The reference frame $K^{\prime }$ moves in the positive direction of the x axis of the frame $K$ with a relative velocity $V$. Suppose that at the moment when the origins of coordinates $O$ and $O^{\prime }$ coincide, the clock readings at these points are equal to zero in both frames. Find the displacement velocity $\dot{x}$ of the point (in the frame $K$) at which the readings of the clocks of both reference frames will be permanently identical. Demonstrate that $x<V$.

Given: $v$ is relative velocity of the frame $K^{\prime }$ w:r.t $K$

& at t=0 $K$ & $K^{\prime }$ are at origin.

To find :displacement velocity.

i.e. $\frac{dx}{dt}$

Concept: We'll be using Lorentz Transformation for this problem which is.

$t^{\prime }=\frac{t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$

Here, $(x,t)$ & $(x^{\prime },t^{\prime })$ are values of two inertial frames where x is displacement and $t$ is time.

Solution:

Let us imagine that both $K^{\prime }$ & $K$ are at origin at t=0

at t=0

after moving for a distance x or displacement x.

$K$ will have time $t=t_1$

& $K^{\prime }$ will take time $t=t_2$

Now according to Lorentz Transformation:

$t_2=\frac{t_1-\frac{xv}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$

and according to question we've been asked to find that $\frac{dx}{dt}$ for $t_2$=$t_1$=$t$(Let)

So,

$t=\frac{t-\frac{xv}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$

Lets differentiate the above equation w.r.t $t$

$\frac{dt}{dt}=\frac{\frac{dt}{dt}-\frac{dx}{dt}\frac{v}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$

which gives us

$1=\frac{1-\frac{dx}{dt}\frac{v}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$

by rearranging the above equation we get

$\frac{dx}{dt}=\frac{c^2}{v}(1-\sqrt{1-\frac{v^2}{c^2}})$