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Einstein Is Wrong But Einstein's Effect Is Right.

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#35 maheshkhati

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Posted 13 July 2020 - 06:22 AM

dE/dt =v. dp/dt = v. F   this formula is use many times directly in relativity.



#36 maheshkhati

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Posted 14 July 2020 - 02:55 AM

Just consider that one man is at location (x,y)  is pulling Box on friction less platform with rope with force F whose component in x-direction & y-direction are Fx & Fy respe.

Calculation of relativity math in Post 4 suggest that actual force acting on Box is more i.e. Fx+Fmy & Fy+Fmx.

Due to more force , more work done happen.

So, less consumption of energy create more work or more energy in relativity.

This is against consistency of energy in that frame.


Edited by maheshkhati, 14 July 2020 - 03:07 AM.


#37 devin553344

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Posted 14 July 2020 - 03:50 AM

Just consider that one man is at location (x,y)  is pulling Box on friction less platform with rope with force F whose component in x-direction & y-direction are Fx & Fy respe.

Calculation of relativity math in Post 4 suggest that actual force acting on Box is more i.e. Fx+Fmy & Fy+Fmx.

Due to more force , more work done happen.

So, less consumption of energy create more work or more energy in relativity.

This is against consistency of energy in that frame.

You don't appear to understand the physics correctly. Relativity requires more force in the x direction since it increases momentum.


Edited by devin553344, 14 July 2020 - 03:54 AM.


#38 maheshkhati

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Posted 15 July 2020 - 01:28 AM

Thanks, you understand something. You are true. Relativity requires more force in the x direction since it increases momentum but mathematics in post 4 proves that

the additional force is not the applied force but mathematical force. & That additional force is not responsible for acceleration in that direction. 

That is the problem.

OR mathematics in post 4 proves that more rate of change of moment happen than applied force in that direction due to combination of velocity in that direction & acceleration perpendicular to that direction.



#39 maheshkhati

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Posted 20 July 2020 - 01:38 AM

Mathematics in post 4 indicate that

step 1 proves that when applied force in x-direction is 0 then force acting in x-direction  Fx= 0+y3 mo. (ux/c2} uy ay= y3 mo. (ux/c2} uy ay

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STEP 2 proves that applied force in x-direction is -Fx = -y3 mo. (ux/c2} uy ay  then  force acting in x-direction  Fx= -y3 mo. (ux/c2} uy ay +y3 mo. (ux/c2} uy ay =0

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This proves that applied force in x-direction is Fx  then  force acting in x-direction  (Fx+y3 mo. (ux/c2} uy ay)

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This proves that acting force or (rate of change of momentum in x-direction happen) is more than applied force in relativity.

 

Same mathematics indicate by transformation of forces in relativity.

 

Same mathematics indicate that direction of acting force & acceleration are different in relativity.

Even positive direction acting force in x-direction create -ve direction acceleration.


Edited by maheshkhati, 20 July 2020 - 01:39 AM.


#40 devin553344

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Posted 21 July 2020 - 12:04 PM

Mathematics in post 4 indicate that

step 1 proves that when applied force in x-direction is 0 then force acting in x-direction  Fx= 0+y3 mo. (ux/c2} uy ay= y3 mo. (ux/c2} uy ay

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STEP 2 proves that applied force in x-direction is -Fx = -y3 mo. (ux/c2} uy ay  then  force acting in x-direction  Fx= -y3 mo. (ux/c2} uy ay +y3 mo. (ux/c2} uy ay =0

----------------------------------------------------------------------------------------------------------------------------------------

This proves that applied force in x-direction is Fx  then  force acting in x-direction  (Fx+y3 mo. (ux/c2} uy ay)

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This proves that acting force or (rate of change of momentum in x-direction happen) is more than applied force in relativity.

 

Same mathematics indicate by transformation of forces in relativity.

 

Same mathematics indicate that direction of acting force & acceleration are different in relativity.

Even positive direction acting force in x-direction create -ve direction acceleration.

This is ridiculous, if force is more, then as a value it would be greater. Yet you can't supply any examples with values, which makes me think it's fake math.



#41 maheshkhati

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Posted 22 July 2020 - 01:37 AM

In relativity force in X-direction is mathematically calculated as

Fx  =d/dt (m ux) = dm/dt . ux + m . du/dt = dm/dt . ux + m . ax

Let, consider fighter plane with horizontal velocity ux drop the bomb B 

& Observer is on ground

For observer :- Bomb B will move with constant horizontal velocity ux & accelerate vertically due to gravity.

So, Force applied on Bomb is gravity only in vertical direction.

but one mathematical acting force is created in horizontal direction i.e. Fx= dm/dt .ux + m. (0) =dm/dt .ux 

as mass of Bomb increases due to vertical acceleration.

So, applied force is different that acting force in relativity.

& even horizontal acceleration is zero there is acting horizontal force

I am giving detail mathematics below

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FORCES IN SPECIAL THEORY OF RELATIVITY MAKE SPECIAL THEORY OF RELATIVITY WRONG

 

(Special theory of relativity is very great & close theory. If we prove one thing in this theory is mathematically wrong then whole theory gets collapsed because all mathematics of  the total theory are interlinked. So, If we prove acting force is different than applied force or energy consumed is different than energy produce then whole special theory get collapsed because same mathematics can be used to prove transformation equation of  forces for relativity , same mathematics can be extended & can be used to prove dE= y . dE or dM= y. dMo. etc)   

 

CALCULATION 1:- Force without acceleration, acceleration without force & applied force is less than acting force in SPECIAL THEORY OF RELATIVITY.

 

STEP 1:-This problem can easily be understood by following paradox.

{Before starting this paradox, I want to put one relativity formula’s given in standard book of relativity for example “Page no. 135 of Elements of special relativity” by Dr T.M. Karade, Dr K S Adhav & Dr Maya S Bendre.

In any frame, for force in X-direction by S.R.

      Fx = d/dt( y.  mo. ux)  where y=(1-u2/c2)-0.5

   So, after differentiation

Fx= y. mo. (dux/dt) + y3. mo. {ux/c2}. (u . du/dt)  

 Fx= y. mo. ax + y3. mo. {ux/c2}. (u . a)  -----(A)

We know,  u2=ux2+uy2+uz2    

So, after differentiation

       2 u. (du/dt) = 2.ux (dux/dt) +2.uy (duy/dt) + 2.uz (duz/dt)

      2 u. a = 2.ux ax +2.uy ay + 2.uz az    

       u. a = ux ax + uy ay + uz az    --------(   :cool:

from (A) & (   :cool:

So, Fx=y. mo. ax+y3 mo. (ux/c2} (ux ax+uy ay+uz az) ------(1)}

Now, Consider Paradox:-

On frictionless platform, object is moving with constant velocity ux in X-direction & only magnetic force is acting in Y-direction & there is acceleration in Y-direction only with velocity uy & Fz=0

If we apply eq(1) to this case then result will be

       Fx= y3 mo. (ux/c2} uy ay    ----------   as ax=0  

  Or  Fx=Fay  as this force is form due to ‘ay’ only

Mean’s even there is no magnetic force acting on object from outside in x-direction & no ‘ax’ then also above force will act on object in +ve direction of x-axis due to ‘ay’

Important point (1):-

Mean’s applied magnetic force on object in X-direction is 0 & acting force in X-direction is   Fx= y3 mo. (ux/c2} uy ay+0 or Fay+0=Fay

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STEP 2:-Now, Force acting in X-direction is  Fx= y3 mo. (ux/c2} uy ay or Fay

Now, after this happen, very small magnetic force of same intensity

 -fx = -y3 mo. (ux/c2} uy ay or  -Fay  start acting on object in direction opposite to above force (but velocity is still positive ux) & cancel that above force.

Mean’s equation (1) becomes

  0=y. mo. ax+y3 mo. (ux/c2} (ux ax+uy ay) 

Or  0 =y. mo ax. (1+ y2  {ux2/c2} ) +Fay 

(Here as  Fay= y3 mo. (ux/c2} uy ay)

Mean’s  Fay = y. mo. -ax. (1+ y2. {ux2/c2} )

Mean’s there must be acceleration in –ve X-direction to fulfill above equation of S.R.

Now, see above equation carefully, it is of nature

     0= -fx + Fay

Important point (2):- Mean’s applied magnetic force on object in X-direction is -fx & acting force in X-direction is  -fx + Fay = 0  or 0.

Here, resultant force in X-direction is zero but there is acceleration in –ve direction.

 

STEP3:- same things happen for +ve force in X-direction (for less than Fay or more)

Now, I am generalizing above result.

Step 1 & 2 clearly shows that when we apply any magnetic force (Fmx) in X-direction on the object, actual force acting on object is more & that quantity is (Fmx+Fay)

Similarly,

If we apply any magnetic force (Fmy) in Y-direction on the object then actual force acting on object is more & that quantity is (Fmy+Fax)

This is completely complicated results, which says that applied force & acting forces on objects are different & more in S.R.

STEP4:- Force does work, consume energy, gain energy & we must know that energy cannot be created. It can be transferred only:-

From above setup it must be clear that energy get transfer from magnet to object but if applied force is less than acting force then energy gain by the object will be more than energy loose by the magnet. Means due to more work done by more force for same displacement, more energy get generated.

HERE, more energy (& force) is the problem.

Where this additional energy (or force) does comes from?

There is no answer in S.R. for this problem.

 

THIS MATHEMATICS PROVES THAT THE S.R. IS COMPLETELY WRONG:-

In S.R., force is not related to change in the state of motion or acceleration as Newton consider but with change in moment.

So, even I move towards falling ball,

fx= y3 mo. (ux/c2} uy ay  --------  this force will act on the ball.

&  Direction of applied force is different than acting force.

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If this mathematics is true then if old man pulls the cart on horizontal platform with force f and fx, fy are their components in X & Y direction respectively

Then above calculation says that actual forces acting on the cart are not fx, fy but

Fx=fx+ y3 mo. (ux/c2} uy ay = fx +Fmay

 & Fy=fy+ y3 mo. (uy/c2} ux ax = fy +Fmax

This will create further problem because if F is actual force acting on the cart then

F2= Fx2+Fy2

F=(fx2+fy2+Fmax2+Fmay2+2 .fx. Fmay + 2 .fy. Fmax)0.5

F=(f2+Fma2 +2 .fx. Fmay + 2 .fy. Fmax)0.5

So, here actual force acting cannot be equated to the sum of resultant force applied by old man i.e. f & resultant of additional force created by Fmay & Fmax.

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Above  mathematics proves that

  1. Acting force is different than applied force.
  2. Energy consumed is less than energy produce

Proof:-

Applied force  <  acting force.

So, in this inertial frame

So,  (Applied force X displacement) < (acting force X displacement).

So,  Energy consumed <  energy produce.

This is against  the  law of consistency of energy.

  1. Above mathematics proves that even there is zero ‘Fx’  force acting on object then also body will accelerate in –ve x-direction.

Mathematics of step 2 proves that for applied force 0 to –fx,

Acting force direction is +ve & acceleration direction is -ve

  1. If above calculation is proved wrong then

a)Trnsformation equation of forces in special relativity is wrong.

As same mathematics if extended gives transformation equation in relativity

For example:-

So, if

F’x = d/dt’( y’.  mo. U’x)  where y’=(1-u’2/c2)-0.5  This equation is wrong then

F’x = Fx – ( v/c2 . Fy. Uy)/(1-V .Ux/c2)   ----transformation equation in relativity is wrong because ..

If this differentiation extended by proper transformation equations of frame like putting equations of  U’x, y’  & d/dt’ then we can prove that

F’x = Fx – ( v/c2 . Fy. Uy)/(1-V .Ux/c2)   ----transformation equation in relativity.

So, if F’x = d/dt’( y’.  mo. U’x)  where y’=(1-u’2/c2)-0.5  is   wrong then above transformation equation for force is wrong.

:cool: dE= y . dEo is wrong

Proof:-

As,  F’x = Fx + ( v/c2 . Fy. Uy)/(1-V .Ux/c2) 

 F’y = (Fy/ y ) /(1-V .Ux/c2)   ----transformation equation in relativity.

Now, consider event

Consider ball is falling under gravity in rail cabin with vertical force Fy  only then by above transformation equation of relativity

 for person on platform.

F’x =(Uy.v/c) . Fy    &      F’y = (Fy/ y ) --------(1)

As Fx =0 & Ux =0

Similarly,

d’y = dy ------(2)

d’x = y (dx + v dt)

As dx =0 in rail cabin frame.

d’x = y v dt  ------(3)

Now, Energy consume in this event by observer on Platform :-

d’w = F’ d’s = F’x.d’x+F’y.d’y

put the values (1), (2) & (3)

d’w = y v dt . (Uy.v/c) . Fy    + dy . (Fy/ y )

d’w = Fy  . {y dt . (Uy.v2/c) + dy / y }

d’w = Fy  .dy . {y.v2/c + 1 / y }

d’w = y. Fy  .dy

In Rail cabin force acting is F=Fy & displacement ds =dy only.

So, d’w= y . dwo

So, d’E = y . dEo

So, if above mathematics & force transformation is wrong then above calculation is also wrong because special theory of relativity is interlinking theory. If we prove one thing is mathematically wrong then whole theory is collapsed.

  1. If dE= y . dEo is wrong then

dE/c2= y . dEo /cis wrong.

So, dM= y . dMo is wrong.

 

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Above mathematics is not wrong & not fake.

Fake mathematics can be easily proved wrong & you have not proved anything wrong in above mathematics. 


Edited by maheshkhati, 22 July 2020 - 01:38 AM.


#42 maheshkhati

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Posted 22 July 2020 - 03:06 AM

https://youtu.be/W9yWv5dqSKk

 

This video beautifully explain the particle, wave duality. This also proves that this is only possible when particle (Droplet) moving in some medium then only duality is possible. I consider the space as given in 2nd chapter of my paper https://vixra.org/pdf/2006.0033v1.pdf act as medium through which particle travel. Last chapter explain how electromagnetic vibration happen in particle level.



#43 devin553344

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Posted 22 July 2020 - 11:59 AM

https://youtu.be/W9yWv5dqSKk

 

This video beautifully explain the particle, wave duality. This also proves that this is only possible when particle (Droplet) moving in some medium then only duality is possible. I consider the space as given in 2nd chapter of my paper https://vixra.org/pdf/2006.0033v1.pdf act as medium through which particle travel. Last chapter explain how electromagnetic vibration happen in particle level.

So you're an ether guy without any proof of your idea. Yeah that explains a lot.



#44 maheshkhati

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Posted 23 July 2020 - 02:21 AM

I am very sure, if Einstein develop general theory of relativity first & then see toward local theory of relativity. He will develop it in my way. Big object  not only bend the space but create the space. Vacuum is different than space. Particle is local vibrating charge spike in a space as given in last chapter OR world is the interaction of big charge with local charges.

Many scientist knows that Einstein special theory of relativity is wrong & they try to prove it wrong by experiment but relativity wins because time & mass are the properties of space. They do not remain same. So, we get effect that time slow down due to motion or require more force for same acceleration after velocity gain. 

But this create other problems like dark matter, dark energy, sudden expansion of universe after big bang,  very large unexplained black holes at early universe, collapse of standard model of particle physics, wave particle duality etc. This all problem cab be solve only by revisiting physics again with new thought. Starting is space & vacuum are different & physic's mathematics is get applicable in space only not in vacuum.


Edited by maheshkhati, 23 July 2020 - 02:23 AM.




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