# Using the energy-momentum invariant for a decay process

For a decay process in which

particle A ----> particle B + a photon

in which particle A has mass $m_A$, particle of mass $m_B$ and energy and momentum are conserved.

Show that in the frame in which particle A is initially at rest, the energy of particle B, $E_B$, is given by $$E_B = \frac{(m_A ^2 + m_B ^2) c^2}{2m_A}$$

What I did:

conservation of energy: $m_Ac^2 = E_B + E$, where E is energy of photon

conservation of momentum: $0 = p_B + \frac{E}{c}$, where $p_B$ is momentum of particle B

from these and the e-p invariant, I get $$p_B = \frac{E_B - m_Bc^2}{c}$$

but I am not sure how to proceed after this.

• Are you sure about that last equation? You should be using $E^2=(pc)^2+(mc^2)^2$ – PM 2Ring Apr 14 '18 at 15:55
• @PM 2Ring yes that was the equation I used. Since the momentum initially was 0, the net momentum of particle B and photon is 0, therefore $(E_B + E)^2 = (m_Bc^2)^2$, then sub in E that I got from conservation of energy and momentum, is this right? – Student 1 Apr 14 '18 at 17:27
• $E_B=E_A-E$ so $E_B=m_Ac^2-E$ . – PM 2Ring Apr 14 '18 at 17:44
• @PM 2Ring yes but using what you suggested and the $p_B$ I stated above, I cannot get $E_B$ in the form they are asking for? – Student 1 Apr 14 '18 at 18:01
Your first equation can be written as $m_Ac^2 = E_B + \sqrt{E_B^2-m_B^2c^2}$. Solve this for E_B. This equation represents conservation of energy in the rest system of the decaying particle A. The square root is the energy of the photon, which is equal to its momentum. The photon momentum equals the momentum of particle B, which is given by the square root.