Question #108331
A particle of mass M, initially at rest, decays into two particles with rest masses
m1
and
m2
respectively. Show that the total energy of the mass
m1
is:
M
c M m m
E
2
[ ]
2
2
2
1
2 2
1
 
1
Expert's answer
2020-04-08T10:29:53-0400

E1+E2=m0c2E_1+E_2=m_0c^2


p1+p2=0\overrightarrow{p_1}+\overrightarrow{p_2}=0


E12=m12c4E_1^2=m_1^2c^4 and p12=2m1E1p_1^2=2m_1E_1


(m0c2E1)2=E22(m_0c^2-E_1)^2=E_2^2



(m0c2E1)2=(m_0c^2-E_1)^2=


=(m02c4+m12c42m0c2m1c2)==(m_0^2c^4+m_1^2c^4-2m_0c^2m_1c^2)=


=c4(m02+m122m0m1)==c^4(m_0^2+m_1^2-2m_0m_1)=


=c4(m02+m12)c22m0E1=c^4(m_0^2+m_1^2)-c^22m_0E_1



E22=m22c4E_2^2=m_2^2c^4


m22c4=c4(m02+m12)c22m0E1m22c2=c2(m02+m12)2m0E1m_2^2c^4=c^4(m_0^2+m_1^2)-c^22m_0E_1\to m_2^2c^2=c^2(m_0^2+m_1^2)-2m_0E_1


So, we have


E1=c22m0(m02+m12m22)E_1=\frac{c^2}{2m_0}(m_0^2+m_1^2-m_2^2) Answer.






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