Solution to Problem Set 4

Transcription

1 Solution to Problem Set 4 October 017 Pb 1. 0 pts. There are many ways of doing this problem but the easiest would be â α =â ˆD(α) 0 = â exp ( αâ α â ) 0 = â e α α/ e αâ 0 = α + α e α α/ e αâ 0 = α + α α, (1) where we used the Zassenhauss formula (see full expression in Wiipedia or any advanced quantum textboo), which reduces to e ˆX+Ŷ = e ˆXeŶ e [ ˆX,Ŷ ]/, () if [X, Y ] commutes with both X and Y. The displacement operator can then be expressed as ˆD(α) = e α α/ e αâ e α â. (3) Pb. 80 pts. For this problem, I highly recommended two boos for reference. Sean Carroll Geometry and Spacetime, an Introduction to General Relativity. This may be the standard textboo 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 boo 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 Gree letter α, β, γ,, µ, ν, for spacetime Lorentzian indices 0, 1,, 3 and Latin letter i, j,, for spatial indices 1,, 3. Still we set c = 1. a) Maxwell Equation with Lorentz Indices In this problem we loo at the covariant formulation of classical electromagnetism. As defined in the problem statement, the action S is S = d 4 xl = 1 g d 4 xf µν F µν = 1 g d 4 xf µν g µγ g νκ F γκ, (4) F µν = µ A ν ν A µ = F νµ, (5) 1

2 and we re interested in finding the equations of motion for the electromagnetic 4-potential A µ = (φ, A), which can be determined from the Euler-Lagrange equations for fields, L 0 = β L ( g µγ g νκ ) (F µν F γκ ) = β ( β A α ) A α g ( β A α ) ( g µγ g νκ [ ]) Fµν = β g ( β A α ) F F γκ γκ + F µν ( β A α ) ( g µγ g νκ [( = β δ β g µ δν α δν β δµ α ) Fγκ + ( δγ β δκ α δκδ β γ α ) ] ) Fµν = 1 g (( β g βγ g ακ g αγ g βκ) F γκ + ( g µβ g να g µα g νβ) ) F µν = g ( β F βα F αβ) = 4 g βf βα, (6) and we found two of the Maxwell s equations in vacuum (Gauss electric law & Ampere law), The electromagnetic tensor also satisfies the Bianchi property β F βα = 0. (7) [µ F νγ] = 0, (8) where [] is the alternating sum over all permutations. This is because partial derivatives commute (recall the definition of F µ ν contains derivative). This identity leads to the other two Maxwell equations (Faraday law & Gauss magnetic law). The Maxwell equations with sources are and it means that the source terms should come from meaning the action should be modified as S 1 1 g µ F µν = J ν, (9) L A ν = 4 g J ν, (10) d 4 xf µν F µν 4 g d 4 xa µ J µ. (11) To write Maxwell Equation explicitly, we may tae advantages of anti-symmetric tensor. First we want to define E and B by F i0 = E i and F ij = ij B and inversely we have E i = F i0 and B i = ij F j. Starting with (9), first we set ν = 0 and µ = i = 1,, 3, (9) becomes i F i0 = i E i = ρ (1) which is Gauss s Law for electric field. Then we set ν = j = 1,, 3 and Ampere s Law for magnetic field appears µ F µj = t F 0j + i F ij = Ej ij i B = Ej + ( B) j = j j (13) As we mention above, the other two equations are from Bianchi identity. For simplicity, we rewrite Bianchi identity as µναβ ν F αβ = µναβ ν ( α A β β A α ) = µναβ α β A β = 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 = ij, we have 0 = ij i F j = ij jl i B l = δ i l i B l = B (15) For µ = i = i, j,, calculation is a little bit more complicated but we can still wor out 0 = i0j t F j + ij0 j F 0 + ij0 j F 0 = ij t F j ij j F 0 = B (i) + E (i) = 0 (16) Till now we have derived all four Maxwell equations explicitly.

3 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 S 1 taes the simpler form S = 1 g F ( F ) S 1, (17) A A µ dx µ (18) F F µν dx µ dx ν (19) d dx µ µ (0) where the wedge product satisfies dx µ dx ν = dx ν dx µ as stated in the problem. Now we can find the equations of motion using S. We define the exterior derivative of p-form X as dx = (dx 1 p ) dx 1 dx p = 1 [ 1 X p+1]dx 1 dx p+1, (1) which can be found from the definition of d given in the problem. Some Useful Properties of Differential Form Suppose we have a p-form X X I dx I, where I represent multiple indices. d X = 0, this is because d(x Y ) = dx Y + ( 1) p X dy, we can proof it as d(dx) = µ ν X I dx µ dx ν dx I = 0 () d(x Y ) =( µ X I Y J + X I µ Y J )dx µ dx I dy J =( µ X I Y J )dx µ dx I dy J + ( 1) p (X I µ Y J )dx I dx µ dy J =dx Y + ( 1) p X dy (3) Some Useful Properties of Hodge Operator. At first we will figure out is a ind of self dual. X = X 1 p dx 1 dx p = X I dx I, (4) det gµν X = 1 p p+1 d+1 X 1 p dx p+1 dx d+1, (5) det gµν ( X) = (d + 1 p)! p+1 d+1 det gµν 1 p p+1 d+1 X 1 p dx q1 dx qp, (6) where applying the Hodge star operator twice gives bac a p-form, d + 1 (d + 1 p) = p. We will need the following identity p+1 d+1 p+1 d+1 1 p... Using the identity above we have the following result p+1 d+1 1 p p+1 d+1 =( 1) d+1 p p+1 p+ d+1 =( 1) p(d+1 p) p+1 d+1 = (d + 1 p)!δ q1 1 δ qp p, (7) p+1 d+1 p+1 d+1 1 p 1 p =( 1) p(d+1 p) g q1α 1 g qpα p p+1 d+1α 1 α p g 1β1 g pβp p+1 d+1 β 1 β p = ( 1) p(d+1 p) (d + 1 p)!g q1α 1 g qpα p g 1β1 g pβp δ α1 β 1 δ αp β p = ( 1) p(d+1 p) (d + 1 p)!δ 1 q 1 δ p q p, (8) 3

4 which allows us to simplify ( X) = ( 1) p(d+1 p) det g µν δq 1 1 δq p p X 1 p dx q1 dx qp = ( 1) p(d+1 p) det g µν X 1 p dx 1 dx p = ( 1) p(d+1 p) det g µν X = ( 1) p(d+1 p) X, (9) where det g µν = 1 in our case. Now we focus on the special case ( F ) = ( 1) 4 F = F. (30) Let s loo 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 A B = A B (31) c) Maxwell equation in differential form It is useful to realize that F = da and this is very easy to verify F = F µν dx µ dx ν = 1! ( µa ν ν A µ )dx µ dx ν = da. From (), we can derive second pair of Maxwell equations. df = d(da) = 0 (3) By action principle, we can extremize the action S to get the first pair of Maxwell equation. 0 = δ A S = 1 g [d(δa) ( da) + da ( d(δa)) ] = g = g = g d(δa) ( da) [d(δa ( F )) + δa d( F )] 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 Note that in the presence of sources you would find δa d( F ), (33) d( F ) = 0. (34) df = 0, (35) d( F ) = J. (36) S 3 = 1 g F ( F ) + θ F F, (37) and extremize the action 0 = δ A S 3 = g d(δa) ( F ) + θ d(δa) F = g δa d( F ) + θ δa df, (38) but since df = d(da) = 0, the equations of motion are unchanged: d( F ) = 0. 4

5 d) A particle moving in fields S 4 = 1 g F ( F ) + θ F F + m dtẋ + q A = dtl, (39) The 4-current due to the point particle is I µ = q(1, X ) = q(1, V ) (note: this is not a density), and thus qa = qa µ dx µ = A µ (qdx µ /dt)dt = A µ I µ dt = qφdt qa V. First the canonical momentum of the point particle is P j = dl dẋj = mẋj qa j P = m X q A, (40) where j stands for spatial coordinates. The momentum of the particle is shifted by qâ. Secondly we find the equations of motion of the point particle d L L dt Ẋj X j =mẍj q daj q dφ dt dx j + q d dx j ( A V ) =mẍj q Aj q dφ dx j + q d dx j ( A V ) q dxi da j dt dx i =mẍj q Aj q dφ dx j + q d dx j ( A V da j ) qv i dx i. (41) We find that m [ A X = q + q φ q ( A V ) + q( V ) A = q A ] φ + V ( A) [ ] = q E + V B (4) It shouldn t be surprising that we found the Lorentz force. Then we extremize the action for the EM fields 0 = δ A S 4 = 1 g d(δa) ( F ) + δa ( J) = 1 g δa d( F ) + δa ( J), (43) and we found (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) ) d( F ) = J, (44) where X(t) is the path of the particle (such that d xa µ J µ = A µ I µ x=x which we used before). Then do the Hodge star operation and you get the Maxwell equations with sources. (45) e) Chern-Simons term S 5 = c 1 A da + c A A A (46) and extremize the action 0 = δ A S 5 =c 1 δa da + c 1 A d(δa) + 3c δa A A =c 1 δa da + 3c δa A A (47) F := da + 3c c 1 A A = 0 (48) Here the cubic term is related to the structure constant of gauge group. Notice that in general we have non- Abelian gauge theory (Yang -Mills Theory). For instance, usually we c = 3 c 1 and we define field strength (in explict form) F µν = µ A ν ν A µ + [A µ, A ν ] (49) 5