PHYS606: Electrodynamics Feb. 01, 2011 Homework 1 Instructor: Dr. Paulo Bedaque Submitted by: Vivek Saxena Problem 1 µ Under a Lorentz transformation L ν, a rank-2 covariant tensor transforms as α β A A = L L A (1) µν −→ µν µ ν αβ Symmetric Tensor By definition, Aµν = Aνµ,so α β Aν µ = Lν Lµ Aαβ α β = Lν Lµ Aβα β α = Lµ Lν Aβα = Aµν (2) So, symmetry is preserved under Lorentz transformations. Antisymmetric Tensor By definition, A = A ,so µν − νµ α β Aν µ = Lν Lµ Aαβ = L αL βA − ν µ βα = L βL αA − µ ν βα = A (3) − µν So, antisymmetry is also preserved under Lorentz transformations. Contraction of a symmetric and antisymmetric tensor A Sµν =( A )(Sνµ) (4) µν − νµ = (A Sνµ) − νµ = (A Sµν)(µ and ν are dummy indices) (5) − µν ∵ µν Hence, AµνS =0. 1 -1 Curl of grad, and div of curl f =(eˆ ∂ ) (eˆ ∂ f) ∇ × ∇ i i × j j = ijk∂i∂jfeˆk = ijk∂j∂ifeˆk (∵ f is well-behaved) = ∂ ∂ feˆ − jik j i k = f (6) −∇ × ∇ So, f =0 (7) ∇ × ∇ Also, ( a)=(eˆ ∂ ) [(eˆ ∂ ) (eˆ a )] ∇ · ∇ × i i · j j × k k = δiljkl∂i∂jak = ijk∂i∂jak = ijk∂j∂iak (∵ f is well-behaved) = ∂ ∂ a − jik j i k = ( a) (8) −∇ · ∇ × So, ( 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 1231lm. So, either l =2,m =3so ⇒ that the LHS = +1 or l =3,m = 2 so that the LHS = 1. For i =1,j =2,k = 3, the RHS − = δ2lδ3m δ2mδ3l which equals +1 if l =2,m= 3 and equals 1ifl =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, ijkilm = δjlδkm δjmδkl (10) − Equivalently, if we use ijkijk = 3! = 6 as the starting point, the general form has to be ijkilm = C(δjlδkm δjmδkl) − 1 -2 where C is an appropriate normalization factor. Contracting indices (j, l) and (k, m) on both sides, we get ijkijk = C(δjjδkk δjkδkj) − = 6=6C ⇒ = C =1 ⇒ which establishes Eqn. (10). (b) Using Eqn. (11) with l = j (with a sum on j), we get ijkijm = δjjδkm δjmδkj − =3δkm δkm =2δkm − So, ijkilm =2δkm (11) (c) To Prove: A (B C)=B (C A)=C (A B) · × · × · × A (B C)=(a eˆ ) (eˆ b c ) (12) · × i i · l ljk j k = ijkaibjck (13) B (C A)= b c a (14) · × ijk i j k = b c a (15) − kji k j i = jkibjckai (16) = A (B C) (17) · × C (A B)= c a b (18) · × ijk i j k = c a b (19) − jik j i k = kijckaibj (20) = A (B C) (21) · × Hence, A (B C)=B (C A)=C (A B) (22) · × · × · × 1 -3 To Prove: (A B)=A ( B)+B ( A)+(A )B +(B )A ∇ · × ∇ × × ∇ × · ∇ · ∇ The LHS is (A B)=eˆ ∂ (a b ) ∇ · i i j j = eˆi [(∂iaj)bj + aj(∂ibj)] (23) and the four terms of the RHS are A ( B)=eˆ a ∂ b × ∇ × i ijk klm j l m = eˆ (δ δ δ δ )a ∂ b (24) i il jm − im jl j l m B ( A)=eˆ (δ δ δ δ )b ∂ a (25) × ∇ × i il jm − im jl j l m (A )B = eˆ a ∂ b (26) · ∇ i j j i (B )A = eˆ b ∂ a (27) · ∇ i j j i So, the RHS is the sum of the four terms, given by eˆi [(∂iaj)bj + aj(∂ibj)] (28) which identically equals the LHS. Therefore, (A B)=A ( B)+B ( A)+(A )B +(B )A (29) ∇ · × ∇ × × ∇ × · ∇ · ∇ To Prove: (A B)=B ( A) A ( B) ∇ · × · ∇ × − · ∇ × (A B)=(eˆ ∂ ) (eˆ a b ) (30) ∇ · × i i · k klm l m = ilm∂i(albm) = b ∂ a a ∂ b m mil i l − l lim i m = b ( A) a ( B) m ∇ × m − l ∇ × l = B ( A) A ( B) (31) · ∇ × − · ∇ × So, (A B)=B ( A) A ( B) (32) ∇ · × · ∇ × − · ∇ × To Prove: ( A)= ( A) 2A ∇ × ∇ × ∇ ∇ · − ∇ ( A)=(eˆ ∂ ) (eˆ ∂ a ) (33) ∇ × ∇ × i i × j jlm l m = eˆkkijjim∂i∂lam = eˆkjkijlm∂i∂lam = eˆ (δ δ δ δ )∂ ∂ a k kl im − km il i l m = eˆ ∂ ∂ a eˆ ∂ ∂ a k i k i − k i i k = ( A) 2A (34) ∇ ∇ · − ∇ So, ( A)= ( A) 2A (35) ∇ × ∇ × ∇ ∇ · − ∇ 1 -4 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 γ γ1v1 00 1 − c2 γ1v1 γ1 00 L1 = − (36) 0010 0001 v2 where γ =1/ 1 1 . Likewise, the second boost is given by 1 − c2 γ γ2v2 00 2 − c2 γ2v2 γ2 00 L2 = − (37) 0010 0001 The product of these boosts is γ γ2v2 00 γ γ1v1 00 2 − c2 1 − c2 γ2v2 γ2 00 γ1v1 γ1 00 Lprod = L2L1 = − − 0010 0010 0001 0001 v1v2 γ1γ2 γ1γ2 1+ c2 c2 (v1 + v2)00 − v1v2 γ1γ2(v1 + v2) γ1γ2 1+ 2 00 = − c (38) 0010 0001 Therefore, the product of the boosts is also a boost along the x-direction with speed v3 given by 0 v3 (Lprod) 1 2 = 0 (39) c −(Lprod) 0 γ1γ2 2 (v1 + v2) = c (40) v1v2 γ1γ2 1+ c2 So, v1 + v2 v3 = v1v2 (41) 1+ c2 The boost parameter γ of the combined boost is v v γ = γ γ 1+ 1 2 (42) 1 2 c2 1 -5 Problem 4 ν Let Tµ denote an orthogonal transformation, so that µ α µ Tα T ν = δν (43) Now, under such an orthogonal transformation, µ µ µ β α δ (δ ) = T T δ (44) ν −→ ν α ν β µ α = T αTν µ = δν (using (44)) (45) µ Hence, δν is an invariant tensor under an orthogonal transformation. Similarly, µν µν µ ν αβ g (g ) = T T g (46) −→ α β µ σ ων αβ = T αTω g gσβg α δσ µ σ ων α = T αTω g δσ µ α ων = T αTω g µ ων = δωg (using (44)) = gµν (47) Hence, gµν is an invariant tensor under an orthogonal transformation. Finally, under the orthogonal transformation, µνλρ µνλρ µ ν λ ρ αβωσ ( ) = T T T T (48) −→ α β ω σ =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, δF[ϕ(x)] F [ϕ(x)+δ(x y)] F [ϕ(x)] =lim − − (51) δϕ(y) 0 → Now, δF[x(s)] F [x(s)+δ(s s )] F [x(s)] =lim − − δx(s ) 0 → dF [x(s)] 2 F [x(s)] + δ(s s) + O( ) F [x(s)] =lim − dx − (Taylor expanding) 0 → dF = δ(s s) (52) − dx 1 -6 Also, taking F [x(s)] to be the identity functional, i.e. F [x(s)] = x(s) in (52), we have δx(s) dx = δ(s s) δx(s) − dx = δ(s s) (53) − dx(s) Next, taking F [x(s)] = ds in (51), we have d dx(s) δ dx(s) x(s)+δ(s s) =lim ds { − }− ds δx(s ) ds 0 → dx(s) d dx(s) + δ(s s) =lim ds ds − − ds 0 → dδ(s s ) = − (54) ds δ dy sin([f(y)+δ(y x)]2) dy sin([f(y)]2) dy sin(f 2(y)) = lim − − δf(x) 0 → dy sin([f 2(y)+2δ(y x)f(y)]) dy sin([f(y)]2) =lim − − 0 → dy sin([f(y)]2)+ dy cos([f(y)]2)sin(2δ(y x)f(y)) dy sin([f(y)]2) =lim − − 0 → dy cos([f(y)]2)sin(2δ(y x)f(y)) =lim − (∵ cos(θ) 1 for small θ) 0 ≈ → dy cos([f(y)]2)[2δ(y x)f(y)] =lim − (∵ sin(θ) θ for small θ) 0 ≈ → =2f(x) cos([f(x)]2) where we have used the trigonometric identity sin(A + B)=sin(A) cos(B) + cos(A)sin(B). So, δ dy sin(f 2(y)) = 2f(x) cos([f(x)]2) (55) δf(x) 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) sin([f(y)]2) δf(x) − df (x) d = dy δ(x y) sin([f(y)]2) − df (y) = dy δ(x y)2f(y) cos([f(y)]2) − =2f(x) cos([f(x)]2) 1 -7 Similarly, δ dy sin(f (y)+δ(y x)) dy sin(f (y)) dy sin(f (y)) = lim − − δf(x) 0 → dy sin(f (y)) cos(δ (y x)) + dy cos(f (y)) sin(δ (y x)) dy sin(f (y)) =lim − − − 0 → dy cos(f (y)) sin(δ(y x)) =lim − (∵ cos(θ) 1 for small θ) 0 ≈ → dy cos(f (y))[δ(y x)] =lim − (∵ sin(θ) θ for small θ) 0 ≈ → = dy cos[f (y)]δ(y x) − d = dy δ(y x) cos[f (y)] − − dy df (y) = dy δ(y x)sin[f (y)] − dy = f (x)sin[f (x)] So, δ dy sin(f (y)) = f (x)sin[f (x)] (56) δf(x) 1 -8.