In prime frame, if Fz =0 & ratio Fx/Fy is equal to ( v/c(adsbygoogle = window.adsbygoogle || []).push({}); ^{2}. Uy)/(1-V .Ux/c^{2}) then after transformation in S’ frame F’x becomes F’x = 0 because

F’x = Fx – ( v/c^{2}. Fy. Uy)/(1-V .Ux/c^{2}) ----transformation equation

Let, consider one situation

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/c^{2}. Uy)/(1-V .Ux/c^{2}).

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/{r^{3}. (1-ux. v/c^{2})^{3}} where r =1/(1-v^{2}/c^{2})^{0.5}but there is no force in X'- direction because

as F’x = Fx – ( v/c^{2}. Fy. Uy)/(1-V .Ux/c^{2}) & as Fx/Fy=( v/c^{2}. Uy)/(1-V .Ux/c^{2})

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.

In relativity, is this possible that in some frame, there is acceleration but no force in X-direction.

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# I Acceleration without force is possible in relativity

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