Solutions to problem set 6

Transcription

1 Solutions to problem set 6 Donal O Connell February 3, Problem 1 (a) The Lorentz transformations are just t = γ(t vx) (1) x = γ(x vt). () In S, the length δx is at the points x = 0 and x = δx for all t. In S, we measure the length instantaneously, so δt = 0 = γ(δt vδx). Thus, δx = γ(1 v )δx = 1 δx (3) γ. (b) The charge density ρ in any frame is given by ρ = Q V (4) where Q is the total charge and V is the volume of the region enclosing the charge. If the charge is moving with some velocity v, we can choose the x-direction to be the direction of movement of the charge. Then lengths in the x-direction are contracted according to the length contraction formula, while lengths in the other directions are invariant. Moreover, the charge is a relativistic invariant since you can find the charge in some region by just counting it. The result is that ρ = γq V 0 = γρ 0 (5) where V 0 is the volume of the region enclosing the charge in the rest frame of the charge density. 1

2 (c) The current density is j = ρ v where v is the (three) velocity of the charges. But the four-velocity is u µ = (γ, γ v) so j i = ρv i = ρ 0 γv i = ρ 0 u i. (6) The fourth component of the current 4-vector is just the charge density, so indeed j µ = ρ 0 u µ. Problem (a) The Maxwell tensor F µν transforms as Therefore, F µν = Λ µ ρ Λ ν σ F ρσ. (7) F 01 = Λ 0 ρ Λ 1 σ F ρσ (8) = Λ 0 0 Λ 1 σ F 0σ + Λ 0 1 Λ 1 σ F 1σ (9) = Λ 0 0 Λ 1 1 F 01 + Λ 0 1 Λ 1 0 F 10 (10) = (Λ 0 0 Λ 1 1 Λ 0 1 Λ 1 0 )F 01 (11) = F 01, (1) where in the last step we used the fact that det Λ = 1 for a proper Lorentz transformation. (b) The electric field is E = F 01 = t A 1 x A 0. The potential A µ transforms as a covariant vector, so A 0 = γ(a 0 + va 1 ) (13) A 1 = γ(a 1 + va 0 ). (14) The derivatives also transform as a covariant vector, but it s useful to express the transformation the other way around: The transformation of E is t = γ( t v x) (15) x = γ( x v t). (16) E = ta 1 xa 0 (17) = t(γ(a 1 + va 0 )) x(γ(a 0 + va 1 )) (18) = (γ t vγ x)a 1 (γ x vγ t)a 0 (19) = t A 1 x A 0 = E. (0)

3 3 Problem 3 (Zwiebach problem 10.6) (a) A quick calculation shows that H µνρ is antisymmetric. For example, H µρν = µ B ν ρ + ρ B νµ + ν B µρ (1) = ( µ B νρ + ρ B µν + ν B ρµ = H µνρ. () Next, we are asked to show that H is invariant under the gauge transformation δb µν = µ ɛ ν ν ɛ µ. We calculate: H µνρ = H µνρ + δh µνρ (3) = H µνρ + ρ µ ɛ ν ρ ν ɛ µ + µ ν ɛ ρ ν µ ɛ ρ + ν ρ ɛ µ µ ρ ɛ ν (4) = H µνρ. (b) We just have to show δb = 0 for ɛ = µ λ: (5) δb µν = µ ɛ ν ν ɛ µ (6) = µ ν λ ν µ λ = 0. (7) (c) We want to show that we can always choose ɛ + = 1 (ɛ 0 + ɛ 1 ) = 0 with a suitable choice of λ. We have so we can choose ɛ + = 0 by choosing ɛ + (0) = ɛ + (p) + ip + λ(p) (8) (d) The action is S λ(p) = i ɛ+ p +. (9) d D x ( 16 ) H µνσh µνσ. (30) We will vary this action and set the variation equal to zero in order to find the equation of motion. δs = 0 d D xh µνσ δh µνσ = 0 (31) d D xh µνσ µ δb νρ = 0 (3) d D x µ (H µνσ )δb νρ = 0 (33) µ H µνσ = 0. (34) 3

4 In terms of B this equation of motion is In Fourier space this is just B νρ + µ ν B ρµ + µ ρ B µν = 0. (35) p B νρ + p µ p ν B ρµ + p µ p ρ B µν = 0. (36) (e) In Fourier space, the gauge transformation of B is δb µν = ip µ ɛ ν ip ν e µ. (37) Following the graviton discussion in the book, we want to set B +µ = 0. This is possible since B ++ = 0 by definition, and δb + = ip + ɛ ip e + = ip + ɛ (38) δb +I = ip + ɛ I ip I e + = ip + ɛ I (39) so we can choose ɛ and ɛ I to set B + and B +I to zero, respectively. This completely fixes the gauge freedom ɛ. Now let s look at the implications of these choices on the equations of motion. Taking ρ = + in (36) we find p + p µ B µν = 0 (40) so p µ B µν = 0. Expanding this, we find p + B ν = p I B Iν. (41) Notice that this equation determines B ν. We have already fixed B + so these constraints determine B J where J is a transverse index. In total, we have fixed B +, B +I and B I. All that remains are the transverse degrees of freedom B IJ. There are therefore (D )(D 3) (4) degrees of freedom. Now, since µ B µν = 0 the equation of motion simplifies to p B µν = 0. (43) (f) To quantise the theory, we assume that there exists some vacuum state Ω. This field is like a collection of point particles, one particle for each physical degree of freedom. Therefore there are creation and annihilation operators, labelled by 4

5 the energy-momentum of the state. In light cone gauge, the energy-momentum is determined by p + and p T, so these variable label the creation and annihilation operators. Since there is one particle for each physical degree of freedom, the creation and annihilation operators also carry indices describing which degree of freedom is being excited. In total, the operators must be a IJ and a IJ, where the operator is antisymmetrix under interchange of I, J. A given physical state is a superposition of the states a IJ Ω. (44) Hence, an arbitrary state must be of the form D 1 I,J= The matrix ζ is an antisymmetric matrix. ζ IJ a IJ Ω. (45) 5