Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Looking at the equation for binding energy and mass defect,
$$ B = m_{\text{free}} - m_{\text{bound}} \\ \Rightarrow m_{\text{bound}} = m_{\text{free}} - B, $$ my question is the following. Suppose that the binding energy $B$ becomes larger than the free rest mass $m_{\text{free}}$. It would seem that the mass of the bounded system $m_{\text{bound}}$ can become negative. As this is impossible there must be something wrong in my assumptions but I don't see it.

Take for example the hydrogen atom, the binding energy for the ground state is $13.6 \text{ eV}$. What would happen to the mass of the hydrogen atom if hypothetically the binding energy would become larger than $ \approx 938.783 \text{MeV} $ (which is just the sum of the rest masses of the proton and the electron) by, for example, increasing the EM-coupling constant.

share|cite|improve this question
Of course, the binding energy of the quarks that make up a proton/neutron is far in excess of the rest mass of the quarks, and therefore, most of the mass of any atom is tied up in quark binding energy. – Jerry Schirmer Mar 7 '13 at 17:07
up vote 5 down vote accepted

A good question touching the deep physical concepts. I will try to give an answer in several steps.

  • Before answer your question let me answer another one: what is a free particle? A simple (but complete) answer is that a «particle» is a long-living elementary excitation of a system (a quantum field). «Long-living» means living long enough to be observable, i.e., weakly interacting with other excitations, «elementary» means that the the total spectrum of the system is (approximately) different sums of energies of such excitations.

You correctly noted that the energy of the bound state is less then the energies of its «free» components. Let’s for example consider a positron-muon bounded state (a muonic atom). As it was correctly noticed by John Rennie you can take these two fermions in a continuous spectrum and they can form the muonic atom with some amount of electromagnetic radiation.

  • The interaction of elementary excitations (particles) can be characterized by a some constant usually called a «coupling constant». As I noticed before for a «long-living» excitation the interaction should be week thus the coupling constant should be small.

Our muon and positron can form a bound state because of the interaction with electromagnetic field. The corresponding coupling constant is usually referred as the fine-structure constant and denoted as $\alpha$. If the coupling constant is small enough then the binding energy is proportional to some power the coupling constant, e.g., the binding energy of the muonic atom is approximately $-m_{0}c^2(4\pi\alpha)^2/(2\,n^2)$, where $m_{0}=m_{\mu}m_{e}/(m_{\mu}+m_{e})$ is the muon-positronium reduced mass. For the sake of simplicity let's assume that $m_{\mu}=m_{e}=m$.

Hence your question can be reformulated as follows: what happens if we would increase (by any means) the coupling constant $\alpha$ so that the binding energy becomes to be comparable with (or greater than) $2m$?

  • Here I would like to notice that you can produce muonic atoms from the vacuum. For example, an accelerated conductor (or any media) produces photons, e.g., via the dynamical Casimir effect. Photons can interact and produce electron-positron pairs, muon-antimuon pairs and (if there is enough energy) pairs of muonic and anti-muonic atoms. The quantity which characterize the intensity of such production is usually called a «production cross section».

If you start to increase the coupling constant then the production cross section also grows. Since the muonic atoms become lighter and lighter, the energy threshold for the muonic atom production decreases. Thus your system (universe, whatever) becomes very easy excitable, any process leads to the production of the enormous number of very light excitations. And if you will continue to increase the coupling constant so that the binding energy will be greater than $2m$, then the ground state (vacuum) of your system becomes unstable, e.g., all photons in your universe immediately decay into fermions, these fermions form the bound states (atoms), these bound states fill the whole volume of the universe and form Bose-Einstein condensate. Such scenarios are referred as critical phenomena or phase transitions. The main feature of these transitions is that they lead to a total spectrum transformation. After the transition of a critical point you will find a new ground state (vacuum) and new elementary excitations (particles) with new masses and new coupling constants.

share|cite|improve this answer
Thank you for your answer. Also excellent reformulation of my original question. I still have some questions concerning the phase transition. 1) Am I correct in stating that as the bound states that are created in massive numbers consist of fermions, subjected to Pauli's exclusion principle, this new vacuum cannot become infinitely dense but will stop at a certain moment because Pauli's exclusion will raise the excitation energy of the new vacuum. 2) Can I view the new elementary excitations and masses in a similar fashion as electrons behaving differently in solids (like hole excitations)? – camelthemammel Mar 7 '13 at 21:56
Actually, my answer to your question is equal to the short statement «if you increase the coupling constant, at some moment something bad will happen». So do not take the scenario I described literally. The real dynamics (the first or the second order phase transition, what kind of vacuum is generated etc.) strongly depends on details: long-range or short-range interaction, gauged or not, Abelian or not etc. – Grisha Kirilin Mar 8 '13 at 16:43
For example, the QCD coupling constant strongly depends on the distance, thus it is believed that there is the second order phase transition, e.g., during the cooling down of quark-gluon plasma, so that a non-trivial non-perturbative QCD vacuum appears and it is filled with the quark condensate. The resulting excitations (hadrons) are absolutely non-trivial collective excitations of this vacuum. – Grisha Kirilin Mar 8 '13 at 16:44
For example, pions can be considered as pseudo-Goldstone bosons of the chiral-flavor symmetry breaking. There are plenty of phenomenological models which consider baryons (like proton) as a nontrivial soliton of some effective chiral fields (see, e.g., Skyrme model, Nambu—Jona-Lasinio model etc). – Grisha Kirilin Mar 8 '13 at 16:45
The example of the numerical calculation of the complete proton structure you can find in G. S. Adkins, C.R. Nappi, Nucl.Phys. B233 (1984) 109. Although the real dynamics is still unknown, I would like to stress here that neither baryons nor mesons are the bound states of any number of quarks and gluons (in the conventional sense), they are nontrivial low-temperature collective excitations of the quark and gluon fields. – Grisha Kirilin Mar 8 '13 at 16:45

protected by Qmechanic Mar 21 '13 at 14:10

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.