Lecture 11 8.251 Spring 2007 Lecture 11 - Topics • Static gauge, transverse velocity and string action • Motion of free open string endpoints Reading: Section 6.6 - 6.9 ∂Pτµ + Pσµ = 0 ∂τ ∂σ δS = � dτ [δxµµσ]0 σ1 � τf dτ � σ1 dσ � ∂Pµτ + ∂Pµτ � δxµP − τi 0 ∂τ ∂σ σ σ 1 � τf dτ [δX0(τ, σ1)P (τ, σ1)−δX0(τ, 0)P (τ, σ1)+δX1(τ, σ1)P σ)(τ,σ1 )−δX1(τ,0)Pσ 0 0 1 τi σ ∂LµP = ∂x�µ σ = 0 and σ = σ1 are the endpoints of the string. [τi, τf ] is the time interval that the string is evolving over. String described at end by δxµ(τf , σ) (σ varies from 0 to σ1). How do we make this variation δS vanish? Let σ∗ ∈ {σ = 0, σ1} (so whatever we say about σ applies to both σ = 0 and ∗ σ1). δxµ(τ, σ )∗Impose a Dirichlet BC. Some xµ(τ, σ ) is a constant oas a function of τ.∗∂xµ ∂τ (τ, σ∗) = 0 Then δxµ(τ, σ ) = 0. But can’t impose Dirichelt BC on x0(τ, σ ). Time always ∗ ∗flows. Never a constant. 1Lecture 11 8.251 Spring 2007 Impose Free BCs: δxµ(τ, σ ) arbitrary ⇒ P∗since time flows) σµ(τ, σ ) = 0. Must include P∗=0σµ(τ, σ ) = 0 (again ∗D-Branes D2-Brane: 2 = number of spatial dimensions of the object DP-Brane: P = number of spatial dimensions (with free BCs) where endpoints can move freely BCs for motion of an open string on a D2-brane: x3(τ, σ ) = 0 ∗σ(τ, σ ) = 0 ∗0Pσ(τ, σ ) = 0 ∗1Pσ(τ, σ ) = 0 ∗2Pσ = 0 or σ1∗ D0-Brane: All strings forced to start and end at the point (string looks like a closed string but has different equations of motion because it can’t move away freely) 2Lecture 11 8.251 Spring 2007 D1-Brane: Looks like a string, but not actually one. Can have up to Dd-Branes. Have cartesian coordinates for: 1. String swejpg out spacetime surface by moving through time. 2. Actually matters whether 3Lecture 11 8.251 Spring 2007 Orientation matters. Will see this later. Static Gauge Will enable us to draw lines on the surface (2) from lines on (1) Consider line τ = τ0 or 1 Draw on worldsheet τ(Q) = t(Q) t = τ 4Lecture 11 8.251 Spring 2007 x 0(τ, σ) = ct = cτ Description of Coordinates: xµ(τ, σ) = {cτ, �x(t, σ)} = {cτ, �x(τ, σ)} Remember σ is not the length of the string - it’s a parameter, so σ1 is constant. BUT the string can elongate or shorten, of course. Some useful quantities: x˙µ(τ, σ) = (c, ∂�x/∂t) xµ�(τ, σ) = (0, ∂�x/∂σ) x˙ x� = ∂�x/∂t ∂�x/∂σ · · x˙2 = −c 2 + (∂�x/∂ t)2 x�2 = (∂�x/∂σ)2 Consider static string stretched between x1 = 0 and x1 = a x 1(τ, σ) = x 1(σ) (independent of τ) Plot x1(σ) vs σ Since Nambu-Gotta action, reparam-invar, doesn’t matter what path chosen as long as not e.g 5� � Lecture 11 8.251 Spring 2007 So: x˙µ = (c, 0,�0) ⇒ x˙2 = −c 2 xµ� = (0, dx1/dσ,�0) x�2 = (dx1/dσ)2 ⇒ Nambu-Gotta action: � � σ1 � T0 dx1 S = − c dτ dσ ( ˙x · x1)2 − (−c2)( dσ )2 � � 0 σ1 = −T0 dτ dσ(dx1/dσ) � 0 = −T0 dτ (x 1(σ1) − x 1(0)) tf = dt(−T0a) ti Recall S = (K − V )dt. String not moving so K = 0. For stretched string, V = tension distance. So physically, the tension of the string in constant. T0a:·potential energy of static string stretched to length a. Where did point P go from t to t + dt? No physical answer! So hard to talk about velocites. 6� � Lecture 11 8.251 Spring 2007 But let’s construct a velocity we can all agree on. Construct plane perpendicular to P . Say P �� moves to (P �)�� where P � is the intersection of the plane with the string at t + dt. This is called the perpendicular velocity. (String doesn’t actu-ally neccesarily move perpendicularly, but this well-defined quantity pretends it does). ds = |d�x| → d�x/ds = 1 d�x/ds is a unit vector tangent to the string. ∂�x ∂�x ∂�x ∂�x v� = ⊥ ∂t − ∂t · ∂x ∂s v 2 = (∂�x/∂t)2 − (∂�x/∂t ∂�x/∂s)2 ⊥ · Let’s simplify Nambu-Gotta action: 7� � � � Lecture 11 8.251 Spring 2007 � �2 � � �2�� �2 2 ∂�x ∂�x 2 + ∂�x ∂�x ( ˙x · x�)2 − x˙ x�2 = ∂t · ∂σ − −c ∂t ∂σ � �2�� �2 � �2� = ds ∂�x ∂�x +c 2 ∂�x dσ ∂t · ∂s − ∂t = ds [c 2 − v 2 ]dσ ⊥Nambu-Gotta action knew nothing mattered except perpendicular velocity. No way to tell how a point moves. ds vSo √. . . = c dσ 1 − c2 2 � � σ1 � 2 � ��ds vdt dσ ⊥ dt ds 2 /c2S = −T00 dσ 1 − c2 = −T0 1 − v⊥Recall: L = −m0c2 1 − v2/c2. T0ds: rest energy of small section of string. Consider a totally free open string (no D-branes) Pσµ(τ, σ∗) = 0 σµ T0 ( ˙x x�) ˙xµ − x˙2x�µ · P = − c √. . . � � � � �2� ∂�x ∂�x 2 ∂x ∂xµ ∂t ∂s ∂t ∂s T0 · + c − = − c2 � 2 /c21 − v⊥Note magic with the ds/dσ µ = 0: ∂�x ∂�x Pσµ = − Tc 0 � 1 ∂s −· v∂t /c2 |endpoint = 0 2 ⊥Numerator = 0 = ∂�x ∂�x = 0. So endpoint ∂�x ∂�x .∂s ∂t ∂s ⊥ ∂t · So at endpoint, either: 1. ∂�x/∂t⊥ string or 2. ∂�x/∂t = 0 But (2) can’t be, because: 8� Lecture 11 8.251 Spring 2007 P�σ = T0 1 − v2/c2∂�x/∂s = 0 v 2 = c 2 since ∂�x/∂s = 0 �Motion perpendicular to string always if free.