Friday, August 20, 2010

Week 4 Lex 3: Multipole Expansion II

The multipole expansion arises from the idea that the potential due to a point charge q at a distance r goes as 1/r. If we calculate the approximate potential at a large distance from a physical dipole which is a pair of equal and opposite charges +q and -q at a distance d  where the vector d is from the -ve to the +ve charge is given by a binomial expansion to first order ( higher orders neglected) as 
V=(1/4πε0)[ q d cos θ/r2] and goes as 1/r2 to a first order.


This tells us that it may be possible to obtain the approximate potential at a large distance from an arbitrary charge configuration in terms of a series of contributions in powers of (1/r). This is somewhat like expressing a complex waveform in terms of some simple waves of different frequencies ( fundamentals and harmonics), called a Fourier expansion.. What allows this is the Superposition Principle.


When we expand the denominator of the the expression for the potential at a faraway point r due to an arbitrary volume charge distribution ρ(r') given as
V=(1/4πε0)∫τ ρ(r')dτ'/r
( note that r is the Griffiths script/curly r.
)
in a binomial expansion, we obtain a series of contributions in powers of (1/r) which is the distance between the source point r' and the field point r. The coefficient of each power of (1/r) is an integral that provides the respective moments like the monopole moment ( total charge Q=∫ ρ(r')dτ') then a dipole moment p= ∫τ r'ρ(r')dτ' and
the contribution to the potential is given as Vdip=(1/4πε0)p. r cap /r2 please note that r cap is a unit vector along r. Note that θ or θ' is simply the angle between the position vectors r and r'. It is not the polar angle unless z axis is chosen along the r' or r direction. Note that this is the dipole moment of a PURE DIPOLE


*For a point charge at origin the monopole term is the only non zero term and the potential due to a point charge is EXACT all higher multipoles vanishing. ( Check it out)


*The dipole moment expression works for a line charge λ, surface charge σ distribution replacing the volume distribution ρ and integrating over
a surface or a line.


For a discrete point charge distribution of qi at r'i the pure dipole moment is p=∑i=1n qi r'i. A physical dipole moment is a special case of this for just two charges +q and -q and is given as p=qd where the vector d is the vector from - to the + charge. This is consistent as it gives the correct potential where
for a physical dipole. But note that this potential is approximate because for a physical dipole of two unlike but equal charges the monopole term is zero as total charge Q=0 but higher multipole terms are not zero. They are smaller than the dipole contribution and hence neglected in the approximation that the field point r is far away from the source.


It should be obvious now that for a single point charge away from the origin ( that is r' not equal to 0) will have all possible moments starting from the monopole moment ( total charge Q=q in this case) and all possible such terms in the expression for the potential V at a field point r.


* The physical dipole reduces to a pure dipole in the limit q
goes to infinity and d goes to 0 with qd=p held fixed. So a physical dipole are finitely separated charges and a pure dipole is more like a point. The pure dipole is the second moment in the multipole expansion. For large distances the field
of a pure dipole and a physical dipole are identical but are very different close to the source. ( See the two figures given in Griffiths).


The multipole expansion is dependent obviously on the coordinate system. The example of a point charge q away from the origin is an example of this where all higher multipoles also contribute although the dominant term is the monopole moment.


For a shift in origin the monopole moment Q does not change. The (pure) dipole moment changes due to the shift except when the total charge Q=0 ( algebraic sum). In that case ( for total charge Q=0) the (pure) dipole moment is unchanged. See the example for a shift of origin by a vector a  in Griffiths at the end of the section which was discussed in class.

For quick reference

http://cr4.globalspec.com/blogentry/2842/Multipole-Expansion

14 comments:

  1. So given a certain charge distribution if you are asked to calculate the dipole moment use the formula for the PURE DIPOLE moment. Note that even a single point charge has a dipole ( and all higher)
    pole moments if it is not at the origin of coordinates.

    ReplyDelete
  2. sir,
    its not related to this lecture but i wanted to know if the long problems and derivations of IPSA will come in our mid sems or jus the applications of the concepts....
    also the speed of the course PHY103 is gr8....cnt anything be done in that regard....the theory is that we can revise aftr coming back from class but i dont feel that much confident about the problems bcoz v have not been exposed to the problems othr then DIPA IPSA n solved examples of the textbook....not even the unsolved questions under the heading "problems"
    practically we cant even think of touching them and now the midsems are round the corner...even if we glance at them we can loose our nerves...would revising everything (theory from notes n textbook) n examples wid dipa ipsa are sufficient????
    coz m nt even sure abt these.....

    ReplyDelete
  3. There is no need to get panicked. If you have revised the class notes and worked through the DIPA and IPSA problems and a few related same level problems from Griffiths then you should have no difficulties in exams. Its important to understand the concepts. Exam questions will be of the same level as typically DIPA problems which are slightly harder than IPSA problems.

    Not much can be done about the speed of the course. In fact I am one of the slower instructors
    and in the initial part I have gone even more slowly. There is a syllabus which needs to be completed.

    My advise is to get out of the JEE coaching class mental framework. In the semester system if you are not regular in your academics and are not coming prepared to the lectures then it will be difficult. Its very important to understand the lectures and the concepts.

    Again my advise is not to panic but to go over the class notes using the blog to clear any doubts
    or conceptual difficulties and also you can consult me or your IPSA and DIPA instructors with regard to any difficulty. Work through the IPSA and DIPA problems with clear understanding and the exam should be quite OK.

    ReplyDelete
  4. PLEASE NOTE THAT THERE WAS SMALL ERROR ( NOW CORRECTED) IN THE POST. THE VECTOR d IN THE DIPOLE MOMENT OF A PHYSICAL DIPOLE IS FROM THE -VE TO THE +VE CHARGE LOCATION.

    ReplyDelete
  5. Sir,

    i hv 2 queries (not related to this blog)-
    (a) What is the exact syllabus outline of mid sems?will evrything taught till last day come into the exams?

    (b) Where can i get previous years mid sems' question papers? Would you mind mailing them/ uploading them on brihaspati.

    since there are a large no.of courses, I hope you would reply positively.

    ReplyDelete
  6. (a) This is already announced in the class, i hope you are attending them. Yes the full syllabus is by tradition is always till the last lecture. But obviously more emphasis will be on what has been taught earlier than the last lecture.

    (b) There is no system in the IITK Physics department for sample papers. This is because
    the paper and the Instructors keep changing and
    a sample from previous year may be quite misleading to the students.

    ReplyDelete
  7. multipole expansion consists of a lot of big and somewhat dangerous looking terms , do we have to remember it all or if we just know that it will have a monopole , dipole and higher contributions , it works,, and we have to remember that every system of charges has a monopole and other contributions , right sir

    ReplyDelete
  8. You will have to remember the monopole and dipole contributions which are short and easy.

    ReplyDelete
  9. sir i can't figure out the use of multipole expansion because though the expressions for a point charge and a dipole is clear but that big expansion misleads me and as you said that even a point charge could have dipole moment and higher moments what's the use of couloumb's law......i can't figure out an application of the multipole expansion.it looks as if it has increased our problem of finding electric fields......those integrals looks dangerous

    ReplyDelete
  10. You have to first read it from Griffiths. The expansion is a simple binomial expansion for which i have also given the formula. If you systematically read it all the integrals will become clear.

    Coulombs law is for the Electric field, not the potential. The potential at a point depends on the coordinate system chosen. If your point charge is at the origin only the monopole moment survives and all other higher moments are zero.
    But if its not at the origin then all the other
    higher moments will contribute, however the dominant term is the monopole moment.

    You should try to understand in the class itself and stop me if you are not understanding. I dont think you are doing that. Integrals will look dangerous when you dont understand them or come to me right after the lectures. I do hope that you are attending the lectures and the DIPA and IPSA.

    You dont have to worry about the quadrupole terms
    only monopole and dipole are sufficient at your level.

    Electric fields are very hard to determine except in simple charge distributions. The multipole expansion is a very powerful tool for arbitrary chrage distributions to get an approximate potential as
    a series expansion.

    ReplyDelete
  11. sir, when total charge is zero. then dipole moment doesnt change.. but as given in griffiths, they have changed r prime to r dash prime but havnt correspondingly changed r prime for ρ(r') and dτ' in p= ∫τ r'ρ(r')dτ'. why??

    ReplyDelete
  12. sir, can you please have a look at my interpretation for the difference b/w pure n physical dipole--- i suppose physical dipole is one which have finite charges(of opposite nature) separated by a finite distance while physical dipole is one where there is volume charge distribution and hence charges have zero separation. n in physical dipole, only dipole moment term is there while in physical dipole, monopole, dipole,..... terms r there. basically the funda behind coming of all these terms is that we can assume the formation of infinite no. of monopole and dipole and so onforming inside these charge distribution. since the elementary charges are infinite, so infinite sets of dipole, quadrupole could be formed,, thus leading to summing up i.e. intregation... how much of my interpretations were correct???

    ReplyDelete
  13. The physical dipole is just a special case of the pure dipole. The pure dipole is just one of the terms in the multipole expansion. It will occur for all charge distributions, volume or not. Although in some cases it may be ZERO VALUED as it depends on the geometry of the charge distribution. So even for a discrete charge distribituion there will be all terms in the multipole expansion including the pure dipole term.

    The physical dipole is just a special case of a discrete charge distribution with two equal and opposite charges separated by d. It will have zero monopole contribution but all higher multipoles will be there in general.

    ReplyDelete
  14. They have not changed r' in the integrals as that variable is being integrated over and it will not matter finaly. Because any variable that you do a
    definite integral over disappears from the result of the integration.

    ReplyDelete