In my A-Level physics notes on Nuclear Physics, my teacher states the following two points (taken directly from his notes):

  1. In an unbound system (e.g. 6 individual protons and 6 individual neutrons), the rest mass of the composite system is greater than the sum of the rest masses of the separated particles by an amount equal to the kinetic energy of the amalgamating particles at combination.

  2. In a bound system (e.g. a [here he had the symbol for a C-12 atom] nucleus), the rest mass of the composite system is less than the sum of the rest masses of the separated particles by an amount called the binding energy ∆E.

I understand the concept of the changing mass of the particles related to their potential energy, but my question is about certain words that he used in these statements:

What does ‘unbound composite system’ mean? It seems somewhat of a contradiction.

I’m not entirely sure about the energy difference being described in the first point (is this a theoretical kinetic energy or are the particles actually moving?)

In the second point he appears to compare the combined particles (a nucleus) to the seperated particles, all under the banner ‘bound system’. I don’t understand that.

Many thanks, Hugo


2 Answers 2


What does ‘unbound composite system’ mean?

It just means that the collection of protons and neutrons is being treated as a closed system, but none of the particles are interacting via the nuclear force. The nuclear force is negligible unless the distance between the nucleons is very small (which is why a nucleus is so much smaller than an atom), and you have to overcome a potential barrier to get nucleons to "stick" to one another.

is this a theoretical kinetic energy or are the particles actually moving?

Yes, the particles are actually moving. The protons are all repelling each other due to their positive electric charge. So even if we're working in the centre of mass frame of the collection of particles, the individual protons will all have some velocity and hence some kinetic energy in that frame.

Actually, his point 1 isn't quite correct: as well as the KE, we also have to include the electrostatic potential energy between the protons. You have to perform work to reduce the distance between a pair of protons. So a system consisting of a pair of protons has more energy (in its rest frame) when you reduce the distance between them. FWIW, this is related to the concept of specific orbital energy, except that we're working with electrostatic repulsion instead of gravitational attraction, so the specific potential energy is positive rather than negative.

Neutrons have no electric charge, so those considerations don't apply to them.

  • $\begingroup$ I have doubt about the second last paragraph. You indeed need to perform work to cross the Coulomb barrier of protons, but once they get near enough they fall into nuclear force well, and the energy excess is radiated away. $\endgroup$ Nov 25, 2021 at 2:00

For 1, it presupposes that there is a motion in the particles, described by the four vector of each. Mass, in special relativity is the "length" of the four vector, so when the four vector of many particles is added, from the algebra of four vectors, the combined four vector will have a mass larger than the sum of the individual masses, unless all the particles are at rest.

For 2, In a bound system:In a nucleus, it is an experimental fact that the mass of the nucleus is less than the sum of the individual masses. Using the mass energy formula ,it needs that much energy for the particles to be free. It is one of observations that validates special relativity.


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