Acceleration without force is possible in relativity

[ravi#]

Interesting
You are right. Thiscan be solution if $u_y$ & $a_y$ are in opposite direction but condition in post 1 is
Fx/Fy is always equal to $\frac { \frac {v}{c^2} . u_y}{(1- \frac {v. u_x}{c^2})}$
here, $F_y$ , $a_y$ & $u_y$ are in same direction.
So, this can not be solution .
Point 1:-When I was working on this problem. I interact with one interesting situation.
It is wrong if I will not share it to this important forum. If I am wrong please correct me.
$F'_x$ = Fx -$\frac { \frac {v}{c^2} .F_y u_y}{(1- \frac {v. u_x}{c^2})}$ is the transformation equation
Means if Fx = 0 then also there is -ve force in non prime frame.
as $F'_x$ = 0 - $\frac { \frac {v}{c^2} .F_y u_y}{(1- \frac {v. u_x}{c^2})}$
so, $F'_x$ = - $\frac{ \frac {v}{c^2} .F_y u_y}{(1- \frac {v. u_x}{c^2})}$
Now, just consider that $u_x =v$ or object velocity is same as frame velocity or for object whose velocity in X- direction is zero for non-prime framethen also there is -ve force in non prime frame because
$F'_x$ = -$\frac{ \frac {v}{c^2} .F_y u_y}{(1- \frac {v^2}{c^2})}$
Mean's for observer in train, for object in his cabin where $u_x =v$ for platform frame if there is forced acceleration in Y-direction then
$F'_x$ = - $\frac{ \frac {v}{c^2} .F_y u_y}{(1- \frac {v^2}{c^2})}$
This -ve force will automatically act on it.
Mean's for every forced acceleration of object in train cabin in Y-direction, there is -ve force formation in X-direction as per above formula due to motion with prime frame.

is this possible ?

 

[A.T.]

You only have to let
$$
a_y = - \frac{c^2 a_x}{u_x u_y}\left(\gamma^{-2} + \frac{u_x^2}{c^2}\right)
$$
and $a_z = 0$.

This can be solution if $u_y$ & $a_y$ are in opposite direction

Or if $u_x$ & $a_x$ are in opposite direction

 

[Ibix]

Your original problem specification is in a frame S. In this frame there is a body moving with 3-velocity \vec{u}=(u_x,u_y,0). It is subject to a force \vec{F}=(vu_yF/(c^2-vu_x),F,0), where F is some constant. This produces a 3-acceleration \vec{a}=(a_x,a_y,0). In general, Orodruin's expression from #30 is not satisfied in this frame because the x-component of the force is not zero.

You then transform to a frame S', which moves with velocity v in the x direction relative to S. In this frame, the body is moving with 3-velocity \vec{u'}=(u'_x,u'_y,0). It is subject to a force \vec{F'}=(0,F'_y,0) - note that the x-component is zero because you have carefully chosen the direction of your force in S so that this happens. This produces a 3-acceleration \vec{a'}=(a'_x,a'_y,0). In this frame I do expect Orodruin's expression (using primed quantities and noting that \gamma is the \gamma associated with u') from #30 to be satisfied.

You appear to have noted that Orodruin's expression is not satisfied in S, but do not appear to have checked it in S'. It is the latter that is of interest to you.

As for the rest of your post #31, if I understand correctly, you seem to be worried that a force in the y-direction is not parallel to y'. Isn't that an obvious consequence of the force transformation equations you quoted?

 

[ravi#]

(1)
Yes, Mr A.T. you are right I also think about this solution
but problem remains un-resolve when $u_x$ becomes more than $V$ in prime frame.
See transformation equation for F’x, there is –ve force created due to action of Fy or Fz perpendicular to X-direction. It is not form due to any acceleration & velocity has opposite direction. Following problem will make it very clear. I think that these are problem happen due to force can have without acceleration in that direction.

(2)
My second problem in post 31:- This is very important problem which I put in my previous post 31.
This problem can easily be understood by following paradox.
IMPORTANT NEW PARADOX:- On friction-less platform object is moving with constant velocity $u_x$ in X-direction & only magnetic force $F_y$ is acting in Y-direction & there is acceleration in Y-direction only.
For non-prime observer moving with velocity V in X-direction :-
For this observer by transformation of forces equation in x-direction
here, F'x =0 - $\frac { \frac {v}{c^2} .F_y u_y}{(1- \frac {v. u_x}{c^2})}$
F'x = - $\frac { \frac {v}{c^2} .F_y u_y}{(1- \frac {v. u_x}{c^2})}$
This force will act in –ve direction on object in non-prime frame. (Even there is no Fx in prime frame).
Mean’s in this non-prime frame, this standing force will act in x-direction.
Let, consider that $u_x=V$
Then also this force will not get cancel &
F'x = - $\frac { \frac {v}{c^2} .F_y u_y}{(1- \frac {v^2}{c^2})}$ in X'-direction--------(1)
--------Now, this create problem because as $u_x=V$
Here, object velocity matches with observer velocity
Or There is no relative motion of object in this frame in X’-direction
but there is force F’x present in this non-prime frame by equation (1) of transformation.

--------But we know that in any frame if there is no motion in x-direction then
$F_x = 0$ because differentiation of $\gamma m_0 u_x$ gives
F_x=\gamma m_0 a_x+\gamma^3 m_0 \frac { u_x }{c^2} (u o a) in any frame
& if acceleration & velocity is zero in X-direction
Then $F_x = 0$ -----------(2)
Now case(1) & (2) are completely opposite.
Transformation equation says that there can be force even there is no relative motion in X-direction but above equation says that this is impossible.
Interesting but opposite situation.
Now, somebody will say this situation is not possible in reality but I can give example...

 

[Ibix]

Your problem statement appears rather confused to me. Let me restate it to be clear that I understand.

You have a frame S in which there is a body moving with velocity $(u_x,u_y,0)$. It is subject to a force $(0,F_y,0)$.

You then transform to a frame S' where the object moves with velocity $(0,u'_y,0)$ and is subject to a force $(F'_x,F'_y,0)$.

Your claim is that $F'_x \neq 0$ when derived from the force transformation, yet $dP'_x/dt'=0$. The solution is simple - the second statement is false. In the case that $u'_x=0$, $dP'_x/dt \propto a'_x \propto a_x$, and $a_x$ is not zero, as you can see easily from $0=F_x=dP_x/dt$.

 

[PeterDonis]

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