Question

first picture is the notation on the test paper
Second picture is the question i would like to solve

thanks~

Notation and convention: r x +y The distance from the origin to the point r [x,y,z] + ê: The unit vector along the direction of r-[x, y,z] (a.e,6)-i.j.):m :The orthonormal bases of a Cartesian coordinate system. for dummy indices Einstein convention: Omitting the summation notation (repeated indices). Examples:ab,-a b, ab a b Notice: No dummy index is allowed to be repeated more than twice. You should change the "names" of the dummy indices before taking the product of two summations of the same type. Example: A B A,B, C.D C,D, (A B)(C D) A,B,C,D 4. The Kroneker Delta and Levi-Civita symbols are defined as , (à, xe), a.b.ce(1,2,3) (a) Give the values of 8 , E2and E13 (b) For 3-dimentional vectors A and B, write thej-th component of AxB (using Levi-Civita symbols and Einstein convention). and the result of (b), find the -6 (e) Using the identity S expansion of (AxB) x C. Please express the final result in the vector form, not the component form.

Answer 1

From definitio (a) 623 mutually orthog onal unit vectors, so, Since are S23 0 822 = 1 From depinition Eabc = ea (xej E312 1 e, e1, this implies, е, хе - ез So, eз (е, хе) E312 1, and 1 X E213 1 1, E213 -1 823 = 0, 622 = 1, E312 Hence

Then B mem Let A A e and B (b) х Bm em Ag e A X B Ay Bm ёрхёт A Bm Emn en АХВ S0, as n th component of vector AxB is given Thus, the (АxB), dummy voriable therefore replace 'r' by Emn Ag B, m Since 'n' is a i.e (AXB);= Emj Efmj A Bm

(C) (AxB)x C Substitute in From part (b), АхВ 3D АgBт Eemp en the expression above, to obtain, хс Ag Bm Еgmn en Let C= Cp ep hen, en A Bm Etmn X A Bт Сp Epmn (en х ер) Again by depinition, e x e, = Enpy A BтСр Егmp Епрт By NDw use the ide ntitų Ekij Eklm = Sit Sjm Sim 8ji Enim Eдmn So, Emn Enpy Enim Enp SO,

Substitute in the ezpession FOY (A X BXC (Аxв) xс - А Вm Cp Ermn Enpy Cr - A, Bm Cp Enhm Enpy er Ap Bm Cp ( S4p бmy - бдr 5mp) er - (At Bm Ср бiр Smr A Bm Cp mp Jer Consider, A Bm Cp 6i bmr = Ag(Bm Bmr) Cp Spt) A C By = A BrC - and, AL Вт Сp буу Smp - (A бpr Bm 6mp) Сp Aт Bр Сp 3 ВрСp Ay Clearly From depinition, A B = Aj Bj, therefore A CA.Cand Bp C Thus, A Bm Cp S1p m - (A C) Br and (BC) A A Bm Cp 6y mp Subs titute, to obtain, v2)- = 2x ( gxy (A.C) Bre, - (BC) A, e (AXE)X(A)B - (B-C)