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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.

Thank you for your help.

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    $\begingroup$ Are you sure about that last equation? You should be using $E^2=(pc)^2+(mc^2)^2$ $\endgroup$ – PM 2Ring Apr 14 '18 at 15:55
  • $\begingroup$ @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? $\endgroup$ – Student 1 Apr 14 '18 at 17:27
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    $\begingroup$ $E_B=E_A-E$ so $E_B=m_Ac^2-E$ . $\endgroup$ – PM 2Ring Apr 14 '18 at 17:44
  • $\begingroup$ @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? $\endgroup$ – Student 1 Apr 14 '18 at 18:01
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    $\begingroup$ You have energy conservation, momentum conservation, and the on-shell condition on B. So solve for E and eliminate it in the above condition of @PM2Ring. $\endgroup$ – Cosmas Zachos Apr 15 '18 at 0:55
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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.

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  • $\begingroup$ @ Jerrold Franklin may I ask where you have got the expression inside the square root from? And how have you used the energy-momentum invariant? $\endgroup$ – Student 1 Apr 15 '18 at 9:41

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