Harnessing Zero Point Energy -- Mathematical Physics Perspective

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The topic covered are: 1. A completely mathematical proof of fallacy in Einstein's mass energy derivation. 2. Proof of why classical or quantum mechanical equations cannot even satisfy deflection of billiards balls. 3. The missing factor. 4. Matter Light duality. 5. Speed greater than light. 6. Zero energy non-zero momentum particles. List of related papers: ------------------- 1. Matter-Light Duality and Speed Greater Than Light: http://www.vixra.org/abs/1309.0167 2. Zero Energy Non-Zero Momentum Particles (Zen Particles): http://www.vixra.org/abs/1309.0168 3. Are All Photon Frequencies a Result of Doppler Effect? : http://www.vixra.org/abs/1309.0169 4. Principle of Super Inertia: http://www.vixra.org/abs/1309.0173 5. Fallacy in Einsteins 2 blackboard derivation of mass energy: http://www.vixra.org/abs/1309.0172 -----------------

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<ul><li> 1. 9/28/2013 Priza Technologies Inc. 1 Hi To All Thanks for attending to this seminar The topics we will cover are: 1. A completely mathematical proof of fallacy in Einstein's mass energy derivation. 2. Why classical equations cannot even satisfy deflection of billiards balls. 3. The missing factor. 4. Matter Light duality, Speed greater than light. 5. Zero energy non-zero momentum particles. </li></ul><p> 2. 9/28/2013 Priza Technologies Inc. 2 Let us start with a BANG (2D, 2 particle case) Given 2 particles in different directions and , if the total energy of them is conserved then it implies that momentum is conserved, which means is conserved and is conserved. 1 1 2E E E= + 1 1 2 2cos cosxP P P = + 1 1 2 2sin sinyP P P = + 2 3. 9/28/2013 Priza Technologies Inc. 3 BANG Continued Then not only the quantities in previous slide are conserved but also following are conserved: Shocking but true! We will see proof after some time. 1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 ( ) cos(2 ) cos(2 ) ( ) sin(2 ) sin(2 ) ( ) cos((2 1) ) cos((2 1) ) ( ) sin((2 1) ) sin((2 1) ) x y x y E r E r E r E r E r E r P r P r P r P r P r P r = + = + = + + + = + + + 4. 9/28/2013 Priza Technologies Inc. 4 BANG Continued For the case of 2 billiard balls: 5. 9/28/2013 Priza Technologies Inc. 5 BANG Continued Initially angles are 0, : Final angles are , : Where: 1 2( ) 2 ( ) 0 ( ) 0 ( ) 0 x y x y E r E E E E r P r P r = + = = = = + 1 2 1 2 1 2 1 2 ( ) cos(2 ) cos(2 ( )) cos(2 )(2 ) ( ) sin(2 ) sin(2 ( )) sin(2 )(2 ) ( ) cos((2 1) ) cos((2 1)( )) 0 ( ) sin((2 1) ) sin((2 1)( )) 0 x y x y E r E r E r r E E r E r E r r E P r P r P r P r P r P r = + += = + += = + + + += = + + + += 1 2 1 2 E E E P P P = = = = 6. 9/28/2013 Priza Technologies Inc. 6 BANG Continued For conservation momentum seems ok but for energy: As , if the previous given results are true then the difference is be staggering. 2 cos(2 )(2 ) 0 sin(2 )(2 ) E r E r E 2 E mc = 7. 9/28/2013 Priza Technologies Inc. 7 BANG Continued Statement: A matter particle can be represented by 2 photons going in opposite directions. A photon can be represented by 2 electrons, one going below the speed of light = u, other going above the speed of light = v with uv = 2 c 8. 9/28/2013 Priza Technologies Inc. 8 BANG Continued 9. 9/28/2013 Priza Technologies Inc. 9 BANG Continued 10. 9/28/2013 Priza Technologies Inc. 10 Proof 1 Introduction The derivation is based on looking at conservation of energy of particles from 2 reference frames, one moving at an infinitesimally small speed with respect to the other frame of reference. For the simplicity of equations the derivation is given for the 2D space. The same derivation works well for the 3D or N dimensions as well. 11. 9/28/2013 Priza Technologies Inc. 11 Prerequisites: None I will explain and refresh whatever is required 1. Lorentz law of velocity addition. 2. Taylor series expansion. 3. Relativistic energy equation. 4. Doppler effect. 12. 9/28/2013 Priza Technologies Inc. 12 Lorentz law of velocity addition Galilean law is what appeals to intuition. Relative Speed = Speed of the frame of reference + Speed of the particle 13. 9/28/2013 Priza Technologies Inc. 13 Lorentz law of velocity addition Cont Problem: It is contradictory to the observation that the speed of light is same in all frames of reference. Example: If a train is moving at the speed of 100 miles per hour the speed of light as observed by a observer on the ground is 100 mph + speed of light. This is contradiction. 14. 9/28/2013 Priza Technologies Inc. 14 Lorentz law of velocity addition Cont Lorentz law in one dimension: Approximates to Galilean Law when speeds are small in comparison to the speed of light: 2 ( ) / (1 / )observed frame object frame object lightV V V V V V= + + Correction 2 1 / 1 ( )/ (1) frame object light observer frame light observer frame light V V V V V V V V V + = + = + 15. 9/28/2013 Priza Technologies Inc. 15 Lorentz law of velocity addition Cont Works well when the object is photon: object lightV V= 2 ( ) / (1 / ) ( ) / (1 / ) ( / 1) / (1 / ) observed frame light frame light light observed frame light frame light observed light frame light frame light observed light V V V V V V V V V V V V V V V V V V V = + + = + + = + + = 16. 9/28/2013 Priza Technologies Inc. 16 Lorentz law of velocity addition Cont It is only valid for the inertial frames of reference Lorentz Law for 3D/ND velocity addition Where is the velocity of frame of reference ( )2 2 ( ) ( ) 2 1 / (1 / ) object parallel object perpendicular observed object V h h c V V V h c + + = + h 17. 9/28/2013 Priza Technologies Inc. 17 Lorentz law of velocity addition Cont Magnitude of observed velocity if the frame of reference is moving in x direction and the object is moving in 2D with angle ( ) ( ) ( ) 2 2 cos sin cos sin cos cos 1 / sin (1 cos / object object object object parallel object object perpendicular object object object object object observed object h hi V V i V j V V i V V j V h V h V i hi h c V j V V h = = + = = = + + = + 2 )c 18. 9/28/2013 Priza Technologies Inc. 18 Lorentz law of velocity addition Cont Magnitude of observed velocity if the frame of reference is moving in x direction and the object is moving in 2D with angle ( ) ( ) ( ) ( ) 2 2 2 22 2 2 2 2 2 2 2 2 2 ( cos ) 1 / sin (1 cos / ) cos 1 / sin (1 cos / ) cos 1 sin / 2 cos (1 object object observed object object object observed object object object object o V h i h c V j V V h c V h h c V Speed V V h c V h V c V h Speed V + + = + + + = = + + + = + 2 cos / )bjecth c 19. 9/28/2013 Priza Technologies Inc. 19 Lorentz law of velocity addition Cont Magnitude of observed velocity if the frame of reference is moving in x direction and the object is moving in 2D with angle ( ) ( ) ( ) ( ) 2 2 2 22 2 2 2 2 2 2 2 2 2 ( cos ) 1 / sin (1 cos / ) cos 1 / sin (1 cos / ) cos 1 sin / 2 c object object observed object object object observed observed object object object object observed V h i h c V j V V h c V h h c V V V V h c V h V c hV V + + = + + + = = + + + = 2 os (1 cos / )objectV h c + 20. 9/28/2013 Priza Technologies Inc. 20 Lorentz law of velocity addition Cont Thus the scalar speed function as a transformation is: ( ) ( )2 2 2 2 2 2 2 cos 1 sin / 2 cos , , (1 cos / ) object object object observed object object V h V c hV V V h V h c + + = + 21. 9/28/2013 Priza Technologies Inc. 21 Taylor Series Expansion in 1D A function can be written in the form: If the series sum converges. ( )f x h+ ( , )f x y 2 3 2 3 ( ) 1 ( ) 1 ( ) ( ) ( ) ... 2! 3! df x d f x d f x f x h f x h h h dx dx dx + = + + + + 22. 9/28/2013 Priza Technologies Inc. 22 Taylor Series Expansion in 1D Cont.. Example: ( , )f x y 2 2 2 2 2 2 2 2 ( ) ( ) ( ) ( , ) 2 , 2, 0 3 ( ) 2 2 / 2! ( ) 2 ( ) ( ) n n f x x df x d f x d f x y x n dx dx dx f x h x h x h f x h x xh h f x h x h = = = = + = + + + = + + + = + 23. 9/28/2013 Priza Technologies Inc. 23 Taylor Series Expansion chain rule If there is a function , and itself is a function of some other variables Then the Taylor series can be written in the form: &lt; and higher powers &gt; ( , )f x y 1 2 3 1 2 3 1 2 3 1 g( , , ) ( ) (g( , , )) (g( , , )) y y y df x f y h y y f y y y h y dx + = + + ( )f x x 1 2 3( , , )x g y y y= 2 h 24. 9/28/2013 Priza Technologies Inc. 24 Now the real proof ( , )f x y Let us assume there are n particles in a closed system and there exists a scalar conservation function energy E, which solely depends on the speed of the particles. Let the velocities of the particles be in a reference frame A. Let us take another frame of reference B which has a velocity w.r.t. to A. Let the resultant velocity of the particles in frame of reference B be . 1 2{ , ,..., }a a anV V V h hi= 1 2{ , ,..., }b b bnV V V 25. 9/28/2013 Priza Technologies Inc. 25 Proof Cont ( , )f x y Conservation of energy: Sum of all energy is the closed system remains constant Looking from the frame of reference A Looking from the frame of reference B . ( ) ( ) 1 1 (B) . C n n bk bk b k k TotalEnergy E V E V const = = = = = = ( ) ( ) 1 1 (A) . C n n ak ak a k k TotalEnergy E V E V const = = = = = = 26. 9/28/2013 Priza Technologies Inc. 26 Proof Cont ( , )f x y Writing the total energy in B again: Assume that the scalar function E can be represented as a Taylor series. (This is can easily demonstrated for the Relativistic energy scalar function). . ( ) ( ) 1 1 (B) ( , , ) n bk k n bk ak k TotalEnergy E V E V h V = = = = 27. 9/28/2013 Priza Technologies Inc. 27 Proof Cont ( , )f x y Then we can write it as the Taylors series: &lt; and higher powers&gt; . ( ) ( ) 1 1 (B) ( , , ) (0, , ) ( ) (0, , ) n bk ak k n bk ak ak bk ak k ak TotalEnergy E V h V V V dE V E V V h h dV = = = = + + 2 h . bConst C= = 28. 9/28/2013 Priza Technologies Inc. 28 Proof Cont ( , )f x y Let and be two states of the system both in reference frame A &amp; B. Then based on the law of conservation of energy in the frame of reference B: &lt; and higher powers&gt; &lt; and higher powers&gt; . ( ) ( ) 1 1 (B) ( , , ) (0, , ) ( ) (0, , ) n bk ak k n bk ak ak bk ak k ak TotalEnergy E V h V V V dE V E V V h h dV = = = = + + 2 h ( ) ( ) 1 1 ( , , ) (0, , ) ( ) (0, , ) n bk ak k n bk ak ak bk ak k ak E V h V V V dE V E V V h h dV = = = = + + 2 h 29. Proof Cont 30. Proof Cont 31. 9/28/2013 Priza Technologies Inc. 31 Proof Cont ( , )f x y Also kind note that in the frame of reference B: As h = 0 means there relative speed between A and B is zero, which means no transformation of the speed. . ( ) ( ) ( ) ( ) 1 1 1 1 (0, , ) (0, , ) n n bk ak ak k k n n bk ak ak k k E V V E V E V V E V = = = = 32. 9/28/2013 Priza Technologies Inc. 32 Proof Cont ( , )f x y Then the equations simplify to: &lt; and higher powers&gt; &lt; and higher powers&gt; . ( ) ( ) 1 1 (B) ( , , ) (0, , ) ( ) n bk ak k n bk ak ak ak k ak TotalEnergy E V h V V V dE V E V h h dV = = = =+ + 2 h ( ) ( ) 1 1 ( , , ) (0, , ) ( ) n bk ak k n bk ak ak ak k ak E V h V V V dE V E V h h dV = = = =+ + 2 h 33. 9/28/2013 Priza Technologies Inc. 33 Proof Cont ( , )f x y Similarly in the frame of reference A based on the law of conservation of energy: Subtracting the this equation from earlier equation: &lt; and higher powers&gt; &lt; and higher powers&gt; . ( ) ( ) 1 1 (A) n n ak ak k k TotalEnergy E V E V = = = = 2 h 2 h 1 (0, , ) ( )n bk ak ak k ak V V dE V h h dV = + 1 (0, , ) ( )n bk ak ak k ak V V dE V h h dV = + 34. 9/28/2013 Priza Technologies Inc. 34 Proof Cont ( , )f x y Divide both sides by and taking limit (remember that the proof is based on 2 frames of reference, one moving at infinitesimal with respect to the other) This means is conserved for transition from any state to . 1 1 (0, , ) ( )(0, , ) ( )n n bk ak akbk ak ak k kak ak V V dE VV V dE V h dV h dV = = 0h 1 (0, , ) ( )n bk ak ak k ak V V dE V h dV = h 35. 9/28/2013 Priza Technologies Inc. 35 Proof Cont ( , )f x y Now let us look at at . (0, , )bk akV V h ( ) 2 2 2 2 2 2 2 2 2 1 (2 (1 / sin ) 2 cos )( , , ) 2 1 cos / (1 / sin ) 2 cos ak k ak k bk ak ak k ak ak k ak k h V c VV h V h hV c V h V c hV + = + + + ( ) 2 2 2 2 2 2 22 (1 / sin ) 2 cos ( cos / ) 1 cos / ak ak k ak k ak k ak k V h V c hV V c hV c + + + 0h = ( ) 2 2 2 2 (0, , ) cos / cos / (0, , ) cos 1 / bk ak ak k ak ak k bk ak k ak V V V V V c h V V V c h = = 36. 9/28/2013 Priza Technologies Inc. 36 Proof Cont ( , )f x y So is conserved. This is true for any conserved scalar function Let us take relativistic energy function Then . ( ) 1 2 2 1 (0, , ) ( ) ( ) cos 1 / n bk ak ak k ak n ak k ak k ak V V dE V h dV dE V V c dV = = 2 2 2 ( ) 1 / mc E v v c = 2 2 2 2 3/2 2 2 3/2 ( ) 2 /1 2 (1 / ) (1 / ) ak ak ak ak ak ak dE V mc V c mV dV V c V c = = 37. 9/28/2013 Priza Technologies Inc. 37 Proof Cont ( , )f x y So is conserved. This is true for any conserved scalar function Let us take relativistic energy function Then . ( ) 1 2 2 1 (0, , ) ( ) ( ) cos 1 / n bk ak ak k ak n ak k ak k ak V V dE V h dV dE V V c dV = = 2 2 2 ( ) 1 / mc E v v c = 2 2 2 2 3/2 2 2 3/2 ( ) 2 /1 2 (1 / ) (1 / ) ak ak ak ak ak ak dE V mc V c mV dV V c V c = = 38. 9/28/2013 Priza Technologies Inc. 38 Proof Cont ( , )f x y So is conserved. The above is the X component of the relativistic momentum! Similarly if we take we can prove that Y component of relativistic momentum is conserved. So conservation of energy implies conservation of momentum. . ( ) ( ) 2 2 1 2 2 2 2 3/2 1 2 2 1 ( ) cos 1 / cos 1 / (1 / ) cos 1 / n ak k ak k ak n ak k ak k ak n ak k k ak dE V V c dV mV V c V c mV V c = = = = h hj= 39. 9/28/2013 Priza Technologies Inc. 39 Proof Cont ( , )f x y As is conserved for any choice of conserved function E If is magnitude of the momentum then: Take Both are conserved as a sum of conserved quantities. So . ( )2 2 1 ( ) cos 1 / n ak k ak k ak dE V V c dV = ( ) cos ( ) sin akx ak k ak aky ak k ak P V P P V P = =akP ( ) ( ) i ak ak akx aky ak i ak ak akx aky ak P V P iP P e P V P iP P e + = + = = = 1 1 ( ), ( ) n n ak ak ak ak k k P V P V+ = ( ) ( )2 2 2 2 1 1 ( ) ( ) cos 1 / , sin 1 / n n ak ak ak ak k ak k ak k kak ak dP V dP V V c V c dV dV + + = 40. 9/28/2013 Priza Technologies Inc. 40 Proof Cont ( , )f x y Let us take . As this is conserved so following sums are also conserved: Take , , which is also conserved. is also conserved . ( )2 2 1 ( ) 1 / n iak ak ak k ak dP V S V c e dV + = = ( )ak akP V+ 2 1 2( )S c S iS= + ( ) ( )2 2 2 2 1 2 1 1 ( ) ( ) cos 1 / , sin 1 / n n ak ak ak ak k ak k ak k kak ak dP V dP V S V c S V c dV dV + + = = = 2 2 ( ) 1 / i iak ak ak ak ak mV P V P e e V c + = = 41. 9/28/2013 Priza Technologies Inc. 41 Proof Cont ( , )f x y Then is conserved Similarly we keep doing it again and again to prove that: is conserved for integer r. Similarly is conserved for any integer r . ( ) ( ) 2 3/2 3/22 2 2 2 2 2 2 2 ( ) 2 /1 1 21 / 1 / 1 / 1 / i i iak ak ak ak ak ak ak ak ak ak ak dP V mV V cd e m e e m V dV dV V c V c V c V c + = = = ( ) ( ) 2 2 22 2 2 3/2 2 22 2 1 1 1 1 / 1 /1 / k k k k i in n n i i ak ak k k k akak e m e mc S c V c e E e V cV c = = = == 2 1 k n i r ak k E e = (2 1) 1 k n i r ak k P e + = 42. 9/28/2013 Priza Technologies Inc. 42 Proof Cont ( , )f x y Which means there are infinite such equations. It is easy to prove that these are all independent equations. This leads to a very strict form of change of energy and momentum, which we name as: Principle of Super Inertia. We need to revisit all our earlier derivations which were relying on and using conservation of energy and momentum. . 43. 9/28/2013 Priza Technologies Inc. 43 Fallacy in Einstein Derivatio...</p>