A Lagrangian for the dark side of spacetime

A Lagrangian for the dark
side of spacetime
Angelo Tartaglia
Monica Capone
Dipartimento di Fisica, Politecnico di
Torino
July 20 2005
Paris - Einstein centennial
1
A simple classical problem
Motion of a point massive particle in a
viscous medium
b
S   Ldt
a
1 t  x 2
L e
mx
2
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Euler Lagrange equation
x  x 

2
x  0
2
Non-invariance per inversion of x
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Covariant formulation of
the same problem
s2
    x
S  m  e
s1
ds
2
1 dx     x
L  m 1  2
e
c dt
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Time like vector
   ,  ,  ,  
    3  0
r '   ' , x' ,0,0
2
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Euler-Lagrange equation

x' c 1  v'
2
     c 1  v' v'  0
2 2
2
2
dx'
v'
d
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Reference frame of the fluid
   ,0,0,0
r   , x,0,0
 0
 x 2 
x  c1  2  x  0
 c 
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Dependence on v
d
 '
dv
   v 
 ' 1  v +
2
x
v
c
v
1 v
2
= Ke
 
χ(v) given
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Space time
(4)-spherical symmetry
2

dr
2
2
2
2
2
2
2 
ds  d  a 
 r d  sin d 
2
1  kr


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
k  0,1
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Action integrals
Free
Symm.
2
S   Rd 
1
2


S  6 Vk  a aa  a 2  k d
Simplest “dissip.”
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d   g d 4 x
1
2
S  6 Vk 
1
Vk  4 
r2
r1


r2
1  kr 2

dr
2



e a aa  a  k d
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Euler-Lagrange
  constant
a  2aa  aa  k  0
2
a  constant
k 0
a  a0 e
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K 2 
e
2
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empty and flat
inflating
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Non-constant χ
Naive example
     given
a  2aa      k  0
2
Non trivial solution
a  a0e
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2 
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χ depends on a
   a  given
'
d
da


a  2aa  k  a ' aa  a  k  0
2
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Dependence on
a
   (a )

R   ' R  R   a 2  2aa  k  0
a
   ' a  

'
a
R  a aa  a 2  k 
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More appropriate action
integral
S  e
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g     
Rd 
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No pre-defined symmetry
Euler-Lagrange-Einstein equations

G  R     g  
 
2
      
,
2
   
2
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2
,
2
     
,
,  ;
2 ,
2 ,

;
0

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4-sphere symmetry
Cosmic time
  (  ,0,0,0)
S  6 Vk  e a(aa  a  k )d

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Non trivial if
   a 
   a 
   R 
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Let us consider


   (a)
2a ' a  aa  k  a  k  0
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Divergence free vector


 g

,
0

 
 a3
0

K
 3
a
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Empty space time field
equations
2


K
2
2
 6 6 a  aa  k  a  k  0
a
Non-trivial implicit solution (k=0)
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a  Hae
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
a6
36K 2
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Maximum expansion rate for
aM  6K
6
2
2
6
K
a M  6
H
e
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What about γ’s dynamics?
Nothing more than wild guess
An ‘EOS’ needed
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Trying an approach similar to
the one of field theory
S   e Rd 

 
  g     A
Assume
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

;

;

B

;


;


;

0
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;

Introducing symmetry
2


a 2
2
2
    B   3 2  
a


K
 3
a
Obtain a complicated equation for a
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What could the vector field γ
represent?
Again an analogy:
• point-like defect in space time
•consequent radial displacements field
(one-dimensional strain)
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Four-dimensional point defect
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Conclusion
• Use the guiding idea of motion through a viscous
medium
• Express the problem covariantly in the Lagrangian
formalism
• Assume basic expansion or contraction as an
intrinsic property of space time
• Adopt, by analogy, a ‘dissipative’ Lagrangian for
empty space time depending on a vector field
• When the vector field depends in turn on the
scale parameter end with an accelerateddecelerated expansion
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