Answer to Question #125430 in Quantum Mechanics for Sunil

Question #125430
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

E1= c^(2) [M^(2)+ m1^(2) - m2^(2)] / 2M

c= speed of light
1
Expert's answer
2020-07-09T10:33:43-0400

Let's introduce 4 - vector "p_\\mu = (\\frac{E}{c},\\vec p)", "p^\\mu p_\\mu = \\left(\\frac{E}{c}\\right)^2-(\\vec p)^2 = m^2c^2"

"p^{1}_\\mu = (\\frac{E_1}{c},\\vec p_1) \n;p^{2}_\\mu =(\\frac{E_2}{c},\\vec p_2);P_\\mu =(\\frac{E}{c},\\vec P)"

where "P_\\mu" - 4 - vector of momentum of the body "M"

"p^1_\\mu, p^2_\\mu" - 4 - vector of momentum of the "m_1, m_2"

We know that "\\vec P = 0". Also we know that momentum and energy conserve laws are done in this case. So, we have the system of equations:

"\\vec P = \\vec p^1 + \\vec p^2 = 0;"

"\\left(\\frac{E}{c}\\right)^2 = M^2c^2"

"\\left(\\frac{E_1}{c}\\right)^2-(\\vec p^1)^2 = m_1^2c^2"

"\\left(\\frac{E_2}{c}\\right)^2-(\\vec p^2)^2 = m_2^2c^2"

We can solve this system relatively by the energies "E_1,E_2" and get the answer on this question.

"E_1 = c^2\\frac{M^2+m^2_1-m^{2}_2}{2M}"


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Comments

Sunil
07.07.20, 19:00

Thanks sir for the answer. But which photo I have to follow ? I can't see any photo here

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