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A New Argument for the Non-Existence of Closed Timelike Curves

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A New Argument for the Non-Existence of Closed Timelike Curves IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-5728,p-ISSN: 2319-765X, Volume 7, Issue 4 (Jul. - Aug. 2013), PP 43-47 www.iosrjournals.org A new argument for the non-existence of Closed Timelike Curves Sanath Devalapurkar1 1(Physics, Center for Fundamental Research and Creative Education, India) Abstract: In this paper, we attempt to present a short argument, different from that of the original proofs by that of Hawking, for a theorem stated that no closed timelike curves can exist. In a later paper, we apply this to quantum gravity and relate the curvature of spacetime to this theorem. Also, we present this paper as a preliminary introduction to the complete argument of this, and we also provide a preliminary notion of the concepts which will be narrated in the later papers. We also use this as a starting basis for a true theory of everything for a theory of everything. We use the notation of [1] and of [2]. Keywords: Theoretical Physics, Quantum Gravity, Mathematical Physics, General Relativity, Quantum Mechanics I. The elimination of Closed Timelike Curves in Loop Quantum Gravity Theorem : There exists no closed timelike curve in the physical world. To prove this, we introduce spacetime as (ℳ, (4)푔) where we can define ℳ as a four dimensional manifold and (4)푔 as a Lorentzian metric on it. In the third part, we introduce a ADM 3 + 1 split of the four dimensional (ℳ, (4)푔). II. ADM 3+1 split of classical four dimensional (퓜, (ퟒ)품) In this section, we mainly reconstruct the topics in [1]. To do a 3+1 split of classical four dimensional (ℳ, (4)푔), we need (ℳ, (4)푔) to be homeomorphic to the direct product space formed by ℝ × 훴, where 훴 is a three-manifold representing space and 푡 ∈ ℝ represents time. (ℳ, (4)푔) needs to be globally hyperbolic, and we assume causality, that is, no closed timelike curves (CTC) exist. We define closed timelike curves in the following manner. A curve 휋 such that, we have, in a particular coordinate system 픎 on (ℳ, (4)푔), the following equations satisfied: 휋: 푆1 → (ℳ, (4)푔) 1) And in a shifted coordinate system, 픎1, we have 푔(휋픎1, 휋픎1) < 0 2) This allows us to state that the time function is regular. A particular slicing of spacetime, though, would be a matter of choice. A choice of slicing is equivalent to the choice of a regular function $f$ (i.e., a scalar field on (ℳ, (4)푔)) for which 휕휇 푡 is timelike. Suppose there exist two spaces, so that there is a mapping of some sort between them, defined as Proposition 1: 푓 on maps to ℬ. Then, it could be said that {푓} is a space itself, multiplied (i.e., it forms a product space with) by to yield ℬ. Then we could say that {푓} = ℬ/. I.e., if 푓: → ℬ 3) Then we have {푓} = ℬ/ 4) Proof: If 푓 is defined as 푓: → ℬ 5) Then we need ℬ and to have their points as discrete eigenvalues of some operator corresponding to each space (we wish to call this the soperator). Thus, as 푓 should have eigenvalues corresponding to each soperator, we need 푓 to eigenvalues which exist in both the spaces, which is obviously the union of the two, which is given by www.iosrjournals.org 43 | Page A new argument for the non-existence of Closed Timelike Curves {푓} = ℬ/ 6) ∎ −1 The regular values of 푓 then form 3-manifolds, our 훴, defined by 훴(푡0) = 푓 (푡0). Using the submersion theorem, we can find a local coordinate system {푥휇 } over the open set 푈, where ∀ 푝 ∈ 푈, 푓(푝) = 푓(푥0(푝), 푥1 (푝), 푥2(푝), 푥3(푝)) = 푥0(푝). The 1-form 푑푓 is then 푑푥0 and the intrinsic coordinates of each hypersurface are given by 푥1 , 푥2, 푥3. 휕 Thus the vectors 휕 : = span the target space to each hypersurface. We can express the components of each 훼 휕푥 푎 훼 vector field in terms of a general basis {푦 } as 푒푣 : 휕 휕푥푎 휕 7) = =: 푒훼 휕 휕푥푣 휕푥푣 푦훼 푣 훼 Similar to [1], √4 means the metric dual to 푑푥0 . We have a vector field 0 √4 (4) 0 (4) 0푣 (푑푥 ) : = 푔(푑푥 ,∙) = ( 푔) 휕푣 8) (4) 0푣 휇 Thus the vector field with components 휕푣푓 or ( 푔) is a normal to the hypersurface. If 푛 is the unit 휇 normal to 훴, then on decomposing 휕0 into its components parallel to the hypersurface 푁 and orthogonal to is 푁푛휇 , we obtain 휕0 = 푁푛 + 푁 9) 0 Since we have 푑푥 (휕0) = 1, we see that we must have −푁2 = ((4)푔)00 = −||푑푥0||2 10) III. Loop Part We first look at Definition 1.0.2. So, we can define elements of a curve as Definition : If there exists a curve 훾, then the elements of it are its 푥, 푦, 푧 and/or 푡 components, given 푎 by 훾 , or by 훾푎 . So, using this convention, we have a new definition of a closed timelike curve: Definition : A curve 휋 such that, we have, in a particular coordinate system 픎 on (ℳ, (4)푔), the following equations satisfied: 휋: 푆1 → (ℳ, (4)푔) 11) And in a shifted coordinate system, 픎1, we have (4) 픎1 푎 픎1 푏 ( 푔)푎푏 휋 , 휋 < 0 12) We then define a set of all possible timelike curves defined by 휋: 푆1 → (ℳ, (4)푔). This set we call the timelike loop space of (ℳ, (4)푔) and call it 훺(ℳ, (4)푔). Now, if we consider our coordinate system change, to the system 픎1, we would have the following: 훺 ℳ, 4 푔 = ∅ 13) Therefore, we may define the coordinate shift such that the causal structure makes the timelike loops (that is, curves) to vanish and therefore make 훺 ℳ, 4 푔 = ∅. The 휋픎1 may lie on ℝ or 훴. If 휋픎1 lies on 훴, then we need 휋픎1 to lie on the regular values of 푓. But the introduction of a metric in ℳ, 4 푔 induces a metric on 훴, which in turn causes a causal structure to be formed. Since spacetime is not necessarily Ricci flat, 휋픎1 must, to introduce a causal structure on ℳ, 4 푔 , exist on ℝ. Then we may define ℝ as having no causal structure. The coordinate shift creates the closed timelike curves on the ℝ, and therefore a restriction of vectors tangential to ℝ must not exist. This is wrong. But ℝ is Riemannian, therefore 휋픎1 must not exist. That is, slicing causing closed timelike curves do not exist. This means that all slicings of ℝ × 훴 have no closed timelike curves in them, even on coordinate shifts, and therefore ℳ, 4 푔 is globally hyperbolic, ∀ ℳ, 4 푔 and 4 푔. This is the proof of the proposition. ∎ www.iosrjournals.org 44 | Page A new argument for the non-existence of Closed Timelike Curves IV. Introduction to the preliminaries of Future Papers 훼 훼 푎 훼 Define 푡 = 푁푛 + 푁 푒푎 as the components of 휕0. We then obtain from [1], the following equation: 퐷(푎 푡푏) = 푛(훼 푁;훽) + 푁퐷(훽 푛훼) + 퐷(훼 푁훽) 14) We then obtain the extrinsic curvature as 1 15) 퐾 = (푔 푎푏 − 2푁 ) 푎푏 2푁 (푎;푏) Now we use the Ashtekar approach. The canonical variables (these are 푥휇 ) are components of the inverse 3-metric 푔푎푏 intrinsic to 훴 and the components of the spin connection ∇ on 훴. The spin connection ∇ defines a bundle connection with base space 훴 and the spin space 핊. ∇ is then given by ∇= ∇intrinsic + 푖푲 16) Defining a particular 휇 = 0 would lead to 푓 being a component of 푔푎푏 intrinsic to 훴 and the components of the spin connection ∇ on 훴. For the 3 + 1 split, we say that ℳ, 4 푔 = 훴 × ℝ, and in the spin connections's space, 훴 × 핊. We now define 훴 × 핊 as a five-dimensional space 풟, 5 푔 , with a 5-metric. We may perform a 4 + 1 split of 풟, 5 푔 , by stating that 핊 would have a 1 + 1 split on it, so we can define 핊 = ℝ × 풬. What is 풬? We can define the points of 풬 as below: Definition : 풬 is defined as ∀푝 ∈ 풬, 푝 × ℝ is a vector in 핊. Here ℝ denotes time, and 풬 denotes space. (We can see the reasons for this split later.) Consider the following equation: 풟, 5 푔 = 훴 × ℝ × 풬 17) Look at the first part of the right hand side. It has 훴 × ℝ, which is the 3 + 1 split of gravity! So, this can be reformulated as 풟, 5 푔 = ℳ, 4 푔 × 풬 18) The spin connection is basically a functional derivative‟s component, that is, 휇푣 훿 19) 푔 = 푖ℏ 훿∇휇푣 The metric made is 푔휇푣 = 4 푔휇푣 + 푛휇 푛푣. Therefore, we have a refined equation, that is 4 휇푣 휇 푣 훿 20) 푔 + 푛 푛 = 푖ℏ 훿∇휇푣 For uncontrollable infinities to vanish, we need ∇휇푣 ≠ 0, else the 3-dimensional metric would yield infinite distances between any two (even infinitesimally close) points as ∞. Now we ask ourselves a question. In the above statement, why was “3-dimensional metric” mentioned? Why not “the metric” or “4-dimensional metric”? To answer this, look at the fact that 푔휇푣 = 4 푔휇푣 + 푛휇 푛푣. We need the 4-metric 4 푔 to be finite, so we 휇 푣 휇 4 can introduce 푛 푛 = −∞ + 훿푣 .
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