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This may seem like a silly question, but I believe this to be very fundamental because the Standard Model of particle physics seems based on the axiom or assumption that neutrons and protons exist “as-is” inside atomic nuclei.

Why else would the standard model require a strong nuclear force to hold everything together?

Surely there must be more evidence that this is the case besides the fact that neutrons and protons appear when a nucleus is smashed ?

EDIT: It has been a long time since I asked this question, and looking at it now (dec 5 2017) it seems like I have not mentioned an important reason for asking this question. In any case, this is what I want to add to the question now:

Take for example the Helium nucleus which is postulated to consist four separate baryons that need to be kept together with the strong force in the standard model. I would expect that in that case the total mass of a Helium nucleus would be at least that of the 4 individual baryons added together, and then I would expect to have to add more mass because of binding energy of the strong force.

Instead, the mass of the Helium nucleus is less than the four individual baryons combined. Isn’t that evidence that the Helium nucleus cannot consist of four separate “as-is” baryons?

And if that is the case, what is the evidence that these, what I would call “reduced”, baryons still require a strong force to be kept together? I mean, these baryons have lost some mass in the process of fusing together in a Helium nucleus which means they have changed somehow. Then I wonder, what if this change also changes the repulsive forces between them into attractive forces for example while retaining all the other particle specific characteristics? Would that not be a more elegant explanation than a strong nuclear force?

I mean, it would not change a thing in the released energy levels when fusing two protons and two neutrons together. The only thing that changes is the model. A model that seems just as compatible with the data as the model with a strong nuclear force.

Leon Sprenger
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  • All fusion and fission that occurs is only explainable if atoms consist of protons and neutrons, and you can indeed have nuclei "capture" neutrons to form a heavier isotope. – ACuriousMind Mar 18 '15 at 14:41
  • Probably corrections on atomic spectra should come with the presence of neutrons and protons. There are measurements of the effect of the proton form factor to the alpha ray of the hydrogen, where they estimate the charge radius of the proton. – Hydro Guy Mar 18 '15 at 14:41
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    @DroneScientist you're right, that statement isn't true at all - actually neutrons and protons are consequences of the standard model, not axioms of it - but the essence of the question is still valid. – David Z Mar 18 '15 at 14:50
  • Well, some properties of the nucleons are quite different in the nucleus, like how the neutron is stable. I don't know enough nuclear physics to definitively answer the question, though I don't know what sort of answer the questioner is looking for either. – zeldredge Mar 18 '15 at 14:51
  • To echo @SirElderberry The form-factors of bound nucleons differ from those of their free counterparts. There exist a host of different parameterizations of this for use in Monte Carlos. An early one that still see a lot of use is that by de Forrest. – dmckee --- ex-moderator kitten Mar 18 '15 at 16:46
  • Related: http://physics.stackexchange.com/questions/138941/how-do-we-know-that-nuclear-physics-corresponds-to-low-energy-qcd – dmckee --- ex-moderator kitten Mar 18 '15 at 18:20
  • Neutrons will only be stable in certain configurations of the nucleus though, by themselves they are unstable with a mean lifetime approximately 15 minutes where they decay – Triatticus Mar 22 '15 at 18:08
  • @Drone Scientist: My reasoning was: The strong nuclear force is theorized to explain why the nucleons are held together. And one only needs to keep things together that are actually there. So apparently the (hidden?) assumption seems to be that nucleons exist inside a nucleus.Because the solution to theorize a strong force to solve the problems you get when trying to keep these nucleons together “feels” awkward and clumsy to me I was curious what the explanation is why this is “certain”. – Leon Sprenger Mar 23 '15 at 16:33
  • @ Drone Scientist That is not the complete story. The strong force is also needed to counter the repulsive force between protons. In any case: for me the question “Evidence that nucleons contain quarks?” is equivalent to my question above. – Leon Sprenger Mar 26 '15 at 10:25
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    RE: Your added question - the mass of a helium nucleus is less than the sum of the masses of two neutrons and two protons because of the addition of the (negative) binding energy which holds them together. Any bound state, whether it be electromagnetic, gravitational, or nuclear, exhibits this property. – J. Murray Dec 06 '17 at 00:39
  • Negative binding energy. Interesting. That would indeed be the only possible consequence. I know I have not right to speak about these things because I am layperson in this field, but this feels very artificial to me. Wouldn’t a more obvious model be to assume that the baryons lose some mass, and that this loss of mass changes some baryon properties that allow them to stick together naturally without a force? An analogy that comes to mind is how soap bubble stick together in a more efficient configuration with less total “surface” area than separate? – Leon Sprenger Dec 07 '17 at 23:41

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Evidence that there are distinct protons and neutrons in nuclei starts with the Pauli term (pairing term) in the semiempirical mass formula of the liquid drop model.

Furthermore, all nuclei with even numbers of protons and neutrons have nuclear spin of zero. This is consisent with shells being filled with spin up and spin down pairs of nucleons, each pair resulting in net zero spin.

More generally, that experimental data are consistent with the Nuclear Shell Model is evidence that distinct protons and neutrons exist in the nucleus.

Also, the protons and neutrons are held together by exchange of pions. The exchange can result in the proton becoming a neutron and a neutron becoming a proton, so it is not that they exist entirely "as is".

See A reappraisal of the mechanism of pion exchange and its implications for the teaching of particle physics for furthur discussion of pion exchange.

DavePhD
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Short answer: We can measure their energy and momentum distribution functions in the nucleus.

We do this by interacting with them individually, either knocking them out of a nucleus left otherwise undisturbed (quasi-elastic scattering) or by exciting them to higher energy states inside the nucleus (many inelastic scattering reaction backed up by data from various capture and production reactions).

Quasi-elastic Scattering

The quasi-elastic scattering route is a reaction I know well because I did my dissertation work on Color Transparency using $A(e,e'p)$ as the probe. A well characterized electron beam is scattered from a fixed, nuclear target and the products measured with two spectrometers positioned and tuned to detect the scattered electron and proton in elastic kinematics (i.e. as if the target had been $^1\mathrm{H}$ rather than a nucleus) to within Fermi-momentum. The only thing that is difficult about the measurement is how small the cross-section gets as the squared momentum transfer $Q^2$ grows.

The measurement gives us a picture of the energy and momentum distribution of the protons inside the nucleus, and for small $A$ these results are quite consistent with results of mean-field computations (which agree with the abstract shell-model for nuclear structure). For larger $A$ they remain qualitatively consistent but the precision of agreement drops a bit.

I want to emphasize that though the reaction here is a knock-out reaction, the thing we're measuring is the energy and momentum that the knocked out nucleon had inside the nucleus.

Excitation

By measuring the energies of the gamma rays released by excited nuclei and the momentum transfer to particles used to excite them we get another probe of the internal structure of the nuclei and this probe is likewise consistent with the shell model. A data set I've done some reading about here concerns the energy levels of $^{17}\mathrm{O}$, which can be probed in situ, by various inelastic collisions and by creation of short-lived, highly excited states in the reaction $^{13}\mathrm{C}(\alpha,n)^{16}\mathrm{O}$.

Here we are measuring the difference in energies between different states occupied by single nucleons. Again, the energy the nucleon had in the nucleus.

Community
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dmckee --- ex-moderator kitten
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  • It looks to me like you are demonstrating that you can make a collection of assumptions including the assumption that a nucleon you knock out of the nucleus had its independent existence in the nucleus before, and you get experimental results which do not contradict those assumptions. Have you instead proven that no other assumptions can be compatible with your results? – J Thomas Jul 01 '22 at 11:40