Particules Élémentaires, Gravitation et Cosmologie
Année 2010-’11
Théorie des cordes: quelques applications
Cours X: 11 mars 2011
Transplanckian scattering in QST:
IV. The strong-gravity regime
11 mars 2011
G. Veneziano, Cours no. X
1
Outline
• Identification of the relevant diagrams
• Resumming classical corrections via an effective action
• Two-body scattering at b ≠ 0: a numerical solution
• The axisymmetric case: comparison with CTS criteria
• Graviton spectra below and near criticality
• What happens above criticality?
D=4 hereafter
11 mars 2011
G. Veneziano, Cours no. X
2
IV: Small-angle graviton (GW) emission
=> Classical corrections to leading eikonal
�
S(E, b) ∼ exp i
Acl
�
�
�
�
Gs
∼ exp −i (logb2 + O(R2 /b2 ) + O(ls2 /b2 ) + O(lP2 /b2 ) + . . . )
�
V: Large-angle inelastic scattering
VI: Collapse?
=> Resumming classical corrections
11 mars 2011
G. Veneziano, Cours no. X
3
b/lP
log-log plot
I: leading
eikonal
IV:grav.
rad.
strong
gravity
regimes
II:tidal
str. exc.
critical line?
(ls/lP)2
V: large angle
VI: black hole
production/evaporation
ls/lP
III: s-channel
string prod.
1
1
(ls/lP)1/2
ls/lP
R(E)/lP
Classical corrections characterized by absence of h.
Not surprisingly, they are related to tree diagrams
once the coupling to the external energetic particles is
replaced by a classical source.
The exponent (the “phase”) is given by connected trees
Power counting for connected trees:
δ(E, b) ∼ G2n−1 sn ∼ Gs R2(n−1) → Gs (R/b)2(n−1)
11 mars 2011
G. Veneziano, Cours no. X
5
If gravitons do not interact we get the leading eikonal.
Next to leading order: the H diagram
∼ G3 s2 = Gs G2 s = GsR2 → Gs (R/b)2
One of the produced graviton’s polarizations (“TT”) is IR-safe
the other (“LT”) is not.
11 mars 2011
G. Veneziano, Cours no. X
6
NNL-order
∼ G5 s3 = Gs G4 s2 = GsR4 → Gs (R/b)4
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G. Veneziano, Cours no. X
7
Reduced effective action & field equations
There is a D=4 effective action generating the leading
diagrams (Lipatov, ACV ‘93) but the resulting equations
are too complicated. Drastic simplification: “freeze”
the longitudinal dynamics => a D=2 effective action.
Neglecting the IR-unsafe (LT) polarization, it contains:
a and a, representing the longitudinal (++ and --)
components of the gravitational field, coupled to the
corresponding components of the E-M tensor;
φ, representing the TT graviton-emission field.
Besides source and kinetic terms there is a derivative
coupling of a, a and φ.
11 mars 2011
G. Veneziano, Cours no. X
8
ansverse component of the gravitational field, respectively. Equation (1) can be easily
neralized
in order
to deal
with two extended
The
action
(generalized
tosources:
extended null sources):
"
#
!
1
A
ur starting point is the
two-dimensional
= effective
d2 x a(x)s̄(x)
+ ā(x)s(x) − action
∇i ā∇i a of [1] (see their e
2πGs
2
!
!
2 !
2
$
%
A
12
(πR)
2 2 φ)2 + 2φ∇2(πR)
2
2 2(2)
d
x
−(∇
H ,
+
= a(b) + ā(0)
−
d
x∇ā∇a
+
d
x(−(∇
φ) +
2
2πGs
2
2
√
2
2
2
H ≡of ∇
a∇
ā − ∇si ∇
, & corresponding
here −∇
the center
mass
energy
provides
the
normalization
factor
2πGs =
eom
j a ∇i ∇
j āoverall
&
&
R2 , while the two sources s(x), s̄(x) are normalized by d2 x s(x) = d2 x s̄(x) = 1.
2 case
2 of 2two extended axisymmetric sources moving in
Let
to the
here
andspecialize
φ=are2(πR)
three
real
representing
longitudinal
∇2 a,
aus+ānow
2s(x)
(∇
afields
∇ φ−
∇i ∇j a ∇i ∇the
= a(b − x)an
j φ),twoā(x)
posite direction
with theofspeed
of
light and undergoing
a central collision.
Using the
ansverse
component
the
gravitational
field,
respectively.
Equation
(1
4
2
2
nventions of [1] we will
denote
by Ei (rai )∇
(in ā
the−following
2 jwill
∇ to
φ deal
= −(∇
∇i ∇jsources:
ai =
∇1,
ā) represent unbarred
i∇
neralized
in
order
with
two
extended
d barred fields/sources respectively) the energy carried by the ith beam below r = r
" solve
#
!
dSemiclassical
define Ri (r) = 4GEapproximation:
that eom
the two
sources
have finite
support
i (r). Let us also assume
and
compute
the
A for r > L2. By going to the overall center1of mass, we may
that Ri (r) = Ri (∞) ≡ R√
i =
d i x a(x)s̄(x)
+ ā(x)s(x)
− ∇
i ā∇
ia
classical
action
on
the
solution.
At
leading
order
in
R/b
2πGs
2
ways choose Ri = R =
2G s.
The axisymmetric case: general consider
$we recover
%
and for s(x) = δ(x), s(x)
= δ(x-b),
the
leading
(πR)
2
2
2
2
d x −(∇ φ) + 2φ∇ H ,
+
.1eikonal.
Simplifications
2
Iterative procedure
possible but not so useful
ne advantage of considering the axisymmetric
case is that there is simply no dependence
√
for
analytic
study!
here
the center
of azimuthal
mass energy
s provides
thetake
overall
normalization
the physics
upon the
angle, hence
no need to
averages
over it. This is
2
2
11 mars 2011
!
G. Veneziano, Cours no. X
&
92
&
R , while
thesimplification
two sourcesthat
s(x),
s̄(x)
normalized
by todsolving
x s(x)ODE.
=
useful
technical
allows
us are
to reduce
the problem
d
Numerical solutions
(G. Marchesini & E. Onofri, 0803.0250)
Solve directly PDEs by Fast FT methods in Matlab
Result: real solutions only exist for:
b > bc ∼ 2.28R
Compare with Eardely-Giddings’s CTS lower bound:
bc > 0.80R
11 mars 2011
G. Veneziano, Cours no. X
10
8
7
data
empirical fit
best fitted slope = 0.445
6
R
crit
5
4
3
2
1
0
0
5
10
15
b
11 mars 2011
G. Veneziano, Cours no. X
11
For an analytic approach we turn to
Axisymmetric Solutions
(J. Wosiek & G.V. ’08)
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G. Veneziano, Cours no. X
12
Central beam-beam collisions
A very rich problem in spite of its restrictive
symmetry:
1. The two beams contain several parameters: total
energy, shapes (same or different) & we can look for
critical surfaces in their multi-dim.al space.
2. The CTS (KV) criterion is simple (see below).
3. Numerical results are coming (see below).
4. Two major simplifications occur in ACV eqns:
• PDEs become ODEs.
• The IR-singular polarization is just not produced.
11 mars 2011
G. Veneziano, Cours no. X
13
The relevant parameters
R1(r)=4GE1(r)
11 mars 2011
r
G. Veneziano, Cours no. X
R2 (r)=4GE2(r)
14
"
$
hoose Ri = R = 2G #s.
2 2
"
2
ρ
=
t
1
−
(2πR)
straightforward
to
rewrite
the
action
the
cas
A
dt(1
ρ̇)
−2 φ̇ 2 ˙ 2(4) for
!−axisymmetric
!
2
dt [a(t)s̄(t)
ā(t)s(t)
−
onal =
integral
over the +
variable
r (2πR)
= 2ρ
x ā≡ȧ]t. Using d2 x = π dt we fi
Gs
!t !
"
parts and 2using π dt s"i (t! ) = Ri (t)/R we arrive at the follow
A and, 2as in [1], we have introduced the fie
a
dot
means
d/dt
dt(1 =
− ρ̇) dt [a(t)s̄(t) + ā(t)s(t) − 2ρā˙ ȧ]
(5)
−
he action:
2
2
(2πR) 2π Gs
'& $
'(
" & #
2
1
R2 (t)
2
21 (t)
2
2 = t 1R
ρ
−
(2πR)
φ̇
dt(1
− ρ̇)
−+ (2πR)
ρthe
ȧfield:
ȧ2 +
(1
1 (t)R
2 (t)
1 +
d, −asρ̇)in −
[1], R
we
have
introduced
2
(2πR)
ρ
2πRρ
2πRρ
Axisymmetric action, eqns (t=r )
#
$
!t !
2
dot
means
d/dt
and,
asand
in φ̇[1],
we
have
ρ that
=by
t follow
1parts
− (2πR)
(6)we
motion
from
(7)
read:
egrating
using
π introduced
dt si (t! ) the
= field:
Ri (t)/R
$
nient form !of the action: #
2
t 1
! R!i (r) ρ = t 1 − (2πR) φ̇
nd usingȧ"i π= −%dt si (t ) = Ri (t)/R we arrive at the &
following
1
1!
R1 (t
2πρ R 2
2
on:
t 1 (t)R
=rating
− 2 by parts
R
+R(2πR)
ρ arrive
ȧ1 + at th
dt1 (1and
− ρ̇)
−
!
2! )(t)
using
π
dt
s
(t
=
(t)/R
we
1
R
(r)R
(r)
4lP ρ̈ = (2πR)2 ȧ ȧ =&
ρ 1 i 2 ' & iρ(0) = 0 ; '(
ρ̇(∞)2πR
=1
1 of the2 action: 1 22 2
R
R2 (t)
nt form
21 (t)
ρ
ȧ2 +
(7)
− "R1 (t)R
2 (t) + (2πR) ρ ȧ1 +
&
'
&
%
ρ
2πRρ
2πRρ
e 1equations
of
motion
that
follow
from
(7)
read:
1
R1 (t)
R2
nd
2order
2 b.
e
to
a
closed
2nd
equation
for
ρ.
We
want
to
look
for
solutio
2
order
ODE
w/
Sturm-Liouville-like
conditions.
− 2
ȧ2 +
dt (1 − ρ̇) − R1 (t)R2 (t) + (2πR) ρ ȧ1 +
2πRρ
2π
P
nh4lthat
follow from
(7)ρ read:
the
following
boundary
conditions [1]:
11 mars 2011
1 R (r)
G. Veneziano, Cours no.iX
i = −
equations of motion that ȧfollow
from (7) read:
15
S-criteria and critical points in the ACV e
l result
CTS criterion
(KVone
gr-qc/0203093)
xisymmetric case,
can construct explicitly a MCTS [
h that (see eq. If
(4.4)
of exists
[6] foranDrc=
4):that
there
such
R1 (rc )R2 (rc ) = rc2
we canthat
construct
and therefore
a BH to of
form.
argue
such aa CTS
condition
impliesexpect
the absence
real s
0.Theorem
Proof: (VW08):
Let us first
notethe
that,
of (11)
whenever
KV because
criterion holds
theand
ACVthe
field
equationsofdor,not
regularσ,(at
ng
functions
theadmit
quantity
asr=0)
wellreal
as solutions.
ρ̇, are increa
r Thus:
any t:
KV criterion ==> ACV criterion
!
but of course not the other way around!
R1 (t)R2 (t)
σ(t) ≤ σ(∞) = 1 , i.e. ρ̇(t) ≤ 1 −
ρ(t)
16
11 mars 2011
G. Veneziano, Cours no. X
0.4
A sufficient criterion for dispersion
0.3
0.2
(P.-L. Lions, private comm.)
0.1
1
If
2
3
4
5
Out[15]=
Ü Graphics Ü
In[14]:=
Plot@2 x ^ 2 ê H3 * Sqrt@3DL, 8x, 0.01, 5<,
PlotStyle Ø [email protected], RGBColor@1, 0, 0D<<D
�
�
��2
2
8
r
1
log(1 + r )
R1 (r)R2 (r) ≤
1
+
1
−
27 (1 + r2 )2
2
r2
4
8
6
the ACV eqns do admit regular, real solutions.
4
2
1
Out[14]=
To summarize
2
Ü Graphics Ü
3
4
5
collapse
if touched
In[16]:= Show@%15, %14D
R1 (r)R
2 (r)
2.5
2
1.5
1
0.5
1
Out[16]=
2
dispersion if below
3
4
5
Ü Graphics Ü
clearly, there is room for improvement...
11 mars 2011
G. Veneziano, Cours no. X
17
Examples
11 mars 2011
G. Veneziano, Cours no. X
18
Example 1: particle-scattering off a ring
b
Can be dealt with analytically:
ρ = ρ(0) + r2 ρ̇(0) , (r < b)
�
ρ̇ =
1 − R2 /ρ , (r > b)
2
R
ρ̈ = 2 Θ(r2 − b2 )
2ρ
Since ρ(0) =0:
�
2
2
2
2
ρ(b ) = b ρ̇(b ) = b 1 − R2 /ρ(b2 )
This (cubic) equation has
positive real solutions iff
11 mars 2011
√
3 3 2
b >
R ≡ b2c
2
2
G. Veneziano, Cours no. X
(b/R)c ~ 1.61
CTS: (b/R)c > 1
19
Example 2: Two hom. beams of radius L.
The equation for ρ becomes
2
2 4
R
R
r
2
ρ̈(r ) = 2 Θ(r − L) +
Θ(L − r)
4
2
2ρ
2L ρ
We can compute the critical value numerically:
� �
R
∼ 0.47
L cr
It is compatible with (and close to) the CTS upper bound
� �
of KV:
R
L
11 mars 2011
< 1.0
cr
G. Veneziano, Cours no. X
20
2σ
2σ
Example 3: Two%different
Gaussian Beams
L
urier transforms
i
si (k) =(GV
2π & J.Wosiek
rdrJ0(kr)si (x)
‘08)
0
0Two
(∞)extended
the problem
reduces
to the
the same
one offixed
the point-ring
(two ho
sources
with
total energy
and Gaussian profiles centered at r=0 with different
interesting example is the one of gaussian sources concentrated
widths
ond
to: L1 and L2:
"
#
"
#
1
t
Ri (t)
t
si (t) =
exp − 2
,
= 1 − exp − 2
2
2πLi
2Li
R
2Li
mentum
space line
corresponds
The critical
in the (Lto
can
be compared with
1 , L2)
#
"plane
2 2
k
Li
2
the one coming from
the
CTS
criterion.
si (k ) = exp −
2
11 mars 2011
G. Veneziano, Cours no. X
8
21
4
3
Dispersion
L2 2
1
Collapse
0
0
1
2
L1
3
4
ure 3: Two gaussian sources at the origin. The critical line in the (L1 , L2 ) plane. Th
phase lies below this line. We also show (dash-dotted line) a lower bound on the curv
m the CTS criterion (13).
1.1
1
11 mars 2011
G. Veneziano, Cours no. X
22
Dispersion
1
0
Collapse
-1
Log L2
CTS
-2
-3
-4
-4
11 mars 2011
-3
-1
-2
Log L1
0
G. Veneziano, Cours no. X
1
23
An amusing coincidence?
In 0908.1780 Choptuik & Pretorius analyzed a “similar”
situation numerically (relativistic central collision of two
solitons of fixed mass and transverse size).
BH formation occurs at a critical γc (i.e. Rc) which is a
factor 2 or 3 below the naive CTS value.
*************
The above results are encouraging but real control over
the different approximations is lacking, in particular on
the freezing of longitudinal dynamics.
This is probably at the origin of some puzzles we find in
connection with gravitational radiation at b >> R.
11 mars 2011
G. Veneziano, Cours no. X
24