Schiff moment in classical electrostatics

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Hello! I just found out about the Schiff moment. This is the paper where I encountered it, equations 3 and 4. The paper covers other things, too, that are not related to my question.

The main question I have is that, it seems like the derivation from equation 1 to 4 is purely classical (one can ignore the shielding term if necessary). They look at the next term after the electric quadrupole, which is equation 3.

As far as I remember from my electromagnetism (EM) class, the next term would be the octupole. However, they seem to get, beside the octupole (eq 6), another term, the Schiff moment (SM) (eq 5). I am pretty sure I didn't hear about the SM in my EM class, so I was wondering what is happening here? How was it hidden in the derivations in the EM classes (honestly I haven't heard of SM in general, even when googling stuff online for the class, not only in my book).

Also I thought that the normal expansion in dipole, quadrupole, octupole etc. is complete i.e. no other terms are needed. Can someone help me understand a bit this SM term, and why is it missing in classical books derivations (assuming I didn't just missed in during that class)? Thank you!
 
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  • #2
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A) The link points to an "Authorization Required" page, with no clue as to what the underlying paper is.
B) Why is this high energy physics?
 
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A) The link points to an "Authorization Required" page, with no clue as to what the underlying paper is.
B) Why is this high energy physics?
A: Sorry, it seems like the PRL link is not working. I replaced it with the arXiv one.
B: It is not high energy physics. The paper I mentioned above is a nuclear physics paper, so I assumed this is the best place to post the question. I am not sure where it would fit better.
 
  • #4
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I think you should look at ref 9 of the paper.
 
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I think you should look at ref 9 of the paper.
I looked at that reference. That is the derivation as to why one can't see the EDM of a nucleus (in the case of a point like nucleus). But my questions is purely classical, and it has to do only with the derivation between equations 1 and 5. It seems like the term in equation 5 appears naturally (and it comes from a classical EM derivation, not QM involved), but I really don't remember seeing it in my EM textbook (e.g. Jackson, Griffiths) derivations. But just to make sure I am not missing something, what equations form the reference 9 are you referring to?
 
  • #6
vanhees71
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In classical electrodynamics you cannot have a Schiff moment and other T- or P-odd contributions, because classical electrodynamics is P-, T-, and C-invariant. Only with the additional term involving a possible electric dipole moment of the nucleus you get such P- and T-odd contributions, and that's calculated in detail in Eqs. (1)-(5).

It's of course on the one hand convenient to work with Cartesian multipoles, but you always have to take care separately to expand it into irreducible parts (under rotations). This is more convenient using the spherical multipoles (spherical harmonics). The most convenient way to do this expansion in terms of irreducible tensors is also indicated in the paper, i.e., to use the multipole expansion of ##1/|\vec{r}-\vec{r}'|## for ##r'<r## around ##\vec{r}'=0##. This leads to the irreducible homogeneous polynoms of order ##\ell## in the components of ##\vec{r}'##, which are just the spherical harmonics ##\text{Y}_{\ell m}## in Cartesian form.
 
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In classical electrodynamics you cannot have a Schiff moment and other T- or P-odd contributions, because classical electrodynamics is P-, T-, and C-invariant. Only with the additional term involving a possible electric dipole moment of the nucleus you get such P- and T-odd contributions, and that's calculated in detail in Eqs. (1)-(5).

It's of course on the one hand convenient to work with Cartesian multipoles, but you always have to take care separately to expand it into irreducible parts (under rotations). This is more convenient using the spherical multipoles (spherical harmonics). The most convenient way to do this expansion in terms of irreducible tensors is also indicated in the paper, i.e., to use the multipole expansion of ##1/|\vec{r}-\vec{r}'|## for ##r'<r## around ##\vec{r}'=0##. This leads to the irreducible homogeneous polynoms of order ##\ell## in the components of ##\vec{r}'##, which are just the spherical harmonics ##\text{Y}_{\ell m}## in Cartesian form.
Thank you for your reply. I am a bit confused about this statement: "Only with the additional term involving a possible electric dipole moment of the nucleus you get such P- and T-odd contributions, and that's calculated in detail in Eqs. (1)-(5)." If I don't missunderstand it, they state in the same paper (basically between equations 10 to 15), that the Schiff moment can exists even if there is no electric dipole moment. So if my understanding is right, going back to equations 1 to 5, one can obtain a Schiff moment in equation 5, even if d=0 (the second term dissappear, but the overall stuff is not zero). However, if d=0, equation 1 is exactly the classical formula for a charge distribution. What confuses me is that is seems like even if you start from a classical charge distribution (i.e. formula 1 with d=0), you still obtain a Schiff moment (equation 5 with d=0).

Basically you start with equation 1 (d=0), you do the octupole expansion (equation 3 with d=0), then you get the Schiff moment (equation 5 with d=0). But as far as I can tell this is pure classical electrostatics (i.e. it follow naturally just by expanding equation 1). I just simply don't remember seeing equation 5 (with d=0) before in the derivations from my EM books. What am I missing here?
 
  • #8
vanhees71
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Good point. Now I'm confused myself :-((.
 
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  • #9
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Let's start with the Schiff Theorem (and let's not misuse "SM" to mean "Schiff Moment":

"For a non-relativistic system made up of point, charged particles which (should be "that") interact only electrostatically with each other and an arbitrary electrostatic field, the shielding is complete ."

This most certainly does have a classical analog - the electrons act as if they are a conductor and they completely screen the nuclear charge and its dipole moment.

So a non-zero EDM requires at least one of these loopholes:
  1. Relativistic effects (go as β2)
  2. Finite size effects (go as r)
  3. Magnetic interactions
  4. Exotic new physics
"Schiff moments" is a term used to describe the non-cancelling terms. Typically they go as qr2: qr provides the moment, and the other power of r is from finite-size effects.
 
  • #10
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Let's start with the Schiff Theorem (and let's not misuse "SM" to mean "Schiff Moment":

"For a non-relativistic system made up of point, charged particles which (should be "that") interact only electrostatically with each other and an arbitrary electrostatic field, the shielding is complete ."

This most certainly does have a classical analog - the electrons act as if they are a conductor and they completely screen the nuclear charge and its dipole moment.

So a non-zero EDM requires at least one of these loopholes:
  1. Relativistic effects (go as β2)
  2. Finite size effects (go as r)
  3. Magnetic interactions
  4. Exotic new physics
"Schiff moments" is a term used to describe the non-cancelling terms. Typically they go as qr2: qr provides the moment, and the other power of r is from finite-size effects.
Thank you for this! So I read a bit more about it and I actually realized that I am confused about the point 2. i.e. finite size effects. In the Schiff original paper, when he writes down the Hamiltonian of the system, equation 1, he uses ##\rho(r)## to describe the nuclear charge density (and a smiliar notation for the dipole density). He mentions at a point that his derivation (i.e. the cancelation of the EDM) assumes that the two densities are the same, however I don't see in his derivation (equations 1-6) the assumption that the nucleus is pointlike (which is what lots of papers claim that the theorem is based on). Shouldn't the charge density be replace by a delta function, if we assume a point like nucleus? He doesn't seem to use that at any point in his derivation.

Then going back to my original question (I will read a bit more literature about this meanwhile). As I mentioned in the last post, the derivation in the paper I mentioned in the original post, from equations 1 to 5 is purely classical (we assume for now that d=0). You basically start with a charge distribution ##\rho(r)## do the math as it is done in that paper and end up with the term in equation 5 (with d=0). Regardless of whether you call it Schiff moment or not, that terms arises naturally, just by doing the math, without any quantum mechanics reference or assuming any shielding (we assumed d=0). One of my main confusions is, when the multipole expansion for the same ##\rho(r)## was made in my EM textbooks, I don't remember seeing that term (in equation 5). But in both cases you start from ##\rho(r)##, without any other assumptions (again, here I am asking a purely EM question). Why is that term in equation 5 not mentioned in classical EM books (e.g. Griffiths, Jackson)? Thank you again for the help!
 
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  • #11
vanhees71
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Let's start with the Schiff Theorem (and let's not misuse "SM" to mean "Schiff Moment":

"For a non-relativistic system made up of point, charged particles which (should be "that") interact only electrostatically with each other and an arbitrary electrostatic field, the shielding is complete ."

This most certainly does have a classical analog - the electrons act as if they are a conductor and they completely screen the nuclear charge and its dipole moment.

So a non-zero EDM requires at least one of these loopholes:
  1. Relativistic effects (go as β2)
  2. Finite size effects (go as r)
  3. Magnetic interactions
  4. Exotic new physics
"Schiff moments" is a term used to describe the non-cancelling terms. Typically they go as qr2: qr provides the moment, and the other power of r is from finite-size effects.
But in purely classical electrostatics there's just the usual multipole expansion without Schiff moment. It's just simply due to the decomposition of the Green's function of ##-\Delta##, i.e.,
$$\frac{1}{4 \pi |\vec{r}-\vec{r}'|} = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} \frac{1}{2l+1} \frac{r_<^l}{r_>^{l+1}} \text{Y}_{ml}^*(\vartheta',\varphi') \text{Y}_{ml}(\vartheta,\varphi).$$
For ##l=3## this boils down to Eq. (3) of the discussed paper by Auerbach et al. with ##\vec{d}=0##. On the other hand there's a contribution to the Schiff moment for ##\vec{d}=0##, but shouldn't it simply boil down to the octupole contribution in classical electrodynamics (for arbitrary static charge distributions), i.e., the standard-textbook octupole moments given in Eq. (6) of said paper? I think that was the question in the OP, and as I said, I'm now also confused about this addition Schiff-moment contribution to a classical charge distribution.
 
  • #12
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I would suggest that thinking about "classical" atoms is more likely to be confusing than enlightening, especially since the classical model of atoms doesn't work.
 
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  • #13
vanhees71
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Then the paper seems to be misleading...
 
  • #14
Vanadium 50
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You've never written a confusing paper? :wink:
Try it - it's fun!
 
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  • #15
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I would suggest that thinking about "classical" atoms is more likely to be confusing than enlightening, especially since the classical model of atoms doesn't work.
I am not sure why the fact that we have atoms or not matters. As I said my confusions starts from ##\rho(r)##. I could have just posted this question (even in the classical mechanics forum) from the beginning by writing down equations 1-5 in the original paper (with d=0), without mentioning atoms or Schiff moment. Imagine my question was made of only these 5 equations (without any other information or paper links) and the questions was how does the term in equation 5 appears (compared to the derivation in classical EM)? I am not sure at which step of that derivation the fact that we have an atom involved matters at all (again, assuming d=0).
 
  • #16
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But in purely classical electrostatics there's just the usual multipole expansion without Schiff moment. It's just simply due to the decomposition of the Green's function of ##-\Delta##, i.e.,
$$\frac{1}{4 \pi |\vec{r}-\vec{r}'|} = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} \frac{1}{2l+1} \frac{r_<^l}{r_>^{l+1}} \text{Y}_{ml}^*(\vartheta',\varphi') \text{Y}_{ml}(\vartheta,\varphi).$$
For ##l=3## this boils down to Eq. (3) of the discussed paper by Auerbach et al. with ##\vec{d}=0##. On the other hand there's a contribution to the Schiff moment for ##\vec{d}=0##, but shouldn't it simply boil down to the octupole contribution in classical electrodynamics (for arbitrary static charge distributions), i.e., the standard-textbook octupole moments given in Eq. (6) of said paper? I think that was the question in the OP, and as I said, I'm now also confused about this addition Schiff-moment contribution to a classical charge distribution.
Exactly! This is my confusion. So did Vanadium 50 answer help you understand? Could you explain it to me in more details? I am not sure I see how the fact that we have an atom matter (assuming d=0)?
 
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  • #17
vanhees71
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It's my confusion too, and this has not too much to do with making it quantum or not. The argument, however makes some sense. The point is that they have split the reducible octupole moment
$$\tilde{Q}_{ijk} = \int_V \mathrm{d}^3 r r_i r_j r_k \rho(\vec{r})$$
into the irreducible parts, and thereby they encounter an expression
$$\vec{\nabla}_R \Delta_R \frac{1}{|\vec{R}|}=-4 \pi \vec{\nabla} \delta^{(3)}(\vec{R}).$$ I'm still not through with my analysis of it...
 
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It's my confusion too, and this has not too much to do with making it quantum or not. The argument, however makes some sense. The point is that they have split the reducible octupole moment
$$Q_{ijk} = \int_V \mathrm{d}^3 r r_i r_j r_k \rho(\vec{r})$$
into the irreducible parts, and thereby they encounter an expression
$$\vec{\nabla}_R \Delta_R \frac{1}{|\vec{R}|}=-4 \pi \vec{\nabla} \delta^{(3)}(\vec{R}).$$ I'm still not through with my analysis of it...
Right! To be honest I didn't do the math of it yet (trusted them with the separation of terms). Could it be that in classical EM books they just, somehow, keep the reducible tensor, without doing this separation (I should also go back in more details in my EM derivation, too). Please let me know if you get any insight!
 
  • #19
vanhees71
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Well, let's see. We need the traceless part of the polynom ##q_{ijk}=r_{i} r_{j} r_k##. Now you have
$$q_{ii k}=r^2 r_k, \quad q_{iji}=r^2 r_j, \quad q_{ijj}=r^2 r_i$$
So you get the trace less part by subtracting
$$r_{ijk}=\frac{1}{5} r^2 (\delta_{ij} r_k + \delta_{ik} r_j + \delta_{jk} r_i)$$
So you get
$$\tilde{q}_{ijk}=q_{ijk}-r_{ijk},$$
leading to the reducible octupole moment
$$\tilde{Q}_{ijk}=\int_{V} \mathrm{d}^3 r \tilde{q}_{ijk} \rho(\vec{r}).$$
Thne you have
$$\varphi^{(3)}_{\text{oct}}=-\frac{1}{3!} \tilde{Q}_{ijk} \partial_i \partial_j \partial_k \frac{1}{R}.$$
The other piece reads
$$\varphi_{\text{Schiff}}^{(3)}=-\frac{1}{5\cdot 3!} \partial_i \partial_j \partial_k \frac{1}{R} \int_{V} \mathrm{d}^3 r r^2 (\delta_{ij} r_k + \delta_{ik} r_j + \delta_{jk} r_i)\rho(\vec{r})= -\frac{1}{10} \partial_i \Delta \frac{1}{R} \int_V \mathrm{d}^3 r r^2 r_i \rho(\vec{r}).$$
Since now
$$\Delta \frac{1}{R}=-4 \pi \delta^{(3)}(\vec{r}),$$
you get indeed Eq. (4) with the definitions in Eq. (5) and (6) (with the obviously missing fractor ##r^2## in the definition of the irred. quadrupole moment reinstalled).

So this extra distribution-valued term comes from the expansion in terms of irreducible Cartesian multipole moments rather than reducible ones. In the usual multipole expansions in textbooks one does not take these singular contributions into account, because one is interested in the fields far away from the charge distribution, where these ##\delta##-distribution contributions of course don't matter, but they play a role in quantum mechanics, because there you evaluate matrix elements involving integrals over ##\mathbb{R}^3##, and there the singular contributions must be taken into account.
 
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  • #20
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Well, let's see. We need the traceless part of the polynom ##q_{ijk}=r_{i} r_{j} r_k##. Now you have
$$q_{ii k}=r^2 r_k, \quad q_{iji}=r^2 r_j, \quad q_{ijj}=r^2 r_i$$
So you get the trace less part by subtracting
$$r_{ijk}=\frac{1}{5} r^2 (\delta_{ij} r_k + \delta_{ik} r_j + \delta_{jk} r_i)$$
So you get
$$\tilde{q}_{ijk}=q_{ijk}-r_{ijk},$$
leading to the reducible octupole moment
$$\tilde{Q}_{ijk}=\int_{V} \mathrm{d}^3 r \tilde{q}_{ijk} \rho(\vec{r}).$$
Thne you have
$$\varphi^{(3)}_{\text{oct}}=-\frac{1}{3!} \tilde{Q}_{ijk} \partial_i \partial_j \partial_k \frac{1}{R}.$$
The other piece reads
$$\varphi_{\text{Schiff}}^{(3)}=-\frac{1}{5\cdot 3!} \partial_i \partial_j \partial_k \frac{1}{R} \int_{V} \mathrm{d}^3 r r^2 (\delta_{ij} r_k + \delta_{ik} r_j + \delta_{jk} r_i)\rho(\vec{r})= -\frac{1}{10} \partial_i \Delta \frac{1}{R} \int_V \mathrm{d}^3 r r^2 r_i \rho(\vec{r}).$$
Since now
$$\Delta \frac{1}{R}=-4 \pi \delta^{(3)}(\vec{r}),$$
you get indeed Eq. (4) with the definitions in Eq. (5) and (6) (with the obviously missing fractor ##r^2## in the definition of the irred. quadrupole moment reinstalled).

So this extra distribution-valued term comes from the expansion in terms of irreducible Cartesian multipole moments rather than reducible ones. In the usual multipole expansions in textbooks one does not take these singular contributions into account, because one is interested in the fields far away from the charge distribution, where these ##\delta##-distribution contributions of course don't matter, but they play a role in quantum mechanics, because there you evaluate matrix elements involving integrals over ##\mathbb{R}^3##, and there the singular contributions must be taken into account.
Ohhh I see! Thank you so much! So this has nothing to do with QM, atomic physics, EDM etc. That Schiff term is supposed to be there (and it arises naturally) in classical EM, too. The reason why it is ignored and I didn't notice it, is that it is zero anywhere except at the origin but in classical EM we look far away from origin, so that term is actually zero there. Is my understanding right?

I have one more question (which came along the way while trying to look at other papers). I found this paper which tries to generalize the formula of the Schiff moment potential to the case of a non-pointlike nucleus. They claim that the natural generalization of equation (1), which is the one you derived above, is equation (4) and the derivation is in section V A (the rest of the paper gives a relativistic generalization of the Schiff moment, which is not the cause of my confusion for this question). So my confusion is this: they claim that the old formula is for a point like nucleus and one needs to "introduce a finite-size Schiff moment potential" in order to get more accurate numerical results. I don't understand why the original derivation is for a point like nucleus. It wasn't at any point in the derivation where we assumed that. I guess this would imply that ##\rho(r)=\delta(r)##, but we never used a delta function (as far as their and your derivation is concerned, it seems like it holds for any ##\rho(r)##). Moreover, if we assume a point-like nucleus, the formula for the Schiff moment (not potential, just the S vector) would be automatically zero, so I thought that all this Schiff moment idea starts from the assumption of a non point-like nucleus. Do you have any idea what do they mean in this new paper? What do they mean by the fact that the initial derivation is for a pointlike nucleus? Thank you!
 
  • #21
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I came across the Schiff theorem and I am a bit confused about cases in which it doesn't apply. The statement for the theorem I found in all the papers/slides I read is of the form: "For a non relativistic system made up of point, charged particles which interact electrostatically with each other and with an arbitrary external field, the shielding is complete" (e.g. page 12 from here). Also in the same presentation and other papers, it is mentioned that if the particles are not point-like, we can get a Schiff moment, which basically says that this theorem applies only to point particles. However I went to Schiff's original paper and in the derivation he provides (eq. 1-6), the only things he assumes is that the distribution of the charge and dipole moment of each particle are the same. However, he never claims in his derivation that the particles are point-like. He even states this directly: "there is no term in the interaction energy that is of first order in the electric dipole moments, regardless of the magnitude of the external potential. This is true even if the particles are of finite size, provided that the charge and dipole moment of each have the same spatial distribution.". Of course being point-like is a particular case of having the same distribution, but it is not necessary to be pointlike. Am I missing something? How come no one mentions this and they all state the theorem as for pointlike particles (which again it is just a particular case). Also, their claim that if we take into account the size of the particles the theorem doesn't hold anymore seems clearly wrong. If the particles have the same charge and dipole distribution, the theorem still holds, despite the size of the particles. I am also aware that all these paper can't be all wrong, so I was wondering if someone can help me understand why this claim about similar charge and dipole distributions seem to be totally ignored in the literature? Thank you!

[Moderator's note: Result of a merger.]
 
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  • #22
vanhees71
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Yes, and of course if you look at a continuous charge distribution the singular terms are not there at all. It's just that you approximate the charge distribution entirely as a point-like object getting only "structure" with all the multipole moments. Then you have to take into account these singular contributions.

It's most easy to see in the more systematic spherical multipole expansion, which is most easily understood as solving the Poisson equation
$$\Delta \phi=-\rho$$
for the electrostatic potential. The Green's function of the negative Laplace operator is
$$G(\vec{r}-\vec{r}')=\frac{1}{4 \pi |\vec{r}-\vec{r}'|} = \sum_{\ell=0}^{\infty} (2l+1) \text{Y}_{\ell m}(\vartheta,\varphi) \text{Y}_{\ell m}^*(\vartheta',\varphi') \frac{r_{<}^{\ell}}{r_{>}^{(\ell+1)}}.$$
The ##\text{Y}_{\ell m}## are the spherical harmonics and ##r_{<}=\text{min}(r,r')## and ##r_{>}=\text{max}(r,r')##.

The solution of the Poisson equation then is
$$\phi(\vec{r})=\int_V \mathrm{d}^3 r' G(\vec{r}-\vec{r}') \rho(\vec{r}').$$
Then you see that there's no singularity at the origin if ρ is a sufficiently smooth function since in the volume integral over ##\vec{r}'## has to be done considering that
$$\frac{r_{<}^{\ell}}{r_{>}^{\ell +1}} = \Theta(r-r') \frac{r^{\prime \ell}}{r^{\ell+1}} + \Theta(r'-r) \frac{r^{\ell}}{r^{\prime (\ell+1)}}.$$
So there's no singularity at the origin when integrating over a smooth charge distribution.
 
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  • #23
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Yes, and of course if you look at a continuous charge distribution the singular terms are not there at all. It's just that you approximate the charge distribution entirely as a point-like object getting only "structure" with all the multipole moments. Then you have to take into account these singular contributions.

It's most easy to see in the more systematic spherical multipole expansion, which is most easily understood as solving the Poisson equation
$$\Delta \phi=-\rho$$
for the electrostatic potential. The Green's function of the negative Laplace operator is
$$G(\vec{r}-\vec{r}')=\frac{1}{4 \pi |\vec{r}-\vec{r}'|} = \sum_{\ell=0}^{\infty} (2l+1) \text{Y}_{\ell m}(\vartheta,\varphi) \text{Y}_{\ell m}^*(\vartheta',\varphi') \frac{r_{<}^{\ell}}{r_{>}^{(\ell+1)}}.$$
The ##\text{Y}_{\ell m}## are the spherical harmonics and ##r_{<}=\text{min}(r,r')## and ##r_{>}=\text{max}(r,r')##.

The solution of the Poisson equation then is
$$\phi(\vec{r})=\int_V \mathrm{d}^3 r' G(\vec{r}-\vec{r}') \rho(\vec{r}').$$
Then you see that there's no singularity at the origin if ρ is a sufficiently smooth function since in the volume integral over ##\vec{r}'## has to be done considering that
$$\frac{r_{<}^{\ell}}{r_{>}^{\ell +1}} = \Theta(r-r') \frac{r^{\prime \ell}}{r^{\ell+1}} + \Theta(r'-r) \frac{r^{\ell}}{r^{\prime (\ell+1)}}.$$
So there's no singularity at the origin when integrating over a smooth charge distribution.
Thank you for this (and sorry for the late reply)! So in this derivation if we take the limit ##r_{<} \to 0## we would get the results we got originally? However, I am still confused by this statement: "It's just that you approximate the charge distribution entirely as a point-like object" (which is similar to the one found in most of the papers). I am not sure I see at which point in the original derivation, we used the fact that the charge distribution is point like. The formula for the Schiff moment (eq. 5 in that paper, as well as your derivation) has the nuclear density explicitly appearing i.e. ##\rho(r)##. If we assume that the nucleus is point like, shouldn't the density be ##\rho(r)=Z\delta(r)## i.e. a delta function, whose integral is the total charge? I don't see this fact (that ##\rho(r)## is a delta function) used in the initial derivation and also the formula has ##\rho(r)## in it, suggesting that any form of the nuclear density (eg. hard sphere, gaussian) can be assumed for actual numerical calculations of the Schiff moment. So I guess my confusion is, at which step do we assume that the nuclear density is point like? And if we assume that, why do we have a ##\rho(r)##in the final formula, and not a delta function instead? Thank you!
 
  • #24
vanhees71
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No! Pointlike doesn't mean to restrict yourself to the monopole term, which indeed is a point-charge when you neglect the finite extension of the nucleus.

You just go on describing the charge distribution with the entire set of multipole-moments,
$$\rho_{\ell,m}=\int_V \mathrm{d}^3 r' r^{\prime \ell} \text{Y}_{lm}^*(\vartheta',\varphi') \rho(\vec{r}')$$
and then neglecting the finite extension of the charge using the expansion for the electrostatic potential

[EDIT: corrected a typo, i.e., inserted the before forgotten multipole moments]

$$\varphi(\vec{r}) \equiv \varphi(r,\vartheta,\varphi)=\sum_{\ell=0}^{\infty} \sum_{m=-\ell}^{m=\ell} \frac{2 \ell+1}{r^{\ell+1}} \rho_{\ell,m} \text{Y}_{\ell m}(\vartheta,\varphi)$$
everywhere. Then you have all these singular contributions in the origin with the ##\delta##-distribution and its derivatives.
 
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  • #25
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No! Pointlike doesn't mean to restrict yourself to the monopole term, which indeed is a point-charge when you neglect the finite extension of the nucleus.

You just go on describing the charge distribution with the entire set of multipole-moments,
$$\rho_{\ell,m}=\int_V \mathrm{d}^3 r' r^{\prime \ell} \text{Y}_{lm}^*(\vartheta',\varphi') \rho(\vec{r}')$$
and then neglecting the finite extension of the charge using the expansion for the electrostatic potential
$$\varphi(\vec{r}) \equiv \varphi(r,\vartheta,\varphi)=\sum_{\ell=0}^{\infty} \sum_{m=-\ell}^{m=\ell} \frac{2 \ell+1}{r^{\ell+1}} \text{Y}_{\ell m}(\vartheta,\varphi)$$
everywhere. Then you have all these singular contributions in the origin with the ##\delta##-distribution and its derivatives.
I am really sorry for being so dumb, but I am still a bit confused. In this formula: $$\rho_{\ell,m}=\int_V \mathrm{d}^3 r' r^{\prime \ell} \text{Y}_{lm}^*(\vartheta',\varphi') \rho(\vec{r}')$$

what is ##r'## referring to and what is the integration volume? I would normally assume that the volume is over the charge distribution of the nucleus, but isn't the nucleus pointlike? How can you have a volume integral over a point? Also, if the nucleus is a point how can one have ##\rho(r) \neq 0## for ##r>0##. Wouldn't that imply that the charge distribution of the nucleus is more than a point? (Sorry for not getting it...)
 

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