PHYS606: Electrodynamics Feb. 01, Homework 1. A νµ = L ν α L µ β A αβ = L ν α L µ β A βα. = L µ β L ν α A βα = A µν (3)

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1 PHYS606: Electrodynamics Feb. 01, 2011 Instructor: Dr. Paulo Bedaque Homework 1 Submitted by: Vivek Saxena Problem 1 Under a Lorentz transformation L µ ν, a rank-2 covariant tensor transforms as A µν A µν = L α µ L β ν A αβ (1) Symmetric Tensor By definition, A µν = A νµ,so A νµ = L ν α L µ β A αβ = L ν α L µ β A βα symmetry is preserved under Lorentz transformations. Antisymmetric Tensor By definition, A µν = A νµ,so = L µ β L α ν A βα = A µν (2) A νµ = L ν α L µ β A αβ = L ν α L µ β A βα antisymmetry is also preserved under Lorentz transformations. Contraction of a symmetric and antisymmetric tensor = L µ β L ν α A βα = A µν (3) A µν S µν =( A νµ )(S νµ ) (4) = (A νµ S νµ ) = (A µν S µν ) ( µ and ν are dummy indices) (5) Hence, A µν S µν =0. 1-1

2 Curl of grad, and div of curl f =(ê i i ) (ê j j f) = ijk i j fê k = ijk j i fê k ( f is well-behaved) = jik j i fê k = f (6) Also, f =0 (7) ( a) =(ê i i ) [(ê j j ) (ê k a k )] = δ il jkl i j a k = ijk i j a k = ijk j i a k ( f is well-behaved) = jik j i a k = ( a) (8) ( a) =0 (9) Problem 2 (a) The LHS is antisymmetric in (j, k) and (l, m), so it suffices to consider the case j = k and l = m (as otherwise, the LHS is zero). Substituting j = 2,k = 3 ( = i = 1) the LHS equals 123 1lm. either l = 2,m = 3 so that the LHS = +1 or l =3,m = 2 so that the LHS = 1. For i =1,j =2,k = 3, the RHS = δ 2l δ 3m δ 2m δ 3l which equals +1 if l =2,m= 3 and equals 1 ifl =3,m= 2. So the LHS = RHS for this permutation of indices. Similarly, we can show that the LHS = RHS for every permutation of the indices [1, 2, 3]. This establishes the identity, ijk ilm = δ jl δ km δ jm δ kl (10) Equivalently, if we use ijk ijk = 3! = 6 as the starting point, the general form has to be ijk ilm = C(δ jl δ km δ jm δ kl ) 1-2

3 where C is an appropriate normalization factor. Contracting indices (j, l) and (k, m) on both sides, we get which establishes Eqn. (10). (b) ijk ijk = C(δ jj δ kk δ jk δ kj ) = 6=6C = C =1 Using Eqn. (11) with l = j (with a sum on j), we get ijk ijm = δ jj δ km δ jm δ kj =3δ km δ km =2δ km ijk ilm =2δ km (11) (c) To Prove: A (B C) =B (C A) =C (A B) A (B C) =(a i ê i ) (ê l ljk b j c k ) (12) = ijk a i b j c k (13) B (C A) = ijk b i c j a k (14) = kji b k c j a i (15) = jki b j c k a i (16) = A (B C) (17) C (A B) = ijk c i a j b k (18) = jik c j a i b k (19) = kij c k a i b j (20) = A (B C) (21) Hence, A (B C) =B (C A) =C (A B) (22) 1-3

4 To Prove: (A B) =A ( B)+B ( A)+(A )B +(B )A The LHS is and the four terms of the RHS are (A B) =ê i i (a j b j ) the RHS is the sum of the four terms, given by which identically equals the LHS. Therefore, = ê i [( i a j )b j + a j ( i b j )] (23) A ( B) =ê i ijk klm a j l b m = ê i (δ il δ jm δ im δ jl )a j l b m (24) B ( A) =ê i (δ il δ jm δ im δ jl )b j l a m (25) (A )B = ê i a j j b i (26) (B )A = ê i b j j a i (27) ê i [( i a j )b j + a j ( i b j )] (28) (A B) =A ( B)+B ( A)+(A )B +(B )A (29) To Prove: (A B) =B ( A) A ( B) (A B) =(ê i i ) (ê k klm a l b m ) (30) = ilm i (a l b m ) = b m mil i a l a l lim i b m = b m ( A) m a l ( B) l = B ( A) A ( B) (31) (A B) =B ( A) A ( B) (32) To Prove: ( A) = ( A) 2 A ( A) =(ê i i ) (ê j jlm l a m ) (33) = ê k kij jim i l a m = ê k jki jlm i l a m = ê k (δ kl δ im δ km δ il ) i l a m = ê k i k a i ê k i i a k = ( A) 2 A (34) ( A) = ( A) 2 A (35) 1-4

5 Problem 3 Suppose the boosts are performed along the x-axis, and the (transformed) x -axis (which is parallel to the x-axis). Then, the first boost is given by L 1 = γ 1 γ 1v γ 1 v 1 γ where γ 1 =1/ 1 v2 1. Likewise, the second boost is given by The product of these boosts is L prod = L 2 L 1 = = L 2 = γ 2 γ 2v γ 2 v 2 γ γ 2 γ 2v γ 2 v 2 γ (36) (37) γ 1 γ 1v γ 1 v 1 γ γ 1 γ 2 1+ v 1 v 2 γ 1γ 2 (v 1 + v 2 ) 0 0 γ 1 γ 2 (v 1 + v 2 ) γ 1 γ 2 1+ v 1 v 2 c (38) Therefore, the product of the boosts is also a boost along the x-direction with speed v 3 given by The boost parameter γ of the combined boost is v 3 = (L prod) 0 1 (L prod ) 0 0 (39) γ 1 γ 2 (v c = 1 + v 2 ) γ 1 γ 2 1+ v 1 v 2 (40) v 3 = v 1 + v 2 1+ v 1v 2 (41) γ = γ 1 γ 2 1+ v 1v 2 (42) 1-5

6 Problem 4 Let T µ ν denote an orthogonal transformation, so that Now, under such an orthogonal transformation, T α µ T α ν = δ µ ν (43) δ µ ν (δ µ ν ) = T µ αt ν β δ α β (44) = T µ αt ν α = δ µ ν (using (44)) (45) Hence, δ µ ν is an invariant tensor under an orthogonal transformation. Similarly, g µν (g µν ) = T µ αt ν βg αβ (46) = T µ αt ω σ g ων g σβ g αβ δ α σ = T µ αt ω σ g ων δ α σ = T µ αt ω α g ων = δ µ ωg ων (using (44)) = g µν (47) Hence, g µν is an invariant tensor under an orthogonal transformation. Finally, under the orthogonal transformation, µνλρ ( µνλρ ) = T µ αt ν β T λ ωt ρ σ αβωσ (48) =det(t ) µνλρ (49) =+1 µνλρ (if T is a proper orthogonal transformation.) (50) Hence, µνλρ is also an invariant tensor under a (proper) orthogonal transformation. If the orthogonal transformation matrix has a determinant 1, then the components acquire a minus sign. This shows that µνλρ is a pseudotensor. Problem 5 By definition, Now, δf[x(s)] δx(s ) δf[ϕ(x)] δϕ(y) F [ϕ(x)+δ(x y)] F [ϕ(x)] F [x(s)+δ(s s )] F [x(s)] F [x(s)] + δ(s s ) df [x(s)] dx + O( 2 ) F [x(s)] (Taylor expanding) = δ(s s ) df dx (51) (52) 1-6

7 Also, taking F [x(s)] to be the identity functional, i.e. F [x(s)] = x(s) in (52), we have δx(s) δx(s ) = δ(s s ) dx dx = δ(s s ) (53) Next, taking F [x(s)] = dx(s) in (51), we have δ dx(s) δx(s ) d {x(s)+δ(s s )} dx(s) dx(s) + d δ(s s ) dx(s) = dδ(s s ) (54) δ dy sin(f 2 (y)) = lim dy sin([f(y)+δ(y x)] 2 ) dy sin([f(y)] 2 ) dy sin([f 2 (y)+2δ(y x)f(y)]) dy sin([f(y)] 2 ) dy sin([f(y)] 2 )+ dy cos([f(y)] 2 )sin(2δ(y x)f(y)) dy sin([f(y)] 2 ) dy cos([f(y)] 2 )sin(2δ(y x)f(y)) dy cos([f(y)] 2 )[2δ(y x)f(y)] =2f(x) cos([f(x)] 2 ) ( cos(θ) 1 for small θ) ( sin(θ) θ for small θ) where we have used the trigonometric identity sin(a + B) = sin(a) cos(b) + cos(a) sin(b). δ dy sin(f 2 (y)) = 2f(x) cos([f(x)] 2 ) (55) Note that this result could have been obtained more directly by using (52) with F [f(y)] = dy sin([f(y)] 2 ), so that δf[f(y)] d = dy δ(x y) df (x) sin([f(y)]2 ) d = dy δ(x y) df (y) sin([f(y)]2 ) = dy δ(x y)2f(y) cos([f(y)] 2 ) =2f(x) cos([f(x)] 2 ) 1-7

8 Similarly, δ dy sin(f (y)) = lim dy sin(f (y)+δ (y x)) dy sin(f (y)) dy sin(f (y)) cos(δ (y x)) + dy cos(f (y)) sin(δ (y x)) dy sin(f (y)) dy cos(f (y)) sin(δ (y x)) ( cos(θ) 1 for small θ) dy cos(f (y))[δ (y x)] ( sin(θ) θ for small θ) = dy cos[f (y)]δ (y x) = dy δ(y x) d dy cos[f (y)] = dy δ(y x)sin[f (y)] df (y) dy = f (x)sin[f (x)] δ dy sin(f (y)) = f (x)sin[f (x)] (56) 1-8