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There is a recent paper in PRL, also on the ArXiv that claims the discovery of a fifth force, mediated by a 17 MeV boson, which explains an anomaly in nuclear decays.

Is this a real effect? Could the claimed particle be some sort of composite particle such as a pion?

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    $\begingroup$ I would stop the horses right at "Recently, studies of decays of an excited state of 8Be to its ground state..." and "The discrepancy from expectations may be explained by as-yet-unidentified nuclear reactions or experimental effects...". My money is on one of those, not on the fifth force, especially if you start looking at where the data comes from... no offense to the folks who did this, but I would want, at the very least, a second independent experiment to confirm this. $\endgroup$ – CuriousOne May 27 '16 at 3:02
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    $\begingroup$ nature.com/news/… $\endgroup$ – user83548 May 27 '16 at 3:16
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    $\begingroup$ @brucesmitherson: No offense, but a six page arxiv preprint wasn't quite what they used to do in the past to test a new hypothesis against all previous data. That used to take years and involve a significant fraction of the entire community. The more I am looking at this, the shadier it becomes. $\endgroup$ – CuriousOne May 27 '16 at 3:40
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    $\begingroup$ Basically, such a thing could be possible. I wouldn't give it too much emphasis till this is cross-checked e.g. by direct e+e- collisions, which should give a small bump at the correct energy. There is naturally the question of where would this new boson get its mass from. One could guess that if other SM particles are millicharged under this new symmetry, the Higgs would also have a non-zero charge (should be micro-charged to get to the right ballpark) and that would explain the smallness of the mass. $\endgroup$ – Janus Boffin May 27 '16 at 7:52
  • $\begingroup$ @Rodriguez, read the paper. They claimed it is a boson. $\endgroup$ – Janus Boffin May 27 '16 at 7:54
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I wouldn't say that either of the linked papers "claims the discovery of a fifth force." What we have here is an subtle experimental observation with a very interesting interpretation that can't yet be ruled out, plus a path forward to search in data from upcoming experiments.

The reaction chain of interest in the PRL by Krasznahorkay et al. is \begin{align} \rm^7Li + p &\to{} \rm^8Be^* \\ \rm^8Be^* &\to{} \rm^8Be + \gamma \to{} ^8Be + e^+e^- \\ \rm^8Be &\to 2\alpha \end{align} The gamma rays in the second step are unusually energetic, about 18 MeV. There is some small chance that the gamma ray emitted in the relaxation of the excited nucleus $\rm^8Be^*$ will interact with the electric field around that nucleus to produce an electron-positron pair. These lepton pairs are overwhelmingly likely to leave the nucleus traveling nearly parallel to each other, so that the angle between their momenta is nearly zero. For the $\rm e^+e^-$ pair to be emitted back-to-back, this interaction between the relaxation photon and the nuclear electric field would have to take place in the rest frame of the nucleus. The ratio of pairs separated by 40° to pairs separated by 180° is about a factor of twenty.

The subtlety is that, for a narrow range of proton energies, there are perhaps 30% extra $\rm e^+e^-$ pairs separated by about 140°. It's not impossible this is some acceptance effect in the detectors, since the detector geometry has a feature at 150°. However one would naïvely expect the same sorts of acceptance issues at the other special angles in the detector, at 60°, 90°, and 120°; those are absent. Further arguing against a geometrical detector anomaly, the excess is present at some proton energies but absent at lower and higher energies. However a model where second line in the reaction chain is \begin{align} \rm^8Be^* &\to{} \rm^8Be + \mathit X \to{} ^8Be + e^+e^- \end{align} has a most probable $\rm e^+e^-$ angle which depends on the mass of the $X$. The authors find a best fit to their data a small contribution from $X$ decays for $m_X \approx 17\rm\,MeV$. This is also in the right range for resonant production of the $X$, which would explain why the extra angular correlations are only present at some energies. The branching ratio for $X$ production is quite small.

Krasznahorkay et al. also cite previous observations of this phenomenon, anomalous angular distributions in internal pair production from high-energy nuclear gamma decays, from another group in the era 1996--2001, so the phenomenon is not a bolt from the blue. The abstracts for these older papers suggest a quite different mass for the putative particle; I haven't read to see what is the difference.

A follow-up paper by Feng et al. (now also in PRL) compares this putative particle with known, constrained, and unconstrained particles. Since production of the $X$ seems to proceed from one excited state in $\rm^8Be$ but not another, we know something about its quantum numbers. It can't be a pion, since it's too light by a factor of ten. And it's highly unlikely to be any hadron that's composed of quarks, since the pion is the "massless" Goldstone boson of QCD and the pion mass sets the floor of the strong interaction's mass spectrum. Feng et al. compute the ranges of couplings of the $X$ to electrons, neutrinos, up and down quarks, and nucleons which would permit the $X$ to contribute to $\rm^8Be$ decay but have kept it hidden from a list of about a dozen previous experiments. Feng et al. then take those coupling ranges and claim, in their final figure, that the $X$ should be within the sensitivity ranges of at least five upcoming experiments (looks like the Heavy Photon Search would miss it).

Both of these are soundly reasoned papers which survived peer review and have appeared in the flagship journal of the American Physical Society. There is repeatable (possibly already repeated) evidence for an anomalous result in one system which is consistent with a new force-carrying particle, and theoretical guidance on how this particle would look in other, completely different experiments which have not been completed yet. If we were going to find a new fundamental interaction this is exactly how the early stages would look and it's absolutely worth pursuing further. All those positive comments in mind, I give it a 90% chance that it evaporates under further scrutiny; nothing resembles a new effect quite so much as a mistake.

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  • $\begingroup$ +1 particularly for the last phrase . I still remember a mu-pi 4 sigma resonance in a bubble chamber neutrino experiment back in 1977. It did not appear in other parallel experiments. Now I interpret it as a lack of "look elsewhere" effect to enlarge the statistical error. We were naive back then. $\endgroup$ – anna v Aug 17 '16 at 4:57
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Rob's answer is more or less complete for the stage of data we are aware of. New experiments are needed to see if a 17 MeV particle exists in e+e- , as the existing data in the particle data group stop at about 100 MeV.

I want to address the title question:

Is the “protophobic fifth force” observed in nuclear decays a new fundamental force?

the difference between "fundamental" force and force in general in particle interactions.

Force in the framework where quantum mechanics reigns can still be defined as dp/dt. Any two particle scattering will either transfer momentum or be elastic. A force is exchanged between initial state and final state measured particles if there is a momentum transfer. The crossections are calculated using the iconal representation of Feynman diagrams. These are integrals in a perturbative expansion series, each order dependent on coupling constants., in diminishing importance.

The standard model of particle physics , which describes successfully the enormous plethora of data gathered the last fifty or so years,depends on three fundamental forces. They are called fundamental because the lowest order diagrams, contributing the most to the crossections, are dependent on the corresponding couplings of electromagnetic forces, weak, or strong. These lowest order diagrams have as an exchanged particle a photon, a Z/W , a gluon corresponding to the force characterization of the interaction. These are identified with the gauge bosons in the SU(3)xSU(2)xU(1) of the standard model.

There is a hypothesis that the gravitational force, once quantized , will have the graviton as the exchanged particle, but this is still a research project.

The SM model is based on a plethora of data and it is doubtful that it can be upset by the introduction of a "fifth force" when just an exchange feynman diagram with this possible X being a complex quark antiquark particle might explain the interaction given an appropriate format. After all, all particles can be exchanged and transfer a dp/dt without introducing any fundamental forces. See my answer here.

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  • $\begingroup$ If I understand, force in this context becomes mainly a reference to the specific theory, like strong for QCD or weak/EM for the electroweak. $\endgroup$ – user46925 Aug 17 '16 at 10:33
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    $\begingroup$ @igael That is how we use the word "force" in particle physics, thus only talking of fundamental forces. But an exchange of an electron is also a force, the couplings are electromagnetic, but it is not "fundamental". An exchange creating this 17 MeV particle would be a "force", (they imply weak coupling constants) but it need not be fundamental,after all the nuclear forces have been modeled with pion exchanges and rho exchanges. This could just be one more not thought out complex particle exchange. $\endgroup$ – anna v Aug 17 '16 at 10:53
  • $\begingroup$ Nicely put. I would think, though, that a low-mass vector boson composed of quarks with milli-charged couplings to the standard model particles would require "new physics" as interesting as a new gauge boson. The PDG name for a particle with these quantum numbers (isoscalar, $J^\pi = 1^+$) seems to be $h_1$; it's too heavy. I'd expect some really magical destructive interference to be required to make all the matter couplings go away as in the Feng et al. paper. $\endgroup$ – rob Aug 17 '16 at 11:55
  • $\begingroup$ @rob it may be I am biased by my background, but I would not expect a new fundamental force to just be observable in a nuclear interaction. $\endgroup$ – anna v Aug 17 '16 at 12:07
  • $\begingroup$ @annav Nor is that the claim; see the conclusions to the Feng et al. paper. $\endgroup$ – rob Aug 17 '16 at 17:32

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