8/10/2019 Jackson 4 4 Homework Solution

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Jackson 4.4 Homework Problem SolutionDr. Christopher S. Baird

University of Massachusetts Lowell

PROBLEM:

(a) Prove the following theore! "or an ar#itrary charge distri#ution(x) the values of the ($l% &)oents of the first non'vanishing ultipole are independent of the origin of the coordinate aes #ut

the values of all higher ultipole oents do in general depend on the choice of origin. (*he different

oents qlmfor fied ldepend of course on the orientation of the aes.)

(#) + charge distri#ution has ultipole oents q p Qij , with respect to one set of coordinate aes

and oents q- p- Q'ij , with respect to another set whose aes are parallel to the first #ut whose

origin is located at the point R (X YZ) relative to the first. Deterine eplicitly the connections#etween the onopole dipole and /uadrupole oents in the two coordinate fraes.

(c) 0f q1 2 can R#e found so that p- 23 0f q1 2 p1 2 or at least p1 2 canR#e found so thatQ-ij 23

SOL!"O#:

(a) Soe point is located at x (xyz) relative to an initial origin. 0f we choose a new origin that is

located at R (X YZ) then the point in this new coordinate syste will #e at x- x4 R or

(x-y-z-) (x4Xy4 Yz4Z).

0n general the lth set of oents is given #y (we ignore the traceless nature)!

Qij...l=(x , y , z)xix j...xldx dy dz

where i & $ 5 andj & $ 5 and etc. andx&xx$yx5z.

*he oents in this initial coordinate syste in ters of the new coordinate syste are!

Qijk...l=(x -+X , y -+Y , z-+Z)(x -i+Xi)(x -j+Xj)...(x -l+Xl)dx -dy - dz-

6pand out all the ters recogni7ing that the coponents of Rare constant and coe out of the

integral!

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Qijk...l=(x -+X , y -+Y , z-+Z)x -ix -j ...x -ldx -dy -dz-+Xi(x -+X , y -+Y , z-+Z)x - j ...x -ldx - dy -dz-+Xj(x -+X , y -+Y , z-+Z)x -i ...x -ldx - dy -dz-

+...+Xl(x -+X , y -+Y , z-+Z)x -j ...x -l& dx - dy - dz-+XiXj(x -+X , y -+Y , z-+Z)x -k...x -ldx - dy - dz-+...+XiXj ...Xl(x -+X , y -+Y , z-+Z)dx -dy - dz-

8ow recogni7e that the first integral is 9ust the ultipole in the new coordinate syste and the otherters are lower order ultipoles in the new coordinate syste.

Qijk...l=Q -ijk...l+XiQ -jk...l+Xj Q -ik...l+...+XlQ -ijk...l&+XiXj Q -k...l+...+Xl&XlQ -ijk...l$

+...

+XiXj...XlQ -

0f the ultiple oents are to have the sae value in #oth coordinate systes all reaining ters

ust #e 7ero. 0n other words!

Qijk...l=Q -ijk...l only if

2=Xi Q -jk...l+Xj Q -ik...l+...+XlQ -ijk...l&+XiXj Q -k...l+...+Xl&XlQ -ijk...l$

+...+XiXj ...XlQ -

*his can only hold true for all ar#itraryXiXj etc if all ters are independently 7ero!

Qijk...l=Q -ijk...l only if Q -ijk...l&=2) Q -ijk...l$=2 ,

0n words this says that the value of the ultipole oent of order lis independent of origin choice if

and only ifall lower order ultipole oents are 7ero.

(#) *his is 9ust a specific application of the general transforation a#ove.

"or a onopole l 2 the general for!

Qij...l=

(x , y , z)xix j...xldx dy dz

#ecoes!

q=(x , y , z)dxdydz

0n the new coordinate syste this is!

q=(x -+X , y-+Y , z-+Z)dx -dy- dz-

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q=q -

"or a dipole l & the general for reduces down to!

p i=(x , y , z)xi dx dy dz for i & $ 5 such thatp&pxp$pyp5pz

0n the new coordinate syste this is!

pi=(x-+X , y -+Y , z-+Z)(x-i+Xi)dx- dy- dz-

p i=(x -+X , y -+Y , z-+Z)x -i dx -dy - dz-+Xi(x -+X , y -+Y , z-+Z)dx -dy - dz-

p i=pi -+Xi q -

p=p -+Rq -

"or a /uadrupole l $ the general for reduces down to!

Qij=(x , y , z)x ix j dx dy dz

*o follow the convention of the #oo: we use the traceless versions!

Qij=(x , y , z)(5x ixjr$ij )dxdydz

0n the new coordinate syste this is!

Qij=(x-+X , y -+Y , z-+Z)(5(xi+Xi)(x j+Xj)((x -+X)$+(y -+Y)$+(z-+Z)$)ij)dx- dy -dz-

Qij=Q -ij+5X jp -i+5Xi p -j+5XiX jq -R$ij q -$ij Rp -

Qij=Q -ij+5X jp -i+5Xip -j$ij Rp -+(5XiX jR$ij)q -

(c) 0f q1 2 then p- 2 if!

R=pq

"or instance consider the siple charge distri#ution consisting of a point charge %Aatz a a point

charge '& atz 'A and a point charge %$Aat the origin. 0n this coordinate syste the total charge is

q %$Aand the dipole oent isp $Aa. 0f we now place the origin at z a this syste will have no

dipole oent.

0f q 2 p1 2 then Q-ij 2 if!

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Qij=5Xjpi+5Xip j$ij Rp

;riting this out and solving the syste of e/uations leads to!

X&=(Q$5p&+Q&5p $+Q&$p 5)

< p$p 5

X$=(Q$5p&Q&5p$+Q&$p5)