it is on?
This is wrong. Solar neutrinos above resonance do not oscillate, they are subject to adiabatic flavour transitions. You need components in at least two propagation eigenstates to induce oscillations.

Does that mean the fraction of mu neutrinos (or tau neutrinos) that arrive at earth is going to be some constant, that does not depend on the length and neutrino energy? How would I calculate that fraction then? Is it just the square of the corresponding element of the PMNS matrix?

Orodruin
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Does that mean the fraction of mu neutrinos (or tau neutrinos) that arrive at earth is going to be some constant, that does not depend on the length and neutrino energy?
For neutrinos produced with energies above resonance, yes. Things change for very low-energy neutrinos, for which you essentially would observe averaged out vacuum oscillations. You also have an intermediate energy range that interpolates between the two.

How would I calculate that fraction then? Is it just the square of the corresponding element of the PMNS matrix?
Yes.

Orodruin
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See, eg,

(From arXiv:1211.5359)

Note that 1 MeV is actually not really sufficient to enter the above resonance region. You would be in the transition region.

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Orodruin
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For reference, figure 4.3 of my MSc thesis shows the fraction of neutrinos produced in the ##|\nu_2\rangle## state at the radius of peak 8B neutrino production (labelled ##|\hat U_{e2}|^2##). As can be seen, 1 MeV is far from high enough to produce mainly the second state. In fact, it is actually closer to the vacuum regime.

nrqed
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## Homework Statement

Consider solar neutrinos of energy 1 MeV which are formed at the center of the sun in the ##\nu_2## eigenstate. What fraction of it do you expect to arrive at earth as ##\nu_\mu## and what fraction as ##\nu_\tau##? Assume that it evolves adiabaticaly inside the sun.

## Homework Equations

$$\left| \psi(x, t) \right> = \left| \nu_2 \right> e^{-i\phi_2}$$

## The Attempt at a Solution

Because the neutrino is formed in the ##\left| \nu_2 \right>## eigenstate,
$$\left| \psi(x, t) \right> = \left| \nu_2 \right> e^{-i\phi_2} \\ = \left( U_{e_2}^* \left| \nu_e \right> + U_{\mu_2}^* \left| \nu_\mu \right> + U_{\tau_2}^* \left| \nu_\tau \right>\right) e^{-i\phi_2} \\ = \left( c_e \left| \nu_e \right> + c_\mu \left| \nu_\mu \right> + c_\tau \left| \nu_\tau \right>\right) e^{-i\phi_2}$$

Then,
$$P(\nu_e \to \nu_\mu ) = \left| \left< \nu_\mu | \psi(x, t) \right> \right|^2 \\ = \left| c_\mu \right|^2 \\ = \left| U_{\mu_2}^* \right|^2$$

But this probability depends neither on the neutrino energy nor on the length it has traveled. What did I do wrong?

You wrote ##
\left| \psi(x, t) \right>
= \left| \nu_2 \right> e^{-i\phi_2} ##. You meant ##\left| \psi(0,0) \right>##. You need to propagate it to the Earth and find
##
\left| \psi(x, t) \right>## at the Earth's location.

You wrote ##
\left| \psi(x, t) \right>
= \left| \nu_2 \right> e^{-i\phi_2} ##. You meant ##\left| \psi(0,0) \right>##. You need to propagate it to the Earth and find
##
\left| \psi(x, t) \right>## at the Earth's location.

Okay. I'm not sure how to write ##\left| \psi(x, t) \right>## though.

Very generally, my guess is that it would be
$$\left| \psi(x, t) \right> = U_{e1}^* \left| \nu_1 \right> e^{-i \phi_1} + U_{e2}^* \left| \nu_2 \right> e^{-i \phi_2} + U_{e3}^* \left| \nu_3 \right> e^{-i \phi_3}$$
But here, since I'm starting at the ## \left| \nu_2 \right>## state, it would just be
$$\left| \psi(x, t) \right> = \left| \nu_2 \right> e^{-i \phi_2}$$
?

Btw my ##\phi_i = p_i \cdot x_i = (E_i t - { \bf p_i \cdot x })##.

Also what Orodruin said above. But that calculation would be trivial since it's just reading off the PMNS matrix.

EDIT: The energy I'm supposed to use is 10 MeV not 1 MeV. I misread it.

Orodruin
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You wrote ##
\left| \psi(x, t) \right>
= \left| \nu_2 \right> e^{-i\phi_2} ##. You meant ##\left| \psi(0,0) \right>##. You need to propagate it to the Earth and find
##
\left| \psi(x, t) \right>## at the Earth's location.
This is not correct. The original state is ##|\nu_2\rangle##, which acquires a phase ##\phi_2## upon propagation to the Earth.

nrqed
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This is not correct. The original state is ##|\nu_2\rangle##, which acquires a phase ##\phi_2## upon propagation to the Earth.
I did not realize that ##\phi## stood for ## P \cdot X ##, I thought it was some arbitrary constant phase.
I will let you, the expert, answer the questions and keep my mouth shut.