How Does the Nuclear Quadrupole Moment Explain Nucleus Deformation?

[patric44]

Homework Statement

How the nuclear quadruple moment describe the deformation of the nucleus?

Relevant Equations

Q =1/e∫∫∫(3z^2-r^2)dV

hi guys
I have read the other day about how the nuclear quadruple moment descries the deformation of the nucleus, however i can't get my head around how is that!, I am familiar with the multiple expansion in which we can describe the potential of an arbitrary charge distribution by the following expansion
$$
\Phi(r') = \frac{1}{4\pi\epsilon_{o}}\sum_{n=0}^{\inf}\frac{1}{r^{n+1}}\int_{v}(r')^{n}P_{n}(cos\theta)\rho(r')dV
$$
and this potential is then expanded as multiple terms : monopole,dipole,quadrupole..., Now how the quadrupole term in this expansion describes a deformation in the shape of the nucleus? and : isn't quadrupole means "generaly speaking" 4 charges two of them are negative and the other is positive, arranged in a specific way, how could such term describes the only positively charged neucles?
I have many questions surrounding this point, can anyone clarify them or suggest a specific book or a set of lecture notes that would help.
thanks in advance.

 

[PhDeezNutz]

Forgive me if this is wrong but from my VERY cursory understanding we refer to a "deformation of a nuclei as deviation from a spherical shape". From my understanding the Electric Quadrupole Tensor is analogous to a Moment of Inertia Tensor. Remember finding principle axes and moments of the inertia tensor and the different cases of spherical tops, prolate, and oblate? I.e. if all the principle moments have the same value then we have a spherical top but if they are unequal we do not (i.e. deformation or deviation from spherical).

 

[ergospherical]

I did a little bit of research and I found the following reference very informative!
https://photonics101.com/multipole-moments-electric/nuclei-model-deformed-sphere-multipole-moments

Summarising briefly; scattering experiments suggested that atomic nuclei can be modeled by a homogeneous charge distribution with a shape given by an axially symmetric function\begin{align*}
R(\theta) = R_0(1+ \xi P_2(\cos{\theta}))
\end{align*}where $\xi \ll 1$ is a parameter representing the level of deformation. If you calculate the moments of this distribution, as the reference does, the dipole moment is zero due to the presence of mirror-symmetry and the first non-zero moment is the quadrupole moment. The multipole moments go as $b_{2n} \propto \xi^n$, so the quadrupole is dominant.

 

[MathematicalPhysicist]

Try to get a hold on the three volumes of Judah Eisenberg and Walter Greiner's books on Nuclear Models.
I stopped reading somewhere after 10-60 pages of the first volume.

 

[robphy]

isn't quadrupole means "generaly speaking" 4 charges two of them are negative and the other is positive, arranged in a specific way, how could such term describes the only positively charged neucles?

A "[pure] quadrupole" would be an arrangement of charges with zero net-charge (zero monopole) and zero dipole-moment. But a general charge distribution can have nonzero multipole moments... where the monopole term handles the net-charge.

A point charge that is not at the origin has nonzero multipole moments.

Possibly useful: has calculations for point-charge arrangements
http://www.physics.usu.edu/Wheeler/EM3600/Notes12MultipoleMoments.pdf
http://orca.phys.uvic.ca/~tatum/elmag/em03.pdf

 

[PhDeezNutz]

@robphy if I understand your post correctly

1) If a nucleus is deformed from a spherical arrangement the monopole stays the same so we can’t use the monopole moment to gauge deformation.

2) Regarding the dipole moment we can imagine a deforming a sphere to an ellipsoid in such a way that the dipole moment stays the same. Likewise the dipole cannot be used to assess deformation from spherical.

3) A quadrupole is spherical if all principle moments are the same. If they aren’t we’ve deformed from spherical. This is why we use the quadrupole to assess deviation from spherical.

Is that correct?

 

[robphy]

@robphy if I understand your post correctly

1) If a nucleus is deformed from a spherical arrangement the monopole stays the same so we can’t use the monopole moment to gauge deformation.

Yes.

2) Regarding the dipole moment we can imagine a deforming a sphere to an ellipsoid in such a way that the dipole moment stays the same. Likewise the dipole cannot be used to assess deformation from spherical.

I think the dipole moment of a sphere that is centered at the origin is zero.
Deforming the sphere into an ellipse will result in a nonzero dipole moment.
However, the dipole moment doesn't have enough information for what you seek.

3) A quadrupole is spherical if all principle moments are the same. If they aren’t we’ve deformed from spherical. This is why we use the quadrupole to assess deviation from spherical.

Is that correct?

So, the quadrupole moment is what you need.