# Simple Mass Calculation

Hadrons, particles composed of quarks, can be sorted into two different categories: Baryons and Mesons. The proton and the neutron are baryons.

According to the Standard Model, baryons contain three quarks, and no other massive particles. Comay’s model assumes that baryons have a massive core and that the three observable quarks occupy outer energy shell.
With regard to mesons, both models agree that they are composed of quark-antiquark pair and contain no other massive particles. It is further agreed that according to Field Theory, there is a probability of finding additional quark-antiquark pairs inside the baryons and the mesons.

Useful Energy Notions
Electrons within the atom are bound to the nucleus, and the Hadrons are bound states of quarks. In Physics, three different notions relate to the energy of these states:

A bound state can only occur if “attraction forces” apply between the system’s components. The energy derived from theses forces is called “Potential energy”. The potential energy of a bound system has a negative value.

In physics it is known that a bound particle cannot be at rest. For example, Earth can’t be at rest within the sun’s gravitational field. (Quantum Theory considerations lead to the same conclusion for electrons in atoms and for quarks in Hadrons). Bound particles therefore have energy as a consequence of their movement. This energy is called Kinetic Energy. Kinetic Energy is always positive.

One other type of energy is Binding Energy. The system’s total energy, which is the sum of the (positive) kinetic energy and the (negative) potential energy, indicates the bond strength between the system’s components. The difference between the kinetic and the potential energies is negative for a bound state, and its absolute value is called “Binding Energy”

What do we know about the properties of particles in systems subject to electromagnetic interactions?
According to Comay’s model, there is an analogy between the baryonic structure and the electromagnetic attraction of electrons to the atomic nucleus, and between mesons and positronium, which is the electron- positron (anti-electron) bound state. According to this model every quark carries a unit of negative magnetic monopole charge, and the baryonic core carries 3 units of positive magnetic monopole charge.

Calculating the actual quantum states under Strong Interaction is an extremely difficult task and goes way beyond the scope of this article. We’ll therefore use simple arithmetics to examine the analogy suggested by Comay’s Model between atomic systems and the positronium on the one hand, and the baryon and meson systems on the other hand.

With regard to systems with electrons, we know that:
– The binding energy of an electron to a proton (hydrogen atom) is less than half of the binding energy of two electrons bound to two protons (the helium atom). This means that the electron’s bond to the helium atom is stronger than its bond to the hydrogen atom.
– The radius of the electron bound to the hydrogen atom is larger than that of any of the two electrons bound to the two protons of the helium nucleus.
– “Heavier electrons” (for example muons) bound to the proton have a smaller radius and a higher binding energy.
– A multi-electron atom contains several shells and the radius increases accordingly.

So far, the masses of many baryons and mesons have been successfully measured, and a few radii of baryons and mesons were measured as well. In this article we’ll try to judge which of the two models better corresponds to the experimental outcomes.

What determines a particle’s mass
The total of the masses of the nuclear building blocks, the neutrons and protons, usually provides a good approximation of the nuclear mass. By definition, the nuclear mass is equal to the sum of its constituents’ masses minus a mass equivalent to the binding energy between them. This is because it takes energy to separate 2 bound particles. The binding energy of protons and neutrons inside atomic nuclei amounts to less than 1% of their global mass, and one can therefore consider that the mass of protons and neutrons in the nucleus is very similar to that of free protons and neutrons.

The situation in hadrons is much more complex, because their binding energy is much higher, due to the Strong Interactions. Theoretically, separating the quarks from each other would require a huge amount of energy. Since energy is equivalent to mass, adding such a significant amount of energy to the system means greatly increasing the quark’s mass. Therefore, if the quark could exist outside the hadron, its mass would have been much larger. In fact, the bound quarks’ mass would have been dozens of times heavier had they not been bound. This is very different from what we know about electric forces. The binding energy of quarks inside the hadrons is a major component of the composite’s particle mass, on top of the masses of the quarks composing it.

But if we overcome our reluctance to performing calculations, and try to see what happens, we may discover some interesting things. The simple calculations presented below confirm that fundamental quantities like mass and radius of baryons and mesons behave according to well known properties of electromagnetically bound systems and according to Comay’s model where baryons have a core carrying three units of magnetic charge.

Baryons compared to Mesons
Data tables specifying baryons’ and mesons’ masses can be found on the internet.
Let’s try to compare between the masses of baryons and mesons. Usually, a specific quark combination can yield more than one baryon or meson. Therefore, the most stable state of each quark combination, which is the lightest particle, is considered below. Let’s start by comparing a baryon – the proton, to a meson – the pi+ particle. The proton has a mass of 938 MeV (the mass-energy units) and the pi+ has a mass of 139 MeV. The proton contains 3 quarks: uud, and pi+ is composed of the quark u and the antiquark d. Everyone agrees that an antiparticle’s mass is identical to the particle’s mass, and only their charges are opposite to each other.
The mass of the u quark is close to that of the d quark. The mass of the neutron, for example, which contains udd quarks, is 939 MeV. I.e., the neutron, which only differs from the proton by the fact that in the neutron a d quark replaces one of the proton’s u quark, is heavier than the proton by about one thousandth of a nucleonic mass.

We also know that the proton’s radius is slightly larger than that of the pi+.

According to Comay’s Strong Interaction Model, the proton contains a complex core with 3 quarks on the external shell. It is therefore not surprising to find that the proton is heavier than the quark-antiquark state of the pi+. The fact that the proton has closed shells implies that the 3 external quarks are not located in the most inner shell. Therefore, in analogy to an atom with several shells, it makes sense that the proton’s radius is slightly larger than that of the pi+.

Let’s consider another baryon and meson pair. The Sigma+ baryon is composed of the quarks uus and its mass is 1,189 MeV. The K+ meson, composed of a u-quark and an anti-s quark, has a mass of 494 MeV. What does this mean? It turns out that every time an s-quark replaces a u or d quark, the resulting particle is heavier. Therefore it is admitted that the s-quark is heavier than the u and d quarks.

It turns out that the radius of the K+ meson is smaller than that of the pi+. The radius of the baryon Sigma+ has not yet been measured. But the radius of an equivalent baryon, containing the quarks d,d,s, and called Sigma-, is known. We see that the radius of each of the particles mentioned above and containing the heavier s-quark is smaller than the radius of the corresponding particle not containing the s-quark. This is coherent with the rule stating that in an electromagnetic system, a heavier particle is more strongly bound (and is characterized by a smaller radius). It is also coherent with Comay’s model.

Another point is that replacing the d-quark by an s-quark increases the baryon’s mass by 251 Mev (from 938 to 1189), and the meson’s mass by 355 MeV (from 139 to 494). This means that exchanging the d-quark by an s-quark increases the proton’s mass to a much lesser extent than the meson mass. And indeed, just as electrons are more intensely bound to the helium atom than to the hydrogen atom, the heavier s-quark inside the baryon interacts more strongly with the inner core carrying 3 units of magnetic monopole charge, than in the meson, where it is attracted only to one quark carrying only 1 unit of magnetic monopole charge. And since a stronger interaction increases the binding energy and decreases the particle’s mass, this result as well is coherent with general laws of electromagnetically bound systems and with Comay’s model. Mass value and mass difference of two baryons and two mesons (see text).

Is the relative mass discrepancy coincidental?
It turns out that it isn’t. And indeed, the comparison to other Baryon – meson pairs containing heavier quarks, such as Charm Sigma composed of the uuc quarks, compared to D-meson composed of a u-quark and an anti-c quark, or Bottom-Sigma composed of uub quarks compared to B-meson composed of a u and an anti-b quarks, reveals exactly the same phenomenon! Any comparison between a baryon and its corresponding meson shows that the same quark is more intensely bound to the baryon than to the corresponding meson. That is, here too, binding energy is coherent with Comay’s model!

In the precedent paragraph we used simple considerations to try and understand a complex problem. Other parameters play non negligible roles in determining a particle’s mass, such as the probabilities of configurations containing quark-antiquark pairs (it has been shown that such additional pairs exist in baryons and are supposed to exist in mesons as well.) But most of all, a high precision calculation of a system characterized by the Strong Interactions is a very complicated problem. Here we made a qualitative calculation taking into account higher or lower interaction intensities, and similarly, larger or smaller particle radii. Such qualitative considerations provide an interesting approximation of the real situation.

Is the Standard Model coherent with experimental findings?
Standard Model fans may argue, and not without reason, that their laws relative to the Strong Interaction are very different than the electrodynamic laws of the electrons in atoms. But the insistence of the gap between baryons and mesons, and the way masses and radii correspond both to what we generally know about systems bound by electromagnetic forces and to Comay’s model, should ring yet another bell with regard to the validity of the Standard Model.

For example, a well-known feature of QCD, called “Asymptotic Freedom”, states that the Strong Interaction’s intensity increases with the distance between the particles. This is in opposition to what we know about electromagnetism, and it is not clear whether or not it is coherent with the experimental findings relative to the K meson radius being smaller than the pi meson, and that of the Sigma- baryon being smaller than the proton.

Furthermore, given the pi-meson’s very small mass in comparison to the proton, QCD fans should demonstrate that the attraction force of the quark anti-quark pair constituting the pi-meson is significantly stronger than that of the 3 quarks inside the proton. Having resolved this paradox, they would further need to explain why this ratio is inverted for the s-quark, and why the s-quark’s meson bond seems weaker than its baryonic bond.

## 8 thoughts on “Simple Mass Calculation”

1. Rod says:

I understand your father’s conjecture and I believe I understand his reasoning but I admire his attempt to “think outside the box”. I suggest however that he has only stuck his toe out to test the water and I guess is unwilling at this point to actually take the plunge.

Thank you for your kind response.

Good luck and best regards.

2. Rod says:

In the graphic illustration above, the horizontal, mathematical comparison of the Baryons suggested that the down quarks have negative mass and the same is true of the strange quarks in the horizontal comparison of the mesons.

On the other hand, the vertical mathematical comparisons produce inequalities.

I thought this was supposed to be Quantum Mechanics, the study of a universe constructed of quantum units of mass and charge. How can you justify this?

• Hi Rod. Why do you say that the comparison above suggests that down and strange quarks have negative mass?

• Rod says:

Hi Ofer. I refer to the basic algebraic evaluation of the above example.

If quarks are supposed to be elementary particles then theoretically at the very least, quarks of the same flavour should all have the same mass. Therefore in the above example:

The proton is represented by; 2u + 1d = 938 MeV
The pi Meson is represented by; 1u + 1d = 139 Mev .
If we solve for d in the latter we get; 1d = -1u + 139 MeV
Substituting for d in the first equation we get;
2u -1u + 139 Mev = 938 MeV or 1u = 938 – 139 MeV = 799 MeV.
Finally, substituting the value for u into the first equation gives us;
(2*799) + d = 938 MeV .
Solving for d gives us; d = 938 – 1598 MeV or – 660 MeV

The same process yields a value of -201 MeV for the strange quark.

Obviously nothing can have negative mass so your father is right there must be something else present.

Your father is on the right track. He has had the courage to admit that there are more particles in a nucleon than are presently admitted by contemporary Physics theories. If he can find the courage to take the next step, he may discover a whole new universe, a far more elegant and beautiful universe, a truly divine universe. All we have seen so far in contemporary Physics is what men can imagine. God’s imagination on the other hand is far more elegant. Understand that I am not a religious man but the universe is trying to explain itself to us, we only need listen to what it is saying instead of trying to make it conform to our own preconceptions.

The mass of the quarks is very sensitive to their states due to special relativity. For example, the u,d quarks inside pi-meson is different than their mass inside the proton. The reason is that there are inner closed u,d quark shells inside the proton, and the three u,u,d quarks that we observe belong to the outer shell. Therefore, these quarks have higher energy (in the same way that electrons in outer shells have higher energy than electrons in inner electronic shell).

This higher energy is very significant and the mass of the u,d quarks is significantly higher when they are in the proton comparing to their mass in pi-mesons.

This is why the calculation that you did cannot work…

I cannot say anything about the beauty of the universe equations – but I hope you are right and there is a simple demonstration of the physical laws in our universe.

3. Curt says:

You say “By definition, the nuclear mass is equal to the sum of its constituents’ masses minus a mass equivalent to the binding energy between them. Then you say ” The binding energy of quarks inside the hadrons is a major component of the composite’s particle mass, on top of the masses of the quarks composing it. ”

So what now? Does binding energy reduce or increase the proton mass?

• Eliyahu Comay says:

For composite particles that are relatively stable, the binding energy is defined as a positive quantity. It always reduces the mass of the composite particle. Unlike atoms and nuclei, quarks cannot be ionized from the proton but the terminology is the same.

In atoms, the gap between the electronic ground state and their excited states is several eV, which is about 1/100000 of the electron’s mass. In nuclei, this quantity is of the order of magnitude of MeV, which is about 1/1000 of the nucleonic mass. On the other hand, there are excited states of nucleons whose energy is greater that of 2 protons (see http://pdg.lbl.gov/2010/tables/rpp2010-qtab-baryons.pdf).

Hence, if one assumes that free quarks have a mass then a large portion of this mass is “swallowed” by their binding energy in hadrons. The second statement quoted in your comment describes this situation.