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Say, I started with a particle with mass MA at rest, and it decayed into another two particles with rest mass of MB and MC.

Ignoring binding energy, what is the energy of MB and MC (in the original reference frame, i.e. the zero-momentum frame)?

What I've attempted is to make use of the relativistic energy equation: $E^2=(mc^2)^2+(pc)^2$. However, the equation became quite complicated, by which time I'm stuck (since my professor explicitly stated that it shouldn't have lengthy-algebra).

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  • $\begingroup$ $m_a = m_b + m_c$ + the binding energy. There is no general answer; each case will be different. $\endgroup$
    – garyp
    Dec 22, 2016 at 19:39
  • $\begingroup$ Hi, foxielmao! Welcome to Physics.SE. As the policy of the site goes, one is not supposed to show their ideas about the resolution of the question rather than just stating the question. I believe, especially when the question is of a rather algorithmic application of the laws in a situation than about the structure of the laws themselves. :-) $\endgroup$
    – youpilat13
    Dec 22, 2016 at 19:40
  • $\begingroup$ As far as the stated question goes, as @garyp has pointed out, it is not complete enough to yield a unique answer. One has to add parameters not specified in the question to formulate an expression that can be represented as the answer in one piece. $\endgroup$
    – youpilat13
    Dec 22, 2016 at 19:42
  • $\begingroup$ @Electrodynamist Sorry about that, I should have been more specific that binding energy is ignored in this context. I will edit my question (along with adding some of the method I've used to try to solve the question). $\endgroup$
    – The First StyleBender
    Dec 22, 2016 at 19:43
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    $\begingroup$ Your professor was right. From conservation of energy, $M_A-E_B=E_C$, which, squared, yields $M_A^2-2M_A E_B=E_C^2-E_B^2=M_C^2-M_B^2$, so that $E_B=(M_A^2+ M_B^2-M_C^2)/(2M_A)$, and likewise for C, the textbook expression. $\endgroup$
    – Cosmas Zachos
    Dec 24, 2016 at 23:39

1 Answer 1

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Momentum is shared equally in magnitude but opposite in direction (conserved), from the mass lost. Which can lead to widely varying velocities due to mass differences in the two particles.

so you have (with units such that $c=1$)

$$ P_{AorB}^2 =(M_A^2 - M_B^2 - M_C^2 )/4$$

Thus $$ E_B^2 = M_B^2 + P_{AorB}^2 = (M_A^2 + M_B^2 - M_C^2 )/4$$ $$ E_C^2 = M_C^2 + P_{AorB}^2 = (M_A^2 - M_B^2 + M_C^2 )/4$$

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  • $\begingroup$ You are right it should be /4. Which now I retrospectively edited in. Good spot. $\endgroup$
    – JMLCarter
    Dec 23, 2016 at 13:48
  • $\begingroup$ But... At threshold, P =0, for $M_B=M_C$, the daughter particle mass is still larger than half of the parent... Can your first relation be sound? $\endgroup$
    – Cosmas Zachos
    Dec 23, 2016 at 16:30
  • $\begingroup$ if I put P=0 into the first equation I get $M_A^2=M_B^2+M_C^2$ its the energy that must equate, which is the square of mass? $\endgroup$
    – JMLCarter
    Dec 23, 2016 at 16:40
  • $\begingroup$ So with your $M_B=M_C$ the daughter particle energy is exactly half the energy of the parent. What is the issue you have? $\endgroup$
    – JMLCarter
    Dec 23, 2016 at 16:42
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – JMLCarter
    Dec 23, 2016 at 16:48

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