# Is the change in mass during alpha decay a change in rest mass?

Energy is released in alpha decay as kinetic energy as the sum of the masses of the nuclei formed is less than the mass of the parent nucleus. However as the nuclei formed have kinetic energy their mass is greater than their rest mass according to $E=mc^2$, so is it just the sum of rest masses that decreases rather than their actual mass?

When we are in particle decays we are in the realm of quantum mechanics and of special relativity. .

In classical mechanics position of masses is described by three vectors, having components (x,y,z), and time is an independent parameter . Kinetic energy is described as $1/2mv^2$ , where $v=dr/dt$, r the vector in space, and masses are fixed and additive.

In special relativity time and space form a four vector, and a corresponding four vector is given by energy and momentum describing a particle. The "length" of this fourvector is invariant under relativistic transformation , it characterizes the particle uniquely , and it is called the invariant mass. It is equal to the energy of the system when the particle is at rest by the equation:

$$\sqrt{P\cdot P}=\sqrt{E^2-(pc)^2}=m_0c^2$$

Thus when Energy is released in alpha decay as kinetic energy as the sum of the masses of the nuclei formed is less than the mass of the parent nucleus. the algebra to use is the addition of the fourvectors, it is energy and momentum which are conserved, not the masses.

However as the nuclei formed have kinetic energy their mass is greater than their rest mass according to $E=mc^2$ .

The $m$ in your formula above is the inertial mass, the equavalent of $F=ma$ and plays no role in the energy balances of the reaction, which has to obey the rules of fourvector addition. This formula is not used in nuclear and particle physics because of the confusion introduced. It is only usefull if one wants to calculate how much fuel is needed to have a spaceship reach a measurable fraction of the velocity of light

• This is a purely classical question. No quantum mechanics is required. – user4552 Apr 2 '18 at 19:44
• @BenCrowell as the question is naive, I just wanted to set the framework for the thinking processes of the questioner. The specific question does not need quantum mechanics, but decays are in the framework of qm. – anna v Apr 3 '18 at 4:01