The frame `K^'` moves with a constant velocity V relative to the frame K. Find the acceleration `w^'` of a particle in the frame `K^'`, if in the frame K this particle moves with a velocity `v` and acceleration w along a straight line
(a) in the direction of the vector V,
(b) perpendicular to the vector V.

Explanation

`--underset(vecv)overset(t)-----underset(vecv+vecwdt)overset(l+dt)---(K)`
In K the velocities at time t and `t+dt` are respectively v and `v+wdt` along x-axis which is parallel to the vector `vecV`. In the frame `K^'` moving with velocity `vecV` with respect to `K`, the velocities are respectively,
`(v-V)/(1-(vV)/(c^2))` and `(v+wdt-V)/(1-(v+wdt)(V)/(c^2))`
The latter velocity is written as
`(v-V)/(1-v(V)/(c^2))+(wdt)/(1-v(V)/(c^2))+(v-V)/((1-(vV)/(c^2)))(wV)/(c^2)dt=(v-V)/(1-v(V)/(c^2))+(wdt(1-V^2/c^2))/((1-(vV)/(c^2))^2)`
Also by Lorentz transformation
`dt^'=(dt-Vdx//c^2)/(sqrt(1-V^2//c^2))=dt(1-vV//c^2)/(sqrt(1-V^2//c^2))`
Thus the acceleration in the `K^'` frame is
`w^'=(dv^')/(dt^')=(w)/((1-(vV)/(c^2))^3)(1-V^2/c^2)^(3//2)`
(b) In the K frame the velocities of the particle at the time t and `t+dt` are respectively
`(0,v,0)` and `(0,v+wdt,0)`
where `vecV` is along x-axis. In the `K^'` frame the velocities are
`(-V, vsqrt(1-V^2//c^2),0)`
and `(-V,(v+wdt)sqrt(1-V^2//c^2),0)` respectively
Thus the acceleration
`w^'=(wdtsqrt((1-V^2//c^2)))/(dt^')=w(1-V^2/c^2)` along the y-axis.
We have used `dt^'=(dt)/(sqrt(1-V^2//c^2))`