A First Course in String Theory.
Solutions for problems in Part II.
Chapters 14, 15, and 16.
The following pages contain the solutions for all the problems to be found in Chapters 14,
15, and 16 of Part II of the textbook A First Course in String Theory. The handwritten
solutions are due to Jeffrey Goldstone. The rest of the solutions have been typeset. They
were mostly written by me. In chapters 14 and 15 the solutions of problems related to string
models were based on those of Daniel Gulotta and Clayton Featherstone.
Barton Zwiebach
MIT
Cambridge, MA
September 2004

Quick Calculations 14.4 to 14.7.
QC14.4 ^(^1,^2) = ~2. The intersection points on the torus are @,0) and (|, |). £2 is rotated
by —vr/2 with respect to £\ (clockwise rotation), so the sign is as expected.
QC14.5
QC14.6 The points with integer coordinates (lattice points) on the torus С are copies of the
point @, 0) on the unit torus, so by the cited 'known' fact, there must by a total of /
lattice points on C. One lies on the corners of the parallelogram. Since the vectors £\
and £2 are reduced, no further lattice points lie along the edges of the parallelogram.
Therefore the other / — 1 lattice points must lie on the interior of the cell C.
Viewed as lines on C, both £\ and £2 each contain the lattice point represented by @, 0).
They do not contain additional lattice points. The same facts hold for any copy of £\
or £2, because copying maps lattice points to lattice points. It follows that all copies of
£\ and £2 go through interior lattice points. Since each copy contains just one lattice
point on C, one copy of £\ and one copy of £2 goes through each lattice point on the
interior of C.
QC14.7 Here is a systematic solution. Associated with the vector £ = (m,n), with m and n
relatively prime integers, we consider the set of lines parallel to £ that go through lattice
points. Such lines have slope n/m, and if they pass through the lattice point (p, q) are
written as nx — my = np — mq. Since n and m are relatively prime, as we vary p and
q, np — mq can take on any integer value (this is a theorem in arithmetic). So the lines
in question are of the form nx — my = k, for arbitrary к G Z. The intersection points
of the lines £\ = (mi,ni) and £2 = (m2,n2) are found by solving the simultaneous
equations n\x — m\y = k\ and П2Х — 1П2У = k2, for arbitrary k\ and &2, and selecting
the solutions (ж, у) inside the unit cell.
For our lines C,2) and A,2), the equations are 2x — 3y = k\ and 2x — у = /с2, which
gives x = {—k\ + 3/^2)/4 and у = {—k\ + &2)/2. Plugging in a few values for k\ and &2
we obtain, as expected, four intersections: @,0), (\,^), (^,0), (§,?;)•

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14.7 Dl-branes at an angle.
Figure 1: Two Dl-branes, one horizontal and one at an angle 7.
Since the open string under consideration begins on the horizontal brane, at a — 0 the
string coordinate X3 vanishes. The beginning endpoint is free to move along x2, so the
associated string coordinate X2 satisfies a Neumann boundary condition. Thus,
Х2'(т,0)=Х3(г,0) = 0.
For the a — it the endpoint lies on the line obtained from the horizontal by a counterclockwise
rotation with angle 7. To study this we use a rotated pair of axes (x'2, xri) such that the x'2
axis is aligned with the rotated brane and the ж'3 axis is orthogonal to the rotated brane.
From the figure,
cos 7
— sin 7 cos 7/
x2 cos 7 + x:i sin 7 \
—x2 sin 7 + x3 cos 7/
At a = 7Г the string satisfies the Dirichlet boundary condition ж'3 = О, so we write
-X2(t, 7r)siii7 + X3(r, tt)cos7 = 0.
Additionally, the endpoint is free to move along the x'2 coordinate, so the corresponding
string coordinate satisfies a Neumann boundary condition
X2'(t,
-0.

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14.10 Oblique tori and orientifold action.
The fundamental domain of the torus can be represented as
ь
a
Figure 1: Fundamental domain of the rectangular torus.
X
(a) The action (j\ y) —>• (ж, —у) is well defined. How could it fail to be so? It could fail if
the action mapped a point to two different points. This can easily happen because we have
given a formula ici: the action in terms of coordinates, and a point in the torus is represented
by sets of equivalent coordinates. It can happen that the action takes equivalent coordinates
to coordinates corresponding to inequivalent points. So we test
■о., у)
(x + a,-y) ~ {x,-y),
0, -y - b) ~ (x, -y),
(x,y) ~
(x,y) ~ ir,y
showing that the action respects the equivalence relations.
A point represented by (ж, у) is a fixed point if
(x,y) ~ (x,-y)
This is clearly satisfied by у — 0. But given the vertical identification, it is also satisfied if
(ж, у) — (ж, —у + nb),
This gives fixed lines
п €
2
For n even these are all equivalent (by the torus identification) to the у — 0 line. For n
odd these are all equivalent (by the torus identification) the the у = 6/2 line. This is the
additional fixed line on the torus.
Consider a point (x. y). Reflection around the у — 6/2 line takes у —> у' where y — \ = \ — y\
that is y' — b — y. So the reflection takes (x, y) —> (x,b — y). Meanwhile, reflection by the

original orientifold action gives (x, —y). Under toroidal identifications, however, we have
(x, ~y) rsj (x,b — y), so that the two reflections are equivalent.
(b) The identifications produce the oblique torus with fundamental domain V shown here:
If
V
/
a
a
l
Figure 2: Fundamental domain of the oblique torus.
The fundamental domain of the torus when a^ — b/2 is displayed below
V
b
b.
2
у
в
A
/
y = 0
Figure 3: Oblique torus with <ii — b/2.
To test that the orientifold action is well defined it is easiest to look at the corners of V■ We
know that @,0) ~ (аьаг)- The orientifold action leaves @,0) invariant, so we must also
have (oi, a^) ~ (ai, —CL'i). This is possible if аг = —Q-2 + nb, with n an integer. With n — 1
we get U2 = b/2, which is the case we are interested in (it is possible to show that for other
values of n one actually obtains the same torus).
2

To prove that the orientifold action is well defined we simply note that the horizontal
identification is compatible with the action, and the oblique identification requires the last
equivalence in
(x, у) ~ (ж + аъу + a2) -* (x + ab -y - a2) ~ (x, -y)
Indeed,
(x + au-y - n-i) ~ {x + аг, -у - a2) - (a1: a2) + @, b) - (z, -y - 2a2 + 6)
and for a2 = 6/2, the last coordinate pair is just (x, —y).
As before, we have a fixed line у = 0, which presented inside V is the lim- у = b lahHrd
В in the figure. The fixed line у — 6/2 is labeled A. These two lines are joined at their ends
and form one closed line on the torus. This can be seen as follows: The right end of line A
is identified with the left end of line В since all the corners of the fundamental region are
identified. Similarly, the left end of the line A is identified with the right end of the line A
under the shift x —^ x + ai and у —> у + 6/2.
(с) We have
а — Ъ — (ai, a2) — @, 2a2) — {a\, — a2) = a*.
This means that a* is just the image under the orientifold action of a. Note also that 6—^—6
under the orientifold action. Hence, the orientifold takes
tia + t2b -+tia* -t2b = t\{a- b) -t2b = t:a- (^
(d) The fact that the intersection number remains the same follows from the fact that oblique
tori are equivalent to square tori under an invertible linear transformation of coordinates.
To explain this, note that the linear transformation
1 1 1
x-+—x, у -*-— x + ~ y,
aj 2ai b
maps the oblique torus V to the square torus with sides of length one. Indeed, the corners
go to each other (check that @, 0) ->• @, 0), @, b) -> @,1), {alya2) -* A,0)) and since lines
go to lines, the edges of V are mapped to the edges of the unit torus. It follows that a line
t — (m, n) on V will map to a line also denoted by (m, n) on the unit torus. Since the linear
transforEiiibtion is one to one, the intersection number of any two lines is preserved under
the map. So we can use the formula we had for square tori to compute intersection numbers
on V.
(e) We have that the orientifold action takes n —* — m — n and leaves m alone. Hence,
~2
[r, s\ = [m,n + —\ -> [m, -m - n + — = [m, -n - — \ = [r, -s\.
(f) The interdiction is computed usm£ tli*: (- . ■) bsusis. *ч we invert tlit- ba^is rhan.^1 uj L!iirn
that [ris Sj] ~ (гг, Si — Ti/2) for i — 1,2. Hence the intersection number is given by
=*t[?
si_?ri/2 S2:

Using the result of part (e), we have
r2(-Sl) = -
t2 ■
(g) Consider the figure
Figure 4: Orientifold line on the oblique torus.
By shifting the line В down by x —> x — b to give the line J3' and also shifting the line
A by ж — h h о to give the line A'i we have represented the fixed line as a single closed line
stretching fmm the origin to the point 2a — b. Hence the fixed line can be reprrsrntfd л>
B, -1) which is equivalent to [2,0].

Quick Calculations 15.4 to 15.14.
QC15.4 q+q'+ + q_q'_ = ±[(9l + q2)(q'1 + q'2) + (qi - q2)(q'1 - q'2)] =
q2q'2
QC15.5
M =
\
l
f
2
"ч/б
ч/2 ч/З
0 Тз у
,мтм =
QC15.6 Quantum field theory requires that the charge of the right-handed particle is opposite to
the charge of the left-handed antiparticle. Therefore, the right-handed and left-handed
particles do not have the same charges (chiral fermion) if and only if the left-handed
anti-particles and left-handed particles do not have opposite charges.
QC15.7 The gluons that begin and end on the same brane each have weight @,0) since they
have qi = q2 = Яз = О. The eight gluon charges are @,0), @,0), B,-1), (-2,1),
(-1,2), A,-2), A,1), and (-1,-1).
QC15.8 uR ~ C,lJ/3, d
R
,l)_b ueR ~ A, lH, and
A,2I/2-
QC15.9 uL has h = \ and Y = \
uR has /з = 0 and Y = |
dL has /3 = — I and Y = |
с?д has /3 = 0 and У = — |
e^ has /3 = — I and У = — |
e^ has /3 = 0 and Y = — 1
z/eL has /з = | and Y = \
veR has /3 = 0 and У = 0
QC15.10 ul has electric charge 0 — | = — |
J^ has electric charge 0 + | = |
z/eL has electric charge \ — \ = 0
Pei has electric charge 0 + 0 = 0
e^ has electric charge — | — | = — 1
e^ has electric charge 0 + 1 = 1
Since the electric charges of left-handed particles and antiparticles are opposite, elec-
tromagnetism does not couple chirally.
1

QC15.11 /12 = (A)A) - B)(l))((l)(-2) - (-1)A))(A)E) - (-2)(-l)) = 3
- (-2)(-l)) = -3
= 0
= 0
= -3
/23 = (A)A) - (l)(l))((l)(O) - (-2)A))((-1)E) - E)(-l)) = 0
/м = (A)B) - A)A))(A)A) - (-2)(-l))((-l)(l) - E)A)) = 6
/и = (A)B) - A)A))(A)A) - (_2)(-l))((-l)(-7) - E)B)) = 3
/26 = (A)A) - (l)(l))((l)(-4) - (_2)C))((-l)(-5) - E)A)) = 0
/34 = (A)B) - A)A))(A)A) - (O)(-l))((-l)(l) - E)A)) = -6
/35 = (A)B) - A)A))(A)A) - @)(-l))((-l)(-7) - E)B)) = -3
/зб = (A)A) - (l)(l))((l)(-4) - @)C))((-l)(-5) - E)A)) = 0
/45 = (A)B) - B)A))((-1)A) - (l)(-l))((l)(-7) - A)B)) = 0
/46 = (A)A) - B)(l))((-l)(-4) - (l)C))((l)(-5) - A)A)) = 6
/56 = (A)A) - B)(l))((-l)(-4) - (l)C))(B)(-5) - (-7)A)) = 3
It is not difficult to see from the diagram that the string directions in the diagram are
the correct directions for left-handed strings.
QC15.12 Y =
[24]:
[34]:
[46]:
[25]:
[35]:
[56]:
— 3Ц/1 ~ '2Q2 ~
Y = -(-l) =
Y = 0
Y = -W)~
Y = -(-!)-
Y = -1
-Qs-Q
1
2
1
(-i) = l
1 = 0
QC15.13 Nq (right Y = 0) 9 incoming quarks from N\ C families and 3 colors). 6 outgoing
neutrinos to N4 and 3 outgoing strings to N5.
N3 (right Y = — 1) 9 incoming quarks from JVi C families and 3 colors). 6 outgoing
positrons to N4 and 3 outgoing antineutrinos to N5.
N4 (leptonic Y = 0) 6 outgoing lepton doublets for 12 total strings to N2. 6 incoming
positrons from N3 and 6 incoming antineutrinos from Л^.
N5 (leptonic Y = — 1) 3 outgoing strings doublets for 6 total strings to N2. 3 antineu-
antineutrinos incoming from Л^з and 3 incoming strings from Л^.
QC15.14 If S is a compact space with no boundary: dS = 0. By the divergence theorem,

is~.i n I- i
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1чаи- I— "Jourc/oue-rvS ^ ^*л*е— -гв^>*Дал-
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CL>O
я or—
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—>
15.3. Properties of the string charge Q.
(a) We are told to evaluate
Q = J dtx? = J <?x\ f da S (x - X{tQ,a)) X'(to,a).
To perform the integral we first change the order of integration:
Q = i I da X'(t0, a) f ddx S (x - X(t0,
The integral over 1Z above is just a function f(a) of a. The function f(a) is either zero or
one. We have f(a) = 1 if X(t0, a) G 1Z and f(a) = 0 otherwise. As a result, the a integral
is constrained to the interval a G [сг^сгу] for which the string is inside TZ
da X'(to,a) = ^ (x{to,<Tf) - X(to,a^ = ±{xf - xt).
For a closed string contained entirely in TZ, the a integral is over the whole range a G [0, 2тг]:
-. 1
daX'(to,a). A)
z Jo
Since the integrand is a a derivative of a well-defined coordinate X (recall that we assume
no compactification of space), we have
4 =0.
(b) Let g(x) be an arbitrary function of x. Consider a volume 1Z which encloses the entire
region where j° ф 0 so that j° = 0 on the boundary dTZ of 1Z. By the divergence theorem,
we have
in
Hence we also have
ddxV ■ (gj°) = f dd-1xg(n-j°) = 0.
Jon
0= / ddxV-(gj°)= / ddx \(Vg)-j° + g(V-j0)} = / ddx(Vg)-j°,
Jn Jn ^ } Jn
since by assumption j° is divergenceless. Setting g = x\ we have Vg = e*, the unit-vector
in the ith direction, and we obtain 0 = jn ddx (j°)%. Since this holds for all values of i:
Q= [ ddxf = 0.
Jn
1

(с) We can still apply equation A) from part (a)
- 1 f27T
Q = - da X'(t0, a).
z Jo
Let x1,... xd, denote the list of spatial coordinates. Assume x1,..., xd~1 are not compactified,
and that xd = x is the coordinate curled up into a circle of radius R. It follows that the string
coordinates X1,..., Xd~x are all single-valued while Xd = X is multivalued (X ~ X + 2irR).
By the same argument used in (a) the first d—1 components of Q vanish. The string charge
therefore takes the form
w
here
Г daX'(to,a)
Jo
The above integral is unchanged under possible reparameterizations of a. We can therefore
evaluate it using the simple parameterization X = Ra. This parameterization indeed rep-
represents a wrapped closed string: as a goes from 0 to 2vr the coordinate X changes by 2ttR.
Despite the fact that the coordinate increases in value throughout the string, because of the
identification, the endpoint coincides with the original point. Since X' = R, the integral
gives
Q = - ■ 2vr • R = ttR.
(It is also possible evaluate the integral by writing X(to,2vr) — X(to,0) = 2irR.) The string
charge Q is proportional to the length of the string. This is similar to the fact noted in
A5.72): the Maxwell charge associated to a wrapped D-brane is proportional to the volume
of the brane. We will see in Problem 17.3 that a string wrapped on a circle represents an
electrically charged state for a lower-dimensional observer.

15.4. Kalb-Ramond field of a string.
We recall equation A5.36)
[ BH-d£ = к2 I f -da,
Jt Js
and A5.38), which gives the only nonvanishing component of j° for an infinite string stretched
along the x axis:
l
Consider a circular contour Г of radius r (= \/y2 + z2) lying on the x = 0 plane and centered
at у = z = 0. Since the current is along the x axis, the "magnetic" field Вн is tangential
to the circle and has a magnitude Вн(г) which is constant on the circle. The two equations
above then give
2irrBH(r) = к2 / dydz j01 = к2 - -► BH(r) =
4vrr
The direction of Вн is given by the right hand rule: the current is along the positive x axis
so the lines of Вн circulate from the positive у axis to the positive z axis. The cartesian
components of Вн are therefore
Using Hokl = eklmBHm (A5.33)) we find
)ггл /l тт о —
4vrr2
rrO12 _ 123 R _ R _ У ^2
П — € £>яз — £>яз — К
rrO13 _ 132 R _ R _
n — e nH2 — —-DH2 —
Z
rrO23 _ 231 R _ D _
4тгг2
Я1
The above, together with Hljk = 0 complete the specification of H^vp. We can confirm that
equation A5.14),
1 d
— д = J^ ,
has been fully satisfied. When both /j, and v are space indices the right hand side vanishes
and the left hand side vanishes because all H fields are time independent and H%^k = 0.
When either /j, or v is zero, the equation is of the form A5.32):
= .ok
к2 dxl J

Although we solved this equation using the magnetostatic analogy, let's check our answers.
The right-hand side vanishes for к = 2 and for к = 3. For к = 2 the left-hand side is
QH
02l
because H023 = 0 and no field has x dependence. For к = 3 the left-hand side is
8Hm дН031 дН032
дх1 дх ду
Finally, for к = 1 we have
1 /ая012 дн013
Equivalently
dBHz дВнЛ
= 0.
This is explicitly the x component of "Ampere's law" for Вц. То see how the left-hand side
gives the delta function that appears on the right-hand side we use
„-. ,. B)
ду \4тг / dz \4тг /
The left-hand side of A) becomes
C)
^ + ^)^lnr Vlnr,
dy2 dz2 J Air Air
where V2 is the two-dimensional Laplacian. It is a familiar result that in two dimensions
V2 lnr = 2irS2(x) (you can prove this by integrating both sides of this relation over a small
disk centered at the origin, using V2/ = V • (V/), and the divergence theorem). It follows
that the last term in C) is equal to the right-hand side of A). This concludes our explicit
verification that all equations have been satisfied by the solution.
We search for a B^v that is independent of time. We assume Blj = 0, so the only
nonvanishing components are Bok = —Bk0. We then have
Hm = d°Bkl + dkB10 + dlBok = -dkB01 + dlBok . D)
We now further assume that only B01 = —B10 is nonvanishing and that it is x independent.
With these conditions D) gives nontrivial information if either к or I, but not both, are 1:
tf012 = dyB01 = ^~:K2, tf°13 = 8zBm = ^K2.
Airr2 Airr2

These equations are solved by
»' i. E)
лт nrreO
B
This solution can be understood through a magnetostatic analog. We introduce an analog
vector potential Ah by
and equation D) becomes
H = tklmBHm = 9 Ан — д Ан .
This equation states the familiar fact that the magnetic field can be written as the curl of a
vector potential: BH = V x AH. The solution in E) corresponds to a vector potential whose
only nonvanishing component is A1:

15.5. Explicit checks of current conservation.
Since the only x dependence in the formula for the current jtl(x) appears on the delta
function, taking a derivative we find
dT —8»(х-х(т)) ^-^. A)
To evaluate this and show it is equal to zero we first note that for an arbitrary function h of
a single variable:
This result holds when h is any function, even a delta function, so we can rewrite A) as
a^) = -^/*[^°(*-*(r))]^. B)
For any quantity А(х(т)) we have
dA dA dx^r)
dr дх^(т) dr
Hence the integrand in the right-hand side of B) is simply a total r derivative:
d^(x) = -qcj dr^-5D{x - x(t)) .
It is reasonable to have a parameterization where r G (—00,00), so
d^f{x) = -qc[5D{x - ж(оо)) - 5D(x - ж(-оо)) .
Since time is parametrized by r we have ж°(т = ±oo) = ±00. It follows that 8{cb — ж°(±оо))
vanishes for any finite t. As a result, the above right-hand side vanishes. We thus conclude
that d^j^ = 0 for any finite time and arbitrary x. The current is conserved.
We now perform a similar analysis for the string current A5.11):
= - drda — -8 (x — X(t, a))\ — —
2 J 1дХ^(т,а) V V ' "\ V дт да дт да)
да дт дт да
1
дХ"д'5°(х-Х(т,*)).

Note that
:v д dXv д
да дт дт да) \дт да да дт
where the partial derivatives J-: and ^ act on everything to the right. The extra terms in
the right-hand side cancel
d2Xv d2Xv
дтда дадт
Hence, we can write
and the total derivatives give
The first term on the right-hand side vanishes for any finite time (Х°(т + ±oo,cr) = oo).
The second term vanishes for closed strings, since cr/ and ai represent the same point on the
string. Thus for closed strings we have d^j^^x) = 0.
For open strings the second term on the right-hand side of C) vanishes when v is an
index for a coordinate orthogonal to the D-brane where the string endpoint lies. The term
does not vanish when v is an index for a coordinate along the brane. The current is only
conserved when we include the contribution to the string current from the Maxwell fields on
the D-brane (see the discussion at the end of section 15.3). We do not attempt to prove this.

КС
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15.8. Intersection numbers and formal variables.
(a) Consider two lines if = (m°;,n°;) and i\ = (m^, tt|), and the product
{тагхг + пагУг){т\хг + пьгУг) = тагтьг хгхг + тагп\хгУг + пагт\угхг + пагп\угУг.
With the stated rules for products of the formal variables Xi and г/i, we have
{тагхг + п«уг)(тьгхг + пьгУг) = тагт\ ■ 0 + тагп\ ■ 1 + т\паг ■ (-1) + пагп\ ■ 0 .
Therefore,
(тагхг + пагуг)(т\хг + пьгУг) = тагп\ - тьгпаг = #(^а) ,ф). A)
Consider now two D6-branes on T6, each defined by three lines
/a) — (ma a\ p{a) _ / a a\ p{a) _ / a a\
if] = (mb1}n\)} ф = (mb2,nb2), ф = (тъ3,пъ3).
We now compute the product of the polynomials Па and Щ associated with the branes
Ua(xi,yi)Ub(xi,yi) = (т^х1+п^У1)(т^х2 + п^У2)(т^х3 + п^Уз)
■ {m\ xx + n\ yi) {m\x2 + n\y2) {m\х3 + пьзУз).
Since the x^'s and yi's commute with the Xj's and г/j's for г ф j, the factors can be grouped
as follows
Па(х{,у{)Пь(х{,у{) = (т*х1+па1у1)(тъ1х1 + пъ1у1)
(m%x2 + n2 y2)(mb2x2 + nh2y2)
{ml x3 + n% y3) (mb3 x3 + n\y3).
Using equation A) and A4.103) we finally have
Ча\-Ьг, Уг)ЧЬ\-1<г, Уг) — W\*-l il\ )tF\12 il2 )tF{13 ) l3 ) — Iab ■
(b) Equation A5.128) states that ^2aNaIla = 0. Multiplying by Щ from the right gives
NaUaUb = Y, NJab = 0 .
Na is the number of branes of type a and \1аь\ is the number of times that each of the a
branes intersects any b brane. The terms in the sum over a with 1аЪ > О gives the number
of left-handed states from strings that originate at any b brane. The terms in the sum with
Iab < 0 gives minus the number of left-handed states from strings that end at any b brane.
The sum over all a is therefore the number of outgoing minus the number of incoming strings
at any b brane. Since the sum is zero, the two quantities are equal. This also holds for the
full collection of b branes.
1

15.9. An intersecting brane model with the particle content of the
Standard Model.
(a) We recall that (m,n) <-> [m,n + y], so we immediately derive that [r,s] <-> (r,s — |).
This gives: if = A, 0), t® = A,1), t® = @,1), if = A, -2).
(b) In order to have 1ц* = 0 it suffices that one of the three lines that defines the brane i
is not changed by the operation of sign reversal of the second entry. For the first brane we
have £\ = £\ , for the second brane £2 = £2 , for the third brane £2 = £2 , and for the
fourth brane £{=£{. Therefore none of the branes intersects its mirror image.
(c)
/12 = (A)(-1) - @)@)) (E)@) - A)A)) (A)(|) - (i)(l)) = 1
/12. = (A)A) - @)@)) (E)@) - (-1)A)) ((l)(f) - (-i)(l)) = 2
/13 = (A)C) - @)D)) (E)@) - A)A)) (A)A) - (I)@)) = -3
/is* = ((l)(-3) - @)D)) (E)@) - (-1)A)) (A)(-1) - (-i)@)) = -3
f(|)A)) = 0
(|) - (^)A)) = 0
/23 = (@)C) - (-1)D)) ((l)@) - @)(l)) (A)A) - (f )@)) = 0
/23* = (@)(-3) - A)D)) ((l)@) - @)(l)) (A)(-1) - (f )@)) = 0
f
/24 = (@)@) - (-1)A)) (A)A) - @)(l)) (A)(^) - (|)A)) = -3
/24* = (@)@) - A)A)) (A)(-1) - @)(l)) ((l)(f) - (^)A)) = 0
/34 = (D)@) - C)A)) (A)A) - @)(l)) (@)(^) - A)A)) = 3
/34* = (D)@) - (-3)A)) (A)(-1) - @)(l)) (@)(|) - (-1)A)) = -3.
These intersection numbers are indeed reflected by the numbers and orientations of strings
in Figure 15.8.
(d) We begin by computing the polynomials that enter into the relation:
П1 = x1Ex2 + У2)(х3 + \уг)
> Hi + П1* = \^ххх2хз + x1y2y3
Пр = x^bx-2 — у2)(хз — \уз)
По = -1
П2 + П2* = — Ъу\Х2
П2* =ШЖ2(ж3-|уз)

П3 = Dxi + 3yi)x2y3
П3 + П3* = §yix2y3
П3* =
П4 = х1(х2
" П4 + П4* =
П4* =хх(х2-у2)(хг
Collecting the above results
■■■&
М{(Т1{ + Т1{*) = 3A0ж1Ж2ж3 + х1у2у3) + 2(-Зу1х2у3)
= 4П
Об •
This is what we wanted to show.
(e) Examining the uL quark at the intersection of N\ and N2, we see that it leaves N2 and
goes to N\, hence, Q2 = —1 and Q\ = 1. So Q = Q\ — 3(<5з + Qa) = 1- But, we know that
Y = 1/6 for a ul quark, so we expect Y = Q/6. Indeed, the Q charges are:
ul and cLl have charge 1 — 0=1
uL has charge — 1 — 3 = —4
dL has charge —1 + 3 = 2
ueL and el have charge — 3 — 0 = —3
veL has charge 3 — 3 = 0
e\ has charge —3 — 3 = —6
For all these particles Q is six times the hypercharge.

15.10. Symmetry breaking by recombination of intersecting branes.
(a) The first brane is represented by a straight line from @,0) to (mi,ni). In the process
of recombination the beginning of the second brane is joined to the end of the first brane.
Therefore, the second brane, initially represented by the straight line from @,0) to (т2,п2)
can be viewed as the straight line from (m-i, щ) to (mi + m2, щ -\-n2). The combined branes
are represented by a composite line beginning at @,0) and ending at (mi + т2,п1 + n2),
with a corner at (m1,n1). This line can be deformed into a straight line from the origin @, 0)
to (mi +m2,ni +n2)
The line from @, 0) to (mi + m2, щ + n2) with a corner at (m2, n2) can also be deformed
into the straight line from @,0) to (mi + m2, щ + n2). Thus attaching brane one to the end
of brane two gives the same final result. In both cases the final brane is represented by the
vector addition of £^ and £B\ The order does not matter: vector addition is commutative.
(b) Below are five frames showing the recombination of the branes A,0) and @,1). In the
first frame we show the both branes, with a little piece of the end of the A,0) brane (the
first brane) removed and a little piece of the beginning of the @,1) brane (the second brane)
removed. The second frame shows a gluing that joins the end of the first brane to the
beginning of the second brane. The third and fourth frames show successive deforemations.
The fifth and final frame shows the brane A,1) resulting from the recombination.
Figure 1: Recombination of A,0) and @,1) branes.
(c) We consider the branes defined by (mi,ni) and (m,2,n2), the recombined brane E =
(mi + m2,ni + n2), and a brane % defined by (гп{,щ) that intersects the previous branes.
Using the triangle inequality we then have
mini - mi7ii\ + |m,;n2 - m2n;|
— mini + mi
i + n2) — (mi
This is what we wanted to show.

16.1. Review of statistical mechanics.
(a) Suppose the system is in the state a with energy Ea(V). The work done by the system
when changing the volume by dV is padV, where pa denotes the pressure of the system in
the state a. This work is equal to minus the change in energy of the system:
padV = -dEa = —T^ydV -> pa = —
dv ra dv
The pressure p is the average of pa over the equilibrium distribution of states
ldlnZ
1
dV Z ^ dV ZBdV ^ В dV
(b) We use F = E — TS. Since we know E in terms of derivatives of Z, we need a formula
for S in terms of Z. As hinted, we consider
= -d{/3 Ё) + P(dE + pdV)
= -d{{3 Ё) + {3dQ ,
where we used the first law of thermodynamics. Since dQ = TdS, we find
dlnZ(P,V) = -d(PE) + ydS ->■ jdS = d(]nZ(/3,V)+/3E).
К К V /
By integration, we obtain
y = lnZ + [3E. A)
к
There is no integration constant, as can be verified by taking aT^O limit and checking
that A) correctly gives S/k = lnfio, where Qq is the number of ground states (assumed to
be separated by a gap from the excited states). Finally, using A)
1 S 1
F = E - TS = E --- = -- In Z ,
/3 k /3
as we wanted to prove.

jrrlU^ IG , 2~
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с
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S ="
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S •
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11
ЛтригиХя pvociud:
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/

16.6. Partition function of the relativistic particle.
We are asked to evaluate the integral A6.83) for the partition function:
ddu
Z(m2) = Vmd I
_ — вту1+u2
Since the integrand only depends on the length и of u, we can use
ddu =
With this formula
Using the following change of variables and definitions
the integral I(z) becomes
We then have
M = sinht, z = /Зт , и =—(d — 1)
I(z)= e-zcosht(smhtJucoshtdt.
Jo
d f°° 9U d
— / e-*^t(8inhtJ"dt = -T(d/2)-=—
dz Jo \J7Г dz
where we used the integral representation cited in the statement of the problem. Back to
Z(m2), the gamma functions cancel and with a small rearrangement we get:
We evaluate the term in parenthesis using the expansion (see correction web-page)

A computation of a few lines gives
d
Back into A) we get the result
, d(d
w
hich gives the first correction to the low temperature partition function A6.89).

16.7. Corrections to the temperature/energy relation in the micro-
canonical ensemble.
Equation A6.58) gives us
lnp24(N) ~ Air^N - Ц- In N.
For high energy E ~ \jN/a! so N = a'E2 and
l(N) AV^E
Щу/оЕ). A)
We then compute
1 d /S\ r- 27 1 1 27 1
We see that 7= < 7^-, so we have the unexpected T > T#. Solving for the temperature,
кТи
kT =
1 _ 2ZMi
2 E
This can only be trusted for high energy. We see that for E —> 00 we have T —> Тя. For
E = 13.5/сТя, which is an energy comparable to the rest mass of the first excited states
of the string (see 16.70), the temperature diverges. This is plausibly an artifact of the
approximation.
For very large energies we have
kTH , 27(fcTHJ
fcT = ^j Wkf^ - кТн +  ^— ' for
The specific heat С is readily found at high energies:
c -(—Y'-f
rfT V^7 V 2 £2 У 27 \kTH,
A negative specific heat is not completely unfamiliar. A Schwarzchild black hole has negative
specific heat: as the energy (mass) increases its temperature goes down.
An alternative approach uses the relation Q(E)dE = p24:(N)dN to find
dN
lnQ(E) = lnp2A(N) + In-— = lnp2A(N) + InE + constant.
dE
With S = klnu(E) we get
S
— = lnp24{E) + InE + constant,
к
1

where the first term on the right-hand side is the previously calculated entropy (see A)).
With this entropy, the temperature result in B) and the specific heat result in C) are changed
into
1 ' "^ С =-±k (£■)'■ D)
кТ кТн 2 E ' 25 \кТн
For open strings on a Dg-brane П(Е) ~ E~Jех.рDтг\<а'Е) with 7 = B5 — q)/2. We then
have
A short computation then gives
Our earlier result in D) corresponds to open strings attached to a DO-brane.

> \
U
>
^
л
ir

fcJL
<U
U
T
?
J
&б~ ^
ь
с и =
R =
H
f^XauS, «*
"
м >
н
U
H -
0
Ю
1
•= \о

16.9. continued
(b) For caution, we note that the Schwarzschild radius R of the black hole of mass M is
so for couplings that are small, so that we can trust the string model, the black hole is much
larger than £s. For such black holes the string model is not expected to be applicable.
(c) As we dial down the coupling the entropy is constant and must match the entropy
S ~ Mla ~ N1'2 of a free string:
g:
(
mpJ \mp
We calculate the size of this string
i
A)
ТПр
We are told to evaluate this in terms of the Schwarzschild radius R and the original value g
of the string coupling. With G ~ 1/mp, we have
^ r\
^= ~ r\ ~ .
WG gta g
For g = 0.1 the size of the free string is ten times the Schwarzschild radius of the original
black hole. The length L of the string would be even larger. Since L =
Rstr mP
For the black hole at the center of the galaxy M = 2.6 x lO6Msun. Moreover, Msun
2 x 1030 kg and mP ~2x 10~8 kg. Thus equation A) gives
N ~ t2£ X Ш X 28X 1QY ~ B.6 x 10")' ~ 4.6 x