Calculating Mass of Particle A After Decay Using Relativity Equations

[alfredbester]

A particle of rest mass Ma, decays into two massles particles of B abd C of energy Eb and Ec respecitvely. The momenta of particles B and C are separated by an angle $$\theta$$. Calculate the combinded momentum and combined energy of B and C and hence show that particle A has a mass given by,
Ma = 1/$$c^2$$ . [sqrt(2EbEc(1- cos $$\theta$$)]

Pa = 0 = Pb + Pc = $$\gamma$$MbVb + $$\gamma$$McVc

E = Ea = Eb + Ec
= $$\gamma$$Ma$$c^2$$

I know that Eb and Ec can be easily found using the $$E^2$$ formula, but am not sure how to take the equations and find Ma.

 

[Norman]

A particle of rest mass Ma, decays into two massles particles of B abd C of energy Eb and Ec respecitvely. The momenta of particles B and C are separated by an angle $$\theta$$. Calculate the combinded momentum and combined energy of B and C and hence show that particle A has a mass given by,
Ma = 1/$$c^2$$ . [sqrt(2EbEc(1- cos $$\theta$$)]

Pa = 0 = Pb + Pc = $$\gamma$$MbVb + $$\gamma$$McVc

E = Ea = Eb + Ec
= $$\gamma$$Ma$$c^2$$

I know that Eb and Ec can be easily found using the $$E^2$$ formula, but am not sure how to take the equations and find Ma.

The idea here is just what you started. But use 4-momentums I think.

Using 4-momentums (I will use capital P for a 4-momentum and a lower case p for 3 momentums so there is no ambiguity)

Please note I am using units in which c=1
$$(P_A)^2 = (P_B + P_C)^2$$
From the LHS of the equation:
$$(P_A)^2 = m_A^2$$
From the RHS of the equation:
$$(P_B + P_C)^2 = m_B^2 + m_C^2 + 2 P_B \cdot P_C$$


$$p_B = (E_B,\vect{p}_b)$$
$$p_C = (E_C,\vect{p}_c)$$

So:
$$(P_B + P_C)^2 = m_B^2 + m_C^2 + 2 (E_B E_C - p_B p_C cos(\theta))$$

But:
$$m_B = m_C = 0$$

Which implies that:
$$p_B = \sqrt{E_B^2 + m_B^2} = E_B$$
$$p_C = \sqrt{E_C^2 + m_C^2} = E_C$$

So therefore:
$$m_A^2 = (P_B + P_C)^2 = 2 E_B E_C(1-cos(\theta))$$

Now just go back and put your c's in!