Both Newton & Einstein Are Wrong.

Vmedvil2 · March 16, 2020

Newton was great scientist. He is completely true in local inertial frame of reference...
but Einstein's special theory of relativity is completely wrong because in that theory we can prove that applied force is different than acting force.
For example:-
Consider one military plane dropped the Bomb A from some height on earth with horizontal constant velocity ux
then Bomb will accelerate in vertical direction due to gravitational force & move horizontally with constant
velocity ux
Now, for observer on ground, force applied on Bomb A is gravity but
Spe relativity create one mathematical additional horizontal force
fx= y3 mo. (ux/c^2} uy ay in horizontal direction which not applied by anyone in this frame.
(Explain in chapter 1)
This will create a situation where applied force on object is different than acting force.
OR energy consumed is different than energy produce.
This horizontal mathematical force Fx is created due to (constant horizontal velocity ux) X (rate of change of mass of bomb A due to vertical acceleration)

Pure Crackpottery!

maheshkhati · March 16, 2020

Then prove mathematically where am I wrong?. I am putting basic mathematics

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

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

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

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.

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

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.

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

Above mathematics proves that

Acting force is different than applied force.
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.

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

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)^-0.5This 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)^-0.5is 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/c²) . 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/c²) . Fy + dy . (Fy/ y )

d’w = Fy . {y dt . (Uy.v²/c²) + dy / y }

d’w = Fy .dy . {y.v²/c² + 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.

If dE= y . dEo is wrong then

dE/c²= y . dEo /c²is wrong.

So, dM= y . dMo is wrong.

Edited March 16, 2020 by maheshkhati

Vmedvil2 · March 16, 2020

Pure Crackpottery!

Then prove mathematically where am I wrong?. I am putting basic mathematics
-----------------------------------------------------------------------------------------------------
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
-------------------------------------------------------------------------------------------------------------------------------
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.
--------------------------------------------------------------------------------------------------------------------------
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.
--------------------------------------------------------------------------------------------------------------------------------------
Above mathematics proves that
Acting force is different than applied force.
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.
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
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)^-0.5This 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)^-0.5is   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/c²) . 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/c²) . Fy    + dy . (Fy/ y )
d’w = Fy . {y dt . (Uy.v²/c²) + dy / y }
d’w = Fy .dy . {y.v²/c² + 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.
If dE= y . dEo is wrong then
dE/c²= y . dEo /c²is wrong.
So, dM= y . dMo is wrong.

Pure Crackpottery!

Thoth101 · March 17, 2020

Pure Crackpottery!

:lol: :lol: :rofl:

maheshkhati · March 17, 2020

I have use mathematics of special theory of relativity to prove it wrong.

It is accepted fact in special theory of relativity that direction of force & acceleration may not be same. This happen due to additional non applied force is created in special relativity & that force do not responsible for acceleration.

Vmedvil2 · March 17, 2020

I have use mathematics of special theory of relativity to prove it wrong.
It is accepted fact in special theory of relativity that direction of force & acceleration may not be same. This happen due to additional non applied force is created in special relativity & that force do not responsible for acceleration.

You haven't proven anything you aren't even using the correct equations for special relativity, PURE CRACKPOTTERY!

maheshkhati · March 17, 2020

I have use correct equation of force in relativity

Fx = rate of change moment in x-direction = d/dt( y. mo. ux) where y=(1-u²/c²)^-0.5

is true.

Please check it.

maheshkhati · March 17, 2020

This definition of force was first used by Lewis (Phil. Mag. xvi. p. 705 (1908)). In Einstein‘s later treatment of the principle of relativity, Jahrbuch der Radioktivität, iv. p. 411 (1907), he defines force by the equations

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

Fy = d/dt( y. mo. uy) , where y=(1-u²/c²)^-0.5

Vmedvil2 · March 17, 2020

This definition of force was first used by Lewis (Phil. Mag. xvi. p. 705 (1908)). In Einstein‘s later treatment of the principle of relativity, Jahrbuch der Radioktivität, iv. p. 411 (1907), he defines force by the equations
Fx = d/dt( y. mo. ux) , where y=(1-u²/c²)^-0.5
Fy = d/dt( y. mo. uy) , where y=(1-u²/c²)^-0.5

Put this crank in a straight jacket.

OceanBreeze · March 17, 2020

He has already been moved to Alternative Theories. He goes wrong right at the start because he does not understand that energy is frame dependent. In fact, he doesn't seem to understand reference frames.

maheshkhati · March 17, 2020

If you read my above calculation then it is clear that I am accepting "in Einstein special theory of relativity, Kinetic energy is frame dependent."

That is reason, I prove

d’w= y . dwo

So, d’E = y . dEo

by using same mathematics.

My problem is with forces in special relativity.

Edited March 17, 2020 by maheshkhati

devin553344 · April 10, 2020

If you read my above calculation then it is clear that I am accepting "in Einstein special theory of relativity, Kinetic energy is frame dependent."
That is reason, I prove
d’w= y . dwo
So, d’E = y . dEo
by using same mathematics.
My problem is with forces in special relativity.

Force and Energy are the same basically. Force is a type of pressure, but so is energy depending on how it's applied. At least that's how I interpret those items. Check your units and math, it sounds like you've done something wrong. I'm not totally following this though. But I doubt everyone got the wrong math and ideas except now you are the only one who can get it right. I don't find that as a normal for the physics that exist today.

I theory may be incorrect if an alternate exists. But usually the math and ideas are well established and verified by top minds in their fields.

Or maybe you're basing your math and ideas from an article that is incorrect. You're supposed to pick up on errors though and fix them yourself. For instance here are some unit errors that I found on the MIT website: http://web.mit.edu/8.334/www/grades/projects/projects08/EvangelosSfakianakis/6.htm

You'll notice he has a displacement x in a place it doesn't belong which destroys the units of frequency. Maybe you have something like that here? Or did I get something wrong? See:https://en.wikipedia.org/wiki/Plasma_oscillation

Edited April 10, 2020 by devin553344

maheshkhati · April 24, 2020

I am sorry. I can not comment on your post timely because I was away from city.

I know, I can be wrong. So, first I visited to mathematics professor of university & presented my above calculation to him. He told me that my mathematics is not wrong.

Here also, No one prove my mathematics wrong.

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

Work done dw is equal to energy loss in that frame but work done is frame dependent & that will create some problem. For example, one old man is applying force F on the heavy box & pulling it in direction of train motion on Platform with very slow average velocity V against friction.

Now, If velocity of train is also V then observer in the train will find very interesting thing.

He find that there is no displacement in his frame in above event.

i.e. Work done = F. dx = F . 0 = 0.

but he find that old man is tired & loosing his energy.

So, I say that we will get correct mathematics only when observer is in primary inertial frame which is created by local space (Here is platform or earth). Space is ether present around every mass created by balance electromagnetic flux present around every substance. When we accelerate in that space (or ether) that space opposes this acceleration. This create the concept of inertia or mass. So, we require energy to accelerate in that local space.

This local space present every where. It is rail cabin or It is space inside substance or It is space around Earth or It is space around Moon.

This concept can be used to solve problem of dark matter.

We consider our existence end at our body edge. I consider in other way. Our existence start from our body edge.

I want to write book on this concept.

World is still very unknown to us.

Vmedvil2 · June 12, 2020

I am sorry. I can not comment on your post timely because I was away from city.
I know, I can be wrong. So, first I visited to mathematics professor of university & presented my above calculation to him. He told me that my mathematics is not wrong.
Here also, No one prove my mathematics wrong.
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Work done dw is equal to energy loss in that frame but work done is frame dependent & that will create some problem. For example, one old man is applying force F on the heavy box & pulling it in direction of train motion on Platform with very slow average velocity V against friction.
Now, If velocity of train is also V then observer in the train will find very interesting thing.
He find that there is no displacement in his frame in above event.
i.e. Work done = F. dx = F . 0 = 0.
but he find that old man is tired & loosing his energy.
So, I say that we will get correct mathematics only when observer is in primary inertial frame which is created by local space (Here is platform or earth). Space is ether present around every mass created by balance electromagnetic flux present around every substance. When we accelerate in that space (or ether) that space opposes this acceleration. This create the concept of inertia or mass. So, we require energy to accelerate in that local space.
This local space present every where. It is rail cabin or It is space inside substance or It is space around Earth or It is space around Moon.
This concept can be used to solve problem of dark matter.
We consider our existence end at our body edge. I consider in other way. Our existence start from our body edge.
I want to write book on this concept.
World is still very unknown to us.

Still Pure Crackpottery!

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