Solutions for Mathematical Physics 1 (Dated: April 19, 2015)

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1 Solutons for Mathematcal Physcs 1 Dated: Aprl 19, Usng the vectors P ê x cos θ + ê y sn θ, Q ê x cos ϕ ê y sn ϕ, R ê x cos ϕ ê y sn ϕ, 1 prove the famlar trgonometrc denttes snθ + ϕ sn θ cos ϕ + cos θ sn ϕ cosθ + ϕ cos θ cos ϕ sn θ sn ϕ. 2 One can use the fact that the angle between P and Q s θ + ϕ. From the dot product and cross product of these two vectors, one can obtan Prove the Jacob s dentty for vector product: snθ + ϕ Q P sn θ cos ϕ + cos ϕ sn θ, Q P cosθ + ϕ P Q cos θ cos ϕ sn θ sn ϕ. 3 Q P Let s re-defne the vectors a, b, and c as a b c + b c a + c a b. 4 Then, one can rewrte Eq. 4 as x 1 a, x 2 b, x 3 c. 5 a b c + b c a + c a b 1 2 ɛ jkx x j x k 6 The repeated ndces mply the summaton. The α-th ndex of the vector gven n Eq. 6 s gven as [ 1 2 ɛ jkx x j x k ] α 1 2 ɛ jkɛ αβγ x β x j x k γ 1 2 ɛ jk ɛ αβγ ɛ γµν x β x j µ x k ν 1 2 ɛ jk δ αµ δ βν δ αν δ βµ x β x j µ x k ν 1 2 ɛ jk x β x j α x k β x β x j β xk α ɛ jk x βx j αx k β 7 Note that the fact that the dummy ndces can be freely exchanged each other has been employed. We can easly show that the last lne becomes zero by usng ant-symmetrc property of the tensor ɛ jk : ɛ jk x βx j αx k β 1 [ ɛjk x 2 βx j αx k β ɛ kj x βx j αx k 1 [ β] ɛjk x 2 βx j αx k β ɛ jk x k βx j αx β] 1 2 ɛ [ jk x β x j αx k β x k βx j αx ] β. 8

2 Three vectors A, B, and C are gven by A 3ê x 2ê y + 2ê z, B 6ê x + 4ê y 2ê z, A 3ê x 2ê y 4ê z. 9 Compute the values of A B C and A B C, C A B and B C A. A B C 12, A B C 6ê x 4ê y + 5ê z, C A B 24ê x + 88ê y 62ê z, B C A 36ê x 48ê y 12ê z An electrc charge q 1 movng wth velocty v 1 produces a magnetc nducton B gven by B µ 4π q v 1 ˆr 1 r 2, 11 where ˆr s a unt vector that ponts from q 1 to the pont at whch B s measured Bot and Savart law. a Show that the magnetc force exerted by q 1 on a second charge q 2, velocty v 2 s gven by the vector trple product F 2 µ q 1 q 2 4π r 2 v 2 v 1 ˆr. 12 F 2 q 2 v 2 B µ q 1 q 2 4π r 2 v 2 v 1 ˆr Let x and y be column vectors. Under an orthogonal transformatons S, they become x Sx and y Sy. Show that x T y x T y, a result equvalent to the nvarance of the dot product under a rotatonal transformaton. x T y can be rewrtten as The -th components of the vector x and y are gven by x T y x y. 14 Enterng ths to Eq. 14 gves x S j x j, y S j y j. 15 From the defnton of the orthogonal transform matrx, we fnd x T y S j S k x j y k. 16 where I s dentty matrx. By enterng ths to Eq. 14, one fnd S j S k S T j S k S 1 j S jk I jk, 17 x T y I jk x j y k x j x j x T y. 18

3 Verfy that the Euler angle rotaton matrx, Eq. 3.37, s nvarant under the transformaton α α + π, β β, γ γ π 19 cos γ cos β cos α sn γ sn α cos γ cos β sn α + sn γ cos α cos γ sn β sn γ cos β cos α cos γ sn α sn γ cos β sn α + cos γ cos α sn γ sn β sn β cos α sn β sn α cos β cos γ cos β cos α sn γ sn α cos γ cos β sn α + sn γ cos α cos γ sn β sn γ cos β cos α cos γ sn α sn γ cos β sn α + cos γ cos α sn γ sn β sn β cos α sn β sn α cos β cos γ cos β cos α sn γ sn α cos γ cos β sn α + sn γ cos α cos γ sn β sn γ cos β cos α cos γ sn α sn γ cos β sn α + cos γ cos α sn γ sn β. 2 sn β cos α sn β sn α cos β a Fnd a vector perpendcular to the surface x 2 + y 2 + z at the pont 1, 1, 1. The gven surface can be defned as the set of the pont satsfyng fr, where fr x 2 + y 2 + z Let s pck two nfntesmally close pont on the surface r and r + dr, then the followng must be satsfed fr + dr fr + dr fr dr fr. 23 Snce the nfntesmal vector dr can have any drecton defned on the surface, fr becomes the perpendcular vector to the surface. fr s gven by Unt vector parallel to the f1, 1, 1 can be found as fr 2xê x + 2yê y + 2zê z. 24 êx + ê y + ê z / b Derve the equaton of the plane tangent to the surface at 1, 1, 1 The plane shares the perpendcular unt vector wth the surface at 1, 1, 1. The functon defnng the plane s gven by gr a x + b y + c z d. 26 Employng the condton that the perpendcular vectors of each surfaces are parallel each other gves The condton g1, 1, 1 gves a b c d 3. 28

4 If a vector functon F depends on both space coordnate x, y, z and tme t, show that In general, df dr F + F dt. 29 t By replacng Y wth F and X wth x, y, z and t, we get dy dx Y X. 3 Ths can be rewrtten as F dx F x + dy F y + dz F z + dtf t dr F + F dt. 31 t For a partcle movng n a crcular orbt r ê x r cos ωt + ê y r sn ωt: a Evaluate r ṙ, wth ṙ dr/dt v. df dr F + F dt. 32 t ṙ ê x rω sn ωt + ê y rω sn ωt. 33 b Show that r + ω 2 r wth r dv/dt. r ṙ ê z ωr Show, by dfferentatng components, that a r ω 2 ê x r cos ωt + ê y r sn ωt ω 2 r. 35 b d da A B dt dt B + A db dt. 36 d dt A B d dt A B da dt B db + A da dt dt B + A db dt. 37 d da A B dt dt B + A db dt. 38 [ ] d da j A B ɛ jk dt dt B db k k + ɛ jk A j dt [ da dt B + A db ]. 39 dt Wth the ad of the result of Exercse , show that f two vectors a and b commute wth each other and wth L, that s, [a, b] [a, L] [b, L], show that

5 5 [a L, b L] a b L. 4 [a L, b L] [a L, b j L j ] a [L, b j L j ] a b [L, L j ] ɛ jk a b j L k a b L A rgd body s rotatng wth constant angular velocty ω. Show that the lnear velocty v s solenodal. Snce we treat a rgd body, we can assume that length of r s constant, whch s equvalent to By the defnton, the angular velocty s gven n form of By takng outer product on left and rght sdes of Eq. 43 wth r, we can get d dr r r 2r. 42 dt dt ω r dr dt. 43 r ω r r dr dr dt dt r2 dr dt r2. 44 Snce the norms of r and ω are constants, we can conclude that dr/dt s constant. Next, take outer product on left and rght sdes of Eq.43 after takng tme dervatve, whch gves We can rewrte ths equaton as 2 r 2 d2 r dt 2 r r d2 r dr r ω 2 dt 2 r dt r 4 r. 45 Snce r ω 2 /r 6 s constant, Eq. 46 becomes solenodal equaton As an alternatve to the vector dentty of Example 3.6.4, show that d 2 r ω 2 r dt2 r 6 r. 46 A B A B + B A + A B + B A. 47 The -th component of A B s Aj B j A j Bj + A j B j A j Bj B j A j + B j A j + A j B j A j B j + A j B j δ a δ jb δ ja δ b Bb a A j + B A + δ a δ jb δ ja δ b Ab a B j + A B ε jk ε abk B b a A j + A b a B j + B A + A B A B + B A + A B + B A Verfy Eq by drect expanson n Cartesan coordnates. V V V, 49

6 6 The -th component of V s gven by ε jk j ε klm l V m Show that any of the equaton δ l δ jm δ jm δ m j l V m j V j j j V V V. 5 automatcally satsfes the vector Helmholtz equaton A k 2 A 51 and the solenodal condton 2 A + k 2 A, 52 Hnt Let operate on the frst equaton. Lke the Hnt suggests, take on left and rght sdes of Eq. 51 to get A. 53 Next, just expand the Eq. 51 to get k 2 A Usng the Paul matrxes σ of Eq. 2.28, show that A 2 A k 2 2 A k σ a σ b a b 12 + σ a b. 56 Here, a and b are ordnary vectors, and 1 2 s 2 2 unt matrx. Eq. 56 can be expended as σ ê x σ 1 + ê y σ 2 ê z σ 3, 57 σ a σ b a b j σ σ j. 58 One may need the followng property of Paul matrces. σ σ j 1 2 Enterng ths nto Eq. 58, one can get 1 σ σ j σ σ j + σ σ j σ σ j δj ε jk σ k a a b j σ σ j a b ɛ jk σ k a b k. 6 v ω r ωρê z ê ρ ωρê φ. 61

7 7 b v 1 r ê ρ ρê ϕ ê z ρ ϕ z ρ ρω 2ωê z 2ω Vρ, ϕ 1 r ê ρ ρê ϕ ê z [ ρ ϕ z V ρ ρ, ϕ ρv ϕ ρ, ϕω 1 ê ρ ρ z ρv ρ + ê ϕ z V ρρ, ϕ + ê z ρ ρv ϕρ, ϕ ] ϕ V ρ 1 [ ρêz ρ ρv ϕ ] ϕ V ρ We know that n sphercal coordnate system, the vector coordnate {x, y, z} becomes Combnng ths wth the Exercse , one can get x y y x {x, y, z} {cos ϕ sn θ, sn ϕ sn θ, cos θ}. 64 r sn 2 θ cos ϕ sn ϕ r + sn θ cos θ sn ϕ cos ϕ θ + cos2 ϕ ϕ r sn 2 θ cos ϕ sn ϕ r sn θ cos θ sn ϕ cos ϕ θ + sn2 ϕ ϕ ϕ Agan, combnng Eq. 64 wth Exercse , we can get and L x r sn ϕ sn θ cos θ r sn ϕ sn2 θ θ r sn ϕ sn θ cos θ r sn ϕ cos2 θ cos ϕ cot θ θ ϕ sn ϕ + cos ϕ cot θ, 66 θ ϕ L y r sn θ cos θ cos ϕ r + cos2 θ cos ϕ θ cos ϕ sn ϕ cot θ θ ϕ sn ϕ cot θ ϕ r sn θ cos θ cos ϕ r + cos ϕ + sn2 θ θ 67 From these relatons, we can get ψ can be wrtten n sphercal coordnate system as L x ± L y e ϕ ± θ + cot θ. 68 ϕ From ths, we can get ψ p cos θ. 69 4πε r2

8 8 [ E ψ ê r r + ê 1 θ r θ + ê 1 ϕ r sn θ The bass of Hlbert space satsfy followng relaton ] ψ ϕ p 4πε r 3 [2ê r cos θ + ê θ sn θ]. 7 If there exst more than one seres of fx, we can say dxϕ mxϕ n x δ n,m. 71 fx a n ϕ n x b n ϕ n x. 72 n n By ntegratng above equaton n whole space after multplyng ϕ m x, we can get a m b m. 73 Snce ths volates the prevous assumpton, we can conclude there are only unque seres of fx In vector notaton, F x can be rewrtten as F a n ϕ n. 74 n and a n can be determned by Then the mean square error becomes a n ϕ n F. 75 b [ F x a n 2 m c n ϕ n x] wxdx m [a n ϕ c n ϕ ] [a n ϕ n c n ϕ n ] + a n 2 ϕ n ϕ n n m a n c n 2 + n n>m n>m a n Ths can be mnmzed only f c n a n L x u x, snce u x s normalzed already; By Gram-Schmdt scheme, L 1 x can be wrtten as dxe x L 1 x A 1 u 1 x u x u L 1, 78 where u L 1 dxe x x 1. 79

9 9 and A 1 s normalzed constant. By substtutng Eq. 79 nto Eq. 78 we get L 1 x have to be normalzed; L 1 x A 1 x 1. 8 dxe x [L 1 x] 2 A 1 2 dxe x x 2 2x From ths, we can check A 1 1. L 2 x can be also calculated where L 2 x A 2 u 2 x u x u L 2 u 1 x u 1 L 2, 82 u L 2 u 1 L 2 dxe x x 2 2 dxe x x 2 x A 2 also can be found by normalzed process of L 2 x. From ths, we can conclude 1 dxe x x 2 4x A