Questions To Special Relativity

[TonyYuan2020]

That is the Newtonian version of gravity which was made incorrect by the Theory of General Relativity the gravitational field is actually, 

 

Gravitational field or space-time curvature is a scientific mathematical model.

The phenomena that can be explained by general relativity can be explained by classical physics.

Edited March 15, 2020 by TonyYuan2020

[TonyYuan2020]

It does not slow down the speed of light however it changes the size of space that it travels through.

 

Gravitational field or space-time curvature is a scientific mathematical model.

The phenomena that can be explained by general relativity can be explained by classical physics.

[ralfcis]

The light doesn't slow. This is not refraction through a medium. Time doesn't slow. Space doesn't contract. The light thinks it's going in a straight line. From our perspective it's following the curvature of spacetime and we explain that as space is contracted and time is slowed so that from our perspective the speed of light remains constant and that if we look at it from the light's perspective, it also remains constant. Even in a black hole the light is going at c away from the black hole. The reality is we don't see it so we come up with relativity to explain this seeming paradox. The truth is the light is unencumbered by gravity. It is travelling in a straight line out of the black hole at c. But there is a relativity of simultaneity between our clock and the black hole's clock for the light's start time. We never see it start. I'd show you the Minkowski diagram but what's the point if no one understands math.

[Vmedvil2]

 

Gravitational field or space-time curvature is a scientific mathematical model.

The phenomena that can be explained by general relativity can be explained by classical physics.

 

No, the Phenomena that are explained by General relativity cannot be explained by classical physics.

[TonyYuan2020]

We can't ignore the influence of gravity however the influence of earth's gravity is much weaker than the more massive object like the sun, it is negligible on the light.

 

It's not. You have to take distance into account. It's like the equation of gravitation

[ralfcis]

I feel like I'm falling into a black hole because I can't keep up with the speed of this thread.

[Vmedvil2]

It's not. You have to take distance into account. It's like the equation of gravitation

Well, I am going to leave this here for now, but Newtonian gravity is wrong, I am done with you for now. Your homework is to read about general relativity, Link = https://en.wikipedia.org/wiki/Introduction_to_general_relativity and https://www.space.com/17661-theory-general-relativity.html

Edited March 15, 2020 by VictorMedvil

[TonyYuan2020]

No, the Phenomena that are explained by General relativity cannot be explained by classical physics.

 

The bending of the light can be explained. The mass energy equation is easier to explain. The Morley experiment can still be explained.

[TonyYuan2020]

Well, I am going to leave this here for now, but Newtonian gravity is wrong, I am done with you for now. Your homework is to read about general relativity, Link = https://en.wikipedia.org/wiki/Introduction_to_general_relativity

 

Well, thank you very much for your discussion. We are good friends in scientific exploration. Thanks.

[TonyYuan2020]

I feel like I'm falling into a black hole because I can't keep up with the speed of this thread.

 

My friend, let's have a rest. We are all explorers who love science. 

[ralfcis]

Well we agree on the facts but not on the explanation for them. There are probably thousands of other explanations.

[maheshkhati]

I have already mathematically proved that Special theory of relativity is also wrong :- Following are calculations

 

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 friction-less 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

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 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.

1. If dE= y . dEo is wrong then

dE/c²= y . dEo /c²  is wrong.

So, dM= y . dMo is wrong.

 

Refer :- site https://vixra.org/abs/1912.0171

Edited March 16, 2020 by maheshkhati

[Vmedvil2]

I have already mathematically proved that Special theory of relativity is also wrong :- Following are calculations

 

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 friction-less 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

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 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.

1. If dE= y . dEo is wrong then

dE/c²= y . dEo /c²  is wrong.

So, dM= y . dMo is wrong.

 

Refer :- site https://vixra.org/abs/1912.0171

All this is crackpottery!

[TonyYuan2020]

All this is crackpottery!

 

We have designed our own LIGO system. As long as the wavelength changes by 0.0001% or the frequency changes by 0.00001hz, we can catch it.

[OceanBreeze]

Moderator note: Since this poster has started two threads on the same topic of Special Relativity, I have merged them into one thread.

 

This will make it easier to move the whole lot into Strange Claims.

[Vmedvil2]

Moderator note: Since this poster has started two threads on the same topic of Special Relativity, I have merged them into one thread.

 

This will make it easier to move the whole lot into Strange Claims.

I am ready send this whole thread to the crank house.

[ralfcis]

 No way to check out this math because it's in some kind of weird format and there's almost no verbal explanation of what it's doing.