# FeaturedA Possible explanation for muon g-2 anomaly: Gravity?

1. Feb 2, 2018

### Staff: Mentor

For many years, the measurements of the Landé g-factor of the muon have been puzzling, as the experimental value and the theoretical predictions showed some disagreement - 3.6 standard deviations for the last years. Experimental and theoretical uncertainties have a similar size, so work on both sides helps.
Muon g-2 at Fermilab is currently taking more data to improve the experimental result, while CERN is studying a muon-electron scattering experiment to improve some experimental values that go into the theory calculations. But a few days a completely new idea was made public.

The difference might come from the gravitational field of Earth. This sounds absurd - something you learn early as particle physicist is that gravity is completely negligible (unless you design detectors!). But we are talking about extremely precise measurements - parts per billion. Three authors investigated if gravity can have an effect - and found a possible contribution that is of the size of the observed discrepancy. Adding this to the theoretical prediction reduces the discrepancy from 3.6 standard deviations to 0.1 standard deviations. If the calculations are correct, theory and experiment are actually in excellent agreement.

The authors uploaded three papers to arXiv:
Theoretical framework
Effect on electrons
Effect on muons

The effect on electrons is very important: Our measurements there are a factor 1000 more precise, and they agree nicely. The authors calculate that gravity doesn't influence the electron measurements due to the different measurement principles - the measurements are done with slow electrons, while the muons are relativistic.

Other theorists are checking these calculations now. If they can confirm the results, the muon g-2 anomaly is gone. On the positive side, we get a direct influence of curved spacetime on particle physics measurements - an interesting effect to study in more detail.

Blog article covering the topic

Something I don't understand: If the effect scales with $1-\frac{3GM}{rc^2}$ for g as given in the abstract and the conclusion, then the Sun should have an effect a factor 14 larger. Yet it is not mentioned at all.

2. Feb 2, 2018

### Jimster41

Do you have any ideas as to why they didn't mention it or consider it? Is there an inertial frame argument based on the idea the sun would be affecting the experimental apparatus and the muon in a way that would almost completely obscure the sun's effect on the muon?

I have no idea, but you seem to be saying they didn't consider something totally obvious.

3. Feb 2, 2018

### websterling

It's the gravitational potential, not the gravitational field. Since the Earth and the apparatus are in orbit (free fall) around the Sun, the equivalence principle says the Sun's field has no effect on the local measurement. Unless, of course, I'm wrong

4. Feb 2, 2018

### ohwilleke

Mentor note: This post was originally in this thread and was moved to keep the discussion together.

The question in the original post is no longer hypothetical. With this latest theoretical value adjustment, the experimental value and the theoretical value are within one sigma of each other.

Post-Newtonian effects of Dirac particle in curved spacetime - I : magnetic moment in curved spacetime" (January 30, 2018).

Takahiro Morishima, Toshifumi Futamase, Hirohiko M. Shimizu, "Post-Newtonian effects of Dirac particle in curved spacetime - II : the electron g-2 in the Earth's gravity" (January 30, 2018).

Takahiro Morishima, Toshifumi Futamase, Hirohiko M. Shimizu, "Post-Newtonian effects of Dirac particle in curved spacetime - III : the muon g-2 in the Earth's gravity" (January 30, 2018).

Last edited by a moderator: Feb 6, 2018
5. Feb 2, 2018

### Staff: Mentor

Why would the gravitational potential of Earth (instead of the local acceleration) matter then?

6. Feb 2, 2018

### Haelfix

I haven't had the time to look at this much, but just glancing at the paper, they seem to be working in some sort of post newtonian (weak field) approximation using the Schwarschild background. If you included the sun, you wouldn't have the Schwarschild metric as a valid solution, and you would need to work numerically (not unlike what they do for binary star mergers). To even get some sort of to first order solution for the combined system would be an absolute mess, and I don't know what you could expand around to even attempt that calculation.

In any event, the suns contribution wouldn't scale like ~ 1 - Mg/r.

7. Feb 2, 2018

### Staff: Mentor

What is wrong with using the Sun with Schwarzschild metric?
MG/r is literally the expression they have in the papers. Multiplied by 3 it is the 2*10-9 discussed.

8. Feb 2, 2018

### PAllen

I've looked at the paper just a little, and I am also confused. If the effect were related to the christoffel symbols, that would make clear why the sun was irrelevant, but as mfb says, all expressions they end up with are based on GM/r (which, of course, is the deviation of metric from Minkowski). The role of the post newtonian approximation just seems to be to express the isotropic SC metric in an algebraically convenient form up to the precision they care about.

9. Feb 2, 2018

### Haelfix

The Schwarschild metric is a model for one large spherically symmetric system. You can't use it to model the composite effects of another large mass some distance away and then superpose solutions like it was a central force problem. The MG/r in the papers is referring to the Earths mass. I don't see how it would make sense to refer to the sun at all in this context. Maybe I am mistaken, do the papers talk about the effect of the sun at all explicitly?

10. Feb 2, 2018

### PAllen

No, that is the question. It appears the effect from the sun should be larger, but it is not mentioned. That is, if you ignored the earth and used the same method for the sun, you would get a larger anomaly.

Last edited: Feb 3, 2018
11. Feb 2, 2018

### Staff: Mentor

If the potential would be the important quantity, then they should take the Sun and neglect the Earth - to get an effect a factor 14 larger. They do not mention the Sun at all (I searched for the word).
If the local gravitational acceleration is important, then the Sun is negligible - but why doesn't it appear in the final expression (GM/r2 instead of GM/r) then?

12. Feb 3, 2018

### Arman777

this might be explain why,

'However those effective values in curved spacetime are respectively
different from those of values in the flat spacetime, the gravitational contribution is canceled
in the ratio Eq. (43) and the anomalous magnetic moment in the curved spacetime coincides
with the case in the flat spacetime."

(Page 11, third article)

They are assuming the magnetic moment of muon in the flat space-time when they considering the earth gravity (or solution based on earth's gravitational potential)

If we try to pass the sun we cannot use the magnetic moment of the muon in flat space-time since in the presence of the sun we should use the magnetic moment in the curved space-time value.

But as in the qoute says there will be cancellation. So they will be the same.

13. Feb 3, 2018

### Keith_McClary

My (dim) understanding is that the effect is not due to the potential or the acceleration/force, but to the curvature (or rate of change of the acceleration with distance). The latter would be small for the Sun's gravity.

14. Feb 3, 2018

### PAllen

But that doesn’t jive with the math they show. If they were talking about curvature in r t components, it would be 1/r2 effect, while in tangent components the curvature is 1/r.

The issue remains they really need to mention the sun and justify ignoring it.

15. Feb 3, 2018

### PAllen

I should also add that at the level of post Newtonian approximation you can easily consider the combination of sun and earth - JPL publishes PN equations combining the effects of all the major solar system bodies, for use in high precision ephemeris calculation.

16. Feb 3, 2018

### vanhees71

Because we are not freely falling around the Earth but sit on it, hold by the electromagnetic force and the Pauli principle of the electrons in the molecules of the matter around us and ourselves.

17. Feb 3, 2018

### Staff: Mentor

Where exactly is that considered? They just use the gravitational potential. Not the acceleration of the experiment or anything similar. These two do not have a unique relation.

18. Feb 3, 2018

### vanhees71

But just think about our daily experience: The effects of the gravity of the sun are completely negligible for nearly anything, the only exception are of course the tides of the ocean. To calculate the free fall of the apple hitting Newton's head inspiring him to discover his universal law of gravitation, you indeed don't need to consider the gravity of the sun.

19. Feb 3, 2018

### Arman777

I considered that too but again it comes to the point where PAllen said,

Or am I mistaken ?

20. Feb 3, 2018

### Arman777

I think we should consider the physical interpretation. We all know that muon is a subatomic particle. In this sense, considering the "size" of the muon and earth's gravtiational energy we can possibly define such relationship as described in the article ( magnetic moment of muon affected by earth's potential energy)

The Sun has larger mass yes but in this case, how can we define for a gravitational potential energy for such a small object in so large distance ? Mathematically we can put the distance between eart and the sun, but does that makes sense ?