Derivation of neutrino mass from neutrino scattering...

[Orion1]

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

[math]p_{e} \sin \phi = p_{f} \sin \theta \tag{1}[/math]

[math]p_{e} \cos \phi + p_{f} \cos \theta = p_{i} \tag{2}[/math]

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

Divide equation [math](1)[/math] by equation [math](3)[/math] for an expression for [math]\tan \phi[/math]:

[math]\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}[/math]

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

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

[math]\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}[/math]

Substituting equation [math](6)[/math] into equation [math](4)[/math], and eliminate [math]p_{i}[/math] and [math]p_{f}[/math] in favor of [math]m_{\nu}[/math] alone.
[math]\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}[/math]

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

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

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

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

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

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

[math]m_{n}[/math] - nuclear mass

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

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

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

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

Any discussions and/or peer reviews about this specific topic thread?

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 June 21, 2018 by Orion1
source code correction...

[swansont]

P = E/c only applies to massless particles 

[Orion1]

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 [math](6)[/math], the energy terms for [math]E[/math] 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 [math](12)[/math] and [math](13)[/math] as kinetic energy approaches infinity, should be the rest mass of the neutrino.

[math]\lim_{E_k \to \infty} m_{\nu} = m_{\nu,0}[/math]

 

Edited June 22, 2018 by Orion1
source code correction...

[Orion1]

Derivation of neutrino mass from neutrino scattering:
[math]\;[/math]
[math]\theta[/math] - scattered neutrino angle
[math]\phi[/math] - electron recoil angle
[math]p_{i}[/math] - initial neutrino momentum
[math]p_{f}[/math] - final neutrino momentum
[math]p_{e}[/math] - electron momentum
[math]\;[/math]
Scattered particles rebounding with relativistic momentum have total energy:
[math]E_{e} = \gamma m_{0e} c^{2} = m_{e} c^{2} \; \; \; \; \; \; E_{\nu} = \gamma m_{0\nu} c^{2} = m_{\nu} c^{2}[/math]
Where [math]m_{e}[/math] and [math]m_{\nu}[/math] are the relativistic particle masses.
[math]\;[/math]
Observational measurements of both particle total energy and velocity, it is possible to calculate the rest particle masses [math]m_{0e}[/math] and [math]m_{0\nu}[/math]:
[math]\boxed{m_{0e} = \frac{E_{e}}{\gamma c^{2}}} \; \; \; \; \; \; \boxed{m_{0\nu} = \frac{E_{\nu}}{\gamma c^{2}}} \tag{0}[/math]
[math]\;[/math]
[math]p_{e} \sin \phi = p_{f} \sin \theta \tag{1}[/math]
[math]\;[/math]
[math]p_{e} \cos \phi + p_{f} \cos \theta = p_{i} \tag{2}[/math]
[math]\;[/math]
Isolate [math]p_{e} \cos \phi[/math] from equation [math](2)[/math]:
[math]p_{e} \cos \phi = p_{i} - p_{f} \cos \theta \tag{3}[/math]
[math]\;[/math]
Divide equation [math](1)[/math] by equation [math](3)[/math] for an expression for [math](3)[/math] for an expression for [math]\tan \phi[/math]:
[math]\;[/math]
[math]\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}[/math]
[math]\;[/math]
Acquire a substitution for [math]\frac{p_{i}}{p_{f}}[/math] to eliminate [math]p_{f}[/math].
Use the Compton equation, which can be rearranged to yield [math]\frac{\lambda_{f}}{\lambda_{i}} = \frac{p_{i}}{p_{f}}[/math] in terms of [math]\lambda_{i}[/math] alone.
[math]\;[/math]
[math]\frac{\lambda_{f}}{\lambda_{i}} = \frac{p_{i}}{p_{f}} \tag{5}[/math]
[math]\;[/math]
[math]\frac{\lambda_{f}}{\lambda_{i}} = \frac{p_{i}}{p_{f}} = 1 + \frac{E_{\nu}}{E_{e}} \left(1 - \cos \theta \right) = 1 + \frac{m_{\nu} c^2}{m_{e} c^2} \left(1 - \cos \theta \right) = 1 + \frac{m_{\nu}}{m_{e}} \left(1 - \cos \theta \right) \tag{6}[/math]
[math]\;[/math]
Substituting equation [math](6)[/math] into equation [math](4)[/math], and eliminate [math]p_{i}[/math] and [math]p_{f}[/math] in favor of [math]m_{\nu}[/math] alone.
[math]\;[/math]
[math]\tan \phi = \frac{\sin \theta}{\frac{p_{i}}{p_{f}} - \cos \theta} = \frac{\sin \theta}{1 + \frac{m_{\nu}}{m_{e}} \left(1 - \cos \theta \right) - \cos \theta} = \frac{\sin \theta}{\left(1 + \frac{m_{\nu}}{m_{e}} \right)\left(1 - \cos \theta \right)} \tag{7}[/math]
[math]\;[/math]
Utilizing a trigonometric identity produces the desired result, specifically:
[math]\frac{1 - \cos \theta}{\sin \theta} = \tan \left(\frac{\theta}{2} \right) \tag{8}[/math]
[math]\;[/math]
Substituting this trigonometric identity into equation [math](7)[/math] results in:
[math]\left(1 + \frac{m_{\nu}}{m_{e}} \right) \tan \phi = \cot \frac{\theta}{2} \tag{9}[/math]
[math]\;[/math]
Solve for neutrino mass [math]m_{\nu}[/math]:
[math]\tan \phi + \frac{m_{\nu}}{m_{e}} \tan \phi = \cot \frac{\theta}{2} \tag{10}[/math]
[math]\;[/math]
[math]\frac{m_{\nu}}{m_{e}} \tan \phi = \left(\cot \frac{\theta}{2} - \tan \phi \right) \tag{11}[/math]
[math]\;[/math]
Electron-neutrino scattering neutrino mass:
[math]\boxed{m_{\nu} = m_{e} \cot \phi \left(\cot \frac{\theta}{2} - \tan \phi \right)} \tag{12}[/math]
[math]\;[/math]
Nuclear-neutrino scattering neutrino mass:
[math]\boxed{m_{\nu} = m_{n} \cot \phi \left(\cot \frac{\theta}{2} - \tan \phi \right)} \tag{13}[/math]
[math]\;[/math]
[math]m_{n}[/math] - nuclear mass
[math]\;[/math]
Electron interaction neutrino scattering angle [math]\theta[/math]:
[math]\boxed{\theta = 2 \operatorname{arccot} \left(\frac{\left(m_{e} + m_{\nu} \right) \tan \phi}{m_{e}} \right)} \tag{14}[/math]
[math]\;[/math]
Neutrino interaction electron recoil angle [math]\phi[/math]:
[math]\boxed{\phi = \arctan \left(\frac{m_{e} \cot \frac{\theta}{2}}{m_{e} + m_{\nu}} \right)} \tag{15}[/math]
[math]\;[/math]
Nuclear interaction neutrino scattering angle [math]\theta[/math]:
[math]\boxed{\theta = 2 \operatorname{arccot} \left(\frac{\left(m_{n} + m_{\nu} \right) \tan \phi}{m_{n}} \right)} \tag{16}[/math]
[math]\;[/math]
Neutrino interaction nuclear recoil angle [math]\phi[/math]:
[math]\boxed{\phi = \arctan \left(\frac{m_{n} \cot \frac{\theta}{2}}{m_{n} + m_{\nu}} \right)} \tag{17}[/math]
[math]\;[/math]
Any discussions and/or peer reviews about this specific topic thread?
[math]\;[/math]

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 March 16, 2019 by Orion1
source code correction.

[Orion1]

On 6/21/2018 at 5:36 PM, swansont said:

P = E/c only applies to massless particles 

Relativistic energy-momentum relation total energy: (ref. 1)
[math]E_{t}^{2} = \left(m_{0} c^{2} \right)^{2} + \left(pc \right)^{2} [/math]
[math]E_{t} = \sqrt{\left(m_{0} c^{2} \right)^{2} + \left(pc \right)^{2}}[/math]
[math]\;[/math]
Relativistic mass particle total energy: (ref. 2)
[math]E_{t} = \gamma m_0 c^{2}[/math]
[math]\;[/math]
Relativistic energy-momentum relation and relativistic mass particle total energy identity:
[math]\boxed{E_{t} = \sqrt{\left(m_{0}c^{2} \right)^{2} + \left(pc \right)^{2}} = \gamma m_0 c^2}[/math]
[math]\;[/math]
Is this equation an identity for a relativistic mass particle?
[math]\;[/math]
Any discussions and/or peer reviews about this specific topic thread?
[math]\;[/math]
Reference:
Wikipedia - Energy-momentum relation: (ref. 1)
https://en.wikipedia.org/wiki/Energy–momentum_relation#Special_relativity
Wikipedia - Energy-momentum relation - heuristic approach for massive particles: (ref. 2)
https://en.wikipedia.org/wiki/Energy–momentum_relation#Heuristic_approach_for_massive_particle

 

Edited March 21, 2019 by Orion1
source code correction.

[swansont]

43 minutes ago, Orion1 said:

Relativistic energy-momentum relation total energy: (ref. 1)
E2t=(m0c2)2+(pc)2
Et=(m0c2)2+(pc)2−−−−−−−−−−−−√

Relativistic mass particle total energy: (ref. 2)
Et=γm0c2

Relativistic energy-momentum relation and relativistic mass particle total energy identity:
Et=(m0c2)2+(pc)2−−−−−−−−−−−−−√=γm0c2

Is this equation an identity for a relativistic mass particle?

It only applies if there is rest mass. If that's zero, E = pc

 

[Orion1]

On 3/21/2019 at 3:59 AM, swansont said:

It only applies if there is rest mass. If that's zero, E = pc 

De Broglie relativistic momentum: (ref. 1)
[math]p = \frac{\hbar}{\overline{\lambda}} = \gamma m_0 v[/math]
[math]\;[/math]
Relativistic energy-momentum relation and relativistic mass particle total energy identity:
[math]\boxed{E_{t} = \sqrt{\left(m_{0} c^{2} \right)^{2} + \left(pc \right)^{2}} = \gamma m_0 c^2}[/math]
[math]\;[/math]
[math]\boxed{E_{t} = \sqrt{\left(m_{0} c^{2} \right)^{2} + \left(\frac{\hbar c}{\overline{\lambda}} \right)^{2}} = \gamma m_0 c^2}[/math]
[math]\;[/math]
Is this equation an identity for a relativistic mass particle?
[math]\;[/math]
Any discussions and/or peer reviews about this specific topic thread?
[math]\;[/math]
Reference:
Wikipedia - Matter wave: (ref. 1)
https://en.wikipedia.org/wiki/Matter_wave

 

[swansont]

57 minutes ago, Orion1 said:

 
Any discussions and/or peer reviews about this specific topic thread?
 

Any modern physics textbook.

[Orion1]

On 3/22/2019 at 5:46 AM, swansont said:

Any modern physics textbook. 

Affirmative, revision complete.

Derivation of neutrino mass from neutrino scattering:
[math]\;[/math]
[math]\theta[/math] - scattered neutrino angle
[math]\phi[/math] - electron recoil angle
[math]E_{\nu i}[/math] - initial neutrino total energy
[math]E_{\nu f}[/math] - final neutrino total energy
[math]E_{e}[/math] - electron total energy
[math]E_{\nu}[/math] - neutrino total energy

Scattered particles rebounding with relativistic momentum have total energy:
[math]E_{e} = \gamma_{e} m_{e} c^{2} \; \; \; \; \; \; E_{\nu} = \gamma_{\nu} m_{\nu} c^{2}[/math]
Where [math]m_{e}[/math] and [math]m_{\nu}[/math] are the particle rest masses.
[math]\gamma_{e}[/math] and [math]\gamma_{\nu}[/math] are the Lorentz factors. (ref. 1)
[math]\;[/math]
Observational measurements of both particle total energy and velocity, it is possible to calculate the particle rest masses [math]m_{e}[/math] and [math]m_{\nu}[/math]:
[math]\boxed{m_{e} = \frac{E_{e}}{\gamma_{e} c^{2}}} \; \; \; \; \; \; \boxed{m_{\nu} = \frac{E_{\nu}}{\gamma_{\nu} c^{2}}} \tag{0}[/math]
[math]\;[/math]
[math]E_{e} \sin \phi = E_{\nu f} \sin \theta \tag{1}[/math]
[math]\;[/math]
[math]E_{e} \cos \phi + E_{\nu f} \cos \theta = E_{\nu i} \tag{2}[/math]
[math]\;[/math]
Isolate [math]E_{e} \cos \phi[/math] from equation (2):
[math]E_{e} \cos \phi = E_{\nu i} - E_{\nu f} \cos \theta \tag{3}[/math]
[math]\;[/math]
Divide equation (1) by equation (3) for an expression for [math]\tan \phi[/math].
[math]\;[/math]
[math]\tan \phi = \frac{E_{\nu f} \sin \theta}{E_{\nu i} - E_{\nu f} \cos \theta} = \frac{\sin \theta}{\frac{E_{\nu i}}{E_{\nu f}} - \cos \theta} \tag{4}[/math]
[math]\;[/math]
Acquire a substitution for [math]\frac{E_{\nu i}}{E_{\nu f}}[/math] to eliminate [math]E_{\nu f}[/math].
Use the Compton equation, which can be rearranged to yield [math]\frac{\lambda_{\nu f}}{\lambda_{\nu i}} = \frac{E_{\nu i}}{E_{\nu f}}[/math] in terms of [math]\lambda_{\nu i}[/math] alone.
[math]\;[/math]
[math]\frac{\lambda_{\nu f}}{\lambda_{\nu i}} = \frac{E_{\nu i}}{E_{\nu f}} \tag{5}[/math]
[math]\;[/math]
[math]\frac{\lambda_{\nu f}}{\lambda_{\nu i}} = \frac{E_{\nu i}}{E_{\nu f}} = 1 + \frac{E_{\nu i}}{E_{e}} \left(1 - \cos \theta \right) = 1 + \frac{\gamma_{\nu} m_{\nu} c^2}{\gamma_{e} m_{e} c^2} \left(1 - \cos \theta \right) = 1 + \frac{\gamma_{\nu} m_{\nu}}{\gamma_{e} m_{e}} \left(1 - \cos \theta \right) \tag{6}[/math]
[math]\;[/math]
Substituting equation (6) into equation (4) and eliminate [math]E_{\nu i}[/math] and [math]E_{\nu f}[/math] in favor of [math]m_{\nu}[/math] alone.
[math]\tan \phi = \frac{\sin \theta}{\frac{E_{\nu i}}{E_{\nu f}} - \cos \theta} = \frac{\sin \theta}{1 + \frac{\gamma_{\nu} m_{\nu}}{\gamma_{e} m_{e}} \left(1 - \cos \theta \right) - \cos \theta} = \frac{\sin \theta}{\left(1 + \frac{\gamma_{\nu} m_{\nu}}{\gamma_{e} m_{e}} \right)\left(1 - \cos \theta \right)} \tag{7}[/math]
[math]\;[/math]
Utilizing a trigonometric identity produces the desired result, specifically:
[math]\frac{1 - \cos \theta}{\sin \theta} = \tan \left(\frac{\theta}{2} \right) \tag{8}[/math]
[math]\;[/math]
Substituting this trigonometric identity into equation (7) results in:
[math]\left(1 + \frac{\gamma_{\nu} m_{\nu}}{\gamma_{e} m_{e}} \right) \tan \phi = \cot \frac{\theta}{2} \tag{9}[/math]
[math]\;[/math]
Solve for neutrino rest mass [math]m_{\nu}[/math]:
[math]\tan \phi + \frac{\gamma_{\nu} m_{\nu}}{\gamma_{e} m_{e}} \tan \phi = \cot \frac{\theta}{2} \tag{10}[/math]
[math]\;[/math]
[math]\frac{\gamma_{\nu} m_{\nu}}{\gamma_{e} m_{e}} \tan \phi = \left(\cot \frac{\theta}{2} - \tan \phi \right) \tag{11}[/math]
[math]\;[/math]
Electron-neutrino scattering neutrino rest mass:
[math]\boxed{m_{\nu} = \frac{\gamma_{e} m_{e} \cot \phi}{\gamma_{\nu}} \left(\cot \frac{\theta}{2} - \tan \phi \right)} \tag{12}[/math]
[math]\;[/math]
Nuclear-neutrino scattering neutrino rest mass:
[math]\boxed{m_{\nu} = \frac{\gamma_{n} m_{n} \cot \phi}{\gamma_{\nu}} \left(\cot \frac{\theta}{2} - \tan \phi \right)} \tag{13}[/math]
[math]m_{n}[/math] - nuclear rest mass
[math]\;[/math]
Electron interaction neutrino scattering angle [math]\theta[/math]:
[math]\boxed{\theta = 2 \operatorname{arccot} \left(\frac{\left(\gamma_{e} m_{e} + \gamma_{\nu} m_{\nu} \right) \tan \phi}{\gamma_{e} m_{e}} \right)} \tag{14}[/math]
[math]\;[/math]
Neutrino interaction electron recoil angle [math]\phi[/math]:
[math]\boxed{\phi = \arctan \left(\frac{\gamma_{e} m_{e} \cot \frac{\theta}{2}}{\gamma_{e} m_{e} + \gamma_{\nu} m_{\nu}} \right)} \tag{15}[/math]
[math]\;[/math]
Nuclear interaction neutrino scattering angle [math]\theta[/math]:
[math]\boxed{\theta = 2 \operatorname{arccot} \left(\frac{\left(\gamma_{n} m_{n} + \gamma_{\nu} m_{\nu} \right) \tan \phi}{\gamma_{n} m_{n}} \right)} \tag{16}[/math]
[math]\;[/math]
Neutrino interaction nuclear recoil angle [math]\phi[/math]:
[math]\boxed{\phi = \arctan \left(\frac{\gamma_{n} m_{n} \cot \frac{\theta}{2}}{\gamma_{n} m_{n} + \gamma_{\nu} m_{\nu}} \right)} \tag{17}[/math]
[math]\;[/math]
Any discussions and/or peer reviews about this specific topic thread?
[math]\;[/math]
Reference:
Wikipedia - Lorentz factor: (ref. 1)
https://en.wikipedia.org/wiki/Lorentz_factor
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 April 22, 2019 by Orion1
source code correction.

[Orion1]

On 3/22/2019 at 5:46 AM, swansont said:

Any modern physics textbook.

If a neutrino has zero mass, would the neutrino obey Compton scattering?

Any modern physics textbook answers this question for zero mass photons, however, do they also answer for zero mass neutrinos?

Any discussions and/or peer reviews about this specific topic thread?

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

Edited August 27, 2023 by Orion1

[John Cuthber]

Has anyone ever observed the path of a neutrino?

[swansont]

5 hours ago, Orion1 said:

Any modern physics textbook answers this question for zero mass photons, however, do they also answer for zero mass neutrinos?

There’s nothing inherent in the equations that says the zero-mass particle must be a photon. i.e. it applies to all zero-mass particles.

[Orion1]

On 8/27/2023 at 4:58 AM, John Cuthber said:

Has anyone ever observed the path of a neutrino?

This file has been identified as being free of known restrictions under copyright law, including all related and neighboring rights.

"The first use of a hydrogen bubble chamber to detect neutrinos, on 13 November 1970, at Argonne National Laboratory. Here a neutrino hits a proton in a hydrogen atom; the collision occurs at the point where three tracks emanate on the right of the photograph." (ref.1)

The neutrino track itself is invisible in a bubble chamber due to the neutrino having zero charge and only interacts via the weak nuclear force. However the neutrino path can be deduced by the resultant particle production angles.

Coherent neutrino scattering has been seen with a compact detector. (ref. 2)

In my opinion, two or more of these coherent neutrino detectors would be capable of determining if neutrinos scatter through zero-mass Compton scattering or through a non-zero-mass process that is described by the equations above and capable of determining the neutrino rest-mass with a high degree of precision.

On 8/27/2023 at 5:43 AM, swansont said:

There’s nothing inherent in the equations that says the zero-mass particle must be a photon. i.e. it applies to all zero-mass particles.

This is the correct intuitive answer, although any modern physics textbook would be challenged to provide such a comprehensive answer, beyond a photon scattering process.

Reference:
Wikipedia - Neutrino: (ref. 1)
https://en.wikipedia.org/wiki/Neutrino
Physicsworld - Coherent neutrino scattering:  (ref. 2)
https://physicsworld.com/a/coherent-neutrino-scattering-seen-with-compact-detector/

 

Edited August 29, 2023 by Orion1

[swansont]

4 hours ago, Orion1 said:

The neutrino track itself is invisible in a bubble chamber due to the neutrino having zero charge and only interacts via the weak nuclear force. However the neutrino path can be deduced by the resultant particle production angles.

That tells you where the scattering event occurred. It doesn’t tell you the neutrino path, because you don’t know where the neutrino originated. You might estimate the angle, but how do you know the momentum of the neutrino?

When I was a postdoc, we did an experiment where we knew where the originated, because they came from beta decay in a magneto-optic trap, so the source was localized to a small volume. We couldn’t detect them, though, but could deduce their momentum by detecting the beta, the daughter nucleus, and any orbital electrons that were ionized in the decay process, for decays where all the particles were co- or counter-propagating.