# Are some protons more massive than others due to spinning at a different rate or being in a more excited state?

The old Usenet Physics FAQ, which is a page originally by Matt Austern, says that

all protons have identical rest masses, and so do all electrons, and so do all neutrons; these masses can be looked up in a table. As the particle is accelerated to ever higher speeds, its relativistic mass increases without limit.

However, writing in 2011 in the “Chemistry” section of PNAS, Hammes-Schiffer says:

This work calls into question the traditional definition of proton transfer as the movement of a hydrogen nucleus between a donor and acceptor atom. According to this perspective, a proton transfer reaction is defined in terms of the change in electronic configuration rather than the movement of the hydrogen nucleus. In this case, a proton transfer reaction is defined as the movement of electronic charge density from the donor–hydrogen covalent bond to the acceptor–hydrogen covalent bond. Such a change in electronic configuration without the motion of the hydrogen nucleus is special to optically induced processes, where the optical excitation is much faster than nuclear motion. The proton is in a highly excited vibrational state immediately after optical excitation and subsequently relaxes to its equilibrium state."

If a proton can be in excited state or not, it can have more mass or not, I would have thought, which would contradict the widely held belief that all protons have the same rest mass.

• All protons have the same spin. However, there is similar particle, analogous to the proton, with higher spin (and mass) called the $\Delta^+$. Dec 5, 2021 at 19:59
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– Buzz
Dec 9, 2021 at 2:53

No. Protons are indistinguishable particles; they all have the same mass, exactly.

We like to tell beginning students that the proton is “made of three quarks,” analogous to the way a hydrogen atom is “made of a proton and an electron.” But that analogy is poor. The quarks in a proton interact via the gluon field (or “by exchanging gluons,” if you like the particle picture). But the gluons also interact with each other, and these interactions produce an “ocean” of virtual quark-antiquark pairs. Also the quarks have electrical charge, which interact via the photon field. The weak interaction is lurking in there as well. The inside of a proton is a hot mess.

But it is exactly the same hot mess, inside of every proton. That’s what indistinguishability means.

If you hit a proton hard enough to store additional energy in the strong-interaction fields, you get a different sort of baryon. Excited baryons which come in two charge states (like the proton and neutron, collectively “the nucleon”) are called $$N$$ baryons. Excited baryons which come in four charge states are called $$\Delta$$ baryons. You can think of a $$\Delta$$ baryon as being “made of” a nucleon bound to a pion —— but like “made of quarks,” that model understates the complexity of the system.

Your second quote is talking about proton exchange between molecules, in the context of hydrogen bonding. The reference to a proton in an “excited vibrational state” refers to the proton vibrating with respect to the recipient molecule.

Compare this language with discussions of the excited states of the hydrogen atom, where people say things like “the electron absorbs a photon and enters an excited state.” This isn’t technically correct, either. In an excited atom, the electron wavefunction is different from a ground-state atom. In the approximation that the nucleus is infinitely more massive than the electron, and in the approximation that there’s only one electron whose state changes, it’s not too wrong to say that the electron’s dynamics have changed and the nuclear dynamics have not. But there is no such thing as an excited electron: all electrons are indistinguishable. For that matter, all ground-state hydrogen atoms are indistinguishable. Indistinguishability plays a key role in quantum mechanics.

Note that an excited molecule is more massive than a ground-state molecule. But not by much.

• Only one highly pedantic nit: All protons have the same rest mass. Dec 7, 2021 at 15:31
• I prefer the pair of concepts “invariant mass” and “relativistic energy” over the outdated “rest mass” and “relativistic mass.” When I write about mass, the invariant mass is always what I mean (which I suppose is a pun), so sometimes I forget to specify.
– rob
Dec 7, 2021 at 19:21

The word proton is used differently in these two quotes.

Physicists use proton only for the three-quark baryon itself. Chemists sometimes use the term proton as a shorthand for a chemically bonded hydrogen group. That is the excited state to which the second quote refers, not the state of the proton internally. One way to tell is by its mention of "optical excitation," which means electron-bond energy levels.

Exciting a proton internally -- which again is not what the chemistry quote meant -- requires orders of magnitude more energy. As others have noted, the result is so different from a proton that it has its own name, the $$\Delta^+$$. The $$\Delta^+$$ is unstable, more massive, and has spin $$\frac{3}{2}$$ versus the proton's spin $$\frac{1}{2}$$.

Addendum 2021-12-07.12:16 EST Tue: This is only for folks interested in the in-depth particle physics aspects of this question.

While this is more a matter of opinion in definitions, I would say that the resonance (which means "a short-lived particle") that best qualifies as an excited proton is not $$\Delta^+$$ but the one-positive-electric-charge $$uud$$ $$N(1680)({\frac{5}{2}}^+)$$ resonance. So little emphasis is placed on this resonance that I could not find a standard notation to distinguish between the neutron-like no-electric-charge $$udd$$ version and the proton-like one-positive-electric $$uud$$ variants of this resonance. The plus sign at the end of its designation means "even parity," not a positive charge. That accidental pre-emption of the plus sign likely makes adding an electric-charge "$$+$$" a bit tricky, so perhaps folks just don't bother?

The $$uud$$ $$N(1680)({\frac{5}{2}}^+)$$ resonance is more like a proton because two of its three valence (non-virtual) quarks spin in the same direction while the other has reverse spin, just as in a proton. In sharp contrast, all four $$\Delta$$ baryons (plus their Regge trajectory excitations; if you don't immediately get a page image on that link, try refreshing the page) have valence quark spins in the same direction. The Regge trajectory resonances add spin in units of $$+2$$ to each of these configurations using orbital momentum. If you think of this added orbital momentum of the quarks as the angular momentum of the quarks swinging around each other at higher speeds, bola style, that's actually a pretty good heuristic model for comprehending what's going on.

You can use that mental model because the strong force behaves remarkably similar to a bungee cord, increasing the force of its binding as it gets stretched. That's why protons and neutrons have well-defined surfaces. It also means that, if someone could measure it (not easy!), the higher Regge excitations should all have larger diameters than protons and neutrons. I have no idea if that has ever been tried.

(On a related note: The critical need for a force with a bungee-cord-like behavior to get string-like vibrations is also why what's now called "string theory" (it was at first called superstring theory) should have died after no more than a handful of papers. String theory unavoidably assumes that some variant of gravity is the binding force, and no variant of gravity shows the bungee-cord-like containment needed to make the hypothesized string vibrations physically meaningful. It's really sad that superstring theory didn't get wrapped up quickly for just that reason, since instead, it went on for half a century, eating up enormous funding and wasting research careers on unproductive speculations that predicted nothing.)

Adding orbital angular momentum arguably changes the character of the nucleon far less than flipping one of the valence quark spins relative to the other two valence quarks since the latter operation profoundly changes the nature of the nucleon and enables entirely new (and quite weird) particles such as $$\Delta^{++}$$.

Terminology note: My thanks to user Christoffer for replacing my "canonical quarks" phrase with "valence quarks," a phrase I did not know before. Standard terms are always better!

• Note that, while the lowest-mass $\Delta$ baryon has spin 3/2, there are spin-half delta baryons. The difference between a $\Delta$ baryon and an $N$ baryon is the strong isospin.
– rob
Dec 6, 2021 at 5:31
• +1 and accepted because it is at the level of this layman. Like other laymen, I need jargon to be clarified first. But what exactly is "a chemically bonded hydrogen group" and why would it be referred to by chemists as a "proton"? Dec 6, 2021 at 10:32
• By "a chemically bonded hydrogen group" I mean a hydrogen atom attached directly to, say, a carbon, nitrogen, or oxygen atom in some larger molecule. A ball-and-stick model of methane has one central carbon atom and four hydrogens sticking out in a tetrahedral shape, and any of those hydrogens would be an example. Atoms such as oxygen that attract bond electrons more than hydrogen tend to leave the proton at the center of the hydrogen atom more exposed and inclined to drift off. Such proton donors are better known as acids. The more easily the proton is detached, the stronger the acid is. Dec 6, 2021 at 12:36
• @MatthewChristopherBartsh A Hydrogen atom is an electron and a proton. In chemical bonds, electrons are shared between atoms. When chemical reactions happen, the nucleus of a hydrogen frequently leaves the molecule, but "its" electron is left behind. So when you do the bookkeeping, instead of transferring a full hydrogen atom (electron+proton) you've just transferred a proton. -- Contrast this with "hydride transfer", where a proton and two electrons are transferred, or a "hydrogen radical" where one proton and one electron are transferred. (Chemists think in terms of filled electron shells.)
– R.M.
Dec 6, 2021 at 12:40
• @TerryBollinger Do you mean only the ball and stick model is tetrahedral, and the actual molecule isn't? Why isn't water a proton donor, then? Dec 6, 2021 at 13:27

Protons have no excited states merely by definition. The $$uud$$ baryon does have higher energy states, but the word "proton" is normally only used for the ground state.

In contrast, excited states of, say, a hydrogen atom are called hydrogen. There isn't any fundamental difference between the two cases. The energy levels of hydrogen are much more closely spaced, and therefore more similar to each other, but that's a difference of degree, not kind.

• I don't agree with this point of view; any uud with isospin 1/2 may be called a proton regardless of excitation. That we call it by its other name N^+ (or just N) is simply an indication that we usually don't care about the charge of such particles. (e.g. arxiv.org/pdf/nucl-ex/0210006.pdf, arxiv.org/abs/1109.1294) Dec 6, 2021 at 14:55