Answer to Question #125430 in Quantum Mechanics for Sunil

Question

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

Answer 1

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

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

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.

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