Solutions to problem set 5

Transcription

1 Solutions to problem set 5 Donal O Connell February 16, Problem 1 Zwiebach problem 7.6) Since Xt, σ) = 1 Using ) F t + σ) F t σ) we have X t, σ) = 1 F t + σ) + ) F t σ). F u) = cos γ cos πu ], sin γ cos πu ]) and the formulae for cosa + B) and sina + B) we find where X t, σ) = cos γ sin πt sin πσ 1) ] cos β, sin β) ) β = γ cos πt cos πσ. 3) The string will be horizontal when the y component of X vanishes everywhere on the string, that is, when sin β = 0 for all σ. Hence, so t = /. b) From the definition of X we find X t = 1 cos πt = 0 4) F t + σ) F t σ)). 5) Inserting the definition of F and simplifying we find ] X t = sin γ cos πt cos πσ σ 1 ] sin γ sin πt sin πσ ]. 6) cos γ cos πt cos πσ 1

2 Hence, X t = sin γ sin πt sin πσ ]. 7) Now, since sin takes values in 1, 1], it is clear that if γ < π/ then the velocity never becomes 1. If γ = π/ then the velocity at the midpoint can become 1 when sin πt = 1 t =. 8) We already showed that at this time the string is horizontal. c) Now we take γ = π/. Hence, X t = π πt sin sin sin πσ ]. 9) At t = /4, sin πt/ = sin π/4 = 1/, so ] X π t = πσ sin sin = 1 σ =. 10) At the later time t = /3, sin πt/ = 3/ so X ] t = 3 sin π πσ sin = 1 σ = sin 1 ) 3 π, 11) that is, when σ 0.3 or σ 0.7. Near these points, cos γ sin πt sin πσ ] π cos + ɛ π ] ɛ π 1) Hence, X switches sign around these point. This is the signature of a cusp. d) I made an animated gif of the string motion which you can see by directing your browser to Figures 1, and 3 are pictures of the string at times t = 0, t = /4, and t = /3, respectively.

3 Figure 1: The string at time t = 0. Problem The coordinates transform as Figure : The string at time t = /4. x 0 = γx 0 βx 1 ) 13) x 1 = γ βx 0 + x 1 ) 14) with the other coordinates unchanged. The light cone coordinates therefore transform as x + = 1 x 0 + x 1 ) 15) 1 β = 1 + β x+ 16) Similarly, x = 1 + β 1 β x. 17) Now, if is the rapidity associated with β, then β = tanh. Hence 1 + β 1 β = 1 + tanh 1 tanh = e e = e, 18) 3

4 Figure 3: The string at time t = /3. as required. 3 Problem 3 Zwiebach problem 9.3) We are told that x 0 = x I 0 = 0, and all the oscillators are also zero except 1 = a and 1 3 = 1 3 = ia. a) The mass is given by 1 = M = 1 n=1 I n I n = a. 19) b) We know that X I τ, σ) = x 0 + I 0τ + i n 0 1 n I ne inτ cos nσ. 0) In view of the conditions we are given, it is clear that X I = 0 except for X and X 3 which are given by X τ, σ) = a cos σ sin τ 1) X 3 τ, σ) = a cos σ cos τ. ) Hence, the length of the string is l = 4a. c) From Zwiebach eq. 9.76, the transverse Virasoro modes are given by L n = 1 p= I n p I p. 3) 4

5 In our case, the only non-zero modes is L 0 which is given by Hence, by Zwiebach eq. 9.79, L 0 = 1 a + a + ia ia) + ia)ia)) = a. 4) X τ, σ) = a τ p +. 5) d) Since X 1 τ, σ) = 0, X 0 = X + = X. 6) Therefore, Since X 0 = t this tells us that e) From 7.66, we expect Since M = a, we find in agreement with our previous result. 4 Problem 4 p + τ = a τ p + p+ = 1. 7) t = p + τ. 8) l = E πt 0 = 4 M. 9) l = 4a, 30) a) The closed string is examined in detail in Chapter 13 so I will just state the results here. The oscillator expansion for closed strings in light-cone gauge is X + τ, σ) = p + τ 31) X τ, σ) = x τ + i e inτ n n e inσ + ᾱn e inσ) n 0 3) X I τ, σ) = x I 0 + I 0τ + i n 0 e inτ n I n e inσ + ᾱ I ne inσ). 33) 5

6 Note the presence of a new set of oscillators ᾱ µ n. These exist because on a closed string the left and right moving waves are not coupled by the boundary conditions. Nevertheless, periodicity of the string enforces a level matching condition: p= I p I p = p= ᾱ I pᾱ I p 34) b) The mass of the string is given by the same formula as for open strings, except we must include the contribution of the additional oscillators: M = n=1 ) I n n I + ᾱnᾱ I n I. 35) c) We are told that 1 = a and 3 = b. Their complex conjugates must also be non-zero but we take all the other oscillators to vanish. But we must also satisfy the level matching condition. To do this, we will take ᾱ1 = ᾱ1 = a and ᾱ 3 = ᾱ 3 = b. All other ᾱ oscillators are zero. Therefore, X τ, σ) = a sinτ σ) + sinτ + σ)). 36) X 3 τ, σ) = b sin τ σ) + sin τ + σ)). 37) This describes two standing waves of different frequency on the string. Taking X 1 = 0, we find X + = p + τ 38) so that we see that the motion is a standing wave in X +. 5 Problem 5 Zwiebach problem 10.) a) We use the orthogonality of the functions e i x in the box. We have d d xf x)e i q x = f) d d xe i q) x 39) = f)δ, q V. 40) 6

7 Hence, f) = 1 V d d xf x)e i x. 41) Now, inserting this result into the Fourier series, we find f x) = 1 d d x f x ) ) e i x x ). 4) V Since the delta function is defined by f x) = d d x f x )δ x x ) 43) we see that δ x x ) = 1 V b) The mode expansion for the field φ is so the conjugate momentum is e i x x ). 44) φt, x) = 1 1 ap e iept+i x + a pe iept i x) 45) V Ep Πt, x) = i V Ep Hence, the equal time commutator is φt, x), Πt, x )] = i V ap e iept+i x a pe iept i x). 46) 1 E p ap e iept+i x + a E pe iept i x, p, ] a p e ie pt+i x + a p e ie pt i x. 47) Since the only non-trivial commutator is a p, a k ] = δ p,k this simplifies to φt, x), Πt, x )] = i V ) e i x x ) + e i x x ) = iδ x x ). 48) 7