Classically, the point particle and the string exhibit the same kind of motion. For instance in flat space both of them move in straight lines albeit for string oscillations which occur because it has to obey the wave equation. When we put it in AdS3 space both the point particle and the string move as if they are in a potential well. However, coordinate singularities arise in the numerical computation of the string so motion beyond ρ = 0 becomes computationally inac- cessible. Physically the string should still move beyond this point in empty AdS3 spacetime. This singularity is an artefact because coordinate systems in general are not physical. The behaviour of the string in the vicinity of a black hole background in AdS3 spacetime is well defined a fair bit away from the horizon. It moves in the same manner as in the AdS3 spacetime in the absence of the background. Un- fortunately, when the string approaches the horizon part of the string overshoots into the horizon. The solutions become divergent and the numerical solution fails before we can observe anything interesting.

1. Classical String in curved space calculations
Ismail Abdulaziz Ibrahim
Queen Mary University of London,
Mile End Road, London
E1 4NS
April 21, 2017
Abstract
Classically, the point particle and the string exhibit the same kind of motion.
For instance in flat space both of them move in straight lines albeit for string
oscillations which occur because it has to obey the wave equation.
When we put it in AdS3 space both the point particle and the string move as if
they are in a potential well. However, coordinate singularities arise in the numerical
computation of the string so motion beyond ρ = 0 becomes computationally inac-
cessible. Physically the string should still move beyond this point in empty AdS3
spacetime. This singularity is an artefact because coordinate systems in general are
not physical. The behaviour of the string in the vicinity of a black hole background
in AdS3 spacetime is well defined a fair bit away from the horizon. It moves in
the same manner as in the AdS3 spacetime in the absence of the background. Un-
fortunately, when the string approaches the horizon part of the string overshoots
into the horizon. The solutions become divergent and the numerical solution fails
before we can observe anything interesting.
1

2. Contents
1 Motivation 3
2 Point Particle Dynamics 3
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.2 Analytical Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2.1 Flat Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2.2 2-Sphere in 4D Flat Space . . . . . . . . . . . . . . . . . . . . 9
2.2.3 AdS3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.3 Numerical Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.3.1 Results: Exitrinsic motion in a sphere . . . . . . . . . . . . . 16
2.3.2 Results: AdS3 . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.4 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.4.1 Flat Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.4.2 2-sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.4.3 AdS3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3 String in space 26
3.1 String Sigma Model . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.2 Flat Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.3 Explicit calculation of a string in flat-space . . . . . . . . . . . . . . 33
3.3.1 Simple case: Xµ
L ≡ Xµ
R . . . . . . . . . . . . . . . . . . . . . . 33
3.3.2 More general wave: Xµ
L = Xµ
R . . . . . . . . . . . . . . . . . . 34
3.3.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.4 Numerical Method and Results . . . . . . . . . . . . . . . . . . . . . 39
3.4.1 4D Flat spacetime . . . . . . . . . . . . . . . . . . . . . . . . 39
3.4.2 AdS3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.4.3 Blackhole Background . . . . . . . . . . . . . . . . . . . . . . 46
3.5 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.5.1 Flat Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.5.2 AdS3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.5.3 Blackhole Background . . . . . . . . . . . . . . . . . . . . . . 52
4 Conclusions 52
References 53
2

3. 1 Motivation
String theory is about n−branes in general and a special class of them are strings.
This is an n = 1 brane and it is thought to be the natural generalisation of the
point particle. Classically, we expect that the string and the point particle behave
in a similar way when we put them inside the same space-time system. However,
the string should exhibit some unexpected behaviour in these situations. The string
has a well defined size. Each point in the string can be well defined by parameters
τ and σ. So when we put a string in curved space we expect to see some physical
deformation. This must be true because in the classical framework each point on the
string experiences a Newtonian attraction force. Each point will occupy different
regions of space so we expect tidal forces to come into play. We wish to calculate
the equations of motion of the classical string in different spacetime geometries and
then we will change some properties of the string to see the effect it will have on
the string [2, sect 2 p 17].
2 Point Particle Dynamics
2.1 Introduction
Imagine the motion of a point particle. An idealised object wherein all the mass
is concentrated into a single point is allowed to travel freely in space. We simply
throw the object from an initial height say h0 up into the air from rest and we let
it fall. We start the timer immediately after releasing the object using a suitable
equipment, a stop-watch for instance. Then we plot the motion of the particle using
a suitable space-time graph. Repeating this experiment with a different object we
observe a similar trend – a parabolic trend. In space-time it seems as if nature is
choosing the path in which the unperturbed object is supposed to go. Basically, the
nature examines all different kind of paths that may be suitable, discards the one
that is not, and carries on doing so until a single path is left out. This single path
varies with different configuration of spaces. In physics, this is called the Action
Principle.
Figure 1: Action minimisation principle [4].
The first formulation of mechanics by way of this principle was published by
1746 in a paper called Les lois du mouvment et du repos deduites d’un principe
metaphysique(Laws of motion and rest deduced from a metaphysical principle).
3

4. Pierre Louis Maupertuis had first introduced the principle of least action in optics
in 1744. In 1893 Florian Cajori published the book “A history of Mathematics”
in which Leonhard Euler worked out the theory of the rotation of a body around
a fixed point, established the general equation of motion of a free body and the
general equation of hydrodynamics [1].
∂L
∂xµ
−
d
dt
∂L
∂ ˙x
= 0. (1)
where the action of the particle, we will call this S, will be defined in terms of L,
the Lagrangian named after Joseph-Louis Lagrange who managed to discover the
beautiful Euler-Lagrange equation (1) under the tutelage of Leonhard Euler who
was his academic advisor at the time.
The curvature of space is identified by the Reimann curvature tensor Rβδ
µν.
This tensor fundametally depends on gµν therefore, the form of gµν is going to be
representative of the spacetime configuration of the problem. In flat 4D space for
instance, gµν = ηµν. This is the Minkowski metric and it is a diagonal identity
matrix which involves a certain Lorentz signature, (−, +, +, +). This type of signa-
ture controls the signs in front of each component. We can define an infinitesimal
space-time length element ds0
ds2
0 = −c2
dt2
+ dx2
, (2)
(x ∈ V 3). Using the notion of an Einstein summation, we can make more compact
the above relation (2) by rewriting it in the following way. This form will be
subsequently used throughout and Einstein summation will be assumed,
ds2
0 = ηµνdxµ
dxν
(3)
and furthermore,
ηµν =




−1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1



 . (4)
Lorentz signatures, such as the one used above, on any gµν make sure that space
time distance between any two events are Lorentz invariant. When we take a point
in space-time calling it (Xi, ti) we can apply a Lorentz boost.
˜Xi =
Xi − vti
√
1 − v2
(5)
˜ti =
ti − vXi
√
1 − v2
(6)
In other words, a Lorentz boost is to simply apply the above transformation on to
a spacetime coordinate in other words (X, t) → ( ˜X, ˜t). We determine the spacetime
distance from (2) as
s2
= −(t0 − t1)2
+ (X0 − X1)2
. (7)
When we apply a Lorentz boost to each point we find that the new length is, from
the transformation formula,
˜s2
= −(˜t0 − ˜t1)2
+ ( ˜X0 − ˜X1)2
(8)
= −
t0 − vX0
√
1 − v2
−
t1 − vX1
√
1 − v2
2
+
X0 − vt0
√
1 − v2
−
X1 − vt1
√
1 − v2
2
. (9)
4

5. The quantity ∆s = s − ˜s can be shown to be zero which means that a Lorentz
transformation does not change the length between two points in spacetime. We
have found that the notion of length is the same in spacetime regardless of the frame
of reference that we are in. Every observer will agree on the spacetime distance ds.
The motion of a relativistic particle of mass m in a curved D-dimensional spacetime
can be formulated as a variational problem, an action principle. The action of the
particle should be proportional to the invariant length of the particles trajectory.
In other words,
S = −α ds. (10)
α is some arbitrary constant of proportionality. The nonrelativistic limit of the
action S (26) in flat Minkowski space-time determines the value of the constant α
to be the mass of the point particle. [2, Sect 2 p 18]
In the non relativistic limit the action (26) becomes
S = −α −ηµνdxµdxν = −α dt2 − dx2 (11)
= −α dt 1 − v2 (12)
≈ −α dt(1 −
1
2
v2
+ . . . ). (13)
Comparing the above expansion with the action of a nonrelativistic point particle,
namely
Snr = dt
1
2
mv2
, (14)
gives α = m. In the nonrelativistic limit an additional constant (the famous E =
mc2 term) appears in the above expansion of S. This constant does not contribute
to the classical equations of motion [2, Sect 2 p 20-21]
S = −m ds. (15)
5

6. In general we can write the line element ds as an invariant line element in space
as
ds2
= −gµν(x)dxµ
dxν
(16)
here gµν(x), with [µ, ν] = [0, . . . , D − 1] and D denotes the number of spacial
dimensions. The minus sign has been introduced so that ds is real for a time-like
trajectory. In mathematical formulism, we express the action of a point particle in
space as
S = −m −gµν(x)dxµdxν. (17)
The action (17) has the disadvantage that it contains a square root because it
makes computation a little less trivial. We introduce an auxillary field e(τ) which
we will use to rewrite an action equilavent to the first one without the square root.
We call this ˜S,
˜S =
1
2
dτ
1
e
˙X2 − m2
e , (18)
so ˙X2 = gµν
˙Xµ ˙Xν where gµν is an arbitrary metric [2, Sect 2 p 19].
In order extract the equations of motion for the point particle we will need to
minimise (18). Taking the variation of (18) we see that
δ ˜S =
1
2
dτδ
˙X2
e
− m2
δe (19)
⇒ δ ˜S =
1
2
dτ
δ ˙(X2)e − δe ˙X2
e2
− m2
δe (20)
⇒ δ ˜S =
1
2
dτ
2 ˙Xµδ ˙Xµ
e
−
˙Xµ ˙Xµ
e2
δe − m2
δe . (21)
Then δ ˜S
δe , the variation in the action with respect to e, the auxilary field metric
is:
δ ˜S
δe
= −
1
2
e−2 ˙Xµ ˙Xµ + m2
. (22)
Free to choose a gauge for e we will use e = 1 just arbitrarily. Using the equation
above we find that
˙Xµ ˙Xµ + m2
= 0. (23)
Since pµ ≡ ˙Xµ, a conjugate momentum to Xµ, we simply get the mass-shell con-
dition. This basically means that the equations due to the minimisation of ˜S will
give classical equations that are going to be physical [2, Sect 2 p 23].
Minimising the variation of ˜S with respect to Xµ we obtain the following:
δ ˜S
δ ˙Xµ
= −
d
dτ
(gµν
˙Xµ ˙Xν
) +
1
2
∂λgρλ
˙Xρ ˙Xλ
, (24)
This must be zero because we want to minimise the action. We obtain the following,
⇒ ¨Xµ
+ Γµ
ρλ
˙Xρ ˙Xλ
= 0 (25)
where we have defined a new symbol,
Γµ
ρλ =
1
2
gµα
(∂ρgαλ + ∂λgαρ − ∂αgρλ) , (26)
the Christoffel symbol. This is the geodesic equation for particles in curved space
and it will allow us examine the dynamics of the particle.
6

7. 2.2 Analytical Method
2.2.1 Flat Space
In non-relativistic physics we know that if an object is travelling inside flat space
it will do so in a straight line. In flat space we take it to be the space where there
are no sources for potential energy to the object. Due to Noether, if there are no
sources for potential energy then the momentum of the object will be conserved.
That is in other words
˙p = 0. (27)
In the relativistic limit as we also expect the point particle to be moving in a
straight line in flat space. We will attempt to find the equation of motion for the
point particle in the relativistic limit by taking the minimal value of variation of
the action in flat space in order to determine the dynamics.
Using (17) with the metric gµν = ηµν we see that the action of a particle in flat
spacetime is
S0 = −m −dt2 + d ˙x2 (28)
= m dt 1 − ˙x2. (29)
From the Lagrangian definition of S we determine the equation of motion for
the particle,
d
dt


˙x
1 − | ˙x|2

 = 0 (30)
⇒ c =
˙x
1 − | ˙x|2
(31)
⇒ ˙x = 1 − | ˙x|2 c. (32)
Dotting (32) with itself we obtain
| ˙x|2
≡
|c|2
1 + |c|2
. (33)
Using (33) in (32) we find that
˙x = 1 −
|c|2
1 + |c|2
c (34)
⇒ ˙x =
1
1 + |c|2
c (35)
Solving (35) for x we find that
dx =
1
1 + |c|2
cdt (36)
⇒ x(t) = (At)c + a. (37)
7

8. From (37), x grows linearly with time so that x traces out a straight line. Letting
Ac = v, the constant velocity vector of the particle and a = x0 then (37) reads as
x = tv + x0. (38)
As we can see, (38) describes motion at constant velocity where A ≡ 1
1+|c|2
and v is a constant vector in time. We find that the motion of the relativistic point
particle in flat space is also along a straight line as expected.
8

9. 2.2.2 2-Sphere in 4D Flat Space
We will now attempt to derive the equations of motion for when a point particle
moves along a 2-sphere. A 2-sphere is a sphere in two dimensions. It is not a solid
ball but a 2-sphere is a 2-d surface. It appears to be 3 dimensional since, we have
3 coordinates x, y, z that determine a point on the sphere, it is embedded in a 3
dimensional euclidian space.
The metric dss for the 2-sphere embedded in Minkowski space can be shown to
be
ds2
s = −dt2
+ R2
(dθ2
+ sin2
θdφ2
) (39)
and
gµν =


−1 0 0
0 R2 0
0 0 R2 sin2
θ

 . (40)
We can use (25) to deduce the equations for each x = (t(τ), φ(τ), θ(τ)) that
make up the geodesics for our particle in the 2-sphere. The Euler-Lagrange (EL)
equations (1) and the solutions of the geodesic equation (25) for a point particle
give the same outcome.
−1 = gµν(x) ˙xµ
˙xν
. (41)
We will call this the proper time gauge. This will be convenient because it will
remove any factor of −gµν(x) ˙xµ ˙xν that we may encounter. As an example, using
the EL equations on the world-line time coordinate parameterised by τ as t(τ)
∂L
∂t
−
d
dτ
∂L
∂ ˙t
= 0. (42)
The Lagrangian L is −gµν(x) ˙xµ ˙xν = ˙t2 − R2( ˙θ2 + sin2
θ ˙φ2).
⇒
d
dτ
∂
∂ ˙t
˙t2 − R2( ˙θ2 + sin2
θ ˙φ2) = 0, (43)
⇒
d
dτ
˙t
−gµν(x) ˙xµ ˙xν
= 0, (44)
⇒
d˙t
dτ
= 0, (45)
where we have used (41), the proper time gauge. We can say that
˙t = const. (46)
⇒ t(τ) = Aτ + c, (47)
A = E representing the total energy of the system and c = tτ=0,
t(τ) = Eτ + tτ=0. (48)
Choosing a coordinate system whose origin coincides with τ = 0 then t(τ =
0) = 0 and c = tτ=0 = 0. Our final equation for time t is in the form
t(τ) = Eτ. (49)
I now want to take a closer look at this result for t(τ). The equation (49) is
simply the time dilation we experience as a result of relativistic effects. Recall
9

10. one of the consequences of one of the principles in special relativity (namely the
universal constancy in the speed of light). The clock of the inertial observer for
a system involving an object moving relatively close to the speed of light will be
noticeably slower than the clock in the moving frame. In other words t = τγ where
τ is defined as the proper time. So E can take the role of γ in that it dilates the
time t accordingly.
Using equation (25) for µ = φ,
¨φ + Γφ
ρλ
˙Xρ ˙Xλ
= 0. (50)
We can find that ρ = φ and λ = θ such that only Γφ
φθ and Γφ
θφ survive. The
second one exists because of the symmetry in the metric gρλ when we interchange
λ and ρ indicies. Then,
¨φ + 2
cos θ
sin θ
˙φ ˙θ = 0. (51)
Multiplying both sides of (51) by sin2
θ we see that,
¨φ sin2
θ + 2( ˙φ ˙θ) sin θ cos θ = 0, (52)
which is
d
dτ
(sin2
θ ˙φ) = 0 (53)
by the product rule. This is something we expected in that the derivative of quantity
sin2
θ ˙φ with respect to τ is zero. Rotational symmetry in the configuration implies
conservation of angular momentum due to Noether’s Theorem. Hence,
sin2
θ ˙φ = J. (54)
This result that we have obtained is the ˆz component of the particle’s orbital
angular momentum J per unit mass for a unit 2-sphere (a 2-sphere of radius R = 1).
Consider for instance some motion of the particle around the 2-sphere. The orbital
angular momentum in the general case is defined to be
J = mr × ˙r. (55)
Rewriting ˙r to be the tangential vector of the orbit
˙r = rT
˙φ (56)
so that the orbital angular momentum can also be expressed in the following way
J = m(r × rT
˙φ) (57)
= mR ˙φ(R sin θ)ˆr × ˆφ (58)
where we have resolved the tangential vector in to the direction of ˆr. The vector
product ˆr × ˆφ produces a vector pointing out perpendicular to the orbital plane.
In order to get to Jz we will resolve J along ˆz (sometimes referred to as ˆk in other
papers).
Jz = sin θJ, (59)
where θ is the angle measured from the ˆz to the vector J. Then,
Jz = mR2
sin2
θ ˙φˆz. (60)
We have retrieved (54) with m and R2 factors present.
10

11. For our final equation, we will require to solve the θ component of (25). Setting
µ = θ and also α = θ (because our gαµ only exists for α = µ where α is our
free index as in (26) in our definition for the Christoffel symbol). Furthermore,
our metric gµν is independent on φ and t so that constrains λ to φ and ρ to φ.
Clearly only one Christoffel term survives, Γθ
φφ. This particular Christoffel is just
1
2
1
R2 −2R2 sin θ cos θ ≡ − sin θ cos θ. Our final equation for θ now reads
¨θ − sin θ cos θ ˙φ2
= 0. (61)
Using (54) for ˙φ we get
¨θ − J2 cos θ
sin3
θ
= 0. (62)
This is not a trivial second order equation to solve for θ at first hand. We will
instead use the proper time gauge for τ. Using the equations (49) and (54) to
rearrange for ˙θ by eliminating the variables t(τ) and φ(τ) for θ(τ),
˙θ2
+
J2
sin2
θ
=
E2 − 1
R2
, (63)
⇒ ˙θ = ±
sin2
θ(E2 − 1) − (JR)2
(R sin θ)2
. (64)
Integrating this expression for θ(τ),
dθ
R sin θ
(E2 − 1) sin2
θ − (JR)2
= ± dτ. (65)
Making the substitution X = cos θ we see that X2 = cos2 θ so that dX = sin θdθ
and it can also be shown that (E2 −1)(1−X2) = (E2 −1) sin2
θ from the unit circle
trignomotric identity I need not mention. Therefore, we can equilavently write (64)
as:
R
√
E2 − 1
dX
(E2−1)−(JR)2
E2−1
− X2
= ± dτ. (66)
By setting F2 = (E2−1)−(JR)2
E2−1
we can immediately use an identity for the left
hand side (LHS) where I have used C = R√
E2−1
,
C
dx
√
a2 − x2
≡ C arcsin
x
a
. (67)
To proceed
⇒ C arcsin
X
F
= ±(τ + K). (68)
⇒ cos θ = ±F sin
τ + K
C
. (69)
When τ = 0 we want the particle to orbit a circle on the x-y plane. This requires
cos θ(τ = 0) = 0,
sin
K
C
= 0 (70)
⇒
K
C
= nπ (71)
11

12. Take n = 0 (since K is a constant) so that K = 0, which means θ becomes
cos θ = ±
2
3
sin
√
3τ. (72)
I have chosen E = 2, J = 1 and R = 1 as an example in order to demonstrate (69)
and as a reference to for when we attempt to resolve the same solution numerically.
2.2.3 AdS3
The anti-De Sitter or AdS space-time is best described as a kind of asymptotic cone
of constant negative curvature. A kind of motion on this space would resemble an
object inside a potential well. The particle would be attracted to the origin (of the
coordinate system).
The metric for the AdS3 spacetime is:
ds2
= R2
(−dt2
cosh2
ρ + dρ2
+ dθ2
sinh2
ρ) (73)
and gµν can be expressed as
gµν = R2


− cosh2
ρ 0 0
0 1 0
0 0 sinh2
ρ

 (74)
so that ds2 ≡ gµνdxµdxν. We can get the inverse by simply inverting each element
so that − cosh2
ρ → − sech2
ρ.
We begin by formulating an equation for t(τ), the equation for time from the
intertial reference frame. From the geodesic equation
¨Xµ
+ Γµ
ρλ
˙Xρ ˙Xλ
= 0. (75)
Xµ = t so that means the indicies for α and µ reduce to µ = t and α = t be-
cause gµν is diagonal. α is just some free index we used in the formulation of the
Christoffel symbol (26) for the geodesic equation (25). The only proper choices for
our remaining indicies are λ = t and ρ = ρ and also ρ ↔ λ. We will have now only
Γt
ρt as the non-vanishing symbol.
Γt
ρt = gtt
∂ρgtt (76)
=
1
− sech2
ρ
(−2 cosh ρ sinh ρ) (77)
= 2 tanh ρ. (78)
Solving (25) we find that for t we get:
¨t + 2 ˙ρ˙t tanh ρ (79)
For Xµ = ρ, µ = α = ρ and ρ = λ = t or ρ = λ = θ. This time we have two
different symbols contributing to the solution for ρ, Γρ
tt and Γρ
θθ.
Γρ
tt = −
1
2
gρρ
∂ρgtt = cosh ρ sinh ρ (80)
Γρ
θθ = −
1
2
gρρ
∂ρgθθ = − sinh ρ cosh ρ. (81)
Our equation for ρ is
¨ρ + cosh ρ sinh ρ(˙t2
− ˙θ2
) (82)
12

13. Lastly for θ, the indicies in the geodesic equation for θ are reduced to µ ≡ α and
λ = θ.This means ρ = ρ which leads to the only non-vanishing Christoffel symbol
Γθ
φθ =
1
2
gθθ
∂ρgθθ (83)
= coth ρ (84)
so that our last equation for the time evolution of θ can be expressed as:
¨θ + 2 ˙ρ ˙θ coth ρ = 0 (85)
The equations (79) and (85) can be rewritten in the following more convenient
forms:
˙t cosh2
ρ = E (86)
˙θ sinh2
ρ = J (87)
E and J are just arbitrary constants that are supposed to represent the total energy
and the angular momentum about the ˆz axis respectively.
We could try to solve (82) by using results from (86) and (87) but in hindsight
it looks like a non trivial second order differential equation,
¨ρ + cosh ρ sinh ρ
E
cosh2
ρ
2
−
J
sinh2
ρ
2
= 0 (88)
(sinh3
ρ cosh3
ρ)¨ρ + E2
sinh4
ρ − J2
cosh4
ρ = 0. (89)
Reminding ourselves of the proper time gauge approach in the 2-sphere case it
appears to be the easier approach to determine ρ(τ). So from (41) we get:
−
1
R2
= −˙t2
cosh2
ρ + ˙ρ2
+ ˙θ2
sinh2
ρ (90)
substituting (86) and (87) for ˙ρ,
˙ρ2
=
R2(E2 sinh2
ρ − J2 cosh2
ρ) − sinh2
ρ cosh2
ρ
R2 sinh2
ρ cosh2
ρ
. (91)
Then solving for ρ we require that
dρ
sinh ρ cosh ρ
E2 sinh2
ρ − J2 cosh2
ρ − 1
R2 (sinh2
ρ cosh2
ρ)
= ± dτ. (92)
Using the substitution ξ = cosh2
ρ,
LHS =
1
2
dξ
− 1
R2 ξ2 + (E2 − J2 − 1
R2 )ξ − E2
(93)
=
R
2
dξ
(ER)2 −
R(E2−J2− 1
R2 )
2
2
− ξ −
E2−J2− 1
R2 R
2
2
(94)
= D
dX
√
F2 − X2
, (95)
13

14. I have used another substitution X = ξ −
E2−J2− 1
R2 R
2 and also D = R
2 , and
F2 = (ER)2 −
R(E2−J2− 1
R2 )
2
2
. Our intergral is now reduced to a trivial form we
can replace with one of many recognised integral definitions.
LHS =
R
2
arcsin
X
F
(96)
⇒
R
2
arcsin
X
F
= ±(τ + k). (97)
Therefore, we see that
cosh2
ρ = ±F sin
2
R
(τ + k) +
E2 − J2 − 1
R2 R
2
. (98)
14

15. Choosing the arbitrary initial condition for ρ, ρ(τ = 0) = 1. This means that
the phase difference in cosh2
ρ must be
k = −
R
2
arcsin
2C2
0 − E2 − J2 − 1
R2 R
2 (ER)2 −
R(E2−J2− 1
R2 )
2
2
(99)
where C0 = cosh 1 and S0 = sinh 1. Recognising that F is a square root we need F
to be real for a given set of real numbers [E, J, R] ∈ R+. We have defined E to be
the total energy, J to be the total angular momentum about the ˆz direction and R
to be the radius of curvature of AdS3. For F to be real we require R belonging to
the set [E = 3, J = 1, R] to satisfy
R >
(C0S0)2
9S2
0 − C2
0
. (100)
RHS is roughly 0.575.
15

16. 2.3 Numerical Method
For what will follow, I intend to show the power of numerical method by showing
that they will match my analytic solution. Numerically, its possible to reproduce
for example the θ, t and φ functions and time-evolutions. All we require are a set
of initial conditions in this case t(τ = 0), φ(τ = 0) and θ(τ = 0) and also initial
conditions for the τ derivative of each function. This is denoted by V µ(τ = 0). It is
an arbitrary velocity field dXµ
dτ ≡ ˙Xµ ≡ V µ. We use a first order approximation to
the ordinary differential equations. This is called the Euler’s Method and it follows
the most basic algorithm to a numerical solution for differential equations.
Xµ
(τ + ∆τ) ≈ Xµ
(τ) + ∆τ ˙Xµ(τ). (101)
The above will be used to approximately determine Xµ(τ). This uses a first order
approximation to the gradient function V µ. In simple terms, it takes the gradient
of a function at a point τi to extrapolate the value of the function at the point
τi+1 = τi + ∆τ. That is the gradient at τi is taken to be the gradient over the
region ∆τ. ∆τ is the interval in the domain which separates two points τi and τi+1.
When ∆τ → 0 the gradient over ∆τ approaches the true gradient of the function
Xµ over the interval. Therefore, we take ∆τ to be in the order of a thousandth.
This recursive relation defined by (101) will require knowledge of V µ over the
region τ ∈ [0, N∆τ]. N is the total number of divisions. The interval division ∆τ
will be small. This requires N to be very large. When it is reasonably large (say
N > 100, 000), ∆τ ≈ 10−4 (in the order of ten thousandths), we should have some
result which almost agrees with the analytical solution. We define as
=
τf − τi
N
. (102)
is the division width ∆τ, τi is the starting point and τf is the last point in the τ
domain.
2.3.1 Results: Exitrinsic motion in a sphere
As an example the recursive relation for φ for τ > 0 would look something like this,
φ(τ + ) = φ(τ) + V φ
(τ), (103)
but V φ is dependent on θ therefore we need also need to define θ and ˙θ = V θ. Using
(64) we can find this to be
V θ
=
E2 − 1
R2
−
J2
sin2
θ
. (104)
So we will also need ˙V θ in order for V θ to be defined for all τ > 0. This we can
obtain from (62), our second order ODE for θ,
˙V θ =
J2 cos θ
sin3
θ
. (105)
16

17. We can now complete our valuation of all variables φ, t, θ by the following equations
coupled with their initial conditions;
V θ
(τ + ) = V θ
(τ) +
J2 cos θ(τ)
sin3
θ(τ)
(106)
θ(τ + ) = θ(τ) + V θ
(τ) (107)
φ(τ + ) = φ(τ) +
J2
sin2
θ(τ)
(108)
t(τ + ) = t(τ) + E (109)
17

18. I have arbitrarily chosen the set (E = 2, J = 1, R = 1) as the arbitrary constants
along with N = 500, 000 which is the number of divisions. I have imposed the
following initial conditions
V θ
(τ = 0) =
E2 − 1
R2
− J2 θ(τ = 0) =
π
2
(110)
φ(τ = 0) = 0 t(τ = 0) = 0. (111)
2 4 6 8 10
Τ
1.0
1.5
2.0
Θ
(a) Numerically approximated θ(τ)
2 4 6 8 10
Τ
1.0
1.5
2.0
Θ
(b) Analytically calculated θ(τ)
Figure 2: Comparing analytic method with the numerical method in the range 0 < τ < 10
18

19. 2 4 6 8 10
Τ
1.2
1.4
1.6
1.8
2.0
VΦ
(a) N = 10, 000
(b) N = 500, 000
2 4 6 8 10
Τ
1.2
1.4
1.6
1.8
2.0
VΦ
(c) Analytical result
Figure 3: Plots of ˙φ versus τ in the range 0 < τ < 10.
19

20. (a) N = 10, 000
Figure 4: [x, y, z] → [R cos φ sin θ, R sin θ sin φ, R cos θ] in the range 0 < τ < 10.
2.3.2 Results: AdS3
I have chosen the following initial conditions
t(τ = 0) = 0, ρ(τ = 0) = 1, θ(τ = 0) = 0, (112)
V ρ
(τ = 0) =
E
cosh ρ(τ = 0)
2
−
J
sinh ρ(τ = 0)
2
−
1
R2
. (113)
20

21. 2 4 6 8 10
Τ
2
4
6
8
Ρ
(a) τ vs cosh2
ρ, where N = 200, 000 and (E = 2, J = 1, R = 0.58)
(b) τ vs θ, where N = 200, 000 and (E = 2, J = 1, R = 8)
2 4 6 8 10
Τ
5
10
15
t
(c) τ vs t, whereN = 200, 000 and (E = 3, J = 1, R = 0.58)
Figure 5: Plots of cosh2
ρ, θ, t
21

22. 2 4 6 8 10
Τ
1.5
2.0
2.5
3.0
3.5
4.0
cosh2 Ρ
(a) (E = 2, J = 1, R = 0.58)
5 10 15 20 25 30
Τ
4
2
2
4
Θ
(b) (E = 2, J = 1, R = 8)
2 4 6 8 10
Τ
2
4
6
8
10
12
14
t
(c) (E = 3, J = 1, R = 0.58)
Figure 6: Plots obtained analytically where a)cosh2
ρ vs τ and b)θ vs τ and c) t vs τ.
22

23. (a) (E = 2, J = 1, R = 0.58)
(b) (E = 2, J = 0, R = 0.45)
Figure 7: Space-time plot, [x, y, z] → [ρ cos θ, ρ sin θ, t]
23

24. 2.4 Discussion
2.4.1 Flat Space
The result we found was an expected one. The point particle in relativistic 4D
Minkowski space moves on a straight line at a constant velocity.
2.4.2 2-sphere
The point particle moves around along the 2-sphere along greater circles. This had
to be the case since we found that (72) changes as a sin and also that angular
momentum is conserved in the system.
(72) is clearly 2π periodic. When we take the negative part of the solution, at
τ = 2π the point particle would have completed the lower half of its orbit around
the 2-sphere. The positive part of the equation describes motion above the x, y
plane (the upper half of the 2-sphere). This is because θ(τ = 0) = π
2 and θ is an
angle measured from the z axis anticlockwise increasing. Below the x, y plane its
greater than π
2 and thats when cos θ is negative. So when τ ∈ [0, 2π] for when cos θ
is described by the positive part the point particle traverses the upper half of the
orbit.
Our numerical results for θ, φ and t matched with each other very well. However,
the accuracy had to be quite high. When N = 10, 000 the function V φ did not agree
with the exact solution. Peaks started to grow. We increased N to the order of
105 in order to eliminate the problem, this meant that had to be roughly 0.00002.
The solutions were reliable so long as epsilon was in the order of ten thousandths.
In polar coordinates, the point particle traces out a circle with radius R. We set
the z axis to be the time t(τ), a linearly increasing function in τ. Since the particle
is just going around the circle parallel to the x − y plane the end result is a spiral
path along spacetime.
2.4.3 AdS3
The ˙θ and ˙t functions with respect to τ (86) and (87) are both oscillating functions.
When we integrate these functions with respect to τ, the time t for instance is
increasing. The gradient is seen to fluctuate periodically. These arise due to cosh2
ρ
changing as a sin. Physically, oscillations arise due to the point particle moving
along AdS3 spacetime geodesics. AdS3 spacetime seems to confine the particle to
move around the origin of the system so whenever ρ goes beyond a certain point
it falls back in again. Time is shifted because the particle accelerates toward the
origin and the dilation factor γ effect on time becomes more noticeable. When
J = 0, an absence of angular momentum means that we have removed momentum
along ˆθ. The point particle just oscillates back and forth from ρ = ρ0 to ρ = −ρ0,
as we can see from the plots. When J = 0 the end result is a spiral path because
˙θ is non zero. This implies that we have a particle with a well defined rotational
motion in AdS3 spacetime.
The analytical results generally agree with the numerical results except for θ.
V θ is infinite at τ = 17. Recall that V θ ∝ 1
sinh2
ρ
. Since sinh2
≡ cosh2
−1 when
cosh2
ρ → 1, V θ → ∞. This happens at earlier τ when we increase R the radius of
curvature of AdS3 spactime.
Physically what is going on is when the radius of curvature is roughly in the
same order as E and J or less, there is some considerable curvature of spacetime.
24

25. The particle is going around and at every turning point it has to change direction
sharply along constant ρ. This change occurs in the ˆθ direction.
(a)
Figure 8: AdS3 negative curvature resembles the geometry of a saddle [5]
25

26. 3 String in space
The action for a relativistic string must be a functional of the string trajectory. Just
as a particle traces out a n + 1 surface for the n-brane. Strings are n = 1 branes
so they sweep a two dimensional surface in spacetime. This will vary with the
spacetime configuration. We refer this surface as a world-sheet, the generalisation of
the world-line we previously encountered for the point particle.The lines of constant
X0 in these surfaces are the strings at different τ. Recall for the point particle when
we defined the invariant distance/length in space time (2) and that this was Lorentz
invariant. For strings we will define the Lorentz invariant area of a world sheet in
space time.
The relativistic string action, the Nambu-Goto action will be proportional to
the world sheet area. That is to say that the string action minimises the area of its
world sheet in space time. A surface will require two parameter variables σ0 and
σ1. We take σ0 ≡ τ throughout.
A point on a world sheet is a function of these two variables Xµ ≡ Xµ(τ, σ)
where σ ∈ [0, 2π]. In the parameter space our τ and σ ranges span an area d2σ ≡
dτdσ. The world-space configuration of the parameter space is mapped by the
function Xµ(τ, σ) and this produces the two dimensional surface. This surface is
usually called the physical surface embedded in a target space, since this describes
the string in space-time. A closed-string traces out a tube. This is represented by
periodicity in Xµ, Xµ(τ, σ = 0) = Xµ(τ, σ = 2π).
Parametrisation of a surface allows us to write the area element in an explicit
form. This area should be independent of the parameter we choose. This is repa-
rameterisation invariance. Since we will choose to equate the relativistic action to
some notion of area it - the string action, must also be invariant when the action
is repameterised. This ultimately means that we are free to choose the most useful
parameter without changing underlying physics [3, Chap 6 pp 100-103].
It follows that we can write the area functional as;
SNG = −T d2
σ
∂X
∂τ
·
∂X
∂σ
2
− ˙X2X 2, (114)
The Nambu-Goto string action [3, Sect 6.4 p 111]where T: string tension units [m
L ]
in the natural units c = 1, A · B = gµνAµBν and ˙X = ∂Xµ
∂τ , X = ∂Xµ
∂σ .
3.1 String Sigma Model
Even though this is a nice way to understand the area of the string world-sheet, the
presence of the square root makes it hard to perform string action computations.
An action that is equilavent to the Nambu-Goto action at the classical level, because
it gives rise to the same EOM, is the string sigma model action, Sσ. An additional
degree of freedom is introduced into the action. This is the auxillary world-sheet
metric hαβ(τ, σ).
Sσ = −
T
2 A
d2
σ
√
−h∂αX · ∂βX. (115)
This is specific to n = 1 branes, we are integrating over a surface area A and
h ≡ dethαβ, hαβ ≡ (h−1)αβ. A more general action for n = 1 branes requires
an additional cosmological constant term. In many ways (114) is really equilavent
to (115), but only at the classical level [2, Sect 2.2 p 29]. One can see this by
eliminating the world sheet metric hαβ by applying the EL relations (1) and solve
for the metric, just as we did with the auxillary point particle field e(τ).
26

27. So in order to solve for the equations which govern the dynamics of the string
in curved space we must consider solving the following
δSσ
δXµ
= 0 (116)
δSσ
δhαβ
= 0. (117)
These arise from minimising the string sigma action (115). From (117) we find
δSσ
δhαβ
= −
T
2
√
−h(∂αX∂βX −
1
2
hαβhαβ
∂γX∂δX), (118)
⇒ Tαβ = 0 (119)
where Tαβ ≡ − 2
T
1√
−h
∂αX∂βX − 1
2hαβhαβ∂γX∂δX, the energy-momentum tensor
[2, Sect 2.2 p 27]. The strings equation of motion implies the vanishing of the
energy-momentum tensor. This point is very important and we will exploit this
result later in the paper.
3.2 Flat Space
The worldsheet metric hαβ has three independent components
hαβ =
h00 h01
h10 h11
.
h01 = h10 because the world-sheet metric must remain a symmetric matrix.
Due to some (local) symmetries in hαβ we are allowed to choose the values for
each component belonging to h. We can gauge fix the metric to be the Minkowski
metric[2, Sect 2.3 p 31] Reparameterisation invariance is when the action remains
the same under the transformations
σα
→ fα
(σ) = σ α
(120)
hαβ =
∂fγ
∂σα
∂fδ
∂σβ
hγδ(σ ). (121)
We have two functions fα and fβ to play with. The choice for f that we make
is completely up to us because the metric is invariant in this type of transforma-
tion. Therefore, this leaves one independent component for hαβ. The other local
symmetry, Weyl transformation symmetry
hαβ → e2φ(σ)
hαβ (122)
suggests that we can choose φ arbitrarily because its an arbitrary function of σ.
This will allow us to choose the last independent component of the worldsheet
metric h that is suitable for a Minkowski metric. Our new String Sigma Model
action [2, Sect 2.3 p 31] in Minkowski space can be written in the following form
Sσ =
T
2
d2
σ( ˙X2
− X 2
), (123)
where in (123) we have defined ˙X = ∂τ X and X = ∂σX and T is the tension in
the string.
27

28. Suppose for the open or closed string, σ ∈ [0, π]. We demand the string action
Sσ be invariant under the shift Xµ → Xµ + δXµ. Under this transformation the
boundary terms in the action must vanish. Consider the variation of the Polyakov
action [7, Sect 1.3 p 18]
δSP = T
τf
τi
dτ
π
0
dσ∂αX · ∂α
δX (124)
Integrating dσ part of (124) by parts
δSP = [∂α · δX]π
0 −
π
0
(∂α
∂αX) · δXdσ, (125)
integrating over τ ∈ [τi, τf ] and also note that [∂α ·δX]π
0 is a total derivative in δSP
so this can be ignored,
δSP = T
τf
τi
dτ
π
0
dσ(∂α
∂αX) · δX (126)
= T
π
0
dσ ˙X · δX
τ=τf
τ=τi
− T
τf
τi
dσX · δX
π
0
. (127)
The first term of (127) will always be zero because the variation is zero at the end
points τ = τf and τ = τi and furthermore,(127) must vanish so that the action is
minimised which means the second term must vanish also.
The second term will only vanish when
∂σXµ
δXµ = 0, σ = 0, π
From this we find two different types of boundary conditions for the open-string.
The closed string automatically causes the boundary terms to vanish because of
the periodicity condition X(τ, σ = 0) = X(τ, 0), it is a closed loop (of string).
What we can gather are two different types of boundary conditions. One of
which is described by,
∂σXµ
= 0, σ = 0, π. (128)
This is called the Neumann boundary condition. In this type of constriction, the
endpoints of the string are completely free to move but no momentum can flow
transverse to the string (or world-sheet) at σ = 0, π.
The other boundary condition arises when
Xµ
= Cµ
, σ = 0, π. (129)
This is called the Dirichlet boundary condition. Above µ is simply an index, which
we will be using throughout. This is a number wherein µ ∈ [0, . . . , D − 2] and
D denotes the number of space-time dimensions. In this kind of constraint, the
ends of the string are fixed positionally in space-time so that they cannot move at
all at τ > 0. Generally, the Neumann boundary condition is used. The Dirichlet
boundary condition sounds very unphysical because the whole string should be
able to move in space that is to say its centre of mass should move. The modern
interpretation however, is that these two fixed points represent the positions of Dp-
branes. A Dp-brane is a special type of p-brane on which a fundamental string can
end. The presence of a Dp-brane breaks Poincare (transform) invariance unless it
is space-time filling (p = D − 1) [2, Sect 3.1 p 31], [7, Sect 1.3 p 18].
28

29. In order to determine the equations of motions for the string it is important
to note that the Euler-Lagrange equations need to be modified to account for an
additional parameter. The original definition (30) was for the point particle and its
world-line parameter τ. It can be shown that the new EL equations for the string
with world-sheet parameters σ and τ can be written in the following form:
∂L
∂Xµ
−
d
dτ
∂L
∂ ˙Xµ
+
d
dσ
∂L
∂X µ = 0. (130)
We apply (130) to our definition of Sσ in (123), taking L ≡ T
2 ( ˙X2 − X 2) and see
that
∂α∂α
Xµ
= 0, (131)
the Xµ equation of motion is the wave equation [2, Sect 2.3 p 34].
Since the metric hαβ was gauge fixed, the vanishing of the Energy-Momentum
tensor, Tαβ = 0, must be imposed as an additional condition. This is the Virasoro
constraint, on the string. In the gauge hαβ = ηαβ the components of Tαβ are
T01 = T10 = ˙X · X = 0 (132)
(133)
T00 = T11 =
1
2
( ˙Xµ2
+ X 2
) = 0 (134)
TrT = ηαβTαβ ≡ T11−T00 by the Weyl Transformation invariance. Since T00 = T11,
a consequence to Tαβ is that it will always be traceless [2, Sect 2.3 p 34].
It is now convenient to introduce world sheet light cone coordinates. This is
defined by the following equation
σ±
= τ ± σ. (135)
In these coordinates the derivatives and the two dimensional Lorentz metric take
the form [2, Sect 2.3 p 33]
∂± =
1
2
(∂τ ± ∂σ) (136)
(137)
ηγδ =
1
2
0 1
1 0
. (138)
The wave equation (131) in light cone coordinates becomes
∂+∂−Xµ
= 0 (139)
and the Virasoro constraint becomes,
∂±Xµ
∂±Xµ = 0 (140)
while T+− = T−+ = 0 expresses the vanishing of the trace, which is automatic [2,
Sect 2.3 p 34]. The solution to (139) being a wave equation is a sum of a left moving
and right moving wave (Xµ
L and Xµ
R respectively).
Xµ
= Xµ
L(σ+
) + Xµ
R(σ−
) (141)
or
Xµ
= fµ
(τ + σ) + gµ
(τ − σ). (142)
29

30. Here fµ and gµ are just arbitrary functions of a single argument σ±.
We can impose the Neumann boundary condition on an open string. At σ = 0
this gives us
∂Xµ
dσ τ,σ=0
=
1
2
(fµ
(τ) − gµ
(τ)) = 0, (143)
(144)
⇒ fµ
− gµ
= cµ
. (145)
This is possible since the operator ∂σ is linear. Now reabsorbing cµ into fµ,
Xµ
(τ, σ) =
1
2
(fµ
(τ + σ) + fµ
(τ − σ)).
At σ = π,
∂Xµ
dσ τ,σ=π
=
1
2
(fµ
(τ + π) − fµ
(τ − π)) = 0, (146)
which must hold for all τ. Therefore we notice that fµ is periodic in σ with period
2π. This enables us to define fµ as a Fourier series
fµ
(τ) = fµ
1 +
∞
n=1
(aµ
n cos nτ + bµ
n sin nτ). (147)
Integrating (147) with respect to τ we get the expression fµ [3, Chap 9]
fµ
(τ) = fµ
0 + fµ
1 τ +
∞
n=1
(Aµ
n cos nτ + Bµ
n sin nτ). (148)
As you can see new constants Aµ
n and Bµ
n are really the integration constants reab-
sorbed into the previous Fourier series constants aµ
n and bµ
n.
Substituting (148) back into our definition of Xµ:
Xµ
(τ, σ) = fµ
0 + fµ
1 τ +
∞
n=1
(Aµ
n cos nτ + Bµ
n sin nτ) cos nσ. (149)
This is called a mode expansion for the Xµ coordinates in the open string case.
Using appropriate variables it can be shown that for the closed string Xµ can be
expressed by the following mode expansion
Xµ
= xµ
+ α pµ
τ + i
α
2
m∈Z
1
m
(αµ
me−iσ−
+ ˜αµ
me−iσ+
) (150)
pµ = 2
α αµ
0 . This is a necessary definition because the sum in (150) is indeter-
minable for when m = 0. The zeroeth mode is always defined in terms of the
momentum pµ.
˜α are called the mode coefficients for the left moving part of the string and α
are the mode coefficients for the right moving part of the string. We get an open
string when α = ˜α and a closed string when α = ˜α For a string (open or closed)
we need to satisfy the Virasoro constraints initially at τ = 0.
ηµν∂±Xµ
∂±Xν
= 0. (151)
30

31. In terms of mode coefficients, (151) is equilavent to
Lm =
n∈Z
ηµναµ
m−nαν
n = 0. (152)
A similar relation holds for ˜Lm and ˜αµ
n. The constraints (152) are very difficult to
handle in general because it is quadratic in αµ
m. However, in a light cone gauge this
can be easily solved. First we define the light cone coordinate and the metric in
space time as
X±
=
1
√
2
Xµ=0
± Xµ=1
(153)
ds2
= −2dX+
dX−
+ dXi
dXi
. (154)
Our Virasoro constraints (151) then become
−2∂±X+
∂±X−
+ (∂±Xi
)2
= 0. (155)
In terms of mode coefficients,
Lm =
m∈Z
(−2α+
m−nα−
n + αi
m−nαi
n) = 0. (156)
This is still not so easy to solve, so at this point we introduce the light-cone gauge.
That is to set X+ to the following form, a differeomorphism if you will,
X+
(τ, σ) = x+
+ τ. (157)
This means that
p+
=
1
α
(158)
where
α =
l2
s
2
. (159)
Also
α+
0 = ˜α+
0 . (160)
In this gauge we can write α−
n in terms of the transverse modes αi
n. The light cone
gauge allows us to equilavently write out the light cone Virasoro contraints (155),
˙X−
± X−
=
1
2p+l2
s
( ˙Xi
± Xi
)2
(161)
(161) can be used to give us the modes for X− in terms of the transverse modes.
The mode expansion for the open string (αµ
m = ˜αµ
m) is much simpler to look at,
this can give us the following expression for the LHS of (161),
˙X−
± X−
=
√
2α
∞
n=1
α−
n e−inτ
(cos nσ ± i sin nσ), (162)
and cos nσ ± i sin nσ ≡ e±inσ.
⇒ ˙X−
± X−
=
√
2α
∞
n=1
α−
n e−in(τ±σ)
, (163)
31

32. and RHS of (161) can similarly be written as,
1
2p+l2
s
( ˙Xi
± Xi
)2
=
1
2p+l2
s
∞
p,q=1
αi
pαi
qe−i(p+q)(τ±σ)
(164)
Therefore, we see that
√
2α
∞
n=1
α−
n e−in(τ±σ)
=
1
2p+l2
s n∈Z
(
∞
p∈Z
αi
pαi
n−pe−in(τ±σ)
), (165)
where n = p + q, so that finally [2, Sect 2.5 pp 48-49]
α−
n =
α
2
D−2
i=1 m∈Z
αi
mαi
n−m. (166)
32

33. 3.3 Explicit calculation of a string in flat-space
3.3.1 Simple case: Xµ
L ≡ Xµ
R
Let us consider a very simple oscillation. We will presume that α = ˜α so that our
string is the most simple case closed string. We turn on only one transverse mode,
namely lets turn on αi=2
m0=1. This can be some arbitrary complex number but for
simplicity lets set it to be
α2
1 = −iA,
A ∈ R.
To make sure that the transverse coordinates Xi (in our case we are just con-
cerned with one of these coordinates because we are only concerned with α2
1) are
real we need α2
−m0
= (α2
m0
)∗. Therefore α2
−1 needs to be turned on as well. All
other αi
m0
are turned off, including αi
0 for all i = 2.
Using Virasoro constraint (166) we see that consequently the only α−
n ’s which
turn on are
α−
0 =
√
2α A2
, α−
±2 = −
α
2
A2
. (167)
From the mode expansion (150),
X2
= x2
+
α
2
∞
m=0
1
m
(α2
me−iσ−
+ ˜α2
me−iσ+
) (168)
So, m = ±1. For the m = ±1 term we get
X2
±1 = x2
+ 2
2
α
A(cos σ+
+ cos σ−
) (169)
Therefore,
X2
(τ, σ) = x2
+
√
2α A(cos σ+
+ cos σ−
). (170)
Now also for X−, from the mode expansion we can also say,
X−
= x−
+ α τ +
α
2
∞
m=0
1
m
(α−
me−iσ−
+ ˜α−
me−iσ+
). (171)
Using the (167) we get,
X−
(τ, σ) = α A2
2τ −
1
2
(sin 2σ+
+ sin 2σ−
) (172)
Full expression for the Xµ now reads,
X+
(τ, σ) = τ, (173)
X−
(τ, σ) = α A2
2τ −
1
2
(sin 2σ+
+ sin 2σ−
) , (174)
X2
(τ, σ) = x2
+
√
2α A(cos σ+
+ cos σ−
). (175)
33

34. 3.3.2 More general wave: Xµ
L = Xµ
R
We can loosen the restriction on the closed string.
α2
m0
= Aeia
; α2
−m0
= Ae−ia
;
α−
0 =
√
2α A2
; α−
2m0
=
α
2
e2ia
;
α−
−2m0
=
α
2
A2
e−2ia
and
˜α2
˜m0
= Aei˜a
; ˜α2
− ˜m0
= Ae−i˜a
;
˜α−
0 =
√
2α A2
; ˜α−
2 ˜m0
=
α
2
A2
e2i˜α
;
˜α−
−2 ˜m0
=
α
2
A2
e−2i˜a
.
where a = ˜a and also m0 = ˜m0 where A ∈ R. This will set all left movers apart
from right movers.
The fields Xµ are;
X+
(τ, σ) = x+
τ, (176)
X−
(τ, σ) = x−
+ α A2
2τ +
1
2
sin 2(m0σ− − a)
2m0
+
sin 2( ˜m0σ+ − ˜a)
˜2m0
, (177)
X2
(τ, σ) = x2
+
2
√
2α A
m0
sin(m0σ− − a)
m0
+
sin( ˜m0σ+ − ˜a)
˜m0
. (178)
34

35. 3.3.3 Results
Below we see worldsheets of the closed string with different arbitrary parameters.
The axes are labelled (x, y, z) → (X1, X2, X0).
(a) A = 1, m0 = ˜m0 = 1 and a = ˜a = −π
2 . (b) A = 1, m0 = ˜m0 = 1 and a = −π
2 , ˜a = −π.
(c) A = 1, m0 = 1, ˜m0 = 2 and a = −π
2 , ˜a = −π.
35

36. Below we see the string τ slices for each worldsheet respectively from top left
clockwise.
0 2 4
x
2
0
2
y
5
0
5
z
(a) τ = 0, A = 0.5, m0 = ˜m0 = 1,
a = ˜a = −π
2 .
0 2 4
x
2
0
2
y
5
0
5
z
(b) τ = 1.07, A = 0.5, m0 =
˜m0 = 1, a = −π
2 , ˜a = −π.
0 2 4
x
2
0
2
y
5
0
5
z
(c) τ = 1.77314, A = 0.5, m0 =
1, ˜m0 = 2, a = −π
2 , ˜a = −π.
0 2 4
x
2
0
2
y
5
0
5
z
(d) τ = 2.3118, A = 0.5, m0 =
1, ˜m0 = 2, a = −π
2 , ˜a = −π.
36

37. 2
0
2
x
2
0
2y
0
5
10
z
(a) τ = 0, A = 1, m0 = ˜m0 = 1,
a = −π
2 and ˜a = −π
5
0
5
x
5
0
5
y
0
5
10
z
(b) τ = 1.07, A = 1, m0 = ˜m0 =
1, a = −π
2 , ˜a = −π.
2
0
2
x
2
0
2y
0
5
10
z
(c) τ = 1.77314, A = 1, m0 =
1, ˜m0 = 2, a = −π
2 , ˜a = −π.
5
0
5
x
5
0
5
y
0
5
10
z
(d) τ = 2.3118, A = 1, m0 =
1, ˜m0 = 2, a = −π
2 , ˜a = −π.
37

38. 5
0
5
x
5
0
5
y
0
5
10
z
(a) τ = 0, A = 1, m0 = 2, ˜m0 =
1, a = −π
2 and ˜a = −π
5
0
5
x
5
0
5
y
0
5
10
z
(b) τ = 1.07, A = 1, m0 =
2, ˜m0 = 1, a = −π
2 , ˜a = −π.
5
0
5
x
5
0
5
y
0
5
10
z
(c) τ = 1.77314, A = 1, m0 =
2, ˜m0 = 1, a = −π
2 , ˜a = −π.
5
0
5
x
5
0
5
y
0
5
10
z
(d) τ = 2.3118, A = 1, m0 =
2, ˜m0 = 1, a = −π
2 , ˜a = −π.
38

39. 3.4 Numerical Method and Results
3.4.1 4D Flat spacetime
The worldsheet of the string can also be numerically reproduced to a very good
degree of accuracy. Take for instance the string in flat space. Its variables are
X+, X−, X2 and their τ derivatives are V +, V − and V 2 respectively. Note that
these are functions of two variables in σ and τ which make it a little less trivial. We
will attempt to reduce the problem to a rather trivial first order ordinary differential
equation that can be solved for X−. By the light cone gauge differemorphism we
will able to limit the number of functions to be determined in the problem to one.
This will be the X− coordinate and the other function X2 will be given. It can be
shown that the EOM in flat space for the string is;
( ˙X−
)2
− (X−
)2
= 0 (179)
The Virasoro constraint need to be satisfied for τ = 0 (and hence for all τ >
0).This is already determined from equation (166) which is the Virasoro constraint
in terms of mode expansion coefficients α.
Consider a more simple case of flat space, where Xµ
L = Xµ
R which produced the
solution for the closed string whose left and right moving parts was indistinguish-
able. The initial conditions make up 6 functions. They all satisfy the Virasoro
constraints.
X+
(0, σ) = x+
, V +
(0, σ) = 1 (180)
X−
(0, σ) = x−
, V −
(0, σ) = 4α A2
sin2
σ (181)
X2
(0, σ) = x2
+ 2
√
2α A cos σ, V 2
(0, σ) = 0. (182)
The time evolution for X+, X−, X2 can be written in the form of first order
differentials equation as follows
V µ
(τ + , σ) = V µ
(τ, σ) + ˙V µ
. (183)
This ignores O( 2), where is the interval width parameter in the τ domain because
this is a first order approximation. Euler’s method for the recursive solution to first
order differential equations (PDE) approaches the analytical solution when → 0
and also simultaneously η → 0. We essentially will need the curve connecting points
connecting three points X(τ, σ − η), X(τ, σ), X(τ, σ − η) to be straight lines. That
is that both η and must be very small so that the curve over two distinct points
is well approximated by a straight line. We will define these parameters to be;
=
τf − τi
N
, η =
σf − σi
n
; (184)
N and n are the number of divisions for τ and σ variables respectively and and η
are the interval widths for τ and σ respectively.
V µ
(τ, σ) =
Xµ(τ + , σ) − Xµ(τ, σ)
, (185)
that is V µ = ∂τ Xµ where µ ∈ [+, −, 2].
By using the equation of motion for the string in flat space as per (179),
˙V µ
= X µ
(186)
39

40. we can define V µ completely for τ, σ > 0,
V µ
(τ + , σ) = V µ
(τ, σ) +
η2
(Xµ
(τ, σ + η) + Xµ
(τ, σ − η) − 2Xµ
(τ, σ)) , (187)
This is the recipie we will use to determine Xµ for all τ > 0 by (185).
The above will reproduce the worldsheet for the closed string when the left
movers are indistinguishable from the right movers by the equations (185), (187)
recursively. For consistency, we will chose α = 1, m0 = ˜m0 = 1, a = ˜a = −π
2 and
A = 1 just like before (analytical result). The number of divisions in the τ and σ
domains are 800 and 100 respectively.
40

41. (a) 0 < τ < 5
Figure 9: The open string worldsheet reproduced using Euler’s numerical recursive tech-
nique.The [x,y,z] axes are X1
, X2
and X0
respectively.
41

42. 3.4.2 AdS3
Euler’s method managed to reproduce the string world sheet in flat space. We now
intend to extend this to curved space. So in AdS3 space time, we need to look at
a more general recipie for the equation of motion for the classical string. It can be
shown that this is,
∂+∂−Xµ
+ Γµ
ρσ∂+Xρ
∂−Xσ
= 0. (188)
We can easily extract the EOM flat space case , when Γµ
ρσ = 0.
From the following christoffel symbols
Γt
ρt = Γt
tρ = tanh ρ, Γρ
tt = cosh ρ sinh ρ, (189)
Γt
φφ = − sinh ρ cosh ρ, Γφ
ρφ = Γφ
φρ = coth ρ. (190)
we extract the following EOM in AdS3,
0 = ∂+∂−t + tanh ρ(∂+t∂−ρ + ∂+ρ∂−t), (191)
0 = ∂+∂−ρ + cosh ρ sinh ρ(∂+t∂−t − ∂+φ∂−φ), (192)
0 = ∂+∂−φ + coth ρ(∂+t∂−φ + ∂+ρ∂−φ). (193)
In τ and σ parameters these equations (191), (192),(193) equilavently are,
0 = ¨t − t + 2 tanh ρ(˙t ˙ρ − t ρ ), (194)
0 = ¨ρ − ρ + cosh ρ sinh ρ(˙t2
− t
2
− ˙φ2
+ φ
2
), (195)
0 = ¨φ − φ + 2 coth ρ( ˙φ ˙ρ − φ ρ ). (196)
By defining V t ≡ ˙t etc, these can be rewritten as partial differential equations
as follows,
˙V t
= t − tanh ρ(V t
V ρ
− t ρ ), (197)
˙V ρ
= ρ − cosh ρ sinh ρ((V t
)2
− t 2
− (V φ
)2
+ (φ)2
), (198)
˙V φ
= φ − 2 coth ρ(V φ
V ρ
− φ ρ ). (199)
This allows us to completely write the equation governing τ evolution of τ, ρ, φ
for each σ as:
t(τ + , σ) = t(τ, σ) + V t(τ, σ), (200)
ρ(τ + , σ) = ρ(τ, σ) + V ρ
(τ, σ), (201)
φ(τ + , σ) = φ(τ, σ) + V φ
(τ, σ), (202)
(203)
and,
V t
(τ + , σ) = V t
(τ, σ) + ˙V t
(τ, σ), (204)
V ρ
(τ + , σ) = V ρ
(τ, σ) + ˙V ρ
(τ, σ), (205)
V φ
(τ + , σ) = V φ
(τ, σ) + ˙V φ
(τ, σ). (206)
The Virasoro constraints govern the initial conditions on the functions Xµ and
V µ. Using the same approach as before we require 6 functions worth of initial data
(because for each t, ρ, φ we need V t, V ρ, V φ as well).
− cosh2
ρ V t
± t
2
+ V ρ
± ρ
2
+ sinh2
ρ V φ
± φ
2
= 0. (207)
42

43. If we give two functions arbitrarily (namely V φ(0, σ) and φ(0, σ)) and if we
know that we are still in a light-cone gauge, which means another two are predeter-
mined by our differemorphism, then the remaining two functions X− and V − are
determined by the constraints (207).
Now we can introduce new light-cone coordinates X±
X±
=
R
√
2
[(cosh ρ0)t ± (ρ − ρ0)] , (208)
which mean
t(X+
, X−
) =
X+ + X−
√
2R cosh ρ0
; ρ(X+
, X−
) = ρ0 +
X+ − X−
√
2R
. (209)
where ρ0 is some constant. So the point (t, ρ) = (0, ρ0) corresponds to (X+, X−) =
(0, 0).
Using (209) into the AdS3 metric the t, ρ part becomes,
R2
(− cosh2
ρdt2
+dρ2
) = − 1 +
cosh2
ρ
cosh2
ρ0
dX+
dX−
+
1
2
1 −
cosh2
ρ
cosh2
ρ0
((dX+
)2
+(dX−
)2
)
(210)
Note that ρ = f(X+, X−), but we have kept it implicit to avoid clutter. In the X±
coordinate at ρ → ρ0 therefore,
ds2
AdS = −2dX+
dX−
, (211)
which is just ds2
flat. Locally to ρ0, the spacetime must be flat.
Solving for ρ >> ρ0, points that are not in the vicinity of ρ, in which the string
will traverse curved space we find the Virasoro contraint can be expressed in a form
similar to the metric. Simply replacing d → ∂±
0 = − 1 +
cosh2
ρ
cosh2
ρ0
∂±X+
∂±X−
+
1
2
1 −
cosh2
ρ
cosh2
ρ0
((∂±X+
)2
+ (∂±X−
)2
)
(212)
+ R2
sinh2
ρ(∂±φ)2
(213)
and in τ, σ the above can be recast in the following form
0 = −(c2
0 + c2
) V +
± X+
V −
± X−
2
+
1
2
(c2
0 − c2
) V +
± X+
2
+ V −
± X−
2
(214)
+ R2
c2
0s2
(V φ
± φ )2
. (215)
Note: I have used cosh2
ρ0 = c2
0, cosh2
ρ = c2 and sinh2
ρ = s2.
This is just a quadratic equation for (V − ± X− ) (by taking light cone gauge
for X+) so that by the quadratic equation this will yield the following solution
V −
± X−
=
(c2
0 + c2) ± (c2
0 + c2)2 − (c2
0 − c2) (c2
0 + c2) + 2R2c2
0s2(V φ ± φ )2
c2
0 − c2
(216)
43

44. which is just an ODE for X−(0, σ) below,
V −
=
1
2
1
2(c2
0 − c2) + R2c2
0s2(V φ + φ )2
c2
0 + c2 +
√
D+
+
1
2(c2
0 − c2) + R2c2
0s2(V φ − φ )2
c2
0 + c2 +
√
D−
(217)
X−
=
1
2
1
2(c2
0 − c2) + R2c2
0s2(V φ + φ )2
c2
0 + c2 +
√
D+
−
1
2(c2
0 − c2) + R2c2
0s2(V φ − φ )2
c2
0 + c2 +
√
D−
.
(218)
D± = 2c2
0 c2 1 + c
c0
2
− R2(c2
0 − c2)s2(V φ ± φ )2 .
Now we have the complete recipie for a complete ODE which determines X−(0, σ)
from Euler’s Method, given a X−(0, 0),
X−
(0, σ + η) = X−
(0, σ) + ηX−
(0, σ).
This determines X−(0, σ) and V −(0, σ).
44

45. Figure 10: This is the world-sheet of the string in AdS3. We have defined axes as
[X1
, X2
, X0
] corresponding to our [x, y, z] respectively.
45

46. 2 0 2
2
0
2
10
5
0
5
10
(a) τ = 0
2 0 2
2
0
2
10
5
0
5
10
(b) τ = 0.37
2 0 2
2
0
2
10
5
0
5
10
(c) τ = 0.59
2 0 2
2
0
2
10
5
0
5
10
(d) τ = 1.02
Figure 11: Plots of the closed string at different τ.
3.4.3 Blackhole Background
Now we place a blackhole background inside AdS3 space. The string should as we
saw previously move toward the origin.
We set the horizon of the black hole to be the boundary containing the zone of
singularity. At the boundary we expect some strange behaviour. The string should
not disintegrate, that is because the string is in some sense still a particle in that it
is indivisible like the electron. The energy of the string in the frame of the falling
string increases with time as it gets closer, till practically infinity* This has serious
effect on the string manifestly in the physical appearance of the string.
The metric for the black-hole background in AdS3 [6] is usually called the BTZ
(Ba˜nados-Tietelboim-Zanelli) blackhole can suitably be written in the following
form;
ds2
L = R2
−(r2
− r2
H)dt2
+
dr2
r2 − r2
H
+ r2
dφ (219)
where R is the radius curvature represenative of AdS3. The horizon is at r = rH.
We obtain the following christoffel symbols;
Γt
tr = Γt
rt =
r
r2 − r2
H
, (220)
Γr
tt = r2
(r2
− r2
H), Γr
rr = −
r
r2 − r2
H
, Γr
φφ = −r2
(r2
− r2
H), (221)
Γφ
rφ = Γφ
φr =
1
r
, (222)
46

47. from which we can calculate the equations governing motion of the string by the
geodesic equation (188). Considering Γt
tr symbol we can extract the following EOM
0 = ∂+∂−t + Γt
tr [(∂+t∂−r) + (∂+r∂−t)] , (223)
0 =
1
4
(¨t − t ) +
1
4
[(∂τ + ∂σ)t(∂τ − ∂σ)r + (∂τ + ∂σ)r(∂τ − ∂σ)t]. (224)
(∂τ + ∂σ)t(∂τ − ∂σ)r + (∂τ + ∂σ)r(∂τ − ∂σ)t = (˙t + t )( ˙r − r ) + ( ˙r + r )(˙t − t )
(225)
= ˙t ˙r + t ˙r − ˙tr − t r + ˙r ˙t + r ˙t − ˙rt − r t
(226)
= 2( ˙r ˙t − t r ). (227)
So that for the first geodesic equation we can extract
0 = ¨t − t +
2r
r2 − r2
H
( ˙r ˙t − t r ). (228)
When µ = r we need to look at Γr
tt, Γr
rr and Γr
φφ symbols.
0 = ∂+∂−r + Γr
tt(∂+t∂−t) + Γr
rr(∂+r∂−r) + Γr
φφ(∂+φ∂−φ) (229)
0 =
1
4
¨r − r + Γr
tt(˙t2
− t 2
) + Γr
rr( ˙r2
− r 2
) + Γr
φφ( ˙φ2
− φ 2
) , (230)
0 = ¨r − r + Γr
tt(˙t2
− t 2
) + Γr
rr( ˙r2
− r 2
) + Γr
φφ( ˙φ2
− φ 2
) (231)
which leads to the second extracted equation for the string,
0 = ¨r −r +r2
(r2
−r2
H)(˙t2
−t 2
)−
r
r2 − r2
H
( ˙r2
−r 2
)−r2
(r2
−r2
H)( ˙φ2
−φ 2
). (232)
Finally when µ = φ we get the last equation of motion, and from which all three
now completely define the motion of the string about the black hole horizon, by a
very similar calculation, which I need not explicitly express, we get that
0 = ¨φ − φ +
2
r
( ˙φ ˙r − φ r ). (233)
47

48. In order to be able to fully define the variables t, r and φ in the blackhole
background we need to specify the initial conditions. These cannot be arbitrary as
we saw earlier, however some could be arbitrarily chosen like X2 in order to reduce
the Virasoro constrain so that X− is the only constrained variable. Using a similar
approach to the one in the empty AdS3 case we begin by looking at the new X±
relations:
X±
=
R
√
2

 r2
0 − r2
Ht
r − r0
r2
0 − r2
H

 , (234)
such that
X+
+ X−
=
√
2R r2
0 − r2
Ht , (235)
X+
− X−
= −
√
2R
r − r0
r2
0 − r2
H
, (236)
from which we obtain the following inverse relations:
t(X+
, X−
) =
X+ + X−
√
2 r2
0 − r2
H
(237)
r(X+
, X−
) = r0 −
1
R
r2
0 − r2
H
2
(X+
− X−
). (238)
The (t, r) part of the metric becomes as a result,
R2
−(r2
− r2
H)dt2
+
dr2
r2 − r2
H
= R2
−
1
2R2
r2 − r2
H
r2
0 − r2
H
(dX+
)2
+ (dX−
)2
+ 2dX+
dX−
(239)
+ R2 1
2R2
r2
0 − r2
H
r2 − r2
H
(dX+
)2
+ (dX−
)2
− 2dX+
dX−
(240)
=
1
2
(dX+
)2
+ (dX−
)2 r2
0 − r2
H
r2 − r2
H
−
r2 − r2
H
r2
0 − r2
H
(241)
− dX+
dX− r2
0 − r2
H
r2 − r2
H
+
r2 − r2
H
r2
0 − r2
H
. (242)
As a check, at r = r0,
ds2
L =
1
2
(dX+
)2
+ (dX−
)2
(0) − 2dX+
dX−
(243)
It reduces to flatspace metric ds2
flat = −2dX+dX−. Now in the X± coordinates
the Virasoro constraint can be expressed in the following form,
− ∂±X+
∂±X− r2
0 − r2
H
r2 − r2
H
+
r2 − r2
H
r2
0 − r2
H
(244)
+
1
2
(∂±X+
)2
+ (∂±X−
)2 r2
0 − r2
H
r2 − r2
H
−
r2 − r2
H
r2
0 − r2
H
+ R2
r2
(∂±φ)2
= 0.
(245)
Multiply both sides by (r2
0 − r2
H)(r2 − r2
H),
− ((r2
− r2
H)2
+ (r2
0 − r2
H)2
) − ∂±X+
∂±X−
+
1
2
(−(r2
− r2
H)2
+ (r2
0 − r2
H)2
) (∂±X+
)2
+ (∂±X−
)2
(246)
+ R2
r2
(r2
0 − r2
H)(r2
− r2
H)(∂±φ)2
= 0, (247)
48

49. We define V µ = ∂τ Xµ, X µ = ∂σXµ and further use the light cone differemor-
phism X+(τ, σ) = x+ + τ. The initial conditions on these hence become
X+
(0, σ) = x+
, X +
(0, σ) = 0, V +
(0, σ) = 1.
The Virasoro constraint above becomes
0 = −
1
4
((r2
− r2
H)2
+ (r2
0 − r2
H)2
)(V −
± X −
) +
1
8
(−(r2
− r2
H)2
+ (r2
0 − r2
H)2
)(1 + (V −
± X −
)2
)
(248)
+
1
4
R2
r2
(r2
0 − r2
H)(r2
− r2
H)(V φ
± φ )2
(249)
=
1
2
A(V −
± X −
)2
− B(V −
± X −
) +
1
2
A + C± (250)
where A = −(r2 − r2
H)2 + (r2
0 − r2
H)2, B = (r2 − r2
H)2 + (r2
0 − r2
H)2, C± = R2r2(r2
0 −
r2
H)(r2 − r2
H)(V φ ± φ )2.
We solve the quadratic equation in V − ± X − and find that
V −
± X −
=
B ± B2 − A2 − 2AC±
A
(251)
V −
± X −
=
B −
√
D±
A
, (252)
where D± = B2 − A2 − 2AC±. The sign in front of
√
D± was chosen so that if
r = r0 and thus A = 0 it gives
V −
± X −
=
C
B
.
So, we get the following ODE’s
V −
=
1
2
B −
√
D+
A
+
B −
√
D−
A
, (253)
X −
=
1
2
B −
√
D+
A
−
B −
√
D−
A
. (254)
Now just as in the flat space we take φ to be an oscillating function
φ(0, σ) =
2
√
2α A
m0
cos m0σ, V φ
(0σ) = 0.
We have, D+ = D− = D and therefore X −(0, σ) = 0 meaning that,
X−
(0, σ) = const = x−
.
The blackhole problem has been specified sufficiently at τ = 0. We can proceed
to evolve the system by the three dynamic equations for t, r and φ for τ, σ > 0. We
will arbitrarily choose rH = 1, r0 = 2, m0 = 1, α = 1, A = 0.1 and R = 1. R is the
radius of curvature of the spacetime and A corresponds to the size of the string.
Regarding the precision of the recursive algorithm, we will attempt to compute the
solution at N = 3000, n = 1000, ≈ 0.000333, η ≈ 0.00628 and τ ∈ [0, 1]. (N, )
and (n, η) correspond to the number of divisions and the interval widths in the τ
and σ domains respectively.
49

50. 1.0
1.5
2.0
1.0
0.5
0.0
0.5
1.0
0
20
40
(a) τ = 0.
1.0
1.5
2.0
1.0
0.5
0.0
0.5
1.0
0
20
40
(b) τ = 0.40000
1.0
1.5
2.0
1.0
0.5
0.0
0.5
1.0
0
20
40
(c) τ = 0.75000
1.00
1.05
1.10
1.15
1.20
1.0
0.5
0.0
0.5
1.0
0
20
40
(d) τ = 0.78700
1.00
1.05
1.10
1.15
1.20
1.0
0.5
0.0
0.5
1.0
0
20
40
(e) τ = 0.79160
1.000
1.005
1.010
1.015
1.020
1.0
0.5
0.0
0.5
1.0
0
20
40
(f) τ = 0.79233, zoomed in.
1.000
1.005
1.010
1.015
1.020
1.0
0.5
0.0
0.5
1.0
0
20
40
(g) τ = 0.79320, zoomed in.
Figure 12: Plots of the closed string near the black-hole at different τ. The leftmost wall
represents the event horizon.
50

51. 3.5 Discussion
Our flat space calculations laid the grounds for the AdS3 and blackhole calculations.
We kept X2(τ, σ) the same and we did not add any additional mode which was
not already turned off when we derived the Xµ for the string in flat space. This
produced a trivial straight line motion for the string. The left and right modes
were both indistinguishable and so the closed string in AdS3 and in the black hole
background looked like an open string. What we saw is some oscillation in the form
of stretches and contractions like that of a stretched spring.
3.5.1 Flat Space
In flat space the motion was very trivial. It produced the same straight line motion
like the classical point particle. The only difference between the point particle and
the string was that the string was seen to stretch and contract periodically. There
was no deformation because the string then and again returned back to its original
shape.
The numerical result for the string worldsheet when a = ˜a, m = ˜m produced
the same distinctive shape as the analytically obtained worldsheet. When we lifted
the restrictions on the string bit by bit the string started open up. See the last one
for instance, when a = ˜a and also m = ˜m. The string clearly looked like a loop of
string. This was because opening up restrictions on the string make the right and
left moving modes different from one another.
3.5.2 AdS3
When we introduced curved spacetime we needed to derive a new recipie for the
equations of motion for the string. The Christoffel symbol in (188) was non van-
ishing. We had more and more terms appearing in the EOM because the string
in curved space is not trivial. A new recipie for Virasoro constraints on the string
had to be determined. We assumed that the light cone gauge coordinates for X±
was going to work for the string in curved space. This would not produce the
right initial conditions for the string because those X± would not reproduce a flat
space metric in a zone local to the string. Locally we expect that, irrespective of
the curvature of space, a flat space. By making a few adjustments to the original
definition for X± we obtained a new set of light cone coordinates - which we then
tested.
In AdS3 the string started at ρ0 and it was seen falling toward the point ρ = 0.
The centre of mass was seen as falling the quickest and the two ends of the string
was behind trying catch up. This produced an interesting V shape. Also the string
did not appear to oscillate. This was perhaps the most interesting change in the
physical appearance of the string in the curved and flat spaces.
However, when the string reached the point ρ = 0 the string solution broke
down. We plotted the τ evolution of the string within a polar coordinate frame
work. The problem with polar coordinates is that ρ is never zero. So solutions that
take ρ past zero and onto the the negative axes cannot be numerically implemented
by the computer because ρ = 0 is seen as a computational boundary, like division
by zero for instance. The solution rapidly becomes divergent and out of control
after passing the maximum allowed computation number. We know that this is
not a physical problem because the coordinates which we choose are not physical
themselves.
51

52. 3.5.3 Blackhole Background
Coordinate singularity is partly also responsible for divergences at the point r = rH.
We expected the string to make its way around the horizon and become much more
longer in length. This cannot happen because there is a boundary at r = 0. When
the string makes its way toward the horizon, the COM of the string arrives first
and the two ends of the string play catch up. Nearer to the horizon the COM
points upward as it wants to avoid the horizon see figure 3 e). Some time later a
part of the string makes its way beyond the horizon, see figure 3 f). Part of the
string overshoots through and into the horizon and when we allow it to move a
little further the solution collapses. Divergences take over and we get overflow.
4 Conclusions
Classically, the point particle and the string exhibit the same kind of motion. For
instance in flat space both of them move in straight lines albeit for string oscillations
which occur because it has to obey the wave equation (131). Furthermore, in parts
where we had analytical solution to the differential equations we found that they
were consistent with the numerical equilavent.
When we put it in AdS3 space both the point particle and the string move
as if they are in a potential well. However, coordinate singularities arise in the
numerical computation of the string so motion beyond ρ = 0 becomes computa-
tionally inaccessible. Likewise, when we put the string in a Black hole background
in AdS3 spacetime the behaviour of the string is well known outside the horizon.
It exhibits the same behaviour as in the free AdS3 case. Unfortunately, when the
string approaches the horizon part of the string overshoots. The solutions become
divergent and the numerical solution fails. We do expect the string to show off
some unique behaviour, in that it is uniquely different to the point particle. The
string was observed to stretch and deform in AdS3 free spacetime for example. The
coordinate singularity in the AdS3 means that the string motion could not be fully
determined by the computer. Similarly, in the black hole background the string
could not exhibit the full motion near the horizon.
We can resolve this issue by introducing a new coordinate system with a new set
of coordinates. A coordinate system is unphysical so we expect a new coordinate
system like the Eddington-Finkelstein coordinate system for example to resolve the
problem. This would not stop the overshooting that happens to the COM of the
string near the the point r = rH. This is caused by a lack of precision in , the
interval width. When it is too big the numerical solution over/under estimates the
gradient of the function Xµ. We would need therefore a much smaller , and a much
higher N. Unfortunately, attempting this kind of computation was too taxing for
the processor so we would need to use a computer with more processors to show
that this does indeed solve the overshooting problem.
52

53. References
[1] Wikipedia https://en.wikipedia.org/wiki/Pierre Louis MaupertuisLeast action principle.
Visited 23/03/17.
[2] Katrin Becker, Melaine Becker, John H. Schwarz String theory and M-theory:
A modern introduction. Cambridge, 2007
[3] Barton Zwiebach A first course in String theory. Cambridge, 2009
[4] Wikipedia Commons https://commons.wikimedia.org/wiki/File:Least action principle.svg
Visited 22/03/17.
[5] Inspire: HEP https://inspirehep.net/record/1223647/plots Visited 22/03/17.
[6] M. Banados, C. Tietelboim and J. Zanelli, The Black hole in three-dimensional
space-time Phys. Rev. Lett. 1894 (1992) doi:10.1103/PhysRevLett.69.1849 [hep-
th/9204099].
[7] http://www.damtp.cam.ac.uk/user/tong/string/one.pdf page 18, Visited
22/03/17.
53