# 1.3.6. Questions about Halo Problem¶

## 1.3.6.1. Differential Cross Section¶ Fig. 1.188 Scattering of neutrinos off electrons.

Giunti derived the general form of differential cross section for all possible neutrino scatterings off electrons, equation 5.19 in his book [Giunti2007]. We plugin differential forms of Mandelstam variables to get the differential cross section related to neutrino scattering angles,

$\frac{d\sigma}{d ( 2 E_\nu E_\nu' \cos \theta_\nu)} = \frac{G_F}{\pi} \left( g_1^2 + g_2^2 \left( 1 - \frac{E_\nu' \cos \theta_\nu}{m_e} \right) - g_1 g_2 \frac{E_\nu' \cos\theta_\nu}{ E_\nu } \right),$

where $$E_\nu=p_\nu$$ and $$E_\nu'=p_\nu'$$ are the energies of neutrinos before and after scattering.

Applying conservation of four momentum, we have

$p_\nu' = \frac{ m_e (m_e + p_\nu)}{ m_e + p_\nu - p_\nu \cos \theta_\nu }.$

We are interested in the energy scale that $$p_\nu \gg m_e$$. The energy of scattered neutrinos equations simplify to

$p_\nu' = \frac{m_e}{ 1-\cos\theta_\nu }.$

Plugin this into the differential cross section and keep only 0 orders of $$m_e/E_\nu$$, we have

$\frac{d\sigma}{d\cos \theta_\nu} = \frac{2 E_\nu m_e}{(1-\cos\theta_\nu)^2} \frac{G_F}{\pi} \left( g_1^2 + g_2^2 \frac{ 1 -2\cos\theta_\nu }{1 -\cos\theta_\nu} \right)$

For neutral current, Giunti shows that the values of $$g_i$$ are [Giunti2007]

$\begin{split}g_1 &= g_2 = -0.27 \\ \bar g_1 &= \bar g_2 = 0.23,\end{split}$

where bar indicates the values for antineutrinos.

Giunti’s formula

The differential cross section formula 5.29 in [Giunti2007] shows

$\frac{d\sigma}{ d\cos \theta_e} = \sigma_0 \frac{ 4E_\nu^2 (m_e + E_\nu)^2 \cos \theta_e } { [ (m_e + E_\nu)^2 - E_\nu^2\cos^2\theta_e ]^2 } \left[ g_1^2 + g_2^2 \left( 1 - \frac{ 2 m_e E_\nu \cos^2\theta_e }{ (m_e+E_\nu)^2 - E_\nu^2 \cos^2\theta_e } \right)^2 - g_1 g_2 \frac{ 2 m_e^2 \cos^2\theta_e }{ ( m_e + E_\nu )^2 - E_\nu^2 \cos^2\theta_e } \right].$

We are interested in supnernova neutrinos whose energy is usually larger than mass of electrons. We set $$m_e/E_\nu \to 0$$.

$\frac{d\sigma}{ d\cos \theta_e} = \sigma_0 \frac{ 4 E_\nu^2 \cos \theta_e } { [ 1 - \cos^2\theta_e ]^2 } .$

We need a relation between $$\theta_e$$ and $$\theta_\nu$$. The way to derive it is to use conservation of four momentum.

$\begin{split}E_\nu' + \sqrt{p_e'^2 - m_e^2} &= m_e + E_\nu \\ p_\nu' \sin\theta_\nu + p_e' \sin\theta_e &= 0\\ p_\nu' \cos\theta_\nu + p_e' \cos\theta_e &= p_\nu.\end{split}$

We imediately notice that for backward scattering, $$\theta_e = \pi + \theta_\nu$$.

 [Giunti2007] (1, 2, 3) Guinti & Kim, Fundamentals of Neutrinos and Astrophysics

## 1.3.6.2. MISC¶

1. Reflection coefficients for neutrinos and anti-neutrinos are different.

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