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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. |
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Answer» SOLUTION :`--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))` |
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