Transcript Slide 1

Utrecht University
Gerard ’t Hooft
Utrecht
April 15, 2008
A gravitating particle in
2 + 1 dimensions:
x
A
x
A moving particle in
2 + 1 dimensions:
A′
x  vx t
x  x'
 t t'
A
A′
A many-particle universe in 2+1 dimensions
can now be constructed by tesselation
(Staruszkiewicz, G.’tH, Deser, Jackiw, Kadar)
There is no local
curvature; the only
physical variables
are N particle
coordinates, with
N finite.
2 + 1 dimensional cosmology is finite and interesting
Quantization is difficult.
CRUNCH
BANG
How to generalize this to 3 +1 (or more) dimensions ?
 0
straight strings are linelike grav. defects
0
Calculation deficit angle:

T  t  2 ( x)
t33  t00  
 2 ( x) 
1
 2
 (  r )
R  r  2 ( x) ,
r11  r22  8 G 
In units such that R11  R22  1 :
2
Deficit angle →   2 (1  cos  )    8 G
A static universe would contain
a large number of such strings
But what happens when we make
them move ?
One might have thought:
But this cannot be right !
Holonomies on curves around strings:
Q : member of Poincaré group:
Lorentz trf. plus translation
C
SO(3,1)  SL(2, C )
Static string: pure rotation, Q  U  SU (2)
Lorentz boost:
Q  Q†  V
Moving string:
Tr U  real ,
Q  V U V 1
Tr U  2
All strings must have holonomies that are
constrained by
Tr Q  real ,
Tr Q  2
1
2
1
2
2
Q1 Q21
Q1
Q1
1
C
Q21
Q21 Q1
1
1
2
1
1
2
1
1
1
2
C  (Q Q ) QQ
 Q Q2 QQ
1
1
In general, Tr (C) = a + ib can be anything. Only if
the angle is exactly 90° can the newly formed
object be a string.
What if the angle is not 90°?
Q21
Can this happen ?
a
d
c
1
1
b
1
2
C  Q Q2 Q1Q  Qa Qc
This is a delicate exercise in mathematical physics
Answer: sometimes yes, but sometimes no !
Q2
Q1
Qa
Qd
Qc
Q4
Qb
Q3
Notation and conventions:
Q1Q2Q3Q4  1 ,
Qb  QaQ2 ,
Q3  Q1Q21Q11 ,
Qc  QbQ3 ,
Q4  Q31Q11Q3
1
1
Qd  QaQ
.
Q2
Q1
Qa
Qd
Q4
Qc
Qb
Q3
The Lorentz group elements have 6
degrees of freedom.
Im(Q) = 0 is one constraint for each
of the 4 internal lines →
We have a 2 dimensional space of
solutions.
Strategy #1: write in SL(2,C)
 a1  ib1
Qa  
 a3  ib3
a2  ib2 
a4  ib4 
Im(Tr(Q)) = 0 are four
linear constraints on the coefficients a, b .
|Re(Q)| < 2 gives a 4 - dimensional hypercube.
Two quadratic conditions remain: Re (det(Qa )) = 1
Im (det(Qa )) = 0
It is not clear whether or when these two
quadratic curves have points within our hypercube.
but we can do better.
1. Choose the Lorentz frame where string #1 is
static and in the z direction:
 i1
e
Q1  1  
 0
0 
i1 
e 
Im(Tr(Qa ))  0 ; Im(Tr(Qd ))  0 ; Det(Qa )  1 
b1  ib2 
 1  ia2
1
2
2
2
Qa   
;


1

a


(
b

b
2
1 1
2)

 1 (b1  ib2 ) 1  ia2 
In the frame
1  1
string a is static as well.
We must impose:
1  0
because
is the boost giving the velocity of the joint.
 1
0 
Vz  

 0 1/ 1 
2. next, go to the frame where string #3 is
static and in the z direction:
ei3
Q3  3  
 0
Q2
Q1
Qa
0 
i3 
e 
There, one must have:
Qd
Q4
Qc
Qb
Q3
q1  iq2 
 p1  ip2
Qc   


(

q

iq
)
p

ip
2
1
2
 3 1
If one would choose  1 and  3 , then this
gives 4 new conditions, of which only 3 are new,
since we knew already that the determinant is real.
One would expect this to fix the coefficients
a, b, p and q.
However, something else happens ... !
We found that for all initial string configurations
there are four coefficients, A, B, C, D, depending only
on the external string holonomies, such that
A  B1
3 
C  D1
So one cannot choose 3 freely !
Instead,
the coefficients a, b, ... are still underdetermined.
3. This leaves us the option to choose 2 instead.
One finds (from the symmetry of the system):
D  C 2
4 
B  A2
If and only if all μ ’s are positive, the strings and
the joints between the strings ar all moving with
subluminal velocities. This is necessary for
consistency of the scheme. In
A  B1
3 
C  D1
we must demand that this is consistent.
Consequently, the case
( A, B, C, D)  (, , , ) or (, , , )
must be excluded. But we found that this will
happen if strings move fast !
In those cases, one might expect
N
2
1
where we expect
free parameters.
6( N  1)  (3N  2)  3N  4
We were unable to verify wether such solutions
always exist
Assuming that solutions always exist, we have
a model.
1. When two strings collide, new string segments form.
2. When a string segment shrinks to zero length,
a similar rule will give newly formed string segments.
3. The choice of a solution out of a 2- or more
dimensional manyfold, constitutes the
matter equations of motion.
4. However, there are no independent gravity
degrees of freedom. “Gravitons” are composed of
matter.
5. it has not been checked whether the string constants
can always be chosen positive. This is probably
not the case. hence there may be negative energy.
6. The model cannot be quantized in the usual
fashion:
a) because the matter - strings tend to generate
a continuous spectrum of string constants;
b) because there is no time reversal (or PCT)
symmetry;
c) because it appears that strings break up,
but do not often rejoin.
7. However, there may be interesting ways to
arrive at more interesting schemes. For instance,
by limiting ourselves to a discrete subgroup
of the Poincaré group: a (relativistic) crystal group.
Then, space - time is viewed as a crystalline
lattice inbedded in a curved Riemann space.
Such models may well be quantizable, if we
allow ourselves a deterministic quantization
scheme.
In crystalline gravity, this lattice has defect lines
(and perhaps also defect planes) that propagate
according to locally causal laws of physics.
The details of such rules are difficult to formulate,
but this appears to be a technical rather than a
fundamental problem.