# Derivation of neutrino mass from neutrino scattering...

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Derivation of neutrino mass from neutrino scattering:
$\theta$ - scattered neutrino angle
$\phi$ - electron recoil angle
$p_{i}$ - initial neutrino momentum
$p_{f}$ - final neutrino momentum
$p_{e}$ - electron momentum

$p_{e} \sin \phi = p_{f} \sin \theta \tag{1}$

$p_{e} \cos \phi + p_{f} \cos \theta = p_{i} \tag{2}$

Isolate $p_{e} \cos \phi$ from equation $(2)$:
$p_{e} \cos \phi = p_{i} - p_{f} \cos \theta \tag{3}$

Divide equation $(1)$ by equation $(3)$ for an expression for $\tan \phi$:

$\tan \phi = \frac{p_{f} \sin \theta}{p_{i} - p_{f} \cos \theta} = \frac{\sin \theta}{\frac{p_{i}}{p_{f}} - \cos \theta} \tag{4}$

Acquire a substitution for $\frac{p_{i}}{p_{f}}$ to eliminate $p_{f}$.
Use the Compton equation, which can be rearranged to yield $\frac{\lambda_{f}}{\lambda_{i}} = \frac{p_{i}}{p_{f}}$ in terms of $\lambda_{i}$ alone, noting that $p = \frac{E}{c}$.

$\lambda_{f} - \lambda_{i} = \frac{h}{m_{e} c} (1 - \cos \theta) \tag{5}$

$\frac{\lambda_{f}}{\lambda_{i}} = \frac{p_{i}}{p_{f}} = 1 + \frac{E_{\nu}}{E_{e}} (1 - \cos \theta) = 1 + \frac{m_{\nu} c^2}{m_{e} c^2} (1 - \cos \theta) = 1 + \frac{m_{\nu}}{m_{e}} (1 - \cos \theta) \tag{6}$

Substituting equation $(6)$ into equation $(4)$, and eliminate $p_{i}$ and $p_{f}$ in favor of $m_{\nu}$ alone.
$\tan \phi = \frac{\sin \theta}{\frac{p_{i}}{p_{f}} - \cos \theta} = \frac{\sin \theta}{1 + \frac{m_{\nu}}{m_{e}} (1 - \cos \theta) - \cos \theta} = \frac{\sin \theta}{\left(1 + \frac{m_{\nu}}{m_{e}} \right)(1 - \cos \theta)} \tag{7}$

Utilizing a trigonometric identity produces the desired result, specifically:
$\frac{1 - \cos \theta}{\sin \theta} = \tan \left(\frac{\theta}{2} \right) \tag{8}$

Substituting this trigonometric identity into equation $(7)$ results in:
$\left(1 + \frac{m_{\nu}}{m_{e}} \right) \tan \phi = \cot \frac{\theta}{2} \tag{9}$

Solve for neutrino mass $m_{\nu}$:
$\tan \phi + \frac{m_{\nu}}{m_{e}} \tan \phi = \cot \frac{\theta}{2} \tag{10}$

$\frac{m_{\nu}}{m_{e}} \tan \phi = \left(\cot \frac{\theta}{2} - \tan \phi \right) \tag{11}$

Electron-neutrino scattering neutrino mass:
$\boxed{m_{\nu} = m_{e} \cot \phi \left(\cot \frac{\theta}{2} - \tan \phi \right)} \tag{12}$

Nuclear-neutrino scattering neutrino mass:
$\boxed{m_{\nu} = m_{n} \cot \phi \left(\cot \frac{\theta}{2} - \tan \phi \right)} \tag{13}$

$m_{n}$ - nuclear mass

Electron interaction neutrino scattering angle $\theta$:
$\boxed{\theta = 2 \operatorname{arccot} \left(\frac{(m_{e} + m_{\nu}) \tan \phi}{m_{e}} \right)} \tag{14}$

Neutrino interaction electron recoil angle $\phi$:
$\boxed{\phi = \arctan \left(\frac{m_{e} \cot \frac{\theta}{2}}{m_{e} + m_{\nu}} \right)} \tag{15}$

Nuclear interaction neutrino scattering angle $\theta$:
$\boxed{\theta = 2 \operatorname{arccot} \left(\frac{(m_{n} + m_{\nu}) \tan \phi}{m_{n}} \right)} \tag{16}$

Neutrino interaction nuclear recoil angle $\phi$:
$\boxed{\phi = \arctan \left(\frac{m_{n} \cot \frac{\theta}{2}}{m_{n} + m_{\nu}} \right)} \tag{17}$

Reference:
Wikipedia - Compton scattering - Derivation of the scattering formula:
https://en.wikipedia.org/wiki/Compton_scattering#Derivation_of_the_scattering_formula

Physics 253 - Compton Scattering - Patrick LeClair:
http://pleclair.ua.edu//PH253/Notes/compton.pdf

Orion1 - Neutrino mass from Fermi-Dirac statistics...:
https://www.scienceforums.net/topic/90189-neutrino-mass-from-fermi-dirac-statistics/

Science News - Neutrinos seen scattering off an atom’s nucleus for the first time:
https://www.sciencenews.org/article/neutrinos-seen-scattering-atoms-nucleus-first-time

Edited by Orion1
source code correction...

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P = E/c only applies to massless particles

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3 hours ago, swansont said:

P = E/c only applies to massless particles

This is true for the Compton equation, where a mass-less particle is scattering from a mass particle.

However, in equation $(6)$, the energy terms for $E$ represent the total energy of a mass particle, which includes its rest mass plus kinetic energy. I should have been more clear about that equation description and will include the total energy description in the next revision, hence peer review.

If I understand this inelastic scattering correctly, when a lighter particle scatters from a heavier particle, as total kinetic energy is increased, then more kinetic energy is absorbed by the recoiling heavier particle and less kinetic energy is carried away by the lighter particle.

So, the limit of equations $(12)$ and $(13)$ as kinetic energy approaches infinity, should be the rest mass of the neutrino.

$\lim_{E_k \to \infty} m_{\nu} = m_{\nu,0}$

Edited by Orion1
source code correction...