# Is the applied force less than the acting force in Special Relativity?

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#### Mahesh Khati

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##### Is the applied force less than the acting force in Special Relativity?
« on: 16/05/2016 11:24:22 »
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
Fx = $$\frac { d }{d_t} ( \gamma m_0 u_x)$$
After differentiation, we get
So, $$F_x=\gamma m_0 a_x+\gamma^3 m_0 \frac { u_x }{c^2} (u_x a_x+u_y a_y+u_z a_z)$$ ------(1)}
Now, Paradox:-
On frictionless platform, object is moving with constant velocity $$u_x$$ in X-direction & only magnetic force is acting in Y-direction & there is acceleration in Y-direction only with velocity $$u_y$$
$$(&F_z=0)$$
If we apply eq(1) to this case then result will be
$$F_x=\gamma^3 m_0 \frac { u_x }{c^2} u_y a_y$$
$$F_x=F_a_y$$ as this force is form due to $$a_y$$ only
Mean’s even there is no magnetic force acting on object from outside in x-direction & no $$a_x$$ then also above force will act on object in +ve direction of x-axis due to $$a_y$$
Important point (1):-
Mean’s applied magnetic force on object in X-direction is 0 & acting force in X-direction is
$$F_x=\gamma^3 m_0 \frac { u_x }{c^2} u_y a_y +0$$ or $$F_a_y+0=F_a_y$$
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STEP 2:-Now, Force acting in X-direction is
$$F_x=\gamma^3 m_0 \frac { u_x }{c^2} u_y a_y or F_a_y$$
Now, after this happen, very small magnetic force of same intensity
-fx =-  $$\gamma^3 m_0 \frac { u_x }{c^2} u_y a_y$$ or $$-F_a_y$$ start acting on object in direction opposite to above force (but velocity is still positive $$u_x$$) & cancel that above force.
Mean’s equation (1) becomes
$$0 =\gamma m_0 a_x+\gamma^3 m_0 \frac { u_x }{c^2} (u_x a_x+u_y a_y)$$
Or  $$0 =\gamma m_0 a_x(1+\gamma^2 \frac { u_x^2 }{c^2}) +F_a_y$$
(Here as  $$F_a_y=\gamma^3 m_0 \frac { u_x }{c^2} u_y a_y$$)

Mean’s $$F_a_y$$  =$$\gamma m_0 (-a_x)(1+\gamma^2 \frac { u_x^2 }{c^2})$$
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
$$0= -fx + F_a_y$$
Important point (2):- Mean’s applied magnetic force on object in X-direction is -fx & acting force in X-direction is $$-fx + F_a_y=0$$ 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 $$F_a_y$$ or more)
Now, I am generalising above result.
This 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+$$F_a_y$$)
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+$$F_a_x$$)
This is completely complicated results, which says that applied force & acting forces on objects are different 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 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 does this additional energy (& force) comes from?

This clearly shows that something is seriously wrong in Special theory of relativity.
« Last Edit: 17/05/2016 07:14:40 by chris »