Solution to Problem Set 4 October 2017 Pb 1. 20 pts. There are many ways of doing this problem but the easiest would be ∗ y a^y jαi =^ayD^(α) j0i =a ^y exp αa^y − α∗a^ j0i =a ^ye−α α/2eαa^ j0i ∗ ∗ @ α ∗ y @ α = + e−α α/2eαa^ j0i = + jαi ; (1) @α 2 @α 2 where we used the Zassenhauss formula (see full expression in Wikipedia or any advanced quantum textbook), which reduces to ^ ^ eX^ +Y^ = eX^ eY^ e−[X;Y ]=2; (2) if [X; Y ] commutes with both X and Y . The displacement operator can then be expressed as ∗ y ∗ D^(α) = e−α α/2eαa^ e−α a^: (3) Pb 2. 80 pts. For this problem, I highly recommended two books for reference. • Sean Carroll Geometry and Spacetime, an Introduction to General Relativity. This may be the standard textbook for GR class next semester. You will find a lot of good examples and exercises about Lorentz Indices in the first two chapters. • John Baez etc. Gauge Fields, Knots and Gravity. A very cute book tells about gauge fields and gravity as an aspect of differential geometry. You can learn a lot about the second problem in first two parts. In this problem, we use Greek letter α; β; γ; ··· ; µ, ν; ··· for spacetime Lorentzian indices 0; 1; 2; 3 and Latin letter i; j; k; ··· for spatial indices 1; 2; 3. Still we set c = 1. a) Maxwell Equation with Lorentz Indices In this problem we look at the covariant formulation of classical electromagnetism. As defined in the problem statement, the action S is Z 1 Z 1 Z S = d4xL = − d4xF F µν = − d4xF gµγ gνκF ; (4) g2 µν g2 µν γκ Fµν = @µAν − @ν Aµ = −Fνµ; (5) 1 and we're interested in finding the equations of motion for the electromagnetic 4-potential Aµ = (φ, A~), which can be determined from the Euler-Lagrange equations for fields, µγ νκ @L @L g g @(Fµν Fγκ) 0 = @β − = − @β 2 @(@βAα) @Aα g @(@βAα) µγ νκ g g @Fµν @Fγκ = − @β 2 Fγκ + Fµν g @(@βAα) @(@βAα) gµγ gνκ = − @ δβδα − δβδα F + δβδα − δβδα F β g2 µ ν ν µ γκ γ κ κ γ µν 1 = − @ gβγ gακ − gαγ gβκ F + gµβgνα − gµαgνβ F g2 β γκ µν 2 4 = − @ F βα − F αβ = − @ F βα; (6) g2 β g2 β and we found two of the Maxwell's equations in vacuum (Gauss electric law & Ampere law), βα @βF = 0: (7) The electromagnetic tensor also satisfies the Bianchi property @[µFνγ] = 0; (8) where [] is the alternating sum over all permutations. This is because partial derivatives commute (recall the definition of Fµν contains derivative). This identity leads to the other two Maxwell equations (Faraday law & Gauss magnetic law). The Maxwell equations with sources are µν ν @µF = J ; (9) and it means that the source terms should come from @L 4 ν = 2 J ; (10) @Aν g meaning the action should be modified as 1 Z 4 Z S ! − d4xF F µν − d4xA J µ: (11) 1 g2 µν g2 µ To write Maxwell Equation explicitly, we may take advantages of anti-symmetric tensor . First we want to i0 i ij ijk i i0 i ijk define E and B by F = E and F = − Bk and inversely we have E = F and B = − 2 Fjk. Starting with (9), first we set ν = 0 and µ = i = 1; 2; 3, (9) becomes i0 i @iF = @iE = ρ (12) which is Gauss's Law for electric field. Then we set ν = j = 1; 2; 3 and Ampere's Law for magnetic field appears @Ej @Ej @ F µj = @ F 0j + @ F ij = − − ijk@ B = − + (r × B~ )j = jj (13) µ t i @t i k @t As we mention above, the other two equations are from Bianchi identity. For simplicity, we rewrite Bianchi identity as µναβ µναβ µναβ @ν Fαβ = @ν (@αAβ − @βAα) = 2 @α@βAβ = 0 (14) Of course from (14) we will get an explicitly impression on why Bianchi identity holds for electromagnetic fields. For µ = 0, (14) is the Gauss's Law for magnetic field. Noticing 0ναβ = ijk, we have ijk ijk l i l ~ 0 = @iFjk = jkl@iB = 2δl @iB = 2r · B (15) For µ = i = i; j; k, calculation is a little bit more complicated but we can still work out @B~ (i) 0 = i0jk@ F + ij0k@ F + ijk0@ F = −ijk@ F − 2ijk@ F = 2 + 2r × E~ (i) = 0 (16) t jk j 0k j k0 t jk j k0 @t Till now we have derived all four Maxwell equations explicitly. 2 b) A glance at differential Geometry, Hodge Operator, and other properties on dif- ferential forms. Now we turn to differential geometry. In that formalism the action S1 takes the simpler form 1 Z S = − F ^ (?F ) ≡ S ; (17) 2 g2 1 µ A ≡ Aµdx (18) µ ν F ≡ Fµν dx ^ dx (19) µ d ≡ dx @µ (20) where the wedge product satisfies dxµ ^ dxν = −dxν ^ dxµ as stated in the problem. Now we can find the equations of motion using S2. We define the exterior derivative of p-form X as 1 dX = (dX ) ^ dxk1 ^ · · · ^ dxkp = @ X dxk1 ^ · · · ^ dxkp+1 ; (21) k1···kp p! [k1 k2···kp+1] which can be found from the definition of d given in the problem. • Some Useful Properties of Differential Form I Suppose we have a p-form X ≡ XI dx , where I represent multiple indices. { d2X = 0, this is because µ ν I d(dX) = @µ@ν XI dx ^ dx ^ dx = 0 (22) { d(X ^ Y ) = dX ^ Y + (−1)pX ^ dY , we can proof it as µ I J d(X ^ Y ) =(@µXI YJ + XI @µYJ )dx ^ dx ^ dy µ I J p I µ J =(@µXI YJ )dx ^ dx ^ dy + (−1) (XI @µYJ )dx ^ dx ^ dy =dX ^ Y + (−1)pX ^ dY (23) • Some Useful Properties of Hodge Operator. At first we will figure out ∗∗ is a kind of self dual. k1 kp I X = Xk1···kp dx ^ · · · ^ dx =≡ XI dx ; (24) pjdet g j µν k1···kp kp+1 kd+1 ?X = Xk ···k dx ^ · · · ^ dx ; (25) p! kp+1···kd+1 1 p pjdet g j pjdet g j µν kp+1···kd+1 µν k1···kp q1 qp ? (?X) = q ···q Xk ···k dx ^ · · · ^ dx ; (26) (d + 1 − p)! 1 p p! kp+1···kd+1 1 p where applying the Hodge star operator twice gives back a p-form, d + 1 − (d + 1 − p) = p. We will need the following identity q kp+1···kd+1q1···qp = −(d + 1 − p)!δq1 ··· δ p ; (27) kp+1···kd+1k1···kp k1 kp ... Using the identity above we have the following result kp+1···kd+1 k1···kp =(−1)d+1−pkp+1 kp+2···kd+1 k1···kp q1···qp kp+1···kd+1 q1···qp kp+1···kd+1 =(−1)p(d+1−p)kp+1···kd+1 k1···kp q1···qp kp+1···kd+1 p(d+1−p) kp+1···kd+1α1···αp k1β1 kpβp =(−1) gq1α1 ··· gqpαp g ··· g kp+1···kd+1β1···βp α = − (−1)p(d+1−p)(d + 1 − p)!g ··· g gk1β1 ··· gkpβp δα1 ··· δ p q1α1 qpαp β1 βp = − (−1)p(d+1−p)(d + 1 − p)!δk1 ··· δkp ; (28) q1 qp 3 which allows us to simplify jdet g j p(d+1−p) µν k1 kp q1 qp ?(?X) = − (−1) δ ··· δ Xk ···k dx ^ · · · ^ dx p! q1 qp 1 p jdet gµν j = − (−1)p(d+1−p) X dxk1 ^ · · · ^ dxkp p! k1···kp p(d+1−p)q p(d+1−p) = − (−1) jdet gµν jX = −(−1) X; (29) where det gµν = −1 in our case. Now we focus on the special case ?(?F ) = −(−1)4F = −F: (30) Let's look at the Hodge operator more carefully. The Hodge star operator (?) maps the p-form X to a (d + 1 − p)- form ?X as follows. Therefore, it is easy for us to construct a n−form from two p−forms A and B as A ^ ?B. And we have some such identity Z Z A ^ ?B = ?A ^ B (31) c) Maxwell equation in differential form µ ν 1 µ ν It is useful to realize that F = dA and this is very easy to verify F = Fµν dx ^dx = 2! (@µAν −@ν Aµ)dx ^dx = dA. From (22), we can derive second pair of Maxwell equations. dF = d(dA) = 0 (32) By action principle, we can extremize the action S2 to get the first pair of Maxwell equation. 1 Z 0 = δ S = − d(δA) ^ (?dA) + dA ^ (?d(δA)) A 2 g2 2 Z = − d(δA) ^ (?dA) g2 2 Z = − [d(δA ^ (?F )) + δA ^ d(?F )] g2 2 Z = − δA ^ d(?F ); (33) g2 where we used the identity, for a p-form X and a q-form Y , which is easy to show with what's given in the problem. We just found d(?F ) = 0: (34) Note that in the presence of sources you would find dF = 0; (35) d(?F ) = ?J: (36) Now we consider the action 1 Z Z S = − F ^ (?F ) + θ F ^ F; (37) 3 g2 and extremize the action 2 Z Z 2 Z Z 0 = δ S = − d(δA) ^ (?F ) + 2θ d(δA) ^ F = − δA ^ d(?F ) + 2θ δA ^ dF; (38) A 3 g2 g2 but since dF = d(dA) = 0, the equations of motion are unchanged: d(?F ) = 0. 4 d) A particle moving in fields Now we consider the action 1 Z Z m Z Z Z S = − F ^ (?F ) + θ F ^ F + dtX_ 2 + q A = dtL; (39) 4 g2 2 _ The 4-current due to the point particle is Iµ = q(1; ~X) = q(1; V~ ) (note: this is not a density), and thus qA = µ µ µ qAµdX = Aµ(qdX =dt)dt = AµI dt = qφdt − qA~ · V~ .