- #1
indigojoker
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the mc^2 for a pion and muon are 139.57 MeV and 105.66 MeV respectively. Find the kinetic energy of the muon in its decay from [tex] \pi ^+ -> \mu^+ + \nu_{\mu} [/tex] assuming the neutrino is massless. Here's what I did:
Since [tex]E^2=p^2c^2+m^2c^4[/tex] and that c=1, then E, p and m have same units.
[tex]E^2 = p^2 +m^2[/tex]
[tex](139.57 MeV)^2 - (105.66MeV)^2 =p^2[/tex]
[tex]p=91.19MeV[/tex]
Also consider the case where there is a small neutrino mass:
[tex]E^2 = p^2 +m^2[/tex]
[tex](m_{\pi})^2 - (m_{\mu}+m_{\nu})^2 =p^2[/tex]
[tex]p=\sqrt{(m_{\pi})^2 - (m_{\mu}+m_{\nu})^2}[/tex]
I feel like there is ill logic here. Comments on my work would be appreciated.
Since [tex]E^2=p^2c^2+m^2c^4[/tex] and that c=1, then E, p and m have same units.
[tex]E^2 = p^2 +m^2[/tex]
[tex](139.57 MeV)^2 - (105.66MeV)^2 =p^2[/tex]
[tex]p=91.19MeV[/tex]
Also consider the case where there is a small neutrino mass:
[tex]E^2 = p^2 +m^2[/tex]
[tex](m_{\pi})^2 - (m_{\mu}+m_{\nu})^2 =p^2[/tex]
[tex]p=\sqrt{(m_{\pi})^2 - (m_{\mu}+m_{\nu})^2}[/tex]
I feel like there is ill logic here. Comments on my work would be appreciated.