      # Inconsistent Forces In Special Relativity Which Make It Wrong

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### #1 maheshkhati

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Posted 12 September 2017 - 07:01 AM

We apply force 1 Newton in X-direction & 1 N in Y direction on any moving substance.

Then by special theory of relativity, actual forces get acted on substance are  (1+a) N & (1+ N.

This shows that something is seriously wrong in relativity. Where applied forces is different than acting forces.

I am putting mathematics of SR here

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

This is general  equation of force in X-direction in S.R.

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’ & ‘uy’ 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} ) + y3 mo. (ux/c2} uy ay =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.

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 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.

Edited by maheshkhati, 12 September 2017 - 07:20 AM.

### #2 exchemist

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Posted 12 September 2017 - 07:44 AM

We apply force 1 Newton in X-direction & 1 N in Y direction on any moving substance.

Then by special theory of relativity, actual forces get acted on substance are  (1+a) N & (1+ N.

This shows that something is seriously wrong in relativity. Where applied forces is different than acting forces.

I am putting mathematics of SR here

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 any 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

[snip]

Well you are clearly wrong somewhere, since SR is indisputably correct. So it's a question of where the error comes from.

I'm not an expert in this field but the first thing that strikes me is you are differentiating with respect to something called "t". What is it? Because in SR, "t" means different things to observers in different reference frames.

You might take a look here: https://en.wikipedia...wiki/Four-force

Edited by exchemist, 12 September 2017 - 07:45 AM.

### #3 maheshkhati

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Posted 13 September 2017 - 08:28 AM

1)Nothing is wrong in this calculation, t is time in that frame where you are observing event.

This problem created because in SR, force is the rate of change of momentum in that direction & due to that force is automatically created in X-direction due to acceleration & motion in Y-direction.

For example:- If I am moving toward falling Apple with some constant horizontal velocity Vx  then also horizontal force is created because mass of Apple will change due to acceleration in vertical direction. So, ultimately there is rate of change of momentum happen in horizontal direction which is the force. Mean's even there is no acceleration in horizontal direction & applied force, there is force (created).

This creates problem in relativity.

I am taking this problem to physics community from long time for example.

http://www.sciencech...php?f=2&t=30181

http://www.sciencech...ae9c47a&start=0

& even on this forum

& people accept this problem of relativity but there is no solution from them.

2) Problem is that this additional forces(which is not applied from out side) are get added to forces in X-direction & Y-direction automatically.

and remember forces do work, consume energy. This create problem of additional energy in relativity.

### #4 exchemist

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Posted 13 September 2017 - 08:40 AM

1)Nothing is wrong in this calculation, t is time in that frame where you are observing event.

This problem created because in SR, force is the rate of change of momentum in that direction & due to that force is automatically created in X-direction due to acceleration & motion in Y-direction.

For example:- If I am moving toward falling Apple with some constant horizontal velocity Vx  then also horizontal force is created because mass of Apple will change due to acceleration in vertical direction. So, ultimately there is rate of change of momentum happen in horizontal direction which is the force. Mean's even there is no acceleration in horizontal direction & applied force, there is force (created).

This creates problem in relativity.

I am taking this problem to physics community from long time for example.

http://www.sciencech...php?f=2&t=30181

http://www.sciencech...ae9c47a&start=0

& even on this forum

& people accept this problem of relativity but there is no solution from them.

2) Problem is that this additional forces(which is not applied from out side) are get added to forces in X-direction & Y-direction automatically.

and remember forces do work, consume energy. This create problem of additional energy in relativity.

OK, now I see from the links that you  are just one of these cranks who thinks they have proved relativity wrong. But in reality, the problem is you refuse to use four force notation and that is why nobody can bothered to wade through your tedious maths to find your mistake. I'm certainly not going to waste my time trying.

Edited by exchemist, 13 September 2017 - 08:41 AM.

### #5 maheshkhati

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Posted 15 September 2017 - 05:58 AM

I have not use any new mathematics. Relativity knows that direction of acceleration & force are not same in relativity but does not take it seriously. This is wrong. This is serious matter.  I am giving you a wiki link

https://en.wikisourc...nd_Acceleration

I am using same math of relativity in standard text book but goes further & prove that this create new situation where applied force is different than acting force.

I can give you additional mathematics of standard relativity which proves same thing.

This is not very complicated mathematics but simple differentiation.

### #6 maheshkhati

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Posted 15 September 2017 - 06:07 AM

I am giving you simple example.

I am just moving toward falling Apple of mass mo then also

Fx= y3 mo. (ux/c2} uy g

This force will act on Apple horizontally where ay = g

Math is simple.

I have to apply this force horizontally automatically ( Even physically it is not) because in SR force & acceleration is not require to be move in same direction. This automatic generated mathematical force is problem.

Edited by maheshkhati, 15 September 2017 - 06:38 AM.

### #7 maheshkhati

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Posted 15 September 2017 - 06:30 AM

Now, I am giving frame transformation mathematics which proves that I am true. Here, reverse is happening, there is acceleration in x-direction without force in X-direction.

CALCULATION  2:- Relativity is so wrong. It can be proves that there can be acceleration without force again.

Example is given below

In prime frame,  if Fz =0 & ratio Fx/Fy  is equal to ( v/c2 . Uy)/(1-V .Ux/c2) then after transformation in S’ frame  F’x becomes F’x = 0 because

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

In frame S :- Now, just consider that on magnetic substance on frictionless platform magnetic forces are acting in X-direction & in Y-direction. Magnetic force Fx is so adjusted by software program that ratio Fx/Fy is always equal to

( v/c2 . Uy)/(1-V .Ux/c2).

Then, Forces Fx (very small) & Fy in this frame will create acceleration ax & ay in direction x &  y.

Observer frame S’ is moving with velocity V with relative to frame S then in frame S’ :-

There is acceleration in X’ direction because ax’= ax/{r3. (1-ux. v/c2) 3 }

where r =1/(1-v2/c2) 0.5 but there is no force in X’- direction because

as  F’x = Fx – ( v/c2 . Fy. Uy)/(1-V .Ux/c2)    &  as

Fx/Fy=( v/c2 . Uy)/(1-V .Ux/c2)

So,       F’x =0

Means, in this case in frame S’ there is acceleration in X’-direction but no force is present in X’-direction.

Means, some ghost force will accelerate substance in direction X’ in frame S’.  Can you call this as physics?

### #8 exchemist

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Posted 15 September 2017 - 07:03 AM

Like I said, just another anti-relativity crank. I see the typeface is already getting larger, the fonts more varied, and colours are creeping in. All symptoms of crankhood.  Oh well you are in the right section at least. Have fun.

Edited by exchemist, 15 September 2017 - 08:24 AM.

### #9 maheshkhati

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Posted 16 September 2017 - 05:57 AM

I like this but there is difference. I am opposing with mathematics of same SR.

In SR, force in X-direction is

Fx= rate of change of momentum in X-direction

So,

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

You can check it from any standard special relativity book & other is just simple differential mathematics.

All mathematics given by me in post 1, wiki link in post 5 & transformation equation in post 7 are same interrelated mathematics & can be derived from one another. It is simple.

1) Great Einstein to solve problem of Maxwell wave equation  d2E/dx2= (1/c). d2E/dt2 =0 by considering  that light has constant velocity with all observer & interestingly, it was proved by M-Morley experiment that velocity of light is same in all direction on earth. (This can be explain very well by other way as given in my paper page no 53)

This create completely wrong mathematics & due to that problems of dark matter, dark energy, non symmetry of particle physics, non explanation of duality of light has created. Now, we can not even explain 5 % of matter of world matter correctly.

I was in school my teacher was said that photon is bundle of energy. I just stood up & asked," What is energy?" . Energy is the capacity of doing work it can not be bundle. It is derived term like kinetic energy, potential energy etc.

I know Einstein was great physicist & mathematician but he is not a God.

### #10 maheshkhati

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Posted 21 September 2017 - 09:48 AM

What does the force do?

Answer:-It does the work & increases kinetic energy of substance by increasing momentum of substance in that force direction.

Ball is vertically falling under gravity & observer moving with constant velocity Vx horizontally.

Then , observer see

1)  Momentum of ball :- It increases in horizontal direction as mass of ball increases.

2) Kinetic energy of ball:- Kinetic energy of the ball increases in horizontal direction.

3) Work done :- Kinetic energy only increases when there is work done. Mean's some work done happen in X-direction.

Now, my simple question is Who is doing all this thing in X-direction? Momentum is changing, Kinetic energy is increasing & doing work.

Definitely force

This force can be measure by rate of change of moment in X-direction

i.e. Fx= y3 mo. (ux/c2} uy g

interestingly, this is not applied force but automatically created mathematical force.

Edited by maheshkhati, 22 September 2017 - 08:26 AM.

### #11 maheshkhati

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Posted 22 January 2020 - 06:11 AM

Before two years, I put this mathematics in front of scientific community but no one can solve this problem.

This mathematics is further develop then we will get transformation equation of special relativity.

& using those transformation equation we can get equation E = (1-u2/c2)-0.5  Eo   or M= (1-u2/c2)-0.5  Mo

But If it is proved that in special relativity , acting force is more than applied force,

Whole relativity will collapsed.

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

This is general  equation of force in X-direction in S.R.

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’ & ‘uy’ 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} ) + y3 mo. (ux/c2} uy ay =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.

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 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.

http://vixra.org/abs/1912.0171

Edited by maheshkhati, 22 January 2020 - 06:14 AM.

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