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 = \gammaMbVb + \gammaMcVc

E = Ea = Eb + Ec
= \gammaMac^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 = \gammaMbVb + \gammaMcVc

E = Ea = Eb + Ec
= \gammaMac^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
<br /> (P_A)^2 = (P_B + P_C)^2<br />
From the LHS of the equation:
<br /> (P_A)^2 = m_A^2<br />
From the RHS of the equation:
<br /> (P_B + P_C)^2 = m_B^2 + m_C^2 + 2 P_B \cdot P_C<br />


<br /> p_B = (E_B,\vect{p}_b)<br />
<br /> p_C = (E_C,\vect{p}_c)<br />

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

But:
<br /> m_B = m_C = 0<br />

Which implies that:
<br /> p_B = \sqrt{E_B^2 + m_B^2} = E_B<br />
<br /> p_C = \sqrt{E_C^2 + m_C^2} = E_C<br />

So therefore:
<br /> m_A^2 = (P_B + P_C)^2 = 2 E_B E_C(1-cos(\theta))<br />

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