LAPLACE S EQUATION IN SPHERICAL COORDINATES. With Applications to Electrodynamics



Similar documents
(Ch. 22.5) 2. What is the magnitude (in pc) of a point charge whose electric field 50 cm away has a magnitude of 2V/m?

Orbits and Kepler s Laws

Adaptive Control of a Production and Maintenance System with Unknown Deterioration and Obsolescence Rates

Summary: Vectors. This theorem is used to find any points (or position vectors) on a given line (direction vector). Two ways RT can be applied:

r (1+cos(θ)) sin(θ) C θ 2 r cos θ 2

2.016 Hydrodynamics Prof. A.H. Techet

Angles and Triangles

32. The Tangency Problem of Apollonius.

UNIT CIRCLE TRIGONOMETRY

N V V L. R a L I. Transformer Equation Notes

Mechanics 1: Work, Power and Kinetic Energy

A Note on Risky Bond Valuation

Continuous Compounding and Annualization

Mechanics 1: Motion in a Central Force Field

Basic Principles of Homing Guidance

(1) continuity equation: 0. momentum equation: u v g (2) u x. 1 a

Newton s Law of Universal Gravitation and the Scale Principle

Chapter 3 Savings, Present Value and Ricardian Equivalence

2. TRIGONOMETRIC FUNCTIONS OF GENERAL ANGLES

Curvature. (Com S 477/577 Notes) Yan-Bin Jia. Oct 8, 2015

Chapter 22. Outside a uniformly charged sphere, the field looks like that of a point charge at the center of the sphere.

Skills Needed for Success in Calculus 1

Vector Calculus: Are you ready? Vectors in 2D and 3D Space: Review

Week 3-4: Permutations and Combinations

12. Rolling, Torque, and Angular Momentum

Intro to Circle Geometry By Raymond Cheong

Random Variables and Distribution Functions

Gauss Law. Physics 231 Lecture 2-1

The LCOE is defined as the energy price ($ per unit of energy output) for which the Net Present Value of the investment is zero.

Coordinate Systems L. M. Kalnins, March 2009

PY1052 Problem Set 8 Autumn 2004 Solutions

HUT, TUT, LUT, OU, ÅAU / Engineering departments Entrance examination in mathematics May 25, 2004

9.3 Surface Area of Pyramids

The Casino Experience. Let us entertain you

Lesson 7 Gauss s Law and Electric Fields


Figure 2. So it is very likely that the Babylonians attributed 60 units to each side of the hexagon. Its resulting perimeter would then be 360!

Combinatorial Testing for Tree-Structured Test Models with Constraints

MATH 150 HOMEWORK 4 SOLUTIONS


Chapter 19: Electric Charges, Forces, and Fields ( ) ( 6 )( 6

4a 4ab b (count number of places from first non-zero digit to

Screentrade Car Insurance Policy Summary


Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration

Physics 235 Chapter 5. Chapter 5 Gravitation


Fluids Lecture 15 Notes

Appendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:

Finite-Length Analysis of Low-Density Parity-Check Codes on the Binary Erasure Channel

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )

A r. (Can you see that this just gives the formula we had above?)

Graphs on Logarithmic and Semilogarithmic Paper

G.GMD.1 STUDENT NOTES WS #5 1 REGULAR POLYGONS

The invention of line integrals is motivated by solving problems in fluid flow, forces, electricity and magnetism.

Exam in physics, El-grunder (Electromagnetism), , kl

Symmetric polynomials and partitions Eugene Mukhin

Solar wind speed theory and the nonextensivity of solar corona

Lesson 8 Ampère s Law and Differential Operators

Forces & Magnetic Dipoles. r r τ = μ B r

Valuation of Floating Rate Bonds 1

payments Excess demand (Expenditure>output) r > r Excess demand (Expenditure>output) r > r Excess supply (Expenditure<output) Excess supply

Things to Remember. r Complete all of the sections on the Retirement Benefit Options form that apply to your request.

Vindforsk report Project /V-238

XML Data Integration using Fragment Join

Voltage ( = Electric Potential )

1240 ev nm 2.5 ev. (4) r 2 or mv 2 = ke2

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.

4.1 - Trigonometric Functions of Acute Angles

The force between electric charges. Comparing gravity and the interaction between charges. Coulomb s Law. Forces between two charges

Functions of a Random Variable: Density. Math 425 Intro to Probability Lecture 30. Definition Nice Transformations. Problem

Radius of the Earth - Radii Used in Geodesy James R. Clynch Naval Postgraduate School, 2002

ABSORPTION-FREE SUPERLUMINAL LIGHT PROPAGATION IN A V-TYPE SYSTEM

Graphs of Equations. A coordinate system is a way to graphically show the relationship between 2 quantities.

Financing Terms in the EOQ Model

Questions for Review. By buying bonds This period you save s, next period you get s(1+r)

Chapter 22 The Electric Field II: Continuous Charge Distributions

Words Symbols Diagram. abcde. a + b + c + d + e

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY

Lab M4: The Torsional Pendulum and Moment of Inertia

Derivation of Annuity and Perpetuity Formulae. A. Present Value of an Annuity (Deferred Payment or Ordinary Annuity)

CURVES ANDRÉ NEVES. that is, the curve α has finite length. v = p q p q. a i.e., the curve of smallest length connecting p to q is a straight line.

Chapter 23 Electrical Potential

9 CONTINUOUS DISTRIBUTIONS


Density Curve. Continuous Distributions. Continuous Distribution. Density Curve. Meaning of Area Under Curve. Meaning of Area Under Curve

tools for Web data extraction

Binary Representation of Numbers Autar Kaw

Magnetic Field and Magnetic Forces. Young and Freedman Chapter 27

SR-Phlx Page 39 of 43 NASDAQ OMX PHLX LLC 1 PRICING SCHEDULE THE EXCHANGE CALCULATES FEES ON A TRADE DATE BASIS.

Voltage ( = Electric Potential )

CHAPTER 5 GRAVITATIONAL FIELD AND POTENTIAL

12.1. FÖRSTER RESONANCE ENERGY TRANSFER

It is required to solve the heat-condition equation for the excess-temperature function:

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100

Turbulence simulation in wavelet domain based on Log- Poisson model: univariate and multivariate wind processes

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES

Carter-Penrose diagrams and black holes

4.11 Inner Product Spaces

Transcription:

LALACE S EQUATION IN SHERICAL COORDINATES With Appitions to Eetodynmis We hve seen tht Lpe s eqution is one of the most signifint equtions in physis. It is the soution to pobems in wide viety of fieds inuding themodynmis nd eetodynmis. In you ees s physis students nd sientists, you wi enounte this eqution in viety of ontets. It is impotnt to know how to sove Lpe s eqution in vious oodinte systems. The oodinte systems you wi enounte most fequenty e Ctesin, yindi nd sphei po. We investigted Lpe s eqution in Ctesin oodintes in ss nd just begn investigting its soution in sphei oodintes. Let s epnd tht disussion hee. We begin with Lpe s eqution: ( We n wite the Lpin in sphei oodintes s: ( + (sinθ + ( sinθ sin θ φ ( whee θ is the po nge mesued down fom the noth poe, nd φ is the zimuth nge, nogous to ongitude in eth mesuing oodintes. (In tems of eth mesuing oodintes, the po nge is 9 minus the titude, often temed the otitude. To mke ou initi utions itte simpe, et s ssume zimuth symmety; tht mens tht ou pmete does not vy in the φ dietion. In othe wods, / φ, so we n wite the Lpin in ( bit moe simpy. Assuming zimuth symmety, eq. ( beomes: ( + (sinθ sinθ ( This is the fom of Lpe s eqution we hve to sove if we wnt to find the eeti potenti in sphei oodintes. Fist, et s ppy the method of sepbe vibes to this eqution to obtin gene soution of Lpe s eqution, nd then we wi use ou gene soution to sove few diffeent pobems. To sove Lpe s eqution in sphei oodintes, we wite:

Fist Step: The Ti Soution ( + (sinθ sinθ (4 The fist step in soving pti diffeenti equtions using sepbe vibes is to ssume soution of the fom: R( Θ( θ ( whee R( is funtion ony of, nd Θ(θ is funtion ony of θ. This mens tht we n set: ( ( θ R Θ ; R( ( θ Θ (6 Substituting the etionships in (6 into (4 podues: Θ( ( θ R ( R ( + (sinθ Θ ( θ sinθ (7 If we mutipy eh tem in (7 by nd then divide eh tem by R( Θ(θ, we obtin: d d ( R ( + (sinθ Θ ( θ R( d Θ( θ sinθ dθ (8 Notie tht the deivtes in (8 e no onge pti deivtives. This is beuse the method of sepbe vibes hs podued two tems; one is soey funtion of nd the othe is soey funtion of θ. Seond Step: Septing ibes Eqution (8 ows us to septe Lpe s eqution into two septe odiny diffeenti equtions; one being funtion of nd the othe funtion of θ. As we hve disussed in ss, we eize tht eh tem on the ight hnd side of (8 is equ to onstnt. This mens we n septe (8 into: d ( R( d d R ( ( + nd (sinθ Θ ( θ ( + Θ( θ sinθ dθ (9 We now hve two diffeent odiny diffeenti equtions whih we wi sove. We eize tht the podut of soutions wi ow us to use eq. ( (ong with ppopite boundy onditions to detemine the soution to Lpe s eqution. You my wonde we we hoose to wite the seption onstnt s something s non-obvious s (+.

The eson is tht we hve ed hed in the sipt, nd know tht by witing the seption onstnt in this wy we wi podue we known diffeenti eqution whose soution we edy know. Notie tht seption onstnt is positive in one eqution (the di pt nd negtive in the othe (the ngu pt; this is neessy so tht the sum of equtions is zeo s equied by Lpe s eqution. The di eqution Let s stt by soving the di eqution of eq. (9. We mutipy though by R( nd epnd the deivte to find: d R dr + ( + R ( d d This is fiy simpe empe of Fobenius (see http://www.u.edu/futy/dsvsk/ouses/othe/ssnotes/fobeniusempe.pdf diffeenti eqution. This is so n empe of n Eue (o Cuhy diffeenti eqution. See http://www.u.edu/futy/dsvsk/ouses/othe/ssnotes/eue.pdf fo moe detis bout soving Eue s eqution. Using eithe the method of Fobenius o methods of Eue s equtions, we n find the soution to eqution (: ( + R ( A + B ( whee A nd B e onstnts whih wi be detemined one we ppy speifi boundy equtions. The ngu eqution We sove the ngu potion of eqution (9 by mutipying though by Θ(θ nd epnding the deivtive to obtin: d Θ osθ dθ + + ( + Θ dθ sinθ dθ ( This is tuy diffeenti eqution you e vey fmii with, though pehps not in this et fom. Refe bk to the soutions fo the vey fist homewok set of the semeste, http://www.u.edu/futy/dsvsk/ouses/phys/homewok/physhws.pdf nd eview gin the soutions to pobems 4 nd. You wi see tht the eqution you hve deived in ( is just the we known Legende eqution. We know tht the soutions to the Legende eqution e the Legende poynomis, (os θ. Thid Step: Constuting the ompete soution

Hving septed Lpe s eqution into two odiny diffeenti equtions, we n use the esuts bove to substitute into eq. ( to eize tht the gene soution to Lpe s eqution in sphei oodintes wi be onstuted of sum of soutions of the fom: ( + (, θ ( A + B ( Fom ou epeiene with Lpe s eqution in Ctesin oodintes, we know tht the fu soution wi be onstuted by tking sum of soutions of the fom of (; in othe wods, ou gene soution to Lpe s eqution in sphei oodintes is: ( + (, θ ( A + B (4 Now, we need e boundy onditions to detemine the vues of the oeffiients A nd BB. Appying Boundy Conditions Fist Empe (Bos pp. 647-649 Let s see how we n use (4 s the stting point to detemine soution to Lpe s eqution with speifi boundy onditions. Fo this pupose, et s use the empe in Bos pp. 647-649. Without ny oss of mening, we n use tk bout finding the potenti inside sphee the thn the tempetue inside sphee. So, et s ssume thee is sphee of dius, nd the potenti of the uppe hf of the sphee is kept t onstnt +, nd the potenti of the owe hf of the sphee is hed t. How n we detemine the potenti t ny point inside the sphee? Fist, et s wite the boundy onditions s:, < θ < π / o < osθ <, π / < θ < π o < osθ < Remembe tht the noth poe of the sphee oesponds to θ, nd θ π/ in the equtoi pne. Now, et s ook bit moe osey t (4. We e sked to find the potenti t ny point inside the sphee. This egime inudes, of ouse, the point, nd we n ook t (4 nd eize tht the soution diveges t uness BB. Appying the neessity fo meningfu physi soution to this pobem ows us to set oeffiients B B to zeo, so tht (4 simpifies to: (, θ A (

Now, we use the boundy ondition fo the sufe of the sphee. When, we know tht in the uppe hf sphee nd in the owe hf sphee. This mens we n wite ( s: (, θ A fo < osθ < (6 The epession in (6 shoud ook fmii to us: we e seeking to wite funtion (in this se the funtion equs the onstnt in tems of n infinite seies. We hve seen how to do this using both Fouie seies nd Legende oynomis. We know tht ou funtion n be epnded in seies if nd ony if we n epnd tht funtion in tems of ompete set of othogon funtions. Fouie seies e possibe beuse sin nd os epesent ompete set of othogon funtions on (-π, π; epnsion in tems of Legende poynomis is possibe sine we hve ened tht Legende poynomis e ompete set of othogon funtions on (-,. Thus, we n epnd ny funtion f( on (-, s: whee the oeffiients, e detemined by: f ( ( (7 + f ( ( d (8 We n see tht eqution (7 ppies to eq. (6 with f(, nd A, o A (9 A we hve to do now is detemine the vues of the oeffiients fom (8, substitute these vues into (9 nd then use those vues of A in ( to detemine the ompete soution to the potenti inside the sphee. We n detemine seve of the oeffiients esiy by diet integtion; in ft this is done in Bos on p. 8. Using these Legende oeffiients with f( nd substituting into (6 we obtin n epiit epnsion of ou soution fo (, θ: 7 (, θ [ ( + +...] ( 4 6 nd you n epnd the vious Legende poynomis epiity in tems of osθ if you wish, but thee is ey no need to go beyond the epession s it is witten in (.

Seond Empe Conside sphee of dius tht hs potenti on its sufe given by: (, θ os θ ( nd we e sked to find the potenti t points eteio to the sphee. We go bk to eq. (4 nd begin to ppy boundy onditions. Fist, we eize tht A must go to zeo sine n get vey ge, owing us to simpify (4 s: ( + (, θ B ( Now, we ppy the boundy ondition ( nd obtin: (, ( + θ B os θ ( This is just nothe fom of eq. (7. Hee, the funtion f( is os θ, nd the -(+ oeffiient BB stnds in the pe of. So, ou tsk now is vey fmii: ompute the oeffiients using (8, use these to detemine the vues of B B, nd substitute these vues of BB into ( to find ou ompete soution. Let s begin by finding the oeffiients. We n set os θ; sine θ vies fom to π then vies fom - to, whih is vey onvenient in uting Legende oeffiients sine the Legende poynomis e ompete, othogon set on (-,. With this substitution, we wi ute ou oeffiients fom: + + f ( ( d ( d ( + d (4 The fin integ on the ight is petty esy to do; Legende poynomis e, we, poynomis, nd mutipying them by just podues nothe poynomi whih is esy to integte between these imits. But et s think bit moe nd mke ou ives even esie. We e tht Legende poynomis e even funtions fo even vues of, nd e odd funtions fo odd vues of. This mens tht fo odd vnish sine the integnd in (4 beomes the podut of n even funtion ( nd n odd funtion ( ( fo n odd. This mens the integnd in (4 is odd wheneve is odd, nd the integ of n odd funtion between imits symmeti with espet to the oigin vnishes. Let s ompute oeffiients:

( d d ( ( d ( d You wi find tht highe inde oeffiients vnish; does it mke sense tht this funtion is epessibe in tems of ony ( nd (? Thee e ony two tems whih wi ontibute to the seies epnsion of, nmey the nd tems. We emembe fom befoe tht we use ou vues of to find the vues of BB tht substitute bk into (; eq. ( tes us tht: B ; B, + so B (6 Using these vues of BB in ou gene soution ( gives us the ompete nswe to this pobem: (, θ B + B [( + ( ] (7 Thid Empe Let s sy now tht we wnt to find the potenti outside sphee of dius whose sufe is hed t potenti given by os(θ. We know tht sine we e deing with eteio points ou soution wi be of the fom of eq. (, nd tht we wi hve to find the oeffiients BB. The poess we foow is identi to the empe immeditey bove, eept now f( os(θ the thn os θ. We sw in the empe bove how we oud simpify ou utions by eizing we oud set os θ; we woud ike to epess ou uent f( in tems of os θ, but we wi hve to do itte tig nd geb mnipution to ompish this. Let s stt by witing os(θ os(θ+θ. We now epnd this s: os(θ + θ os θ osθ sin θ sinθ (os θ sin θ osθ sin osθ[ {os θ ( os θ } ( os θ ] 4os θ osθ whee os θ. θ osθ 4 (8 Now it is fiy stightfowd tsk to find the neessy oeffiients to sove ou pobem. We foow the empe of eq. (4, now with f( (4, nd sove fo :

( (4 ( (4 d d 8 ( [ (4 7 ( (4 7 d d These esuts te us tht: 8 ; 4 4 B B nd we use the fom of (4 to wite the fin nswe s: ] (os 8( (os ( [ (os (os, ( 4 4 θ θ θ θ θ B B + +