Why Is Octupole Deformation Measurable in Nuclei but Not Dipole Deformation?

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SUMMARY

The discussion centers on the measurable octupole deformation in atomic nuclei versus the absence of measurable dipole deformation. Participants reference the weak interaction's role in violating parity, which allows for octupole deformation to be observed in nuclei like Radium, while dipole deformation remains undetected. The conversation highlights the distinction between the shape of the nucleus and the charge distribution, emphasizing that the lack of observed dipole moments suggests significant implications for understanding nuclear structure and potential new physics.

PREREQUISITES
  • Understanding of nuclear deformation concepts, specifically octupole and dipole moments.
  • Familiarity with the weak interaction and its effects on parity violation.
  • Knowledge of multipole expansion in nuclear physics.
  • Basic principles of charge distribution within atomic nuclei.
NEXT STEPS
  • Research the implications of weak interaction on nuclear structure and parity violation.
  • Explore the experimental techniques used to measure octupole deformation in nuclei.
  • Study the theoretical frameworks surrounding electric dipole moments in atomic nuclei.
  • Investigate the significance of charge distribution shapes in determining nuclear properties.
USEFUL FOR

Physicists, nuclear researchers, and students interested in nuclear structure, particularly those focusing on the implications of deformation in atomic nuclei and the role of weak interactions in particle physics.

Malamala
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Hello! I don't know much about this, so maybe the answer to my questions follows directly from the math of it, but I was wondering if there is an answer providing more physics intuition to this, not just math: Why can a nucleus have an octupole deformation, as a ground state stationary state (https://www.nature.com/articles/nature12073), but no nucleus so far was found to have dipole deformation. I understand that both type of deformations would have to vanish in a stationary state if parity would not be violated. Given that parity is actually violated by the weak interaction (ignore beyond the SM physics for now), we expect to have (a small) octupole and dipole deformation in some nuclei (probably in all nuclei in principle, but for most of them it is too small to be measured). In all cases I encountered so far in physics, when one makes a multipole expansion, the higher the multipole the lower the given effect. So based on that logic I would expect that a dipole deformation to be bigger than an octupole one. Yet, the octupole one was measured, but no dipole one. Why is this the case? Why can the weak interaction lead to a measurable octupole deformation, but not to a dipole one? Thank you!
 
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You are confusing the shape of the nucleus with the shape of the charge distribution. U-238, for example, is known to be cigar-shaped even though as a spin-0 nucleus the "cigar" has no direction in which to point.
 
Vanadium 50 said:
You are confusing the shape of the nucleus with the shape of the charge distribution. U-238, for example, is known to be cigar-shaped even though as a spin-0 nucleus the "cigar" has no direction in which to point.
But then my questions would be, why didn't we see a dipole shaped charge distribution, if we saw an octupole shaped one?
 
You are confusing the shape of the nucleus with the shape of the charge distribution.
 
Vanadium 50 said:
You are confusing the shape of the nucleus with the shape of the charge distribution.
That could be the case (I don't know much about this), so any help is greatly appreciated. So in that paper, with the octupoled deformed Radium, is that octupole the shape of the nucleus or the charge distribution? And, regardless of which one it is, why didn't we see the same one (charge or nuclear shape, whichever that is), having a dipole? I read in many papers the claim that a nucleus having an electric dipole moment, means signs of new physics (at least given our current detectors). So I am a bit confused. Could you explain that a bit to me?
 
I am confused by your confusion. I have no idea what you are talking about.
 
Vanadium 50 said:
I am confused by your confusion. I have no idea what you are talking about.
Why can a nucleus have an electric octupole moment, but not an electric dipole moment?
 
Malamala said:
Why can a nucleus have an electric octupole moment, but not an electric dipole moment?

Why do you think it does?
 
Vanadium 50 said:
Why do you think it does?
Well if I knew I wouldn't ask here...
 
  • #10
Is actual negative charge needed to have a "electric dipole moment", or not needed? There is the negative charges inside nucleons, of course.
If a nucleus had protons concentrated in one end and neutrons in the other end, would the nucleus then have an electric dipole moment?
 
  • #11
snorkack said:
Is actual negative charge needed to have a "electric dipole moment", or not needed? There is the negative charges inside nucleons, of course.
If a nucleus had protons concentrated in one end and neutrons in the other end, would the nucleus then have an electric dipole moment?
You don't need positive or negative charge to have an electric dipole moment. It is all about the charge distribution. For example there are lots of searches for the electron electric dipole moment. The electron doesn't have any positive charge but some theory beyond the standard model predict it actually has an electric dipole moment if its charge distribution has a certain shape. Same for the nucleus. For a given shape, the nucleus could have an electric dipole, but none has ever been measured. However, the octupole deformation was measured. So my questions is why were we able to measure an octupole but not a dipole deformation, given that both of them (the operators corresponding to the dipole and octupole) violate parity?
 

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