1. Using Einstein Summation notation, prove the identity: = A

Transcription

1 1. Usig Eistei Suatio otatio, pove the idetity: ( B ( B B( + ( B ( B [1 poits] We begi by witig the coss poduct of ad B as: So the ou idetity, C B C ( B C, i ε ik B k We coside ( C ε i ε ik ε iε ik ( ε iε ik ( whee the last step follows as a cyclic peutatio of the idices i. We apply the epsilo-delta elatioship to fid: C ( δ δ k δ δ k ( We epad the deivative via the poduct ule to obtai: ( δ δ k δ δ k ( δ δ k[ + ] δ δ k[ + ] I the fist backeted epessio, we ust have (fo the poduct of Koecke deltas that ad also that k ; i the secod backeted epessio we ust have that ad also that k, akig these subsitutios we get: B + B B B (1 We ecall that i Eistei suatio otatio, epeated idices ae sued, so that i Bi B; i ; i i This eas that the fist te i (1 is ( B ; the secod te is ( B ; the thid te is B( ad the last te is - ( B. Suig these fou tes poves the equested idetity.. Spheical coodiates epeset a eaple of a othogoal coodiate syste. The tasfoatio fo Catesia to spheical coodiates is give by: i

2 siθ cosφ, y siθ siφ, z cosθ a Fid the scale factos fo the spheical coodiate syste. I calculatig these scale factos, you ay take as a give that the syste is othogoal ad you ay ake eplicit use of that fact i you calculatios. If you do so, eplai how you ae eploitig the othogoal atue of spheical coodiates. [1 poits] The key tie savig piece of ifoatio is that you could use eplicitly the fact that the spheical coodiate syste is othogoal. This eas you kow that all coss tes, i.e., d ; ddφ; dφ all su to zeo. We stat by takig deivatives of the tasfoatio equatios: d siθ cosφ d + cosθ cosφ siθ siφ dφ dy si θ siφ d + cosθ siφ + siθ cosφ dφ We su the squaes of each te to fid: dz cosθ d siθ ( d + ( dy + ( dz (siθ cosφd + ( cosθ cosφ + ( siθ siφdφ + (siθ siφd + ( cosθ siφ + ( siθ cosφ dφ + (cosθ d + ( siθ Squaig all tes, usig the Pythagoea theoe whe applicable, ad goupig by diffeetials, we fid: ( d + ( dy + ( dz ( d + ( + si θ ( Fo this, we ca ead off that h 1 h 1; h h θ ; h 3 h φ si θ b If is the positio vecto show that 3 fo spheical coodiates. [1 pts] We use the epessio fo divegece i spheical coodiates (give o the table of esults: 1 F ( F1 hh3 + ( Fh1 h3 + ( F3h1 h h1h h3 q1 q q3

3 I this case, the vecto is the positio vecto. I spheical coodiates, thee is oly a copoet of this vecto, so ou epessio above becoes: ( si ( ( ( si 3 si θ + + si θ θ θ φ θ 3. Coside the fuctio: Calculate: 3 y 3 3 v ( + e z i + ( y + l ta z + ( z + + y k v S ds whee S is the suface of a sphee of adius ceteed o the oigi. [1 pts] The key poit hee is to ecogize this as a eaple of the divegece theoe, so that: V vdτ v ds So that all we have to do is fid the divegece of v ad itegate that fuctio ove the volue of the sphee. While the give fuctio v is quite ugly, it has a vey siple divegece, aely: Ou itegal the becoes: Div v 3( + y + z 3 v dτ 3 ( siθ d dφ 3 V S 4 d siθ dφ The agula pat of this itegal is siply 4π, leavig a siple adial itegal evaluated fo,, so ou fial esult is 1π ( 5 /5. 4. Fo ay odd fuctio f(, fid the value of: 3 1 f ( δ ( d whee δ( is the Diac Delta fuctio. Show you wok clealy ad/o eplai the easoig behid you aswe. [1 pts]

4 We ealize that because of the siftig popety of the Diac Delta fuctio, the value of this itegal is equal to f(. We ae told that the fuctio is odd, ad we ca use a ube of ways of showig that f( fo a odd fuctio is, so that the value of this itegal is zeo. [Fo istace, we ca ague that ay odd fuctio ca be epaded i a Taylo seies aoud ; this seies will have the fo a 1 + a 3 3 +a fo a odd fuctio. Clealy, the value of f( ] 5. Deteie the Fouie seies fo f( + 1 o the iteval (-π, π. Wite the fist 3 o-zeo tes of each suatio. You ay use syety aguets to ease calculatios; if you do so, state you easoig eplicitly. [1] This poble was woked out i its etiety i the solutio set to hoewok set #8, questio o. 6. Please efe to: 6. eage studet of atheatical physics obseves that: e d 1 Fo this, ou studet agues that ay fuctio ca be epaded as a seies of the fo: f ( c e ( whee the c ae suitably chose costats. Is ou studet coect? Ca ay fuctio be eaigfully epaded i a ifiite seies of the fo of eq. (? Eplai you easoig. (You gade will be based o you easoig; a aswe of yes o o uaccopaied by appopiate ustificatio will ea o cedit. [1] Ou eage studet is ot coect. I ode to epad f( i tes of a basis set of fuctios, that basis set of fuctios ust epeset a coplete, othogoal set of fuctios o the iteval i questio. Fouie seies ae based o the othogoality of tig fuctios; epasio i tes of Legede polyoials is based o the othogoality of Legede polyoials o (-1, 1. The poposed basis set above, e - is ot a othogoal set of fuctios. Othogoal fuctios have the popety: f ( f ( d δ

5 whee δ is the Koecke delta. We ca show easily that e e d 1 + δ so that these ae ot othogoal fuctios ad caot seve as the basis set fo epasio. 7. adioactive ucleus is ceated by the euto bobadet of aothe ucleus ad is destoyed via its ow adioactive decay. If the ate of poductio is a costat give by Q, ad the decay costat of the ucleus is give by λ, the followig diffeetial equatio descibes the tie ate of chage of the ube of adioactive uclei: dn( dt Q λn( Whee N( is the ube of uclei at a paticula tie t. a Fid the epessio fo N( if N(. [1] This is a vey staightfowad eaple of a fist ode diffeetial equatio that ca be solved by sepaatig vaiables. We goup siila vaiables o each side of the equatio to fid: dn( Q λn( dt dn( Q λn( 1 λ dt l Q λn( t + C Multiplyig though ad epoetiatig both sides yields: N( ( Q Ce λt / λ We ca evaluate the costat C by usig the iitial coditio that N(, showig that CQ, ad we wite ou solutio as: Q(1 e N( λ λ t b What is the liitig value of N( as t? Eplai why this is what you would epect fo the oigial diffeetial equatio. [5 pts] s t gows vey lage, the epoet i the aswe above goes towad zeo, ad the steady-state populatio of N teds towad N( Q/λ. This is epected fo the oigial

6 diffeetial equatio sice as t gows lage, we epect that dn(/dt goes to zeo i the steady-state (that is the eaig of steady-state. Ude these coditios we ca solve the oigial equatio algebaically ad fid that whe dn(/dt, N( Q/λ as we fid. 8. Usig seies solutio techiques, solve the diffeetial equatio: y y Deive the ecusio elatio, ad use this to wite the fist thee o-zeo tes of each solutio to the diffeetial equatio. [1 pts] This poble was solved i its etiety as poble #1 i hoewok #1. Please efe to: 9. Bessel s equatio is witte as: y + y + ( p y whee p is ay ube (positive, egative, a itege o a factio. a Deive the idicial equatio fo Bessel s equatio ad solve fo the oots of that equatio. [5 pts] This diffeetial equatio satisfies the Fuchsia coditios, so we ca use the ethod of + Fobeius with a tial solutio of y a We substitute this ito the oigial diffeetial equatio ad fid: ( + ( + 1 a + ( + a + a p a (1 The idicial equatio is deteied by settig i the suatios cotaiig the lowest powe of. These ae the fist, secod ad last suatios. Settig i those tes yields: ( 1 + p p ± p b Deive the ecusio elatio; deive the epessios fo a fo eve ad sepaately fo odd. [5 pts]

7 To fid the ecusio elatio, go back to eq. (1 above. We ow ust e-ide the thid su so that all powes of ae equal. We set k+, ad obtai: ( + ( + 1 a + ( + a + a p a Note that the thid su ow has to the sae powe as the othe sus, but ow stats at the lowe liit of. We ca gai useful ifoatio by stippig out the 1 te fo the othe sus. Fist, ecall that we have aleady stipped out the te whe we deived the idicial equatio. If we set 1 ad stip out that te, we fid: Fo values of geate tha o equal to : [ ( + 1 p ] a a ( 1 1 [( + ( ( + p ] a a + a a ( ( + p I the last step, we use the fact that p fo the lage oot of the idicial equatio. We ca look at this ecusio elatio ad ealize that it will geeate two baches; oe fo eve ad oe fo odd. Howeve, we aleady deteied that a 1 ; this coupled with the ecusio elatio tells us that all a odd ; ad the ecusio elatio ( holds fo all eve. c I class we solved this equatio fo p ½; fid the solutio fo p -1/ ad show eplicitly how this is elated to a well kow tig fuctio. [5 pts] Use the ecusio elatio i ( with p -1/ to fid: a a : ( 1 a a ; a a 4 a a ; a ! a ! Usig these coefficiets with - ½, we ca substitute back ito ou tial solutio of + y a ad get the seies solutio: y a 1/ (1! + 4 4! 6 6! +... a 1/ cos ( ube of studets foud the coefficiets coectly, but oitted the -1/ te fo the fial esult. 1. Coside Laplace s Equatio i Catesia coodiates: V

8 ssue a tial solutio of the fo V (, y X ( Y ( y. Substitute this solutio ito Laplace s equatio ad deive the geeal fo of the solutio to this equatio i Catesia coodiates. Without ay bouday coditios you ca poceed o futhe tha witig a geeal solutio. [1 pts] We begi by witig Laplace s equatio i Catesia coodiates (i diesios: V V V + (1 y We substitute the tial solutio V (, y X ( Y ( y ito (1, obtaiig: X Y + XY ( We divide ( by XY to obtai: X Y + X Y (3 s we have discussed i class ad ead i the tet, the fist te is a fuctio of X oly ad the secod te is a fuctio of Y oly. Because of this, we ealize that each te ust equal a costat, ad we call this costat the sepaatio costat. We ca the ewite (3 as two sepaate odiay diffeetial equatios: X k X ad Y k Y (4 These equatios ca be ewitte as: X + k X ad Y k Y These equatios have vey well kow solutios: X cos( k + B si( k Y Ce ky + De Sice ou tial solutio was V (, y X ( Y ( y, we ca ultiply these two solutios to fid ou geeal solutio fo: ky V (, y X ( Y ( y ( cos k + Bsi k( Ce ky + De ky With bouday coditios we could deteie the costats, B, C, D ad k ad fid the seies solutios that solves Laplace s equatio; howeve, without those bouday coditios this is the etet of ou aalysis.