A First Course in String Theory.
(Second Edition).
Solutions for problems in Part II.
The following pages contain the solutions for problems to be found in Part II of A First
Course in String Theory, Second Edition. The handwritten solutions are due to Jeffrey Goldstone. Originally written for the first edition they have not been corrected for numbering.
Typewritten solutions are by the author except when indicated.
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
15
16
17
18
19
20
21
22
23
24
25
26
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Complete
Complete
Almost complete.
Almost complete.
Not available
Solutions by Jordan Rubinstein (not proofread).
Missing two last problems. Solutions by D. Gulotta and C. Featherstone.
Complete.
Complete.
Solutions by Jeffrey Falkenbach (not proofread).
Solutions by Andrew Morten (not proofread).
Not available.
Barton Zwiebach
MIT
Cambridge, MA
May 2009
15.7 D1-branes at an angle.
Figure 1: Two D1-branes, one horizontal and one at an angle γ .
Since the open string under consideration begins on the horizontal brane, at σ = 0 the
string coordinate X 3 vanishes. The beginning endpoint is free to move along x2 , so the
associated string coordinate X 2 satisfies a Neumann boundary condition. Thus,
X 2′ (τ, 0) = X 3 (τ, 0) = 0.
For the σ = π the endpoint lies on the line obtained from the horizontal by a counterclockwise
rotation with angle γ. To study this we use a rotated pair of axes (x′2 , x′3 ) such that the x′2
axis is aligned with the rotated brane and the x′3 axis is orthogonal to the rotated brane.
From the figure,
 ′2 
  2  2

x
cos γ sin γ
x
x cos γ + x3 sin γ
=
=
.
′3
3
2
3
x
− sin γ cos γ
x
−x sin γ + x cos γ
At σ = π the string satisfies the Dirichlet boundary condition x′3 = 0, so we write
−X 2 (τ, π) sin γ + X 3 (τ, π ) cos γ = 0.
Additionally, the endpoint is free to move along the x′2 coordinate, so the corresponding
string coordinate satisfies a Neumann boundary condition
X 2′ (τ, π) cos γ + X 3′ (τ, π) sin γ = 0.
1