Solution to Problem Set 4

October 2017

Pb 1. 20 pts.

There are many ways of doing this problem but the easiest would be

ˆ ˆ eXˆ +Yˆ = eXˆ eYˆ e−[X,Y ]/2, (2) if [X,Y ] commutes with both X and Y . The displacement operator can then be expressed as

∗ † ∗ Dˆ(α) = e−α α/2eαaˆ e−α aˆ. (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 = − + (∇ × 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 = 2∇ · 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 + 2∇ × 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 differential 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.

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 |det g | p(d+1−p) µν k1 kp q1 qp ?(?X) = − (−1) δ ··· δ Xk ···k dx ∧ · · · ∧ dx p! q1 qp 1 p

|det gµν | = − (−1)p(d+1−p) X dxk1 ∧ · · · ∧ dxkp p! k1···kp p(d+1−p)q p(d+1−p) = − (−1) |det gµν |X = −(−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~ . First the canonical momentum of the point particle is dL ˙ P j = = mX˙ j − qAj ⇒ P~ = mX~ − qA,~ (40) dX˙ j where j stands for spatial coordinates. The momentum of the particle is shifted by qAˆ. Secondly we find the equations of motion of the point particle d ∂L ∂L dAj dφ d − =mX¨ j − q − q + q (A~ · V~ ) dt ∂X˙ j ∂Xj dt dXj dXj ∂Aj dφ d dXi dAj =mX¨ j − q − q + q (A~ · V~ ) − q ∂t dXj dXj dt dXi ∂Aj dφ d dAj =mX¨ j − q − q + q (A~ · V~ ) − qV . (41) ∂t dXj dXj i dXi We find that " # ¨ ∂A~ ∂A~ h i mX~ = q + q∇φ − q∇(A~ · V~ ) + q(V~ · ∇)A~ = −q − − ∇φ + V~ × (∇ × A~) = −q E~ + V~ × B~ (42) ∂t ∂t It shouldn’t be surprising that we found the Lorentz force. Then we extremize the action for the EM fields 1 Z Z 1 Z Z 0 = δ S = − d(δA) ∧ (?F ) + δA ∧ (?J) = − δA ∧ d(?F ) + δA ∧ (?J), (43) A 4 g2 g2 and we found d(?F ) = ?J, (44) (where J is a 1-form). Now define the current density use Dira delta functions, J µ = q δ(~x − X~ ), V~ δ(~x − X~ ) (45)

~ R µ µ where X(t) is the path of the particle (such that d~xAµJ = AµI x=X which we used before). Then do the Hodge star operation and you get the Maxwell equations with sources.

e) Chern-Simons term Now we consider the action Z Z S5 = c1 A ∧ dA + c2 A ∧ A ∧ A (46)

and extremize the action Z Z Z 0 = δAS5 =c1 δA ∧ dA + c1 A ∧ d(δA) + 3c2 δA ∧ A ∧ A Z Z =2c1 δA ∧ dA + 3c2 δA ∧ A ∧ A (47)

3c F := dA + 2 A ∧ A = 0 (48) 2c1 Here the cubic term is related to the structure constant of gauge group. Notice that in general we have non- 2 Abelian gauge theory (Yang -Mills Theory). For instance, usually we c2 = 3 c1 and we deﬁne ﬁeld strength (in explict form)

Fµν = ∂µAν − ∂ν Aµ + [Aµ,Aν ] (49)