- #751

- 2,967

- 5

The energy of a particle travelling at this close speed to

*c*is:

[tex]

\begin{array}{l}

\frac{E}{m \, c^2} = \gamma = \left ( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}} \\

= \left[ 1 - (1 - \epsilon)^2 \right]^{-\frac{1}{2}} \\

= \left[ 2 \epsilon \, \left( 1 - \frac{\epsilon}{2}\right) \right]^{-\frac{1}{2}} \\

\sim (2 \epsilon)^{-\frac{1}{2}} \, \left[1 + \frac{\epsilon}{4} + O(\epsilon^2) \right]

\end{array}

[/tex]

Considering the rest energy of neutrinos is of the order of 0.1 eV, this means that the energy of these neutrinos would be of the order of:

[tex]

\frac{0.1 \, \mathrm{eV}}{\sqrt{2 \times 10^{-5}}} \sim 20 eV

[/tex]

which is negligible. Even higher energies would bring the speed of neutrinos so close to

*c*that the difference could not be detectable in any terrestrial experiment.