String theory-Dp Brane II

� Lecture 25 8.251 Spring 2007 Last time considered effects of Cal-Baron field: Action S: S = 1 √2πα� � dτdσ � ( ˙X · X�)2 − ˙X2(X�)2− 1 2 � dτ dσ ∂xµ ∂τ ∂xν ∂σ Bµν (x)− 1 6k2 � dDxHµνρHµνρ First term: String action Second term: Interaction Third term: Coupling to current � Interaction term rewritten as: − dDxBµν (x)Jµν (x) where current Jµν (x)= dτdσ ∂x ∂x δ(x − X(τ,σ))∂τ ∂σ · 1 ∂Hµνρ = Jµν k2 ∂xρ Antisymmetric in ρ and ν. Jµν = −Jνµ ∂ ∂Jµ 0= ∂xµ ⇒ ∂xµ Conservation index µ so we seem to have a collection of conserved currents. Jµν µ: conservation index, ν: labels for various currents J0k: string charge densities (to find charge, integrate over space). B0k couples to J0k. Conservation equation for ν = 0: ∂Jµ0 ∂J0µ ∂J0k =0 =0 =0 ∂xµ ⇒ ∂xµ ⇒ ∂xk 1 � � � � � Lecture 25 8.251 Spring 2007 J�o =(J01,J02,...,J0d). ( J�0 is a vector), so: ∂J0k ∂(J�0)k =0 =0 �· J�0 =0 ∂xk ⇒ ∂xk ⇒ String charge lives only on the string. Magnetic fields help us, but not a perfeec analogy. Perfect analogy: string charge is like stationary electric currents. Remember current conservation from E&M : ∂ρ �· J�+ ∂t =0 ∂ρ Stationary currents have ∂t =0 ⇒ �· J�= 0 e.g. a current on a closed loop, or an infinite wire. A current that, e.g. ends at a capacitor to charge it is not stationary. Open strings are problematic: charge flowing in string accumulates at ends. 1 ∂x0 ∂xk ∂xk ∂x0 J0k(t,�x)= 2 dτdσ ∂τ ∂σ − ∂τ ∂σ ·δ(t − x 0(τ,σ)) · δ(�x − �x(τ,σ)) Static gauge, x0 = τ 1 ∂xk J0k = dσδ(�x − �x(t,σ))2 ∂σ 1 ∂x J0 = dσδ(�x − �x(t,σ))2 ∂σ If we have a closed string, we’ll have a string charge density vector everywhere along the string. Now we know the direction of this vector: along the direction of increasing σ. String charge behaves like stationary electric current. Jik for a static string is 0. It has to do with the velocity of the string. Consider a static string in 3 + 1 dimensions (could be an infinitely long string interrestin and simple case). Must solve: 1 ∂Hµνρ = Jµ κ2 ∂xρ 2 Lecture 25 8.251 Spring 2007 Assume all H’s are time independent with Hijk = 0. Static string, so really two equations (µ,ν are i,j) 1 ∂Hijρ = Jij =0 κ2 ∂xρ ∂Xij0 ∂Hijk + =0 ∂x0 ∂xk Expanding over ρ index, which can take values 0 and spatial. 1 ∂H0νρ = J0ν κ2 ∂x0 Totally antisymmeteric, so can’t have other 0 indices, so all other indices spatial. 1 ∂H0kl = J0k κ2 ∂xl Let’s recast this equation as something more familiar. Let: H0kl = κ2�k0mBm Recall �123 = 1. Any reversal of index order changes its sign. Plug in: 1 ∂ (κ2�k0mBm)= J0k κ2 ∂xl �klm∂lBm = J0k This is the familiar (�× B)k =( J�0)k �× B = J�0 3 � � � � � � � Lecture 25 8.251 Spring 2007 So finding Kalb-Raman field of a magnetic field is mathematically equivalent to finding the electric current. We have Hµνρ whose E&M analogue is B�(the magnetic field). Bµν whose E&M analogue is A�(vector potential). J0k whose E&M analogue is J�(current). Action for coupling to Kalb-Raman field: ∂xµ ∂xν SB = − dτdσ ∂τ ∂σ Bµν (x) 1 = − 2 dτdσ�αβ ∂αxµ∂β x ν Bµν (x) where α,β ∈ 1, 2 and �12 = 1, �21 = −1, and ∂1 = ∂τ , ∂2 = ∂σ. α and β are coordinate indices on the worldsheet. qAµdxµ: coupling of E&M to a point charge. Gauge invariant? Yes! Things made with A have a hard time being gauge invariant. Reason above is gauge invariant. ∂xµ qAµdxµ = qAµ(x(τ))dτ(Let δAµ = δµ�)∂τ ∂�(x(τ)) ∂xµ = q dτ ∂xµ ∂τ ∂� = q (x(τ))dτ ∂τ = q(�(τ = ∞) − �(τ = −∞)) Gauge invariant if � vanishes in ∞ past or ∞ future. Good enough! δBµν = ∂ν Λν − ∂ν Λµ δBµν (x)= ∂∂xΛµν − ∂∂xΛνµ Λ = Λ(x) 4 � � � � � � � � � � � Lecture 25 8.251 Spring 2007 δSB = − dτdσ�αβ ∂Λν ∂αxµ∂β x ν ∂xµ = − dτdσ�αβ �αβ ∂αΛν ∂β x ν = − dτdσ∂α(�αβ Λν ∂β x ν ) = dτdσ ∂ (Λν ∂τ x ν )∂σ δSB = dτ [Λν ∂τ xσν =π − Λν ∂τ xσν =0] Consider now a string ending on a D-brane. What kind of violation of the gauge invar. will we get? aΛν ∂τ x ν =Λm∂τ x m +Λa∂τ x 0 since xa at string ends is constant mmδSB = dτ[Λm∂τ xσ=π − Λm∂τ xσ=0] String cannot end on the D-brane. We’ve accumulated so much evidence that this makes sense, but then we get stuck. Here’s where we need inspiration: 5 � � Lecture 25 8.251 Spring 2007 Possible hints: String conservation, charge conservation. Maybe string ends not so innocent -maybe they’re electrically charged (this would be good -then string theory would include electric charge, a very necessary element of a physicca theory). Approach: Let’s believe string endpoints are charged. Say σ = π has a positive charge, σ = 0 has a negative charge. ∂xm ∂xm S = SB + dτAm(x) dτAm(x)∂τ σ=π − ∂τ σ=0 δBµν = ∂µ∂ν − ∂ν ∂µ δBmm = ∂mΛn − ∂nΛm δAm = −Λm Kalb-Maron parameter is changing the magnetic field. Outrageous, but necessaary Then get gauge invariant. Such a strong gauge transformation, might wonder if our cure is worse than the disease. Not so: S = SB + SEM is gauge invariant But what have we done to Maxwell? Not so severe. Fmn used to be gauge invar. Now? δFmn = δ(∂mAn − ∂nAm)= −∂mΛn + ∂nΛm Not gauge invariant, but δFmn = −δBmn So Fmn + Bmn is gauge invariant. So in string theory with Kalb Raman fields, can’t just use F , must use: 6 Lecture 25 8.251 Spring 2007 (gauge invar.) This looks good. Bmn (gravitational, field of closed string) very small usually. So practically �mn ≈ Fmn most of the time. Before: 1 �mn = Fmn + Bmn SEM = − Fµν F µν 4 Now must write: 1 111 SEM = − 4 �mn�mn = − FmnF mn BmnBmn FmnBmn 4 − 4 − 2 Recall string charge B0kJ0k. So F0k couples to B0k -electric fields now have string charge! Charge at string endpoints create electric fields in the brane that continue to carry string charge.Maybe string is made of electric field lines all bunched up together that can flyin/out on the brane.But strings not just made of field lines ... doesn’t quite work.This model implies that particle-antiparticle annihilation are a closed string goiin off the brane.Quarks and QCD: Consider 3 D-branes (for 3 colors):7 Lecture 25 8.251 Spring 2007 1. Open string ending on the red brane is a red quark. 2. Blue quark. 3. Green quark. 4. Red anti-quark. (going away) Add in weak brane and leptonic branes. Call all these branes D4 branes filling world. 8 Lecture 25 8.251 Spring 2007 But string theory has extra dimensions. But all those branes on a torus. This model found to support all the particles we know and more we don’t. Not a complete story since we need symmetry-breaking. Maybe there’s supersymmettry so need some. Higgs boson from tachyons from intersecting branes. Not a 100% model, but a good first attempt. 9