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# Force without acceleration in S.R. , acceleration without force in S.R. & applied force is less than acting force in S.R.

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Force without acceleration in S.R.

& acceleration without force in S.R.

& applied force is less than acting force in S.R.

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

{Before starting this paradox, I want to put some relativity formulae’s

In any frame for force in X-direction

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

After differentiation, we get

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

Now, 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 & Fz=0

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

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

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 fulfil 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.

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

Now, I am generalising 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.

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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 magnetic 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 (& force) does comes from?

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uy = 0, making Fx = 0

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but in above paradox, I have not consider uy=0.

(& same calculation is true in Y-direction for same conditions.)

Edited by mahesh khati
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but in above paradox, I have not consider uy=0.

Yes, you did. "object is moving with constant velocity ux"

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Ok, I again re-write the paradox with above improvement

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

{Before starting this paradox, I want to put some relativity formula's

In any frame for force in X-direction

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

After differentiation, we get

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

Now, 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 & uy is not zero & Fz=0

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

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

Or Fx=Fay as this force is form due to ‘ay (& uy)’

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.

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

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 magnetic 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 (& force) does comes from?

Edited by mahesh khati
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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 & uy is not zero & Fz=0

How do you achieve a magnetic force in the y direction only?

Anyway, I suspect your mistake is trying to break down the force into components before taking the derivative.

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I can give detail calculation:-

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

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

So, after differentiation

Fx= y. mo. ax + y3. mo. {ux/c2}. (u . a) -----(1)

We know, u2=ux2+uy2+uz2

So, u. a = ux ax +uy ay + uz az --------(2)

from (1) & (2)-------- (you can directly differentiate without this two step also)

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

So, calculation given in post is true.

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Point 1:-

Problem which I raised from above paradox is very serious.

For example,

One light object is falling down under gravity in vertical Y-direction & horizontal air is pushing that object in horizontal direction.

For observer on ground

Let, fx is horizontal force applied by air on object & fy is vertical force applied by gravity on object.

Then above calculation says that

Actual force acting on objects are not fx & fy but

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

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

Means, actual force acting on objects are different (more) than force applied on the object.

This is very serious output & can create very complicated problem in relativity. (if not solve).

Edited by mahesh khati
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I suspect your mistake is trying to break down the force into components before taking the derivative.

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HERE, more energy & force is the problem.

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

Clearly, you have made an error somewhere. Are you asking for help in finding it?

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This is very simple derivative but if you find something wrong in it.

Please,inform me, I will be very much thankful to you.

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

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) -----(1)

We know, u2=ux2+uy2+uz2

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

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

from (1) & (2)-------- (you can directly differentiate without this two step also)

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

So, calculation given in post is true (please find out if anything wrong in it.)

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So, calculation given in post is true

Obviously not because you get the wrong answer.

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please, pointed out where I am wrong.

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please, pointed out where I am wrong.

Why?

Whenever people explain your mistakes you either ignore them or insist that you are right and the rest of the world is wrong (as in this thread). It is a waste of time trying to help you.

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please, pointed out where I am wrong.

I have done this twice. Here's the third time: I suspect your mistake is trying to break down the force into components before taking the derivative.

IOW, find the net force by taking the derivative of the momentum, and then break it down into components.

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The relativistic expression relating force and acceleration for a particle with constant non-zero rest mass moving in the direction is:

From wiki

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Above are not the general formula for force in X-directions but conditional, Now, I am giving general formula's

these are the general formula in S.R. for forces in X, Y & Z-direction given in web site https://en.wikisource.org/wiki/The_Direction_of_Force_and_Acceleration

&

in book element of special relativity (Academic book) by Dr T M Karade, Dr K.S. Adhav & Dr Maya S. Bendre page no 135 gives the same formulas

So,

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

Fx = d/dt( y. mo. ux) where y=(1-u2/c2)-0.5 is true & then simple differentiation

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) -----(1)

We know, u2=ux2+uy2+uz2

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

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

from (1) & (2)-------- (you can directly differentiate without this two step also)

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

So, calculation given in post 1 is true.

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

Mr Ematfaal, above general equation can easily converted in to your conditional equation by just two or three steps.

Edited by mahesh khati
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You posted the solutions page,

is there something you don't understand by this statement?

"Hence in order to accelerate a body in a given direction, we may apply any force in the desired direction, but must at the same time apply at right angles another force whose magnitude is given by equation. 5.

Edited by Mordred
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I like this post Mr Mordred & also think about this solution previously but I have problem. In nature we not apply Fx but we get Fx

for example.

On platform old man applied force perpendicular to velocity of train on cart & it accelerate in Y-direction only (acceleration & force has same direction on platform)

let, train velocity is -ux

Then for observer on train :-

then cart will moves with constant velocity 'ux' in x-direction, it accelerate with some acceleration 'ay' in Y-direction only with velocity 'uy'.

Now, if I apply above equation for this frame

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

then, output will be

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

here, I have not applied any force in X-direction but constant velocity in X-direction creates this force.

As velocity of train is more this force is more & when it is zero, force is zero.

Edited by mahesh khati
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I was smelling this problem from last 3 years because in S.R. force (in X-direction) is not depend on change of state of motion (acceleration) in that direction but depends on change of momentum. If some ball is falling & if I am moving toward it horizontally then also

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

this force will act on the ball in horizontal direction.

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

this problem can be complicated as we want

for example man is pulling cart in any direction on platform with force f let, fx & fy are there components in X & Y direction

Then calculations given in post 1 says that actual forces acting on cart, in x & y directions are

Fx = fx +y3 mo. (ux/c2} uy ay & Fy = fy + y3 mo. (uy/c2} ux ax

& this can be complicated further

Now, I have to stop.

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• 2 weeks later...

After nearly one month, there is no solution to above problem in S.R. It is really difficult because I have not change any calculation in S.R. Only proves that there are two forces in S.R.

One is applied & other is actual acting force on object & that force is more than applied force.

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After nearly one month, there is no solution to above problem in S.R.

More accurately, after nearly one month you still haven't understood your error.

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After nearly one month, there is no solution to above problem in S.R. It is really difficult because I have not change any calculation in S.R. Only proves that there are two forces in S.R.

One is applied & other is actual acting force on object & that force is more than applied force.

!

Moderator Note

This is completely unacceptable. You were shown where your calculations may be wrong, and have done nothing about them, yet you insist you're right and no one has offered a solution to your problem.

We can't help fight ignorance when you purposely ignore learning anything. Thread closed due to insufficient rigor.

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