harnessing zero point energy -- mathematical physics perspective

56
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.

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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 Einstein’s 2 blackboard derivation of mass energy: http://www.vixra.org/abs/1309.0172 -----------------

TRANSCRIPT

Page 1: Harnessing Zero Point Energy -- Mathematical Physics Perspective

9/28/2013 Priza Technologies Inc. 1

Hi To AllThanks 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.

Page 2: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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 meansis conserved and is conserved.

1θ1 2E E E= +1 1 2 2cos cosxP P Pθ θ= + 1 1 2 2sin sinyP P Pθ θ= +

Page 3: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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BANG ContinuedThen 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 rE r E r E rP r P r P rP r P r P r

θ θθ θ

θ θθ θ

= += +

= + + += + + +

Page 4: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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BANG ContinuedFor the case of 2 billiard balls:

Page 5: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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BANG ContinuedInitially angles are 0, :

Final angles are , :

Where:

1 2( ) 2( ) 0

( ) 0( ) 0

x

y

x

y

E r E E EE rP rP r

= + =

=

=

=

π

π θ+θ 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 EE r E r E r r EP r P r P rP r P r P r

θ θ π θθ θ π θ

θ θ πθ θ π

= + + =

= + + =

= + + + + =

= + + + + =1 2

1 2

E E EP P P

= == =

Page 6: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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BANG ContinuedFor 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 Er E

θθ

≠≠

2E mc γ=

Page 7: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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BANG ContinuedStatement: 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 = 2c

Page 8: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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BANG Continued

Page 9: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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BANG Continued

Page 10: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Proof 1 IntroductionThe 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.

Page 11: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Prerequisites: NoneI will explain and refresh whatever is required

1. Lorentz law of velocity addition.

2. Taylor series expansion.

3. Relativistic energy equation.

4. Doppler effect.

Page 12: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Lorentz law of velocity addition

Galilean law is what appeals to intuition.

Relative Speed = Speed of the frame of reference + Speed of the particle

Page 13: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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.

Page 14: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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

21 / 1

( ) / (1)frame object light

observer frame light

observer frame light

V V VV V VV V V

+ × ≈

⇒ = +

⇒ = +

Page 15: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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 VV V

= + + ×

⇒ = + +

⇒ = + +

⇒ =

Page 16: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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

observedobject

V h h c VV

V h c

+ + −=

+ ⋅

h

Page 17: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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

observedobject

h hi

V V i V j

V V i

V V j

V h V h

V i hi h c V jV

V h

θ θ

θ

θ

θ

θ θ

θ

=

= +

=

=

=

+ + −=

+

2 )c

θ

Page 18: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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

observedobject

object object

observedobject

object object object

o

V h i h c V jV

V h c

V h h c VSpeed V

V h c

V h V c V hSpeed

V

θ θ

θ

θ θ

θ

θ θ θ

+ + −⇒ =

+

+ + −⇒ = =

+

+ − +⇒ =

+

2cos / )bjecth cθ

θ

Page 19: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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

observedobject

object object

observed observedobject

object object objectobserved

V h i h c V jV

V h c

V h h c VV V

V h c

V h V c hVV

θ θ

θ

θ θ

θ

θ θ

+ + −⇒ =

+

+ + −⇒ = =

+

+ − +⇒ =

2

os

(1 cos / )objectV h c

θ

θ+

θ

Page 20: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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 objectobject

V h V c hVV V h

V h c

θ θ θθ

θ

+ − +=

+

Page 21: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Taylor Series Expansion in 1DA 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 xf x h f x h h hdx dx dx

+ = + + + +

Page 22: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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 xdf x d f x d f x yx n

dx dx dxf x h x h x hf x h x xh hf x h x h

=

= = = ∀ ≥

⇒ + = + +

⇒ + = + +

⇒ + = +

Page 23: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Taylor Series Expansion chain ruleIf there is a function , and itself is a function of some other variables

Then the Taylor series can be written in the form:

< and higher powers >

( , )f x y

1 2 31 2 3 1 2 3

1

g( , , ) ( )(g( , , )) (g( , , )) y y y df xf y h y y f y y y hy dx

∂+ = + +

( )f x x 1 2 3( , , )x g y y y=

2h

Page 24: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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

Page 25: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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) . Cn n

bk bk bk k

TotalEnergy E V E V const= =

= = = =∑ ∑

( ) ( )1 1

(A) . Cn n

ak ak ak k

TotalEnergy E V E V const= =

= = = =∑ ∑

Page 26: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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

bkk

n

bk akk

TotalEnergy E V

E V h V θ

=

=

=

=

Page 27: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Proof Cont…

( , )f x y

Then we can write it as the Taylor’s series:

< and higher powers>

.

( )

( )

1

1

(B) ( , , )

(0, , ) ( )(0, , )

n

bk akk

nbk ak ak

bk akk ak

TotalEnergy E V h V

V V dE VE V V hh dV

θ

θθ

=

=

=

∂= + +

∑ 2h

. bConst C= =

Page 28: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Proof Cont…

( , )f x y

Let and be two states of the system both in reference frame A & B. Then based on the law of conservation of energy in the frame of reference B:

< and higher powers>

< and higher powers>

.

( )

( )

1

1

(B) ( , , )

(0, , ) ( )(0, , )

n

bk akk

nbk ak ak

bk akk ak

TotalEnergy E V h V

V V dE VE V V hh dV

α α

α α αα α

α

θ

θθ

=

=

=

∂= + +

∑2h

α β

( )

( )1

1

( , , )

(0, , ) ( )(0, , )

n

bk akk

nbk ak ak

bk akk ak

E V h V

V V dE VE V V h

h dV

β β

β β ββ β

β

θ

θθ

=

=

=

∂= + +

∑ 2h

Page 29: Harnessing Zero Point Energy -- Mathematical Physics Perspective

Proof Cont…

Page 30: Harnessing Zero Point Energy -- Mathematical Physics Perspective

Proof Cont…

Page 31: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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 akk kn n

bk ak akk k

E V V E V

E V V E V

α α α

α β β

θ

θ

= =

= =

=

=

∑ ∑

∑ ∑

Page 32: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Proof Cont…

( , )f x y

Then the equations simplify to:

< and higher powers>

< and higher powers>

.

( )

( )

1

1

(B) ( , , )

(0, , ) ( )

n

bk akk

nbk ak ak

akk ak

TotalEnergy E V h V

V V dE VE V hh dV

α α

α α αα

α

θ

θ=

=

=

∂= + +

∑2h

( )

( )1

1

( , , )

(0, , ) ( )

n

bk akk

nbk ak ak

akk ak

E V h V

V V dE VE V h

h dV

β β

β β ββ

β

θ

θ=

=

=

∂= + +

∑ 2h

Page 33: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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:

< and higher powers>

< and higher powers>

.

( ) ( )1 1

(A)n n

ak akk k

TotalEnergy E V E Vα β= =

= =∑ ∑

2h

2h

1

(0, , ) ( )nbk ak ak

k ak

V V dE Vhh dV

α α α

α

θ=

∂+

∂∑

1

(0, , ) ( )nbk ak ak

k ak

V V dE Vh

h dVβ β β

β

θ

=

∂= +

∂∑

Page 34: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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 nbk ak akbk ak ak

k kak ak

V V dE VV V dE Vh dV h dV

β β βα α α

α β

θθ= =

∂∂=

∂ ∂∑ ∑

0h →

1

(0, , ) ( )nbk ak ak

k ak

V V dE Vh dV

θ=

∂∂∑ α β

h

Page 35: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Proof Cont…

( , )f x y

Now let us look at

at

. (0, , )bk akV Vh

θ∂∂

( )

2 2 2

2 2 2 2 2 2

1 (2 (1 / sin ) 2 cos )( , , ) 21 cos / (1 / sin ) 2 cos

ak k ak kbk ak

ak k ak ak k ak k

h V c VV h Vh hV c V h V c hV

θ θθ

θ θ θ

− +∂=

∂ + + − +

( )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

θ θ θ

θ

+ − +−

+

0h =

( )

2 2

2 2

(0, , ) cos / cos /

(0, , ) cos 1 /

bk akak k ak ak k

bk akk ak

V V V V V ch

V V V ch

θ θ θ

θ θ

∂= −

∂∂

⇒ = −∂

Page 36: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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 /

nbk ak ak

k akn

akk ak

k ak

V V dE Vh dV

dE VV cdV

θ

θ

=

=

∂∂

= −

2

2 2( )

1 /mcE vv c

=−

2 2

2 2 3/2 2 2 3/2

( ) 2 /12 (1 / ) (1 / )

ak ak ak

ak ak ak

dE V mc V c mVdV V c V c

−= − =

− −

Page 37: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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 /

nbk ak ak

k akn

akk ak

k ak

V V dE Vh dV

dE VV cdV

θ

θ

=

=

∂∂

= −

2

2 2( )

1 /mcE vv c

=−

2 2

2 2 3/2 2 2 3/2

( ) 2 /12 (1 / ) (1 / )

ak ak ak

ak ak ak

dE V mc V c mVdV V c V c

−= − =

− −

Page 38: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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 22 2 3/2

1

2 21

( )cos 1 /

cos 1 /(1 / )

cos1 /

nak

k akk ak

nak

k akk akn

akk

k ak

dE VV cdV

mVV cV c

mVV c

θ

θ

θ

=

=

=

= −−

=−

h hj=

Page 39: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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

akk ak

k ak

dE VV cdV

θ=

−∑( ) cos( ) sin

akx ak k ak

aky ak k ak

P V PP V P

θθ

==akP

( )

( )

iak ak akx aky ak

iak ak akx aky ak

P V P iP P e

P V P iP P e

θ

θ

+

−−

= + =

= − =

1 1( ), ( )

n n

ak ak ak akk k

P V P V+ −= =∑ ∑

( ) ( )2 2 2 2

1 1

( ) ( )cos 1 / , sin 1 /n n

ak ak ak akk ak k ak

k kak ak

dP V dP VV c V cdV dV

θ θ+ +

= =

− −∑ ∑

Page 40: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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 akak

k ak

dP VS V c edV

θ+

=

⇒ = −∑

( )ak akP V+

21 2( )S c S iS= +

( ) ( )2 2 2 21 2

1 1

( ) ( )cos 1 / , sin 1 /n n

ak ak ak akk ak k ak

k kak ak

dP V dP VS V c S V cdV dV

θ θ+ +

= =

= − = −∑ ∑

2 2( )

1 /i iak

ak ak ak

ak

mVP V P e eV c

θ θ+ = =

Page 41: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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 /1 121 / 1 / 1 / 1 /

ii iak ak ak ak

akak ak ak ak ak ak

dP V mV V cd e me e m VdV dV V c V c V c V c

θθ θ+

− = = − = − − − −

( )( )

2 222 2 2

3/2 2 22 21 1 11 /

1 /1 /

k kk k

i in n ni i

ak akk k kakak

e m e mcS c V c e E eV cV c

θ θθ θ

= = =

= − = = −−

∑ ∑ ∑

2

1

k

ni r

akk

E e θ

=∑ (2 1)

1

k

ni r

akk

P e θ+

=∑

Page 42: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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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.

.

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Fallacy in Einstein Derivation

( , )f x y

The derivation by Albert Einstein was based on elastic collision of two particles, which approach each other head on and then divert at some angle. It was shown in the derivation that if this phenomena was observed from any other frame reference, the conservation still was true. Here we prove using the infinite conservation equations that such an elastic collision is not at all possible thus the derivation is wrong.

r

.

Page 44: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Fallacy in Einstein Derivation Cont…

( , )f x y

Let us take a simple case of 2 particles as in the Einstein’s 1934 2 black board derivation:

r

.

Page 45: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Fallacy in Einstein Derivation Cont…

( , )f x y

Initial energy equations:

Initial momentum equations:

Final energy equations:

Final momentum equations:

.22

2 2 ( ) 2

2 21

(1 ) 21 /

ki rni r

k ak

mc e mc e mcV

rc

θπγ γ

=

= ∈+ =−

∀∑

(2 1)(2 1)

2 21

(1 ) 01 /

ki rni rak

k ak

mV e mV e rV c

θπ

++

=

= + = ∀ ∈−

222 2 2 ( ) 2 2

2 21

( ) 21 /

ki rni r i r i r

k ak

mc e mc e e mc e rV c

θθ θ π θγ γ+

=

= + = ∀ ∈−

(2 1)(2 1) (2 1)

2 21

(1 ) 01 /

ki rni r i rak

k ak

mV e mVe e rV c

θθ π

++ +

=

= + = ∀ ∈−

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Fallacy in Einstein Derivation Cont…

( , )f x y

For momentum equations conservation is satisfied.

Energy equations should satisfy:

BUT IT CANNOT SATISFY ABOVE EQUATION FOR ANY GENERAL ANGLE

Which means there is a fallacy in the derivation.

.

2 2 22 2 i rmc mc e rθγ γ= ∀ ∈

θ

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The real fallout

( , )f x y

What Einstein described is a very commonly seen phenomena of life, an almost elastic collision of billiards ball going away at an angle. For our conservation equations to be satisfied, there is a correction needed of the order of , which is complete mass energy!

So how does it happen that equations are not satisfied but the first order equations still hold true? What provides the energy compensation of the higher order?

.

2mc

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Mass Light Duality

( , )f x y

If there are 2 photons of exactly the same frequency going in exactly the opposite directions. The total energy of the photon pair is: Total Momentum of the photon pair is:

Looking from a frame of reference moving at speed u wrt the stationary frame then:The total energy of the photon pair is:

Total Momentum of the photon pair is:

.

2bE ω ω ω= + =

0bE ω ω= − =

1 2 2 2 2 2

1 / 1 / 1 / c 1 / c 121 / 1 / 1 / 1 /

av c v c v vE E Ev c v c v c v c

ω ω ω ω + − + + −

= + = + = = − + − −

1 2 2 2 2 2

1 / 1 / 1 / c 1 / c // / / 2 /1 / 1 / 1 / 1 /

av c v c v v v cP P P c c c cv c v c v c v c

ω ω ω ω + − + − +

= + = − = = − + − −

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Mass Light Duality

( , )f x y

What is interesting to note is that:

and

A pair of photons going in opposite directions transform energy and momentum as matter with mass Furthermore the frequencies of their interference are:

.

2 21 /b

aEEv c

=− 2 21 /

ba

PPv c

=−

11 / 1 / 21 / 1 /

v c v cv c v c

ω ω γω + −

= + = − +

21 / 1 / 2 /1 / 1 /

v c v c v cv c v c

ω ω γω + −

= − = × − +

22 /m cω=

Page 50: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Mass Light Duality

( , )f x y

If put as we found in earlier slide then 2 wavelengths are:

De-Broglie wavelength

The other wavelength

So the model of a particle as 2 photons going in opposite directions not only works perfectly for momentum and energy but also for matter wave model. Further in case of v = 0 De Broglie hypothesis fails but we have the correct answer that at v = 0 only one frequency exists with wavelength: hc/E

. 2 / 2mcω =

2 22 22 /

2 /c c hcv c m vc P

π πλ π ωγω γ

= = = =×

1 1 2

2 22 /2

c c hccm c E

π πλ π ωγω γ

= = = =

Page 51: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Mass Light Duality

( , )f x y

With a similar derivation we have proven that space-time also transforms as if each point is composed of 2 light rays going in opposite directions.A more interesting part in the paper is the proof that a photon can also be modeled as 2 non-zero rest mass particles, one going below the speed of light and the other above the speed of light.We have also formulated and proven the energy and momentum of above the speed of light as:

.

2 2 2/ / 1aE mc v c= − −2 2/ / 1aP mv v c= − −

Page 52: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Lorentz transformation for V >C

( , )f x y

The modified principle of relativity for V > C: The set of “pure measurements” is same if they are done from a frame of reference A into B or B into A. At speed greater than light a single event appears 2 different events. It is light looking into 2 different mirrors. We have given the modified Lorentz transformation with the new principle of relativity in 1D as:

Where are 2 different space-time points in the frame of reference Bgiven a single point in the frame of reference A. This is like seeing the world in mirror but 2 different mirrors.

.

1 1

1 12 2

2 2

2 2

0 0 1 /0 0 / c 1

1/ / 11 / 0 0/ 1 0 0

b a

b a

b a

b a

x xv cct ctv

v cx xv cct ctv c

− − = − −

1 1 2 2( , ), ( , )b b b bx t x t

Page 53: Harnessing Zero Point Energy -- Mathematical Physics Perspective

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Zero energy non zero momentum (ZEN)

( , )f x y

In the case of speed limiting to infinity

These are in-fact zero energy and finite momentum particles! As they have infinite speed they exist only for a single instant of time, where they are omnipresent.

These particles when present in any system can take away or add momentum without taking away or adding energy to the system. When these take away momentum, they lead to conversion of kinetic energy to mass. When these particles add momentum they lead to conversion of mass to kinetic energy. We conjecture that these particles form space component of space time and also the binding force in the atom nucleus.

.

2 2 2lim / / 1 0a vE mc v c

→∞= − − =

2 2lim / / 1a vP mv v c mc

→∞= − − = −

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Further work not published

( , )f x y

We have modified the Klein Gordon/Dirac equation for the In the case of speed greater than light and proven that there are states of electrons in an atom called as negative excited states, which can be achieve only when an atom gets negative energy photon. From the quantum time evolution point of view, these negative excited states have lower energy but still are not the most probable states BUT to bring an electron from the negative excited state to the ground state positive energy photon is required.

We have also proven that an positron is an electron going at a speed greater than light, which manifests as going into past.

.

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At last zero point energy

( , )f x y

We have been able to theoretically prove that zero point energy or free energy can only be harnessed if the particles are tunneled from speed below light to above light (patent pending).

We have also found a very simple way of tunneling particles from speed below the speed of light to above the speed of light. CERN cannot do it because they do not have the complete theory. Particles cannot be accelerated to beyond the speed of light but can be tunneled.

The easiest way to tunnel the particles/electrons across the speed of light is through solid state devices. Solid state devices/semi-conductors are the place where electrons really move in space.

With our theory there are 100s of ways to build devices to harness zero point energy and in all form factors (nano-scales)!

.

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Experimental Discussion

( , )f x y

Will only speak about it!.