Question

Suppose a relativistic particle A with speed v relative to frame S decays into 2 particles B and C. One of these particles is found to be at rest in frame S. The rest masses of A, B and Care ma, me and mc respectively. (a) Expressing your answer in terms of ma, me and mc (and the speed of light c), what is the kinetic energy of the particle A? (b) Expressing your answer in terms of ma, me and me what is the energy of the particle C not at rest in S?

Answer 1

Total energy of a relativistic particle

E² = p²c² + m²c⁴

pc is the KE energy and mc² is the rest mass energy

particle A has a relativistic speed v ;

$\gamma _a$ = 1/(1-v²/c²)^1/2

(a) KE of the particle A = $\gamma _a$m_Ac² - m_Ac² =  m_Ac² /(1-v²/c²)^1/2 -  m_Ac²

relativistic momentum p_A = $\gamma _a$m_Av

after split particle B is at rest so, it total energy = m_Bc²  

In a relativistic momentum is conserved

p_B =0

p_c = p_A = $\gamma _a$m_Av = $\gamma _c$

E_A = m_Bc² + E_c  

Total energy of the particle C

E_c = E_A - m_Bc² = sqrt( p_A²c² + m_A²c⁴ ) - m_Bc²

KE of the particle C = E_c - mc² = sqrt( m²_Av²c² /(1-v²/c²) + m_A²c⁴ ) - m_Bc² - m_Ac²