Causal set is a partially ordered set defined as: a b if and only if - PDF document

Causal set is a partially ordered set defined as: a ≺ b if and only if one can travel from a to b without going faster than the speed of light Topology is defined by Alexandrov sets α ( p, q ) = { r | p ≺ r ≺ q } Discreteness is defined through local finiteness: ♯α ( p, q ) < ∞ Metric is defined through τ ( p, q ) = ξ max { n |∃ r 1 , · · · , r n − 1 ( p ≺ r 1 ≺ · · · ≺ r n − 1 ≺ q ) } Note: It is max rather than min because of the minus sign in Minkowskian metric. For ( dt ) 2 − | d� x | 2 (sign convention is (+ − −− )) � example, if geodesic is along t -axis, | dt | ≥ 1

Key idea a) Assume smooth manifold and the presence of coordinates b) Re-express coordinate-dependent expressions in a way that coordinates aren’t explic- itly mentioned c) Copy the result for the non-manifold situation (ex: tree-like causal structure, etc) Key difference between causal sets and other discrete theories: In manifold situation, the causal set assumption is Poisson scattering = ⇒ lack of structure, emphasis on statistical properties Key difference between my work and other types of causal set theory: I am trying to re-interpret causal structure, the definition of fields, etc, while still sticking to statistical approach 2

Conventional version of causal set Lagrangian (Sorkin at el) Use − φ ∆ φ instead of + ∂ µ φ∂ µ φ 2D case: � (∆ φ )( p ) = ( φ ( p ) + φ ( r ) − 2 φ ( s )) (1) { ( r,s ) | α ( r,p )= { s }} d dimension � ∆ φ = ( c 0 ( d ) φ ( p ) + c 1 ( d ) φ ( r 1 ) + · · · + c n ( d ) ( d ) φ ( r n ( d ) )) (2) r n ( d ) ≺ r n ( d ) − 1 ≺···≺ r 1 ≺ p NOTE Cancelation only occurs sufficiently far away from the boundary What I don’t like about it: Existence of the boundary = ⇒ Preferred frame = ⇒ invalidation of stated claim of causal set theory 3

Steven Johnston’s propagator Summing over all possible paths 1) Propagators are defined direction, WITHOUT the use of Lagrangians 2) Propagators don’t face the problem of non-locality because of the TWO endpoints Problem: Coupling different propagators to each other during φ 4 -coupling Easy solution: Impose a condition by hand which edges are allowed to be φ 4 -coupled and which aren’t 4

DILLEMA: Locality = ⇒ Finately many neighbors = ⇒ Nearest neighbor = ⇒ Preferred frame MY ANSWER: The price for nearest edge neighbor is violation of Newtons first law INSTEAD OF preferred frame a) the nearest edge-neighbor relates to the fact that geodesic wiggles b) wiggling of geodesic is interpretted as gravity THEREFORE c) nearest edge-neighbor phenomenon is ”explained away” through gravity 5

Conventional thinking � γ µ ( τ ) dτ a ( p, q ) = A µ ( γ ( τ ))˙ (3) γ ( p,q ) where γ is a geodesic connecting p and q φ ( x ) is given My thinking a) Replace A µ ( x ) and φ ( x ) with A µ ( x, p ) and φ ( x, p ) b) Assume A µ ( x, p 1 ) ≈ A µ ( x, p 2 ) and φ ( x, p 1 ) ≈ φ ( x, p 2 ) if the relative velocity of refer- ence frames corresponding to p 1 and p 2 is not too close to c c) Assume that in the reference frames, with resect to which p/ | p | isn’t too close to c , φ ( x, p ) and A µ ( x, p ) are both locally linear d) Define � x µ ( τ ) dτdτ a ( p, q ) = A µ ( γ ( τ ) , ˙ γ ( τ )) ˙ (4) γ ( p,q ) 1 � x µ ( τ ) dτ φ ( p, q ) = φ ( γ ( τ ) , ˙ γ ( τ )) ˙ (5) τ ( p, q ) γ ( p,q ) NOTE: Since path integral is dominated by NON-DIFFERENTIABLE paths, the ssumptions b and c are dropped once we are under the path integral; those assumptions ONLY apply to “well behaved” functions we are thinking of in order to “motivate” our definition of the action. NOTE: a ( p, q ) = − a ( q, p ), BUT φ ( p, q ) = + φ ( q, p ) 6

Setup � L = η scal ( K 1 ( φ ; p, q, r ) − C scal ( d ) K 2 ( φ ; p, q )) (6) α ( p,q ) � � d d r (top − bottom) 2 = d d r ( φ ( r, q ) − φ ( p, r )) 2 K 1 = (7) α ( p,q ) α ( p,q ) � ∂φ � ∂φ �� 2 � 2 � 2 � � � � � � r + ˆ e 0 � � r ˆ e 0 � � � d d r d d r d d r (8) K 1 = φ − φ = = � � 2 2 ∂x 0 ∂x 0 � � α ( p,q ) α ( p,q ) α ( p,q ) 0 0 � 2 � φ ( p, r ) + φ ( r, q ) � � d d r (left − right) 2 = d d r K 2 = − φ ( p, q ) (9) 2 α ( p,q ) α ( p,q ) � ∂φ � 2 � 2 � r 0 � r 0 � � � r � � 2 + ∂φ � � d d r d d r K 2 = φ − φ (0) = = � � 2 ∂r 0 ∂r 0 2 � � α ( p,q ) α ( p,q ) 0 0 �� ∂φ � ∂φ � 2 � � 2 � � � � = 1 d d r ( r 1 ) 2 + 2 ∂φ ∂φ � d d r ( r 0 ) 2 + � � d d rr 0 r 1 (10) � � ∂r 0 ∂r 1 ∂r 0 ∂r 1 4 � � α ( p,q ) α ( p,q ) α ( p,q ) 0 0 � d d r r 0 r 1 = 0 Odd Function = ⇒ (11) α ( p,q ) �� ∂φ � ∂φ � 2 � � 2 � � � � K 2 = 1 d d r ( r 0 ) 2 + � � d d r ( r 1 ) 2 (12) � � 4 ∂r 0 ∂r 1 � � α ( p,q ) α ( p,q ) 0 0 7

Finding C scal ( d ) � ∂φ � ∂φ � ∂φ � 2 � 2 � 2 � � � � � − C scal ( d ) � � t 2 � � + � ( r 1 ) 2 � � = � � � ∂x 0 ∂x 0 ∂x 1 4 � � � 0 0 0 �� ∂φ � ∂φ � 2 � 2 � (1 − C scal ( d ) � − C scal ( d ) � � t 2 � � ( r 1 ) 2 � � � = (13) � � ∂x 0 ∂x 1 4 4 � � 0 0 1 − C scal ( d ) � t 2 � = C scal ( d ) ⇒ 1 = C scal ( d ) 4 � ( r 1 ) 2 � = ( � t 2 � + � ( r 1 ) 2 � = ⇒ C scal ( d ) = � t 2 � + � ( r 1 ) 2 � 4 4 4 � 1 � 1 1 0 ξ k ξ d − 1 dξ 0 ξ d + k − 1 dξ d d + k ⇒ � (1 − t ) k � = ξ = 1 − t = = = = (14) � 1 � 1 1 d + k 0 ξ d − 1 dξ 0 ξ d − 1 dξ d d 1 � t � = 1 − � 1 − t � = 1 − d + 1 = (15) d + 1 2 d d � t 2 � = � (1 − (1 − t )) 2 � = 1 − 2 � 1 − t � + � (1 − t ) 2 � = 1 − d + 1 + d + 2 = = d 2 + 3 d + 2 − 2 d 2 − 4 d + d 2 + d = ( d + 1)( d + 2) − 2 d ( d + 2) + d ( d + 1) = ( d + 1)( d + 2) ( d + 1)( d + 2) = (1 − 2 + 1) d 2 + (3 − 4 + 1) d + 2 2 = (16) ( d + 1)( d + 2) ( d + 1)( d + 2) � 1 � 1 0 ( r d − r d +1 ) dr 1 1 0 r 2 r d − 2 (1 − r ) dr d +1 − � r 2 � = d +2 = = = � 1 � 1 0 ( r d − 2 − r d − 1 ) dr d − 1 − 1 1 0 r d − 2 (1 − r ) dr d d +2 − d − 1 1 ( d − 1) d ( d +1)( d +2) ( d +1)( d +2) = = = (17) 1 d − d +1 ( d + 1)( d + 2) ( d − 1) d ( d − 1) d d − 1 1 d � � r 2 � = � ( x k ) 2 � = ( d − 1) � ( x 1 ) 2 � = ⇒ � ( x 1 ) 2 � = d − 1 � r 2 � = (18) ( d + 1)( d + 2) k =1 4 4 4 4 C scal ( d ) = � t 2 � + � ( x 1 ) 2 � = = = = 4( d + 1) (19) 2 d +2 1 d ( d +1)( d +2) + ( d +1)( d +2) ( d +1)( d +2) d +1 8

Avoiding C ( d ) K 1 ( f ; p 1 , q 1 ) − C ( d ) K 2 ( f ; p 1 , q 1 ) = K 1 ( f ; p 2 , q 2 ) − C ( d ) K 2 ( f ; p 2 , q 2 ) (20) K 1 ( f ; p 1 , q 1 ) − K 1 ( f ; p 2 , q 2 ) = C ( d )( K 2 ( f ; p 1 , q 1 ) − K 2 ( f ; p 2 , q 2 )) (21) C ( d ) = K 1 ( f ; p 1 , q 1 ) − K 1 ( f ; p 2 , q 2 ) (22) K 2 ( f ; p 1 , q 1 ) − K 2 ( f ; p 2 , q 2 ) L = η ( K 1 ( φ ; p 0 , q 0 ) − C ( d ) K 2 ( φ ; p 0 , q 0 )) (23) � K 1 ( φ ; p 0 , q 0 ) − K 1 ( f ; p 1 , q 1 ) − K 1 ( f ; p 2 , q 2 ) � L = η K 2 ( f ; p 1 , q 1 ) − K 2 ( f ; p 2 , q 2 ) K 2 ( φ ; p 0 , q 0 ) (24) � � � K 1 ( φ ; p 0 , q 0 ) − K 1 ( f ; p 1 , q 1 ) − K 1 ( f ; p 2 , q 2 ) �� L = η W ( p 1 , q 1 , p 2 , q 2 ) K 2 ( f ; p 1 , q 1 ) − K 2 ( f ; p 2 , q 2 ) K 2 ( φ ; p 0 , q 0 ) (25) W ( p 1 , p 2 , q 1 , q 2 ) w ( p 1 , p 2 , q 1 , q 2 ) = η (26) K 1 ( f ; p 1 , q 1 ) − K 1 ( f ; p 2 , q 2 ) � � L = w ( p 1 , p 2 , q 1 , q 2 )( K 1 ( φ ; p 0 , q 0 )( K 2 ( f ; p 1 , q 1 ) − K 2 ( f ; p 2 , q 2 )) − � −K 2 ( φ ; p 0 , q 0 ))( K 1 ( f ; p 1 , q 1 ) − K 1 ( f ; p 2 , q 2 )) (27) To define f introduce p 3 and write f p 3 ( s ) = τ ( p 3 , s ) (28) Need both p 3 and q 3 for the electromagnetic field 9

Charged scalar field based on short edges Gauge field on the edge: 1 � s 1 ≺ s 2 = ⇒ φ ( s 1 , s 2 ) = φ ( s ) | ds | (29) τ ( s 1 , s 2 ) γ ( s 1 ,s 2 ) Scalar field at the left: φ ( p, q ) Scalar field at the right: ( φ ( p, r ) + φ ( r, q )) / 2 (Note: φ ( r, q ) = + φ ( q, r )) Gauge field from left to right: ( a ( p, r ) + a ( q, r )) / 2 (Note: a ( r, q ) = − a ( q, r )) Left-right contribution to the Lagrangian: 2 � � 1 + i � φ ( p, q ) − 1 � � d d r � � 2( a ( p, r ) + a ( q, r )) 2( φ ( p, r ) + φ ( r, q )) (30) � � � � α ( p,q ) Scalar field at the top: ( φ ( p, q ) + φ ( r, q )) / 2 Scalar field at the bottom: φ ( p, r ) Gauge field from bottom to top: ( a ( p, r ) + a ( p, q )) / 2 Bottom-top contribution to the Lagrangian: 2 � � � 1 − i � φ ( r, q ) − 1 � � 2( a ( p, r ) + a ( p, q )) 2( φ ( p, q ) + φ ( r, q )) (31) � � � � Total charged scalar field contribution to the Lagrangian: 2 � � � � � 1 − i � φ ( r, q ) − 1 � � d d r L scal = ν scal 2( a ( p, r ) + a ( p, q )) 2( φ ( p, q ) + φ ( r, q )) � � � � α ( p,q ) 2 � � � � � 1 + i � φ ( p, q ) − 1 d d r � � − C ( d ) 2( a ( p, r ) + a ( q, r )) 2( φ ( p, r ) + φ ( r, q )) (32) � � � � α ( p,q ) 10