10
$\begingroup$

When I see the equations of multipolar expansions they "look" to me harder than the original expressions. For example:

Multipole expansion - spherical form, in Wikipedia

I bet that this is not the case, but I want to know why. For example, when used as integrands, are these expressions advantageous for analytical or for numerical integration?

My main concern is about its usage in quantum mechanics (specifically about magnetic multipole expansions). I found many references about the math involved but I can't realize why it is useful. I've read the reference:

Ideas of quantum chemistry, Lucjan Piela (Elsevier, 2014). Appendix X; Google Books link

but it is not clear to me "why there is a loss of the accuracy" (as stated in the first page) by using the original expression and how the use of multipole expansion avoid that.

It seems to me that the advantage is in a interchange of variables but it is not clear to me.

Anticipated thanks to any help.

$\endgroup$
  • 1
    $\begingroup$ I thank both for the answers that were given to me. Both answer were useful and above all interesting to me. I chose the Emilio Pisanty's answer because it suits better to my needs. $\endgroup$ – user1420303 Dec 22 '15 at 17:28
4
+50
$\begingroup$

To me, the main advantages of multipolar expansions are typically conceptual ones, in that they help encapsulate and separate the effects of the shape of the charge distribution and the position of the test charge on the electrostatic interaction between the two.

Consider, for example, the interaction between a test charge at position $\mathbf r$ and a charge distribution $\rho(\mathbf r)$ that's localized inside a sphere of radius $R$ around the origin. The electrostatic potential for such a situation is given, in general, by the integral of the Coulomb kernel over the charge cloud, $$ V(\mathbf r)=\int_\Omega\frac{\rho(\mathbf r')\mathrm d\mathbf r'}{|\mathbf r-\mathbf r'|}. \tag 1 $$ This is a bit of a big, messy object, because it requires you to carry out a long, complicated three-dimensional numerical integral every time you want to know the value of $V(\mathbf r)$.

So how does a multipolar expansion help? I'm going to skip all the details of how we obtain it, but it's a pretty standard result that when $r>R$ you can express the potential in $(1)$ as $$ V(\mathbf r) =\int_\Omega\frac{\rho(\mathbf r')\mathrm d\mathbf r'}{|\mathbf r-\mathbf r'|} =\sum_{l=0}^\infty\sum_{m=-l}^l Q_{lm}\frac{Y_{lm}(\theta,\varphi)}{r^l} \tag 2 $$ where $$ Q_{lm}=\frac{4\pi}{2l+1}\int Y_{lm}^*(\theta',\varphi')r'^l\rho(\mathbf r')\mathrm d\mathbf r' \tag 3 $$ are the multipole moments of the system. At a first glance this looks about as ugly as before, and I've only traded the messy integral for a messy series, which I will still need to sum numerically whenever I want to calculate $V(\mathbf r)$.

The point here is that in multipolar series like $(2)$, it is typically only a few terms that contribute significantly, and this means that I only need to calculate a few of the multipole moments. This is important! It means I can precompute only a few of the full three-dimensional numerical integrals in $(3)$, and still have an excellent grasp of the potential.

The precomputation point is the important one. I have managed to isolate the few quantities of the charge distribution - the multipole moments $Q_{lm}$ - that control the weight of the different multipole components of the field, and after that all I need to do is calculate a few potentials (which are both numerically and conceptually simple) and superpose them to get the full interaction.


In addition to needing to calculate fewer integrals, these integrals tend to also be simpler and more benign numerically, and this ties in with the concern voiced by Piela in the book you link to. In particular, the integrals in $(1)$ often involve lots of terms that very nearly cancel out, which means that you need very high accuracy in both $a$ and $b$ to get only mediocre accuracy in $a-b$.

To ground this in an example, consider two charges $\pm q$ on the $z$ axis at $\pm \delta=1\:\mathrm{mm}$, and their interaction with a test particle on the same axis at $z=1\:\mathrm{km}$. If you want to calculate the electrostatic potential there, the equivalent of the integral in $(1)$ is to just add the two contributions directly: \begin{align} V &=\frac{q}{z+\delta}-\frac{q}{z-\delta} =\frac{q}{1000.001\:\mathrm m}-\frac{q}{999.999\:\mathrm m} \\&=\frac{q}{\mathrm{m}}\left(0.00099999900000999\cdots-0.00100000100000100\cdots\right) \\&\approx -2\times 10^{-9}\frac{q}{\mathrm{m}} \end{align} This means in particular that I need all nine decimals on each of the individual potentials - five significant figures - to get a single significant figure of accuracy in the result. If this is the only way to do things then it's mostly a frown-and-bear it sort of situation, but if I want to calculate the potential in a bunch of different positions then there will be a lot of frowning going on, particularly if instead of simple sums I need to be doing high-accuracy numerical integrals all the time, to get only a couple of significant figures in my potential.

In contrast, if I first calculate the dipole moment of the system and use the dipole field, I know that $$Q_{q,z}=q\delta-q(-\delta)=2q\delta=2q\times 1\:\mathrm{mm},$$ so I get the same single significant figure of accuracy for $V$ while using only a single significant figure of accuracy for $z$ and $\delta$. This kind of numerical tit-for-tat, where the level of desired accuracy in the result is well matched by the level of required accuracy in the calculation, is the hallmark of a desirable numerical scheme. As a bonus, I know that the next term up in the multipole series is on the order of $\delta/z\approx 10^{-6}$ with respect to the dipole term, so if you want a more accurate result then for the first five significant figures or so you know you should concentrate on the dipole term, and it's only after that that you need to include higher-order terms.


This leads me to a separate conceptual advantage of multipolar expansions, and it is the separation of scales that's present in $(2)$: each term up the ladder scales as a higher power of $1/r$ than its predecessor; moreover, the lower-$l$ terms tend to have a very 'fuzzy' angular structure, and it's only in higher $l$ that the angular detail comes in. This is the way to make precise the observation that, from far away, a charged system is pretty much indistinguishable from a single ball of charge, and that the details of the distribution of charge only become important as you close in.

This establishes a hierarchy of approximations that is very important in terms of how we conceptualize and apply electrodynamics in a broad range of contexts - from electrodynamics in continuous media to the interaction of atoms and molecules with radiation to the description of intermolecular forces to the focusing optics of electron microscopes and accelerators to a host of other settings. In all of these settings the separation of electromagnetic interactions into different multipolar orders brings conceptual clarity into the relative importance of different parts of the interaction, and that's its main value.

$\endgroup$
3
$\begingroup$

As an example, I will mostly use electrostatics with it's Poisson equation $\Delta \phi=\rho$ (also, constants omitted). This is easily scalable to stationary magnetism ($\Delta \vec{A}=\vec{j}$) or gravity ($\Delta \Phi = 4 \pi \rho_{\rm matter}$).


The multipolar potentials often emerge "without any effort" from trying to solve vacuum (source-less) field equations by the method of separation of variables. As such, they represent an infinite family of "harmonics" which can completely characterize any vacuum field.

However, an important detail is that these harmonics are not completely source-less, they typically have singularities at which infinite densities of charge (or an analogous kind of source) are located.

For example the spherical harmonics, which are obtained by separation of the Laplace equation in radial coordinates, have the first "monopole harmonic" $\phi_{\rm m}\sim Q/r$. This harmonic has $\Delta \phi_{\rm m}=0$ everywhere except $r=0$ where we don't really know what is happening because the potential diverges. A more sophisticated analysis gives us that $\Delta \phi_{\rm m}\sim Q\delta$, where $\delta$ is the infinitely peaked Dirac-delta function.

The dipole $\sim 1/r^2$ field would then correspond to two peaks in opposite direction next to each other (or two point charges of opposite sign right next to each other) which is characterized by the "Dirac-delta derivative" $\delta'$. It is possible to find a similar characterization in terms of "thin" or "point" sources for every multipolar potential. I.e., the multipoles in fact correspond to idealized infinitely packed sources, in the case of spherical multipoles around $r=0$.


But now consider the following: Since any vacuum field is characterized by harmonics, by understanding harmonics, you automatically understand the typical behaviour of the field. If we place the origin somewhere in the middle of the cloud of charge and only study the field outside of it, we will know that the outside field is represented by a series of functions proportional to $1/r^n$. If we are very far from the cloud ($r$ very large), a $\sim 1/r^{10}$ term is most probably quite negligible with respect to the leading $\sim 1/r$ term. This means that we can quite well approximate the exterior field by including just a few first multipoles.


But how do we compute the coefficients of the multipoles? With growing order of the multipole, this might become increasingly difficult, so let us for now consider the first two.

The leading-order coefficient corresponding to the monopole $\sim 1/r$ is just the total amount of charge in the cloud. If, for example, the total amount of charge in the cloud is zero (such as is the case of a neutral atom or molecule), you will always know that the field far away from the cloud will fall-off at least as $1/r^2$.

To compute the dipolar $\sim 1/r^2$ coefficient, one first needs to compute the dipole vector $$\vec{d} = \sum q_i \vec{r}_i$$ where $q_i$ are charges of individual sources in the cloud and $\vec{r}_i$ their positions. The coefficient of the dipole is then proportional to $\vec{d}\cdot \vec{r}$, where $\vec{r}$ is the point we are evaluating the potential at. I.e., if you have a cloud of a 1000 charges with a zero total charge, it is enough to execute this one sum and you get a faithful approximation of the electrostatic potential far from the cloud. This can prove very useful.


As for the cited usefulness in the Ideas of quantum chemistry, the point is that if we represent numbers in a computer with a finite amount of digits, the subtraction of close numbers reduces the number of significant digits. Consider for instance a 5-digit representation. There, you could say that for instance $1.0000$ in fact represents all the number from $0.99995$ to $1.000049999...$, i.e. when the computer gives you $1.0000$ it is in fact $1.0000 \pm 0.00005$. Now consider the subtraction $$1.0001-1.0000 = 0.0001$$ we should add error bars $$1.0001\pm 0.00005-1.0000\pm 0.00005 = 0.0001\pm 0.0001$$ I.e., the relative error of the result is 100%!!!

Because of this, it can be often useful to work around direct evaluation of close-number subtractions by using various approximations. One of them is using a dipolar potential for the potential of very close and equal charges.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.