# Binding energy converted to kinetic energy/mass loss

My particular question is regarding nuclear fission, but applies to other nuclear processes as well.

I understand that in nuclear fission, the two fragment nuclei have a higher binding energy per nucleon, and hence the binding energy of the products is greater than that of the reactants. I also understand that the increase in binding energy means that the nucleons lose energy, and this loss of energy in turn causes the reduction in mass, i.e. the mass of the products < mass of reactants.

However, multiple sources state that this energy released in the fission process is converted into kinetic energy of the products (of the two fragment nuclei and neutrons).

If this is the case, then surely an increase in kinetic energy in turn causes and increase in mass of the products, and so if all binding energy is converted into kinetic energy, then there should be no mass loss.

The only explanation I can think of is that not all binding energy is converted into kinetic energy of the products, and that some is just released (as photons perhaps?), meaning there will be a net reduction in mass.

Either way, I do not understand why my textbook states that increase in binding energy = mass difference * c^2, if the products gain kinetic energy from this binding energy.

Any help is greatly appreciated!

• "[...]then surely an increase in kinetic energy in turn causes and increase in mass of the products[...]" - why? Kinetic energy does not contribute to the rest mass of a system, which is what we usually mean when we say "mass". – ACuriousMind Feb 14 '17 at 12:46
• @ACuriousMind Ok. Could you see if I'm understanding this correctly? When a nucleus forms and the nucleons lose nuclear potential energy, this causes the mass of the nucleus to be less than the mass of the separated constituents. When fission occurs, the total binding energy increases, so the REST mass of the products decrease by an amount (change in binding energy)/c^2. The products then gain a kinetic energy equal to the change in binding energy. This increases their relativistic mass by an amount = binding energy/c^2, but their rest mass is unchanged ................. – John Feb 14 '17 at 14:13
• ........... So that now the relativistic mass of the products = rest mass of reactants, but the rest mass of products is less than the reactants. – John Feb 14 '17 at 14:15

• "In fact they measured the charge to mass ratio and found that the specific charge of beta particles decreased as the speed of the beta particles increased" While that is the historical way in which the idea was described I think that saying the relationship between the kinetic energy and momentum deviated from the Newtonian value has the advantage of being correct in any relativistic framing, while the historical one depends on you using "mass" to mean $\gamma m$. – dmckee --- ex-moderator kitten Jul 21 '18 at 7:44
• "Is the reference to inertial mass correct?" Yes in a specific way and no in general. A particle in relative motion at velocity $v$ will be measured to have inertia $\gamma(v) m$ if the force is applied transverse to the direction of relative motion (as is the case for Lorentz forces), so it is correct in this case. But if you applied a force along the direction of relative motion you would measure an inertia of $\gamma^3 m$, so treating or naming the "relativistic mass" as the inertia of the particle is not generally correct. (Search for "transverse mass" versus "longitudinal mass".) – dmckee --- ex-moderator kitten Jul 21 '18 at 8:31