Lecturer: Bengt E W Nilsson

Transcription

1 9 4 3 Leturer: Bengt E W Nilsson 8::: Leturer absent, three students present. 8::9: Leturer present, three students. 8::: Six students. 8::8: Five students. Waiting a ouple of minutes to see if more ome. 8:3:: Six students again. 8:4:: So this is the hardore of the lass. 8:4:4: Seven students. Leturer wants to start. 8:5:: Eight students. Today you will see the first sign of one of the really important aspet of string theory: the Virasoro generators. You will see how fantasti it is. Let s reall a little bit what we have done: We had a relativisti string, with an ation: S = T Equation of motion: dτ dσ detγ αβ ) = T dτdσ ẋ x ) ẋ x ) τ P τ µ+ P σ µ=, P τ µ = L ẋ µ = ompliated! We need to simplify this situation not nie). Just to remind you: You have these Poisson brakets [q, p] PB =, and these you turn into [qˆ, pˆ] = i. We want to turn this into string theory, and we need anonial momenta. The above is above what we an do, so we need to simplify it. Turning quantities into operators is alled quantisation, but the proedure is not unique. The real physis is in the quantum mehanis: the lassial situation is a limit of the quantum mehanis. Going in the other diretion is not unique, but we have a guiding proedure quantisation). We use the oordinate freedom in τ and σ to simplify these expressions. ) Stati gauge: x τ, σ) = τ. Neumann boundary onditions the ends of the string move with v = v =. ) ẋ x =. The energy gauge : 3) These we ombine into is now gone. x ± ẋ ) =. Is true for the equations of motion: But there is a quadrati onstraint still, x ) + ẋ =. x = x σ x τ = x ± ẋ ) =. Solving it a new.

2 So what we will do today, that will solve all these problems: 9. Light-one relativisti string Another, even better gauge... this is a very important point, you have to understand why we do this weird step to go to the light one: it is all to avoid the ), an even better gauge is the light-one. Here all relations an be solved expliitly without any square roots. This will give us full insight into the set of independent anonial modes i.e. the full set of pairs p, q)!) To use this light-one gauge we need x ± τ,σ)= x τ,σ) ± x τ, σ) ) Pik the light-one gauge: x + α p + τ τ,σ) = α p + τ open string) losed string) The p + here is not the P that is a funtion of τ and σ, this is the integrated p. Here p + is the entre of mass momentum. It is a non-zero number that we an divide by. We put in the α for dimensional units. Units: From now on we use dimensionless τ and σ. Goal: To show that only the transverse string oordinates x τ, σ), p, q) are the independent variables or modes). Notation: We write n µ x µ = n x = x + using n µ =,,, use the notation for the stati gauge if we take another n µ, n µ =,, Then n xτ,σ)= β α n p)τ, x d τ, σ) plus one pair of, ). Nothing deep here, we an, ).) β = for the open string, β = for the losed string). This is a kind of generalised gauge ondition for τ-parametrisations. Choosing an n breaks Lorentz invariane. 9.: σ-parametrisations Before we used the onstant energy gauge along the string σ-oordinate). P τ, = T ds dσ v ) and the equation of motion τ P τ, = beause P σ =. Now we want to demand τ-independene of the more general quantity n P τ, and also set σ = π, just for onveniene. σ [, π] for all open strings. We also need this n P τ to be onstant along the string, i.e. independent of σ. One way to see that we an always ahieve this for one ombination of the string oordinates, i.e. n xτ,σ), is to note P τµ dσ. σ σ σ) dσ P τµ

3 This is lear from the fat that P τµ ds dσ x, x τ s ). v dσ /dσ is one funtion of σ. n xτ,σ) is one funtion of σ. Choosing one speifi ombination n xτ, σ) we an always let the σ-funtion dσ σ) be the dσ reiproal of n xτ, σ), i.e. n xτ, σ) is independent of both τ and σ. We an put this as follows: n p = πn P τ τ,σ)), p small This is the energy gauge ondition. Chek: n p. π π πn p = π n P τ τ,σ)dσ Equations of motion τ P τµ + σ P σµ =. n P σ is σ-independent. τ n P τ ) + σ n P σ ) = So n P σ is onstant along the string. What is the onstant? For the open string: We want this gauge to generalise the stati gauge. The x µ s involved must have Neumann boundary onditions. This means that n P σ = at the ends, and sine σ-independent we have n P σ = for all σ. For losed strings: Reall n P σ = ẋ x )n ẋ) ẋ n x ) πα ẋ x ) ẋ x ) Also here we have that n P σ = onst, but we do not know if it is =. Let us use the τ gauge ondition: n xτ,σ)=α n p)τ We take a σ derivative on this: n x τ,σ)= Consider ẋ x. n P σ = ẋ x )n ẋ) πα 3

4 Figure. So ẋ x = is possible to hoose at, let s say, σ =. So n P σ = at σ = and being onstant in σ, n P σ = for all σ. Note: The hoie of where on the losed string we set σ = is arbitrary, i.e. there is a remaining freedom of letting σ σ + σ where σ is a onstant. This is a rigid rotation of the oordinate system. It sounds ompletely useless, but it will turn out to be extremely ruial at a later point.) 9.3: Constrains and equations of motion n P σ = for all σ ẋ x = This looks ovariant in spaetime. That s a bit of an illusion. Compare ẋ x =, et. Then Then we dot this into n µ : [τ gauge: n x = β α n p)] P τµ = πα x ) ẋ µ ẋ x ) π n p σ gauge β n P τ = ẋ n ẋ) βα ẋ x ) x = ) ẋ x ) ẋ + x ) = 4

5 Combining it with ẋ x = we get: ẋ ±x ) =. Note: in the stati gauge, x being τ, this beomes ẋ ± x ) =. Also P τµ = πα ẋ µ P σµ = πα x µ ẍ µ x µ = 9.4: Wave equation and mode expansions We do the open ase first: ) Solve x µ =. x µ τ,σ) = f µ τ + σ) + g µ τ σ)) ) Neumann boundary onditions Dirihlet later). σ = : f = g g = f + onstant. x µ τ,σ)= f µ τ + σ) + f µ τ σ)) We let the arbitrary funtion absorb the onstant. 3) σ = π f τ +π)= f τ π) f is π periodi. f u) = f µ + a µ n os nu +b µ n sin nu), a µ n R, b µ n R. n= Integrate: fu)= f µ + f µ u+ A µ n osnu + B µ n sin nu) n= u: τ +σ, τ σ in x µ = fτ +σ)+ fτ σ)) x µ τ,σ)= f µ + f µ τ + n= A n µ os nτ + B n µ sin nτ) osnσ Note π ) P µ = dσp τ π µτ,σ)= dσ πα ẋ µ = π πα f µ = α f µ x µ τ,σ)=x µ +α p µ τ zero modes) + i α µ an e inτ a µ n e inτ) n= os nσ n 5

6 Simplify the notation: n =: α µ α p µ n : α µ n = n µ an, α µ n = n an µ µ ) α n = µ αn Open string x µ τ,σ)=x µ + α µ α τ + i α n n α µ n e inτ os nσ ẋ µ = α n Z 9.5: The use of the light-one gauge We have the onstraint α µ n os nσ e inτ, x µ = i α ẋ ±x ) =. n Z α n µ sin nσ e inτ We need to solve this without getting new square roots. In light-one gauge we avoid this effet! So we pik n µ =,,,, ). x + τ,σ)= β α p + τ p + σ = π β σ dσ P τ+ τ,σ ) n p = πβ n P τ ) Now use this in ẋ ± x ) =. In light-one oordinates, with µ = +,,I, with I =, 3, ) ) ) ẋ ± x ) = ẋ + ± x + ẋ ± x + ẋ I ± x I =, d: Solve for ẋ ±x = ) ẋ + ± x + ) ẋ I ±x I Is ẋ + ± x + a number? Well: x + =, ẋ + = βα p + a positive number). ẋ ± x = ) βα p + ẋ I ± x I Important. We learn: The whole mode expansion of x τ, σ), exept x, an be solved for in terms of x I modes! The independent anonial modes ) x I, p I. ) α n I, n=±, ±, [ I ] I α, α = 6

7 3) x, p +. We want to ompute the mass spetrum: M = p = p + p p I p I Reall ẋ ±x = 4α p + ẋ I ± x I ) = mode exp. = p + n Z L n e inτ ±σ), L n = p Z I I α n p α p L = α p = α α = L p + p Z I α p α I p = α I I α + p I) p= I I α p α p This lassial result M.. M = I α α I p α p p= 7