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-u^{2}/c^{2})^{-0.5}

So, after differentiation

Fx= y. mo. (dux/dt) + y^{3}. mo. {ux/c^{2}}. (u . du/dt)

Fx= y. mo. ax + y^{3}. mo. {ux/c^{2}}. (u . a) -----(A)

We know, u^{2}=ux^{2}+uy^{2}+uz^{2}

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+y^{3} mo. (ux/c^{2}} (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= y^{3} mo. (ux/c^{2}} 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= y^{3} mo. (ux/c^{2}} uy ay+0 or Fay+0=Fay

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**STEP 2:-Now, Force acting in X-direction is ** Fx= y^{3} mo. (ux/c^{2}} uy ay or Fay

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

** -fx = -**y^{3} mo. (ux/c^{2}} 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+y^{3} mo. (ux/c^{2}} (ux ax+uy ay)

Or 0 =y. mo ax. (1+ y^{2} {ux^{2}/c^{2}} ) +Fay

(Here as Fay= y^{3} mo. (ux/c^{2}} uy ay)

Mean’s Fay = y. mo. -ax. (1+ y^{2}. {ux^{2}/c^{2}} )

**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= y^{3} mo. (ux/c^{2}} 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+ y^{3} mo. (ux/c^{2}} uy ay = fx +Fmay

& Fy=fy+ y^{3} mo. (uy/c^{2}} ux ax = fy +Fmax

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

F^{2}= Fx^{2}+Fy^{2}

F=(fx^{2}+fy^{2}+Fmax^{2}+Fmay^{2}+2 .fx. Fmay + 2 .fy. Fmax)^{0.5}

F=(f^{2}+Fma^{2} +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__

__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)*^{-0.5 }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.

** ***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**^{2 }) . 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**^{2 }) . Fy + dy . (Fy/ y** )**

**d’w = Fy . {**y dt .** (Uy.v**^{2}/c^{2 }) + dy / y** }**

**d’w = Fy .dy . {**y**.v**^{2}/c^{2 } + 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/c__^{2}= y . dEo __/c__^{2 }is wrong.

__So, dM=__ y . dMo is wrong.

**Edited by maheshkhati, 25 June 2020 - 03:38 AM.**