Understanding nuclear mass defect

Question

I'm having a bit of trouble understanding mass defect, in the context of nuclear physics. The argument from my class is:

1. The nuclear force is attractive, and so does work on the particles as they are brought from being far apart to being close together. I am happy with this

2. Since the force does work on the particles, the change in potential energy is negative. I wasn't happy with this, but I came across this which helped.

3. $E$ in the relationship $E = mc^2$ includes potential energy. I am happy with this.

4. The change in energy is therefore negative, so the total mass of the nucleons decreases after they are brought together. In other words, the mass of the nucleus is less than the mass of its constituent nucleons, when they are far apart.

It is point 4 I really don't understand. I don't see how it follows from points 2 and 3 at all. In thinking about my problem with this, I was constructing the following argument.

1. Consider two point particles, one of mass m and charge q, the other of mass m and charge -q. The masses here are when the particles are far separated.

2. In bringing the particles together, the change in their potential energy is negative. To me, this seems exactly like point 2 of the argument above.

3. The particles can get arbitrarily close together. If we are modelling the particles as points, then this is true.

4. There is a distance $d$ such that the difference between the initial potential energy of the particles and the potential energy of the particles when separated by distance $d$, $\Delta U$, has $\Delta U \leq -m c^2$.

5. Therefore, the particles end up with zero or negative mass, just by being allowed to attract each other.

Obviously, this is argument is not sound, but I can't see how it differs from the argument about the nucleus mass.

My initial thought in constructing this argument was that the particles gain kinetic energy as they attract each other. In particular, the kinetic energy the particles gain cancels out the potential energy the particles lose, and so there is zero change in energy. This doesn't really help me understand mass defect though, since surely the same could be said of the nucleons as they attract each other?

I think we would have to do some work on the nucleons to stop them from moving. In other words, there is no way to go from the state where the nucleons are at rest and far apart, to the state where the nucleons are at rest and bound together, without doing work on the nucleons.

Can anyone help me make sense of this?

Answer 1

You seem to be bothered that

$$ \rm N + N\text{ (free)} \to NN \text{ (bound)} $$

isn't an energy-conserving reaction. And you're right! In a nucleon-capture reaction, there's also an emitted particle in the final state which carries away the energy. The cleanest example is probably neutron capture on hydrogen,

$$ \rm n + p \to d + \gamma $$

If this interaction takes place with the initial neutron and proton at rest (as it does when, say, milli-eV neutrons are incident on liquid or solid hydrogen), then the energy of the emitted photon is exactly the same as the deuteron's binding energy.

In fact, mass spectroscopy of hydrogen and deuterium ions and diffraction-based measurements of the wavelength (and thence energy) of the $\rm npd\gamma$ photon are how we obtain the best estimate of the neutron's mass.

Answer 2

The other answer covers why the interaction isn't energy conserving. Let me add a few things.

You are assuming the strong nuclear force (residual strong force) is attractive but actually it is distance dependent. At certain, very close distances, it becomes repulsive.

The nuclear potential is attractive at modest distances because the pion interaction gives an attractive potential. It turns repulsive because heavier mesons tend to have unit spin and carry a repulsive force betwen nucleons.

https://physics.stackexchange.com/a/119873/132371

Because of this, the nature of this residual interaction is completely different from the quark-level strong interaction. In particular, note the strong repulsion that happens at distances less than 1 fm (i.e. the diameter of a nucleon). This repulsion, mediated by vector-meson (ρ) exchange, is what keeps protons and neutrons apart.

https://physics.stackexchange.com/a/396054/132371

Nuclear binding energy is the minimum needed to disassemble the nucleus into its constituents. Its constituents are protons, and EM neutral neutrons. There is binding energy between them, and we need to spend energy to move these parts away from each other. The mass of the nucleus is less then the sum of the masses of the free constituent parts, the missing energy is the mass defect.

In your example there are two particles with opposite EM charge, but nuclei work differently, since there are EM neutral neutrons and protons. And the potential energy between them is different from a pair of oppositely charged particles.

A nucleon is a composite object made out of three quarks. The nucleon is color-neutral, so to first order, we expect that a nucleon should not interact with another nucleon at all. This is in fact approximately what we do see, since at large distances the nucleon-nucleon interaction falls off exponentially. But the cancellation is not exact, and at small distances we do get an interaction. This is called a residual interaction, and it's exactly analogous to the residual interaction between two electrically neutral atoms, which is the van der Waals force, often modeled by a Lennard-Jones potential.

And the Heisenberg Uncertainty Principle is determining this interaction at very close distances too.

We do not have any usable way of inferring the correct residual interaction between nucleons from a postulated quark-quark interaction. So instead we make models. Some of these models have a repulsive core, and others don't. In particular, it is not necessary to have a repulsive core in order to explain the sizes of nuclei or the fact that they don't collapse; their sizes are fundamentally set by the Heisenberg uncertainty principle.

https://physics.stackexchange.com/a/128062/132371

So basically the reason nuclei are stable, is because certain forces balance out:

1. strong nuclear force (residual strong force)

2. HUP

3. EM interaction between proton and neutron (comparable to Van der Vaals)

So basically when the constituents are brought together, energy is released (and that is the mass defect), and the forces balance out to keep the nucleus stable. If you want to disassemble the nucleus again, you need to spend the same energy you had to use to move them together.