# What percentage of a Proton's mass is potential/kinetic energy?

So in an hydrogen atom, the total mass of the atom is equal to the masses of the proton, the electron, minus their net binding energy of around 13 eV. Making the total less massive than the sum of its parts by about 1 part in 100 million. As it also turns out, the electron follows the virial theorem, so that binding energy is actually in the form of -26 eV of electromagnetic potential energy and 13 eV of kinetic energy.

In a proton however, the total mass is equal to the masses of the three valence quarks plus the net binding energy, which is not only positive but accounts for 99% of the proton's mass. This is because the protons can never be in a free state, so while this binding energy is still positive, it is the minimum possible binding energy these quarks can have, and so attempting to dissassociate a quark increases the total mass of the system just like in the case for the hydrogen atom.

Now onto the question, what fraction of this total net binding energy can be considered as the potential energy of the gluon field between quarks and the gluons themselves, and what percentage can be considered to be in the kinetic energy of the quarks and gluons? The potential energy must still be negative (since it is still a potential well and the strong potential is attractive at nuclear distances), but the virial theorem no longer holds (because the strong potential doesn't follow an inverse square law), so the kinetic energy can no longer be simply negative half of the potential (and that would lead to a negative net binding energy anyway, but we know from above its positive). The kinetic energy must therefore be some multiple greater than 1 of the potential energy, and I am wondering if one has calculated what that multiple is. No doubt some lattice QCD and supercomputer shenanigans are needed to get the result, but surely it has been done?

TL;DR, the ~929 MeV of the proton, ignoring the ~9 MeV of the valence quarks, is some amount of negative potential and positive kinetic energy, say -200 MeV potential and 1129 MeV kinetic, for example. Those numbers are what I'm looking for.

(Note, I am aware that in actuality most of this energy will be taking the form of virtual quark-antiquark pairs, but these are in constant flux and so I am merely looking for the semi-classical baseline that these quantum fluctuations float around.)

• The virial theorem doesn't just apply to inverse square law forces, but it does take a simple form when the potential is proportional to some power of the interparticle distance. FWIW, Frank Wilczek says it took about 3 months of Cray runtime to roughly compute the proton mass (minus the Higgs component) from QCD, but he wasn't using lattice techniques. You may enjoy Frank Wilczek on "Core" Physics: The Analytic Theory of Matter Oct 21, 2021 at 1:12
• Right sorry, meant to say the Virial theorem only gives that result, that kinetic is half the potential, not that it doesn't apply. Also yes you are right, its any power law not just -2. I would also accept any answer that can just give me the form the virial theorem takes for the strong force in a hadron. I know the potentials for a two-quark system and for the residual strong force (the Yukawa potential), but not for the proton, if there even is one since its a 3-body system. Oct 21, 2021 at 1:16
• Ok, you should edit your question to clarify that point about the virial theorem. Oct 21, 2021 at 1:30
• Duplicate of physics.stackexchange.com/q/474084. Oct 24, 2021 at 19:30

There are a number of subtleties, because in a field theory you must define what you mean by "potential" and "kinetic" energy, and there are issues with any decomposition that are related to gauge invariance and scale dependence.

The issue of the proton mass decomposition has been studied in some detail, and continues to be an active area of research (see, for example, these two papers).

The basic idea is the following. Protons are described by QCD, and the QCD energy-momentum tensor is $$T^{\mu\nu} = \frac{1}{2} \bar\psi i\gamma^{(\mu} \stackrel{\leftrightarrow}{D}\mbox{}^{\nu)}\psi +\frac{1}{4}g^{\mu\nu}F^2 - F^{\mu\alpha}F_{\alpha}^{\nu}$$ where $$\psi$$ is a quark field, $$\stackrel{\leftrightarrow}{D}$$ is a symmetrized gauge covariant derivative, $$(\mu\nu)$$ are symmetrized vector indices, and $$F^{\mu\nu}$$ is color field strength tensor. The Hamiltonian is $$H = \int d^3x\, T^{00}$$ and the proton mass is $$m_p = \langle p(\vec{k}=0) | H | p(\vec{k}=0)\rangle$$ where $$| p(\vec{k}=0)\rangle$$ is a proton state with zero momentum. I can now ask whether it is possible to decompose $$m_p$$ into contributions from various terms in the stress tensor, and whether these terms can be independently measured.

The answer is "yes", modulo some of the ambiguities mentioned above (the total mass of the proton is certainly well defined, but there may be ambiguities in individual terms that cancel in $$m_p$$). Experimental input on individual terms comes from deep inelastic scattering (which measures trace free matrix elements of the quark and gluon energy-momentum tensor), and pion-nucleon scattering (which measures quark mass contributions). Individual matrix elements can also be determined in lattice QCD.

There are many details, discussed in the papers cited above. A typical result is shown in this figure (taken from a report on a future electron ion collider) which shows fractional contributions from quarks and gluons (with quark mass contributions separated out) Postscript: Note that the proton is very different from a typical non-relativistic bound state. A non-relativistic bound state is made from some constituents with total rest mass energy $$E_0=m_1c^2+\ldots + m_Nc^2$$, and the statement that there is a bound state implies that $$E=E_0-B$$, where $$B$$ is biding is a (positive) binding energy. States of this type exist in the heavy quark sector of QCD (at least approximately), and are know as charmomium, bottomonium, etc.

However, the proton is not like that. It is made of approximately massless quarks and exactly massless gluons, but the proton mass is large, $$m_p >> 2m_u+m_d$$. There is no binding energy, the mass of the proton is positive energy of quark and gluon fields. Also, the proton cannot be ionized into quark constiutents. If I excite the proton, all I can do is generate additional quark-anti-quark pairs.

Second postscript: If you are really interested in an estimate, there are models of the nucleon (not quite QCD, but QCD inspired) in which one can talk about binding energy. For example, in the constituent quark model, quarks acquire an effective mass $$m_Q\sim 400$$ MeV from chiral symmetry breaking. Constituent quarks interact (by one gluon exchange, string potentials, or instanton induced fores) to make the proton, with a binding energy $$3\times 400 - 935 \sim 265$$ MeV.

Also, one can attempt to take the gluon field energy and split into an electric part ("kinetic energy") and a magnetic part ("potential energy"). This separation depends on the scale, but the magnetic part does indeed come out negative. A simple analysis (using numbers from the paper cited above) gives $$\frac{1}{2}\langle p| E^2 p\rangle \simeq 850\, {\rm MeV}\quad\quad \frac{1}{2}\langle p| B^2 p\rangle \simeq -525\, {\rm MeV}$$ A similar separation cannot be done for quarks, because there is no gauge invariant way to decompose the covariant derivative $$\bar\psi \gamma\cdot D\psi$$ into a kinetic and an interaction term.

• In my research since asking this question, I've recognized the complexity involved in even asking for a semi-classical approximation, like the Yukawa potential is for the residual Strong force, however of the mass-energy breakdowns I've seen, none have recognized any contribution of a negative potential energy. My guess is that the gluon energy is acting as a net gluon energy, with some negative and positive terms mixed in. If we look at the potential for Chromium, there is a clearly negative component of the potential between the quarks, even if overall the gluon energy is positive. Oct 25, 2021 at 1:08
• I have not been able to find an actual example of this Strong potential with numbers plugged in for the Proton, and I guess this is what I would most like an answer to. My main question really is what is the magnitude in MeV of the negative contribution to the Proton's mass? Even a naive calculation that doesn't take into account the full QCD calculation would be appreciated, I just need an estimate. Oct 25, 2021 at 1:11
• @AnthonyKhodanian added a postscript Oct 25, 2021 at 4:30
• I see your postscript, but surely some portion of the mass-energy is negative. To say otherwise is to admit to their being no attractive forces within the hadron whatsoever, which is clearly not the case. The strong force between quarks still has a couloumb-like term, even if overall the positive terms like color confinement and gluon effects dominate the effective potential at all length scales. At this point just an order of magnitude estimate of what the contribution of that term is to the overal mass-energy would satisfy me. Oct 25, 2021 at 7:16
• And anyway, aren't there just as easily relativistic bound states? I believe the virial theorem in that case states the potential energy is equal and opposite in magnitude to the kinetic energy. Obviously that would be different for the strong potential case, but maybe its still roughly on the same order? Could be that the combined gluon energy nets to 600 or so MeV, but theres a portion from the coulomb-like term thats -300 MeV (to counter the +300 MeV of quark kinetic energy), and the rest of the terms that's +900 MeV, or something of the sort? Or maybe its just a miniscule contribution. Oct 25, 2021 at 7:25

I took this image of the from the very instructive webpage of Matt Strassler. It shows impressive complexity. It illustrates that the sum of the rest masses of the naive quark composition uud, 9.4 MeV/c$$^2$$, is only 1% of the proton mass of 938 MeV/c$$^2$$. The rest of the proton mass, according to Strassler, is quark and gluon kinetic energy and interaction energy. In this sense 99% of the proton mass is kinetic and potential energy.

"if the up and down quark masses were zero (and everything else was left unchanged), the proton mass would barely change from what we observe it to be."

• The final sentence is not quite right. Yes, the sum $2m_u+m_d$ is only 10 MeV, but, as the right figure shows, there are more than three quarks in the proton. The value of the proton mass in the chiral (massless) limit can be determined on the lattice. Alternatively, the matrix element $\langle p|m_q\bar{q}q|p\rangle$ can be extracted from pion nucleon scattering, see arxiv.org/abs/1602.07688 . The contribution from light quarks is pretty well constrained, and equal to about 60 MeV. There is a comparable contribution from strange quarks, poorly constrained. Oct 26, 2021 at 2:21
• I believe you misunderstood the question. The breakdown of that 99% is exactly what I'm looking for. The question is not asking how much is potential and kinetic energy, it's asking how much is potential, and how much is kinetic. You basically just repeated the information already in the original question without adding anything new at all. Oct 26, 2021 at 3:39

9 % of proton mass is potential and 91 % is kinetic energy, see for more details article.

Original publication