Lecturer: Bengt E W Nilsson

Transcription

1 Lecturer: Bengt E W Nilsson From the previous lecture: Example 3 Figure 1. Some x µ s will have ND or DN boundary condition half integer mode expansions! Recall also: Half integer mode expansions in the NS-sector of the superstring: { br I, b s I } = δ r+s,0 δ IJ I p, b 1 p, etc. While in the R-sector we have integer mode expansions: 2 { dm I, d n J } = δ m+n,0 δ IJ { d 0 I, d 0 J } =δ IJ Spectrum: vacuum in the R-sector must be such that this commutation relation can be realised d 0 I γ I = spinor! 1

2 Chapter 16: String charge and electric charge Recall the action for a charged particle interacting with electromagnetism: S = m P ds + q A 1 2 P 4κ 0 d D x F µν F µν where A=A µ dx µ. If we are not in four dimensions, κ 0 2 will be a dimensionful constant. P = the path (the world line). ds = η µν dx µ dx ν F µν = µ A ν ν A µ with gauge invariance δa µ = µ Λ S int = q A = q A µ (x)ẋ µ dτ = q dτ P P P d D x δ D (x X(τ))A µ (x) ẋ µ with A µ (x) independent of τ. Can all this be generalised to the string? Electromagnetism: e : point charge world line A µ (one index) String:? world sheet B µν S int = B Σ 2 B µν dx µ dx ν Σ 2 Note: We use weight one forms and antisymmetric brackets: B 1 2 dxµ dx ν B µν dx µ dx ν 1 2 (dxµ dx ν dx ν dx ν ), where is the tensor product dx [µ dτ dx S int = B µν Σ2 µ dx ν dτ dσ dx ν] dτ 1 ( dx µ dx ν 2 dτ dσ dxν dτ ) dx µ dσ where S string = 1 2πα da Σ 2 Σ 2 B 1 6κ 2 d D x H µνρ H µνρ H µνρ µ B νρ + ν B ρµ + ρ B µν 3 [µ B νρ] Gauge invariant under δb µν = µ Λ ν ν Λ µ. 2

3 Now, in electromagnetism we have ν F µν = κ 0 2 j µ µ j µ =0 conservation of charge Q, Q j 0 d d x. String: ρ H µνρ = κ 2 j µν, where j µν is the current defined from S int = B = B µν ẋ µ x ν dτdσ dτ dσ Σ 2 Σ 2 Σ 2 d D xδ D (x x(τ,σ))b µν ẋ µ x ν d D xb µν j µν We get µ ρ H µνρ = κ 2 j µ j µν = 0 µ j µν = 0. (Compare to the stress tensor µ T µν = 0.) So j µν = ( j 0ν, j iν). j 00 = 0 because of antisymmetry. There are only j 0i in j 0ν. j iν = (j i0, j ij ) =( j 0i, j ij ). So the independent components are j 0i and j ij. j 0i d d x j 0i = Q i (vector charge, conserved) j ij vector (the i index) charge current (the j index) density. Implication of µ j µν =0? Electromagnetism: µ j µ = 0 t j 0 + i j i integrate = 0 7 charge conservation but not in the static case i j i =0 (a divergence-free current). String: µ j µν ν=0 =07 t j 00 + i j i0 = 0, i.e. i j 0i = 0 looks like a current that is divergence free. ν = j: t j 0j + i j ij = 0 vector charge conservation. A current that is divergence free cannot stop. i j 0i = 0 the string must be a loop (closed string), of if open it must be infinite, of if ending on D-branes, something new must happen. What? To understand these stringy currents and charges better, consider the static gauge. j µν (t, x)= dτdσδ(x 0 τ)δ d (x x(τ,σ)ẋ [µ (x ) ν] = j 0i = 1 2 dσ δ D 1 (x x(τ σ)) σ x i (t, σ) dσδ d (x x(t, σ)) dx[µ dt dx ν] dσ i.e. the vector string charge density is directed along the string from σ = 0 to σ = π, that is, B µν fells the orientation of the string. 3

4 Recall: Type I string = unorinented string. Start from a closed oriented string and identify the two orientations. This can only be done if the string we start from is orientation symmetric, i.e. IIB. x L x R, ψ 1 ψ 2. Example. In the bosonic string we can introduce the orientation flip operator Ω: Ωα n I Ω 1 =ᾱ n I Then we can keep only states that are orientation flip symmetric: g µν, B µν, φ. 1) If adding open strings. Type I string. 2) If adding closed heterotic string you get heterotic string. 16.2: Visualising the string charge Electromagnetism Figure 2. E da = ( E)dV = ρ dv = q S 2 = B 3 B 3 B 3 4

5 A magnetic version Figure 3. j da B dl = ( B) da = S 1 B 2 B c 2 = jtot c (through the surface B 2 ) 5

6 This can be taken over to the string: Let s be in D = 3+1 ρ H µνρ = κ 2 j µν Put H 0jk ε jkl B l (H) µν = 0j: ρ H 0jρ = k H 0jk =κ 2 j 0i B (H) = κ 2 j 0 just as in electromagnetism, and the above calculation will give the number of strings N instead of j tot. 16.3: Strings ending on D-branes D-branes has A µ fields on them! Is there some scalar associated to the A µ? Figure 4. Can the end point of the string couple to A µ? How is this related to the string vector charge? Recall from electromagnetism: Gauge invariance δa µ = µ Λ. Is S int = q A 6

7 gauge invariant? Vary A µ δs int = δa µ j µ d D x = ( µ Λ)j µ d D x= Λ µ j µ d D x+boundary term (in τ) =0 if µ j µ = 0. So gauge invariance is closely connected to charge conservation. What happens in the string? S B = dτ dσ xµ τ x ν σ B µν δb µν = µ Λ ν ν Λ µ δs B = dτ dσ τ x µ σ x ν ( µ Λ ν ν Λ µ )= = dτdσ (( τ Λ µ ) σ x µ ( τ x µ ) σ Λ µ )= = dτdσ( τ (Λ µ σ x µ ) Λ µ τ σ x µ σ ( τ x µ Λ µ ) + τ σ x µ Λ µ ) = (drop boundary terms in τ-direction) = dτ[ τ x µ σ=π Λ µ ] σ=0 These bound terms are located in the D-branes x µ x m, x a x m : -brane Neumann boundary conditions x a : -brane Dirichlet boundary conditions δs B = dτ [ Λ m ẋ m + Λ a ẋ a] σ=π σ=0 = dτ Λ m ẋ σ=π m dτλ m ẋ m σ=0 This must be made to vanish to save gauge invariance! How? Let us say that the end points actually have q = { + 1 at σ = { π with respect to A 1 0 µ! S tot =S B + dτ A m ẋ σ=π m dτa m ẋ m σ=0 Then define the gauge invariance as { δbµν = µ Λ ν ν Λ µ δa m = Λ m δs tot = 0. 7

8 F mn on the brane is not gauge invariant under this set of transformations. The gauge invariant field strength is really a new object F mn =F mn +B mn F mn F mn F mn F mn = F 2 + B 2 +FB which means that F is like a current of string charge. Figure : D-brane charge 8

9 If the string has a string charge coupling to B µν is this true also for D-branes? What possible fields do we have? In the bosonic string there are no fields like this (only g µν, B µν, φ). unstable! In the superstring: from the RR-sector (call these C µ,,ν): IIA. C µ C µνρ. C µ D0, C µνρ D2 All D-branes are IIB. C D( 1), C µν D1, C (+) µνρσ D3 Recall: q electric charge Figure 6. Q e = 1 E da = 1 F 0i da i = 1 ( F) ij da ij 4π S 4π 2 S 4π 2 S 2 ( F) ij = 1 2 ε ρσµνf µν = 1 2 ε ij0kf 0k m magnetic charge: [B i = 12 ε ijk F jk, F = 12 F ij dx i dx j ] Example: D0-brane in D = 10, electric charge Q m = 1 B da = 1 F 4π 4π S 2 9

10 Q e = Figure 7. 1 Vol(S 8 F 2 ) S 8 D6-brane. F 2 : F µν = µ C ν ν C µ 1 Q m = Vol(S 2 F 2 ) S 2 Figure 8. 10