when we have a decay process, there are many fragments, we can measure their momentum and energy and construct the 4-momentum

$\vec {P_i} = ( E_i , p_i )$

we use the c = 1 unit as usual.

to find out the mass before the decay, we can use

$\sum E_i^2 - \sum p_i^2 = m_{excite}^2$

the reason for the term “excited mass”, we can see by the following illustration.

consider a head on collision of 2 particles in C.M. frame, with momentum p and energy E1 and E2.

the mass for each one is given by

$m_1 = \sqrt {E_1^2 - p^2 }$

$m_2 =\sqrt {E_2^2 - p^2 }$

but if we use the sum of the 4 momentum and calculate the mass,

$\sqrt { (E_1 +E_2)^2 - (p - p)^2} = E_1+E_2$

which is not equal to

$\sqrt { E_1^2 - p^2} + \sqrt{E_2^2 - p^2 }$

in fact, it is larger.

the reason for its larger is, when using the sum of 4 momentum, we actually assumed the produce of collision is just 1 particles, and the collision is inelastic. Thus, if we think about the time-reverse process, which is a decay, thus, some of the mass will convert to K.E. for the decay product.