Recreation Algorithm: Teleportation Deep Bhattacharjee
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1 Recreation Algorithm: Teleportation Deep Bhattacharjee Recreation Algorithm: Teleportation Deep Bhattacharjee (Hobbyist Theoretical Physicist), Independent Researcher, KOLKATA, WEST BENGAL, INDIA Correspondence should be made at [email protected] Phone: +91-8282923025 ‘Teleportation’ means to teleport an object from one place to other with the help of ‘portals’. This is a ‘2 way’ process involving the dissembling from one point to reassembling to the other point separated by a large distance in space and time. The ‘reassemble’ of one person in the other end is based on an algorithm which is solely a polynomial wavefunctions in correspondence with the ‘Minkowski metric’ and based on this ‘algorithm’ the produced image of the persons or the ‘exact copy’ of the persons being teleported is either very accurate or closed to accurate and all of this is the outcome of transforming the matter to energy by means of the Einstein’s famous mass-energy equivalence called E=mc2 as without the conversion of energy, an object ‘being in a state of matter can’t be teleported. There are certain parameters for the ‘teleportation to happen properly’ and these parameters are as follows… • The object needs to be ‘teleported’ through ‘hyperspace’ thereby enabling the need for ‘higher dimensional forces due to gravity’. • The cross section of the energy packets that will be teleported will be changed due to Fitz-Gerald Lorentz Contractions while moving through ‘hyperspace’. • The ‘time will be dilated infinitely’ for the own reference frame of the ‘packets of the objects’ being teleported as the speed of energy is very close to the speed of light. • There will be a pathway through hyperspace called as ‘link line’ which is the main highway of the ‘objects energy packets’ to move during teleportation from one portal to another. • The energy packets must be equal to 1 at the other end of the ‘link line’ or ‘portal’. That is no fragmentation can be done. All these parameters are very important to analyze in case of ‘teleportation’ and if the parameters are consistent with the theory and mathematics, then the teleportation would happen at the blink of an eye just as the ‘science fiction movies’ and its always instantaneous for small distance (by small, I mean within the range of at least some light seconds) and its always constant time or time has been dilated infinitely due to the ‘speed of the energy packet’ close to that of light’ from the own reference frame of the ‘teleported object’, however, some amount of time will pass for the observers at the other distant ends and that amount of passing time is proportional to the length of the ‘link line’ through hyperspace. The nature of ‘teleportation that I’m going to be discussing here is ‘classical teleportation’ and not ‘quantum teleportation’. Quantum teleportation although has been achieved in ‘laboratory’ however, they are restricted on a very tiny scale of a pair of electrons (up and down spin) or photons (vertical and horizontal polarizations). However, there is great difficulty in achieving this nature of ‘quantum teleportation’ in a very large scale objects. As because a tiny dot (.) contains more than 108 atoms and each atom have a number of electrons in different orbital’s depending on the nature of the element comprising the atoms. So, the number of individual electrons are exponentially large in a large scale cluster of a million atoms when taken to be a large objects. Moreover, in quantum teleportation, there is a need for a ‘classical channels’ to establish the contact as the particles are in ‘Bell’s state or entangled pairs’. So, what I would discuss here is a ‘classical’ approach of teleportation and we will proceed both theoretically and mathematically with the process. A complete metric would be derived in this paper to concatenate the proper channel of teleportation through the help of discrete mapping of both locations and coordinates and then the notion of cylindrical polar coordinates are being used in the metric. Here we will use the ‘Polynomial Functions’ rather than ‘Exponential Functions’ as because, in ‘Polynomial functions’… If we look closely into the object being teleported then, there is the object that is a Human with ‘P’ – factors, which will loop ‘P’ times around then such that it’s P*P or P2…. Now, there is a limit of ‘P’ such that… ‘P’ must be equal to the segment that needs to be teleported. But, the whole energy converted from mass has to teleported as a whole, rather in discreet units. That energy has to be divided only 1 Parts as per the ‘Polynomial Function’, Here 1 is a special number which we will discuss below. Here we will take the ‘spacelike wavefunctions’ denoted by ξ(σ) where σ denotes the 4 Space-Time dimensions in Minkowski metric as dl2 = +c2dt2 – dx2 – dy2 – dz2 = (+1-1-1-1) = -2. I will perform a series of operations to find the ‘Polynomial Gap’ which when accurately leads to ‘-2’ or equal to dl2 then a proper ‘teleportation’ can be made. The operations will be made by following the equation, 푃2 훽 = ( − 푃2) … 퐸푞(1) ξ(σ) ( ) 푃2 Therefore, the ‘Polynomial Spacelike Wavefunction’ is… ξ(σ) = |dl2 + P2| By proceeding further we will divide the wavefunction by P2. This results in … ξ(σ) 푑푙2 + 푃2 = 푃2 푃2 Therefore, ξ(σ) 푑푙2 푃2 = + 푃2 푃2 푃2 Now, we will do 4 computation and examine the answers. 2 Recreation Algorithm: Teleportation Deep Bhattacharjee Computation 1 Putting the value of ‘P’ as 1, we get… ξ(σ) −2 12 ( ) = + 푃2 12 12 This gives, ( ) 푃2 ξ σ 2 = -1 and putting the value of P in ξ(σ) , we get -1 푃2 ( ) 푃2 Next we have to compute the ‘gap’ that is -1 - P2 (12) = -2 which is equal to unitary spacelike Minkowski metric Trace(+1-1-1-1) = -2. Hence we obtained a value equal to dl2. Computation 2 Now, to manipulate further lets divide the ‘energy packets’ to 50 units such that ‘P = 50’. 푃2 2 2 This gives the relation of 퐸푞(1) as ( ξ(σ) − 푃 ) = 2.00160128102 which is not equal to dl . ( ) 푃2 Computation 3 Now, divide the ‘energy packets’ to 200 units such that ‘P = 200’. 푃2 2 2 This gives the relation of 퐸푞(1) as ( ξ(σ) − 푃 ) = 2.000100005 which is not equal to dl . ( ) 푃2 Computation 4 Now, divide the ‘energy packets’ to units such that ‘P = 250’. 푃2 2 2 This gives the relation of 퐸푞(1) as ( ξ(σ) − 푃 ) = 2.00006400206 which is not equal to dl . ( ) 푃2 Here we can see that Computations 2,3,4 cannot give the value accurate to dl2 that is ‘-2’, while Computations 1 gives a value -2 which is accurately equal to dl2. This means that the number of energy packets or ‘P’ must be in one piece or equal to ‘1’ which clearly indicates that the whole body needs to be teleported without any fragments. Now that its necessary to compute the gravitational strength of the ‘link line’ and as the acceleration is quite huge compared to the speed of the light (this again depends on the energy packets sizes), the more the size of the packet or the more the mass, the more the energy produced and the less will be the acceleration and there will be a considerable effect of time dilation in the ‘link line’ such that the dilation factor T’ will be given by, 푇 푇′ = 푣2 √1 − 푐2 Where the time from the perspectives of energy packet will be dilated infinitely and comes to a standstill, however, the time from the perspective of the observer standing at the ends of the portal will experience a finite time and if T’ is the time of the ‘energy packet’ and ‘v’ is the travelling velocity, when 푣 ≈ 푐 the equation clearly shows T’ to be infinite provided the energy packet will be just proportional in magnitude to attain an acceleration close to the speed of the light as, 푇 푇 푇 푇′ = = 푇′ = = 푇′ = = ∞ 푣2 푐2 √1 − 1 √1 − √1 − 푐2 푐2 Any particle going through time is subject to length contraction such that if the initial cross-section is ‘A’ then the change in cross-section is equal to, 퐴 ∆퐴 = 퐿표푔 ( ) 푒 퐴 − 1 The ‘link line’ goes through hyperspace such that the force of gravity will be a generalization of the ‘Newton’s gravity’ in higher dimensions given by the relation, 푚1푚2 퐹 (푟) = 퐺 [1] (4+푛) 푁(4+푛) 푟푛+2 The ‘Teleporter’ or ‘the hypothetical machine’ converting energy to mass back to energy via teleported link line is in essence an advanced ‘1 important relations’ as, 푀푎푠푠 → 퐸푛푒푟푔푦 → 푀푎푠푠 3 Recreation Algorithm: Teleportation Deep Bhattacharjee 3 푖 Therefore, from (computation 1) with P = 1 and 훽 = -2 the assigned metric for the Teleportation ‘Link Line’ provided there will be a 4 Dimensional mapping 푈 ∑푖 푥 to map the source and destination would be expressed in the following line element and this is used to map the spatial coordinates to reconstruct the figure of the object where as the ′훾′ is used to map the destination location of the object for its proper arrival in correct location. 2 2 3 푃2 푃2 푑푠2 = 푈 ∑ 푥푖 훾 ( − 푃2) ∗ − (2⁄( − 푃2) ) 푑푡2 + (푑̂ + 휌푑̂ + 푑̂) 2 ξ(σ) ξ(σ) ( ) ( ) 푖 푃2 푃2 ( ) The pictorial representation of this metric can be best visualized with the help of this following diagram. REFERENCE [1] N. Arkani-Hamed, S. Dimopoulos, and G. R. Dvali, Phys. Rev. D59 (1999) 086004. Further references… D.