# Can a big mass defect make the mass negative?

Can two particles with small masses and a strong attractive interaction have a total negative mass when brought together?

Let $$m_1, m_2$$ be the (rest) masses of two particles when infinitely distant. Let the change in potential energy caused by bringing the two masses together be $$\Delta U < 0$$. The mass of the bound state (with the masses at rest) is then $$M = (m_1 + m_2) + \Delta U/c^2 < (m_1 + m_2).$$ As far as I can tell, this is what we call mass-defect. We see that $$\Delta U < -(m_1 + m_2)c^2 \Rightarrow M < 0.$$ Did I make any mistake? Is there something physical preventing this?

• Isn't this the case for quarks? The energy required to separate quarks is greater than the sum of their rest masses, which is why you end up creating new quark pairs before they are ever separated? Commented Jul 7 at 18:22
• That's interesting but I don't know. I have to imagine that quarks carry a lot of kinetic energy in order to make the total mass positive. Could you provide any resource link backing what you said? Commented Jul 8 at 7:25

## 3 Answers

If you apply no rules to the possible interactions then what you describe is possible. But the negative bound state mass can be avoided if we assume two things:

1. A force field that has always positive energy.
2. The definition of mass for one free particle includes the energy in the field around that particle. (And the starting value without that field cannot be chosen arbitrary negative, see below).

We describe the force by a force field $$\bf F$$, and the total energy in that field is: \begin{align} E_\text{field} =& \int d^3{\bf x}\ \lambda |{\bf F}|^2 \tag{1} \end{align} with $$\lambda$$ some positive constant. If the force is attractive this just means that we can lower the energy in the field by bringing the masses together. For instance if the particles have opposite charge, their fields will have opposite sign, and will partially cancel when they start overlapping if the particles are brought together. So to compute $$\Delta U$$ we just have to compute the field energy for the initial and final field configuration: \begin{align} \Delta U =& \int d^3x\ \lambda |{\bf F}_\text{final}|^2 -\int d^3x\ \lambda |{\bf F}_\text{initial}|^2 \\[6pt] \Rightarrow \Delta U =& \int d^3x\ \lambda |{\bf F}_\text{bound}|^2 -\int d^3x\ \lambda |{\bf F}_\text{1,free}|^2 -\int d^3x\ \lambda |{\bf F}_\text{2,free}|^2 \tag{2} \end{align} where the initial configuration just has two terms with the energy in the field $${\bf F}_\text{i,free}$$ around each particle (since they initially are far apart and the fields don't overlap).

For one free particle we include the field energy in the mass: \begin{align} m_1 =&\ m_{1,\text{bare}} + \frac1{c^2} \int d^3x\ \lambda |{\bf F}_\text{1part}|^2 \tag{3} \end{align} Where the "bare mass" $$m_{1,\text{bare}}$$ is the mass a particle would have without its field. Now our bound state mass $$M$$ is: \begin{align} M =&\ m_1 + m_2 + \frac1{c^2}\Delta U = m_{1,\text{bare}} + \frac1{c^2} \int d^3x\ \lambda |{\bf F}_\text{1,free}|^2 + m_{2,\text{bare}} + \frac1{c^2} \int d^3x\ \lambda |{\bf F}_\text{2,free}|^2 \\[6pt] & \qquad + \frac1{c^2} \Big( \int d^3x\ \lambda |{\bf F}_\text{bound}|^2 - \int d^3x\ \lambda |{\bf F}_\text{1,free}|^2 -\int d^3x\ \lambda |{\bf F}_\text{2,free}|^2 \Big) \\[6pt] =&\ m_{1,\text{bare}} +m_{2,\text{bare}} + \frac1{c^2} \int d^3x\ \lambda |{\bf F}_\text{bound}|^2 \\[6pt] \end{align}

Clearly, if both particles have a positive bare mass then the bound sate $$M$$ will always be positive. If, on the other hand, you allow arbitrary negative values for $$m_{1,\text{bare}}$$ and $$m_{1,\text{bare}}$$ then you can always make $$M$$ end up negative, just by taking the combined bare masses more negative than what the energy in $${\bf F}_\text{bound}$$ gives you.

It should be noted that with a negative bare mass we can also make the mass of one free particle negative and we can create other unphysical behavior. It is also related to the [classical electron radius] problem and the unphysical runaway solutions of the Abraham-Lorentz force, for which we also need the radius of the charge distribution [Jackson. p. 787]. So it is definitely safer to keep the bare masses positive.

• Thanks! Let me ask about the intuition: accounting for the energy in the field while calculating the mass is necessary because accelarating a charge is harder than accelarating an uncharged mass with the same bare-mass; this due to the fact that while accelarating the charge we are also adding energy into the field (Larmor radiation). Being harder to accelerate means being more massive. Do you agree? Commented Jul 7 at 14:28
• Even without radiation, just the Coulomb field will be Lorentz contracted to a "pancake" shape for a moving charge and this gives a higher integral, and also a B field will emerge and give additional energy. Commented Jul 7 at 14:47
• I think the relativistic Larmor formula takes all that into account: contracting the electric field (boosting the electromagnetic field) produces radiation. Commented Jul 7 at 14:58

TL;DR: No, quantum field theory does not allow this.

The general situation you describe is not at all exotic, and in fact this is why we have (meta)stable bound states. What you have termed 'mass-defect' we normally call 'binding energy' $$B = -\Delta U$$.

As the easiest example, the interactions of the proton and the electron via quantum electromagnetism leads to a bound state we call a hydrogen atom which has binding energy $$B \simeq 13.6$$ electronvolts. Note this is absolutely miniscule compared to the mass of the particles $$B / m_e \sim B/(511 \ \rm keV) \sim 2.7\times 10^{-5}$$ or $$B/m_p \sim B/(938 \ \rm MeV) \sim 1.5\times 10^{-8}$$.

But your question is whether there might be any interaction which has such large binding energy that the total energy of the whole bound state will be negative. I'll call this a 'super-extremal binding energy'.

In the case of electromagnetism, binding energies are quite small as the result of fact that the fundamental coupling of electromagnetism, the fine-structure constant, is quite small $$\alpha \sim 1/137$$. In general, at weak coupling one expects that binding energies are smaller than the mass scales of the problem.

But still one may reasonably ask whether there is perhaps some exotic theory of physics (maybe very strongly coupled $$\alpha \gtrsim 1$$) which could produce super-extremal binding energy.

In fact, there are no unitary, Lorentz-invariant quantum field theories that lead to this scenario, as effectively proven by Hartman, Kundu, Tajdini (2016). The super-extremal binding energy you envision would allow you to construct a source of negative stress-energy in a sense that would violate the Average Null Energy Condition, which has been shown to be satisfied in all interacting quantum field theories in four dimensions. Physics is described fundamentally by just such a quantum field theory, so this cannot occur.

• Physics is not founded on mathematical abstractions: it is founded on experiments and observations. If an experiment were to show a contradiction to QFT, the postulates of the theory would need revision. Commented Jul 7 at 16:28
• This is certainly true, and indeed my answer was not a mathematical one but a physical one. It is based upon the enormous amount of evidence from experiments and observations that quantum field theory describes the universe exceedingly well. I like Sean Carroll's discussion in The Quantum Field Theory on Which the Everyday World Supervenes. Of course we should and do continue to search for ways in which this might break down, and indeed experiments and observations give us fantastic constraints on theories that violate QFT. Commented Jul 7 at 16:42
• I agree with you on the evidence, but your final paragraph is a mathematical argument, not a physical one: "proven by", "shown to be", "described fundamentally", "cannot occur". It doesn't mention evidence at all. Commented Jul 7 at 16:57
• I can't say I follow your objection. We agree there is lots of evidence that quantum field theory describes our universe, so if we can use mathematics and theoretical physics to prove something about the physics of all quantum field theories, we learn something about our universe. And something much more general than just how physics works with weakly coupled interactions which have classical limits. Commented Jul 7 at 20:52
• You can't prove anything about physics with mathematics. You can create models that capture and connect the evidence, and make predictions. But nothing is proven that way. The imaginary objects of mathematics are not physical reality. Commented Jul 7 at 20:57

This question leads to deep insights into what constitutes a particle relative to the vacuum. If you were to have a system of particles with binding energy less than the vacuum, the universe would simply fill itself up with this state in order to minimize its energy; that is, the vacuum would decay. In the new vacuum, what you thought were particles would no longer be the correct degrees of freedom to understand physics, and you would have to rework your understanding in terms of new particles. With respect to the true vacuum, what you suggest can never occur.

To give a famous example, the Higgs field has a negative mass term. This causes the entire universe to support a nonzero Higgs, the vacuum expectation value or vev. When you look at the particles relative to this vacuum filled with vev, the Higgs boson and all the other particles interacting with it have a normal positive mass.