# Why is mass defect calculated by the rest mass (energy)?

I'm a high school student, and I have a question about the mass defect. I learned that when nuclear reaction occurs, the total rest mass (or rest energy) of reactants is greater than that of products, and the missing rest energy is released as a form of kinetic energy etc. But considering the fact that those reactants and products are moving, why should the missing mass (energy) be calculated by the rest mass(energy), not the relativistic mass? What would happen if you use the relativistic mass when calculating the mass defect?

• Generally one looks in the center-of-mass frame of reference, so overall system motion goes away. (In the lab frame, of course, one needs to put motion back in.) Sep 28 '20 at 15:54
• @JonCuster Even in that reference each atom is in motion, thus having relativistic mass. If not, the atoms would not have collided(?) in the first place
– DH K
Sep 28 '20 at 16:27
• Yes, they are in motion, and yes, if relativistic, you need to account for that energy. So? Sep 28 '20 at 16:29
• @JonCuster But then why my textbook and other websites only consider the rest mass and then calculate the energy derived from the mass defect? Here's my thoughts. 1) The mass defect accelerates the products, therefore making the relativistic mass heavier. 2) Then if you compare the energy of products with that of ractants, the former has greater kinetic energy, and more mass. Then where did the mass and energy come from?
– DH K
Sep 28 '20 at 20:02
• The mass defect is for the various ins and outs, at rest and isolated with respect to each other. How the experiment can actually be done, and how the energy output is measured (say, kinetic energy of the products) is a different question. Sep 28 '20 at 20:43

But considering the fact that those reactants and products are moving, why should the missing mass (energy) be calculated by the rest mass(energy), not the relativistic mass? What would happen if you use the relativistic mass when calculating the mass defect?

What would happen:

relativistic mass of reactants = relativistic mass of products

Why we do not calculate mass defect that way:

For reaction A the relativistic mass defect = 0, also for a very different reaction B the relativistic mass defect = 0.

If we are given a table of relativistic mass defects of different reactions, we are not given any information.

• So you mean that the relativistic mass is always conserved, whereas the rest mass isn't?
– DH K
Sep 29 '20 at 11:37
• @DHK No. Both are conserved. Cooled nuclear reactor loses mass, heated cooling water gains mass. Oct 2 '20 at 8:04
• @DH K is correct; in a nuclear reaction the relativistic mass is conserved, the rest mass is not conserved. See my Dec 4, 2020 answer. Dec 4 '20 at 7:41

$$E = mc^2$$ where m is the relativistic mass. For evaluating a nuclear reaction we can use
E = $$m_0c^2 + T$$ where $$m_0$$ is the rest mass and T is the kinetic energy. In a nuclear reaction E is the same before and after the reaction and m is the same before and after the reaction. Consider the reaction a + X -> b + Y where a and X are the reactants and b and Y are the products: $$E_{reactants} = E_{products}$$ and $$m_{reactants} = m_{products}$$. We define the Q value of the reaction as the difference between the initial and final kinetic energies: $$Q = T_B + T_Y - (T_a + T_X)$$. So Q is also $$[m_0a + m_{oX} - (m_0b + m_{oY})]c^2$$. If Q is positive the reaction is exothermic ; if Q is negative the reaction is endothermic. For an exothermic reaction the sum of the rest masses of the products is less than the sum of rest masses of the reactants and we have concerted rest mass into kinetic energy.

The sum of the relativistic masses of the products is equal to the the sum of relativistic masses of the reactants; its just that the partitioning between rest mass energy and kinetic energy has changed. Some discussions of a nuclear reaction are sloppy in nomenclature and do not distinguish between the rest mass and the relativistic mass.

The same equations apply to a chemical reaction, but the change in rest mass energies is much less for a chemical reaction than for a nuclear reaction. Classical thermodynamics accounts for the change in rest mass as the "heat of formation" for the reaction.