# Cosmology Lecture 11

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### Massive $\nu$ (Hot Dark Matter) continued

Last time we started talking about (1) and (2) below:

• Lab upper bounds on $m_{\nu }$ .
• Neutrino oscillations: $\Delta m^{2}=m_{1}^{2}=m_{2}^{2}$ :
${\begin{pmatrix}\nu _{e}\\\nu _{\mu }\end{pmatrix}}={\begin{pmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \end{pmatrix}}{\begin{pmatrix}\nu _{1}\\\nu _{2}\end{pmatrix}}\,\!$ Now suppose at time $t=0$ we have a pure $\nu _{e}$ produced. Then:

$|\nu (0)\rangle =|\nu _{e}\rangle =\cos \theta |\nu _{1}\rangle +\sin \theta |\nu _{2}\rangle \,\!$ At time t:

$|\nu (t)\rangle =\cos \theta e^{-iE_{1}t}|\nu _{1}\rangle +\sin \theta e^{-iE_{2}t}|\nu _{2}\rangle \,\!$ Where $E_{1}={\sqrt {p^{2}+m_{1}^{2}}}$ , and $E_{2}={\sqrt {p^{2}+m_{2}^{2}}}$ , assuming a fixed momentum state. We assume this for simplicity, but a full wave-packet-based derivation is done in (Kayser (81) PRD 24,110). Anyway, the probability of finding a $\nu _{\mu }$ state at time t is:

{\begin{aligned}P(\nu _{e}\to \nu _{\mu })&=|\langle \nu _{\mu }|\nu (t)\rangle |^{2}\\&=|(-\sin \theta \langle \nu _{1}|+\cos \theta \langle \nu _{2}|)(\cos \theta e^{-iE_{1}t}|\nu _{1}\rangle +\sin \theta e^{-iE_{2}t}|\nu _{2}\rangle )|^{2}\\&=|-\cos \theta \sin \theta e^{-iE_{1}t}+\cos \theta \sin \theta e^{-iE_{2}t}|^{2}\\&=\cos ^{2}\theta \sin ^{2}\theta |e^{-iE_{1}t}-e^{-iE_{2}t}|^{2}\\&={\sin ^{2}2\theta \over 4}(1-\cos(E_{1}-E_{2})t)\\&=\sin ^{2}2\theta \sin ^{2}\left({(E_{1}-E_{2})t \over 2}\right)\\\end{aligned}}\,\! For relativistic $\nu$ (we don’t really need to make this assumption, it just makes our lives easier), then:

$E_{1}-E_{2}\approx {\frac {1}{2}}{m_{1}^{2}-m_{2}^{2} \over E}\sim {\frac {1}{2}}{\Delta m^{2} \over E}\,\!$ Thus:

{\begin{aligned}{\frac {1}{2}}(E_{1}-E_{2})t&\sim {1 \over 4}{\Delta m^{2}Lc^{3} \over e}\\&=1.27{\Delta m^{2}L \over E}\\\end{aligned}}\,\! Where $L\sim ct$ is the distance $\nu$ travels in meters, $\Delta m^{2}$ is in $eV^{2}$ , and $E$ is in MeV Thus we have our expression for the probability of an ${e^{-}}$ neutrino turning into a $\mu$ neutrino in a vacuum:

${P(\nu _{e}\to \nu _{\mu })=\sin ^{2}2\theta \sin ^{2}\left({1.27\Delta m^{2}L \over E}\right)}\,\!$ • Supernova $\nu$ : (SN1987A: Feb 23 1987) "(a)" Supernovae produce $\nu$ via: ${e^{-}}+p\to n+\nu _{e}$ , which happens within the first (0.1 sec) of final star collapse (note that the n is what makes neutron stars. "(b)" They also produce $\nu$ via: ${e^{-}}+{e^{+}}\to \nu _{i}+{\bar {\nu }}_{i}$ , where $i=e,\mu ,\tau$ . "(c)" Time for $\nu$ to travel from 1987A: If neutrinos are massless, $\mu _{\nu }=0$ , $T_{0}={d \over c}\sim 169,000lyrs$ . However if $\mu _{\mu }\neq 0$ , then $E=\gamma mc^{2}$ , $p=\gamma mv$ , so:
${v \over c}={pc \over E}={{\sqrt {E^{2}-m^{2}c^{4}}} \over E}={\sqrt {1-{m_{\nu }^{2}c^{4} \over E^{2}}}}\,\!$ $T={d \over v}={d \over c}\left(1-{m_{\nu }^{2}c^{4} \over E^{2}}\right)^{-{\frac {1}{2}}}\,\!$ The difference in times of arrival for if neutrinos are massive or not is:

{\begin{aligned}\Delta t&\equiv T-T_{0}\approx {d \over c}{m_{\nu }^{2}c^{4} \over 2E^{2}}\\&=0.0257sec\left({d \over 50kpc}\right)\left({m_{\nu } \over 1eV}\right)\left({10MeV \over E}\right)^{2}\\\end{aligned}}\,\! 