ALL 26 CHAPTERS COVERED
, A First Course in String Theory.
Second Edition.
Solutions for problems in Part I.
The following pages contain the solutions for problems to be found in Part I of the textbook
A First Course in String Theory, Second edition. The handwritten solutions are all due to
Jeffrey Goldstone. They are clear and elegant. If you distribute some solutions to students
please let them know that they are due to Goldstone. The rest of the solutions have been
typeset. They were mostly written by me, with some help by Stefanos Marnerides and Ian
Ellwood. Some of the solutions for the newer problems are due to Alan Guth. Other’s
contributions are indicated when applicable.
This file contains the solutions for all problems in Part I of the second edition, but
some edits remain to be done on the handwritten solutions concerning the numbering of the
problems and the equations they cite in the book (both of which changed from the first to
the second edition).
Barton Zwiebach
MIT
Cambridge, MA
26 May 2009
,Quick Calculation 2.2.
a′0 = γ(a0 − βa1 )
a′1 = γ(−βa0 + a1 )
a′2 = a2
a′3 = a3
b′0 = γ(b0 − βb1 )
b′1 = γ(−βb0 + b1 )
b′2 = b2
b′3 = b3
b′0 = −γ(b0 − βb1 )
b′1 = γ(−βb0 + b1 )
b′2 = b2
b′3 = b3
Then
a′µ b′µ = −γ(a0 − βa1 )γ(b0 − βb1 ) + γ(−βa0 + a1 )γ(−βb0 + b1 ) + a2 b2 + a3 b3
h i
= γ 2 a0 b0 (−1 + β 2 ) + a1 b1 (1 − β 2 ) + a2 b2 + a3 b3
= −a0 b0 + a1 b1 + a2 b2 + a3 b3 = a0 b0 + a1 b1 + a2 b2 + a3 b3
= aµ bµ .
1
, Quick Calculation 2.3.
Since we are leaving the coordinates x2 and x3 invariant, we only consider boosts along the
x1 axis, for which
x′0 = γ(x0 − βx1 ) .
To get
1
x′0 = x+ = √ (x0 + x1 ),
2
we would need
1
γ = √ , and β = −1 ,
2
p
which is not possible given that γ = 1/ 1 − β 2 .
More conceptually, under Lorentz transformations we must have
−(x′0 )2 + (x′1 )2 + (x′2 )2 + (x′3 )2 = −(x0 )2 + (x1 )2 + (x2 )2 + (x3 )2
If light-cone coordinates (x+ , x− , x2 , x3 ) defined a (primed) Lorenz frame we would have
−(x+ )2 + (x− )2 + (x2 )2 + (x3 )2 = −(x0 )2 + (x1 )2 + (x2 )2 + (x3 )2 (not true)
This requires
−(x+ )2 + (x− )2 = −(x0 )2 + (x1 )2 (not true)
Indeed, a short calculation shows that
−(x+ )2 + (x− )2 = −2 x0 x1 6= −(x0 )2 + (x1 )2 .
Light-cone frames are not Lorentz frames.
1
, A First Course in String Theory.
Second Edition.
Solutions for problems in Part I.
The following pages contain the solutions for problems to be found in Part I of the textbook
A First Course in String Theory, Second edition. The handwritten solutions are all due to
Jeffrey Goldstone. They are clear and elegant. If you distribute some solutions to students
please let them know that they are due to Goldstone. The rest of the solutions have been
typeset. They were mostly written by me, with some help by Stefanos Marnerides and Ian
Ellwood. Some of the solutions for the newer problems are due to Alan Guth. Other’s
contributions are indicated when applicable.
This file contains the solutions for all problems in Part I of the second edition, but
some edits remain to be done on the handwritten solutions concerning the numbering of the
problems and the equations they cite in the book (both of which changed from the first to
the second edition).
Barton Zwiebach
MIT
Cambridge, MA
26 May 2009
,Quick Calculation 2.2.
a′0 = γ(a0 − βa1 )
a′1 = γ(−βa0 + a1 )
a′2 = a2
a′3 = a3
b′0 = γ(b0 − βb1 )
b′1 = γ(−βb0 + b1 )
b′2 = b2
b′3 = b3
b′0 = −γ(b0 − βb1 )
b′1 = γ(−βb0 + b1 )
b′2 = b2
b′3 = b3
Then
a′µ b′µ = −γ(a0 − βa1 )γ(b0 − βb1 ) + γ(−βa0 + a1 )γ(−βb0 + b1 ) + a2 b2 + a3 b3
h i
= γ 2 a0 b0 (−1 + β 2 ) + a1 b1 (1 − β 2 ) + a2 b2 + a3 b3
= −a0 b0 + a1 b1 + a2 b2 + a3 b3 = a0 b0 + a1 b1 + a2 b2 + a3 b3
= aµ bµ .
1
, Quick Calculation 2.3.
Since we are leaving the coordinates x2 and x3 invariant, we only consider boosts along the
x1 axis, for which
x′0 = γ(x0 − βx1 ) .
To get
1
x′0 = x+ = √ (x0 + x1 ),
2
we would need
1
γ = √ , and β = −1 ,
2
p
which is not possible given that γ = 1/ 1 − β 2 .
More conceptually, under Lorentz transformations we must have
−(x′0 )2 + (x′1 )2 + (x′2 )2 + (x′3 )2 = −(x0 )2 + (x1 )2 + (x2 )2 + (x3 )2
If light-cone coordinates (x+ , x− , x2 , x3 ) defined a (primed) Lorenz frame we would have
−(x+ )2 + (x− )2 + (x2 )2 + (x3 )2 = −(x0 )2 + (x1 )2 + (x2 )2 + (x3 )2 (not true)
This requires
−(x+ )2 + (x− )2 = −(x0 )2 + (x1 )2 (not true)
Indeed, a short calculation shows that
−(x+ )2 + (x− )2 = −2 x0 x1 6= −(x0 )2 + (x1 )2 .
Light-cone frames are not Lorentz frames.
1
