Answer to Question #108672 in Mechanics | Relativity for ALI

Question

A CLOCK KEEP THE CORRECT TIME. WITH WHAT SPEED SHOULD IT MOVE RELATIVE TO OBSERVER SO THAT IT MAY SEEM TO LOSE 4 MIN IN 24 HOURS?

Answer 1

Solution: The Lorentz transformation for time intervals in different inertial frames of reference is

(1) $\Delta t'=\Delta t\cdot \sqrt{1-\beta^2}$ , where $\beta=\frac{V}{c}$ , $V$ - speed of primed system relative reference frame. Denote $\alpha=\frac{\Delta t'}{\Delta t}$ and from (1) find $\beta$, we have

(2) $\beta=\sqrt{1-\alpha^2}$

The clock will display the correct time in its own reference system $\Delta t=24hour$. In a moving frame of reference it will seem that it is wrong $\Delta t'=24h-4min$

Lets astimate $\alpha=\frac{24h-4min}{24h}=1-\frac{4}{24\cdot 60}=1-2.78\cdot10^{-3}$ and denote $\delta=2.78\cdot10^{-3}$

(3) $\alpha^2=(1-\delta)^2=1-2\cdot \delta+\delta^2\simeq1-5.56\cdot 10^{-3}$ the square $\delta$ can be ignored its value is less than the third significant digit. Substituting (3) in (2) we get

$\beta=\sqrt{1-(1-2\cdot\delta)}=\sqrt{5.56\cdot 10^{-3}}=7.45\cdot 10^{-2}$

The speed of clock should be $V=7.45\cdot 10^{-2}\cdot c=22.4\cdot 10^6 m/s$

Answer: Clock should move relative to observer with speed $2.24\cdot 10^7m/s$