Widdekind Posted January 18, 2012 Posted January 18, 2012 Neutrinos only interact Weakly. Er go, neutrinos are generated into Weak eigenstates, which are "mixtures" of the canonical mass eigenstates; and neutrinos remain in those Weak eigenstates, until detection. For example, electron neutrinos [math]\nu_e[/math] are a mixed Weak eigenstate, representing a mixture of the neutrino mass eigenstates [math]\nu_1, \nu_2, \nu_3[/math], specifically [math]|\nu_e> \approx 0.9 |\nu_1> + 0.5 |\nu_2>[/math] The following analysis employs normalized units [math]c \rightarrow 1[/math]; and assumes that the neutrino mass eigenstates, are also eigenstates, of mass [math]\left( \hat{m} \right)[/math], momentum [math]\left( \hat{p} \right)[/math], & squared-energy [math]\left( \hat{E}^2 \right)[/math]. take 1 How can such a Weak eigenstate possess a well-defined energy & momentum ? For, if [math]m \equiv \alpha m_1 + \beta m_2[/math] [math]E \equiv \alpha E_1 + \beta E_2[/math] [math]p \equiv \alpha p_1 + \beta p_2[/math] then [math]E^2 = m^2 + p^2[/math] [math]\left( \alpha E_1 + \beta E_2 \right)^2 = \left( \alpha m_1 + \beta m_2 \right)^2 + \left( \alpha p_1 + \beta p_2 \right)^2[/math] [math]\begin{array}{c} \alpha^2 E_1^2 + \beta^2 E_2^2 \\ + 2 \alpha \beta E_1 E_2 \end{array} = \begin{array}{c} \alpha^2 m_1^2 + \beta^2 m_2^2 \\ + 2 \alpha \beta m_1 m_2 \end{array} + \begin{array}{c} \alpha^2 p_1^2 + \beta^2 p_2^2 \\ + 2 \alpha \beta p_1 p_2 \end{array}[/math] [math]\therefore E_1 E_2 = m_1 m_2 + p_1 p_2[/math] But w.h.t. [math]E = \gamma m[/math] [math]p = \gamma m \beta = \beta E[/math] So [math]E_1 E_2 = m_1 m_2 + p_1 p_2[/math] [math]\gamma_1 \gamma_2 = 1 + \gamma_1 \beta_1 \gamma_2 \beta_2[/math] [math]\gamma_1 \gamma_2 \left(1 - \beta_1 \beta_2 \right) = 1[/math] [math]\left(1 - \beta_1 \beta_2 \right)^2 = \left(1 - \beta_1^2 \right) \left( 1 - \beta_2^2 \right)[/math] [math]\begin{array}{c} 1 + \beta_1^2 \beta_2^2 \\ - 2 \beta_1 \beta_2 \end{array} = \begin{array}{c} 1 + \beta_1^2 \beta_2^2 \\ - 2 \beta_1^2 \beta_2^2 \end{array}[/math] [math]1 = \beta_1 \beta_2[/math] [math]\therefore 1 = \beta_1 = \beta_2[/math] But, neutrinos have mass, i.e. [math]\beta_{1,2} < 1[/math]. take 2 Are "Weak neutrinos", i.e. neutrinos in Weak eigenstates, "off mass shell" ? For, if: [math]|\nu_e> = \alpha |\nu_1> + \beta |\nu_2>[/math] then is the mass expectation value: [math]<m_{\nu_e}> = \left( \alpha^* <\nu_1| + \beta^* <\nu_2| \right) \hat{m} \left( \alpha |\nu_1> + \beta |\nu_2> \right)[/math] [math]= |\alpha|^2 <\nu_1|\hat{m}|\nu_1> + |\beta|^2 <\nu_2|\hat{m}|\nu_2>[/math] [math]= |\alpha|^2 <m_1> +|\beta|^2 <m_2>[/math] [math]\equiv |\alpha|^2 m_1 +|\beta|^2 m_2[/math] [math]\therefore m_{\nu_e} \approx \frac{3}{4} m_1 + \frac{1}{4} m_2[/math] or is the momentum expectation value: [math]<p_{\nu_e}> = \left( \alpha^* <\nu_1| + \beta^* <\nu_2| \right) \hat{p} \left( \alpha |\nu_1> + \beta |\nu_2> \right)[/math] [math]= |\alpha|^2 <\nu_1|\hat{p}|\nu_1> + |\beta|^2 <\nu_2|\hat{p}|\nu_2>[/math] [math]= |\alpha|^2 <p_1> +|\beta|^2 <p_2>[/math] [math]\equiv |\alpha|^2 p_1 +|\beta|^2 p_2[/math] [math]\therefore p_{\nu_e} \approx \frac{3}{4} p_1 + \frac{1}{4} p_2[/math] or is the squared-energy expectation value: [math]<E_{\nu_e}^2> = \left( \alpha^* <\nu_1| + \beta^* <\nu_2| \right) \left[ \hat{p}^2 + \hat{m}^2 \right] \left( \alpha |\nu_1> + \beta |\nu_2> \right)[/math] [math]= |\alpha|^2 <\nu_1|\left[ \hat{p}^2 + \hat{m}^2 \right]|\nu_1> + |\beta|^2 <\nu_2|\left[ \hat{p}^2 + \hat{m}^2 \right]|\nu_2>[/math] [math]= |\alpha|^2 <E_1^2> +|\beta|^2 <E_2^2>[/math] [math]\equiv |\alpha|^2 E_1 +|\beta|^2 E_2[/math] [math]\therefore E_{\nu_e}^2 \approx\frac{3}{4} E_1^2 + \frac{1}{4} E_2^2[/math] ??? If so, then Weak neutrinos are "off mass shell", i.e. [math]E_{\nu_e}^2 \ne m_{\nu_e}^2 + p_{\nu_e}^2[/math], i.e. [math]m_{\nu_e}^2 + p_{\nu_e}^2 = \left( |\alpha|^2 m_1 +|\beta|^2 m_2 \right)^2 + \left( |\alpha|^2 p_1 +|\beta|^2 p_2 \right)^2[/math] [math]= \begin{array}{c} |\alpha|^4 m_1^2 +|\beta|^4 m_2^2 \\ + 2 |\alpha|^2 |\beta|^2 m_1 m_2 \end{array} + \begin{array}{c} |\alpha|^4 p_1^2 +|\beta|^4 p_2^2 \\ + 2 |\alpha|^2 |\beta|^2 p_1 p_2 \end{array}[/math] [math]= |\alpha|^4 E_1^2 +|\beta|^4 E_2^2 + 2 |\alpha|^2 |\beta|^2 \left( m_1 m_2 + p_1 p_2 \right)[/math] [math]= |\alpha|^2 \left(|\alpha|^2 \right)^2 E_1^2 +|\beta|^2 \left( |\beta|^2 \right)^2 E_2^2 + 2 |\alpha|^2 |\beta|^2 \left( m_1 m_2 + p_1 p_2 \right)[/math] [math]= |\alpha|^2 \left(1 - |\beta|^2 \right)^2 E_1^2 +|\beta|^2 \left(1 - |\alpha|^2 \right)^2 E_2^2 + 2 |\alpha|^2 |\beta|^2 \left( m_1 m_2 + p_1 p_2 \right)[/math] [math]= \left( |\alpha|^2 E_1^2 +|\beta|^2 E_2 \right) - |\alpha|^2 |\beta|^2 \left( \left( E_1^2 + E_2^2 \right) - 2 \left( m_1 m_2 + p_1 p_2 \right)\right) [/math] [math]= E_{\nu_e}^2 - |\alpha|^2 |\beta|^2 \left( \left( E_1^2 + E_2^2 \right) - 2 \left( m_1 m_2 + p_1 p_2 \right)\right) [/math] [math]= E_{\nu_e}^2 - |\alpha|^2 |\beta|^2 \left( \left( m_1^2 + p_1^2 + m_2^2 + p_2^2 \right) - 2 \left( m_1 m_2 + p_1 p_2 \right)\right) [/math] [math]= E_{\nu_e}^2 - |\alpha|^2 |\beta|^2 \left( \left(m_1-m_2 \right)^2 + \left(p_1-p_2 \right)^2 \right) [/math] [math]\approx E_{\nu_e}^2 - \frac{3}{16} \left( \left(m_2-m_1 \right)^2 + \left(p_2-p_1 \right)^2 \right) [/math] If so, then [math]E_{\nu_e}^2 > m_{\nu_e}^2 + p_{\nu_e}^2[/math], i.e. "electron neutrinos are energy rich" (by a few eV ?).
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