Lectures on String Theory

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1
R 3 µσνρx σ x ρ , this are Riemann normal coordinates. So that the deviation of g from
the flat metric η is only second order in x (this is a coordinate system of an observer
in free fall). Suppose we have a coordinate system x with metric expanded around
0,
g µν (˜x) = g µν (0) + ∂ σ g µν (0)˜x σ + 1 2 ∂ σ∂ ρ g µν (0)˜x σ ˜x ρ .
We want to transform this using a coordinate transformation ˜x → x = x(˜x) to
Riemann normal coordinates. We expand the coordinate transformation to third
order (zeroth order is zero)
x µ = ∂xµ
∂˜x ν ˜xν + 1 ∂x µ
2 ∂˜x ν ∂˜x σ ˜xν ˜x σ + 1 ∂x µ
6 ∂˜x ν ∂˜x σ ∂˜x ρ ˜xν ˜x σ ˜x ρ .
We can use the first term to bring g to η. Show that after we have done this there
are 1 ∂xµ
d(d − 1) remaining degrees of freedom left for , where d is the dimension.
2 ∂˜x ν
Where do these remaining degrees of freedom correspond to?
Show that we can bring ∂ σ g µν to zero using the second term of the transformation,
by counting degrees of freedom.
For arbitrary dimension there are not enough degrees of freedom to put ∂ σ ∂ ρ g µν to
zero. Count the number of remaining degrees of freedom and show that it equals
1
12 d2 (d 2 −1). This is precisely the number of independent components of the Riemann
tensor.
Exercise 14. Solve equation of motion □X µ = 0 for the case of open string (take
into account the open string boundary conditions).
where
Exercise 15. Consider solution of the closed string equations of motion
X µ (σ, τ) = X µ L (τ + σ) + Xµ R
(τ − σ)
X µ R (τ − σ) = 1 2 xµ + pµ
(τ − σ) +
i
4πT
∑
√
4πT
n≠0
n≠0
1
n αµ ne −in(τ−σ)
X µ L (τ + σ) = 1 2 xµ + pµ
(τ + σ) +
i ∑
√ nᾱµ 1
4πT
ne −in(τ+σ)
4πT
Derive the Poisson algebra of the variables (x µ , p µ , α µ n, ᾱ µ n) by using the fundamental
Poisson brackets of X µ (σ), P µ (σ) variables.