� 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 per-fect 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 -interesting and simple case). Must solve: 1 ∂Hµνρ = Jµ κ2 ∂xρ 2Lecture 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. q Aµdxµ: 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 q Aµdxµ = q Aµ(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 δB µν (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 m mδSB = 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 phys-ical 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 neces-sary. 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: 6Lecture 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 1 1 1 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