Jump to content

trying to understand neutrino oscillations


Widdekind

Recommended Posts

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 ?).

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.