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

In summary: 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.
  • #1
Malamala
299
27
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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  • #2
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.
 
  • #3
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?
 
  • #4
You are confusing the shape of the nucleus with the shape of the charge distribution.
 
  • #5
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?
 
  • #6
I am confused by your confusion. I have no idea what you are talking about.
 
  • #7
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?
 
  • #8
Malamala said:
Why can a nucleus have an electric octupole moment, but not an electric dipole moment?

Why do you think it does?
 
  • #9
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?
 

1. What is parity violation in nuclei?

Parity violation in nuclei refers to the phenomenon where the laws of physics do not exhibit symmetry under the transformation of a system's spatial coordinates. In other words, the behavior of particles in a nucleus is not the same when viewed in a mirror image.

2. How was parity violation in nuclei discovered?

Parity violation in nuclei was first observed in 1956 by physicists Chien-Shiung Wu, Tsung-Dao Lee, and Chen-Ning Yang in an experiment involving the decay of cobalt-60 atoms. They found that the electrons emitted in the decay were not evenly distributed in all directions, indicating a violation of parity symmetry.

3. What is the significance of parity violation in nuclei?

Parity violation in nuclei is significant because it provides evidence for the violation of a fundamental symmetry in the laws of physics. This discovery led to a better understanding of the weak nuclear force and played a crucial role in the development of the Standard Model of particle physics.

4. How is parity violation in nuclei studied?

Parity violation in nuclei is primarily studied through experiments using particle accelerators and detectors. These experiments involve observing the behavior of particles in nuclear reactions and decays, which can provide insights into the underlying mechanisms of parity violation.

5. Can parity violation in nuclei be explained by the Standard Model?

Yes, the Standard Model of particle physics can explain parity violation in nuclei through the mechanism of the weak nuclear force. This force is responsible for the decay of subatomic particles and is known to violate parity symmetry. However, there are still ongoing research efforts to better understand the extent and implications of parity violation in nuclei.

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