Answer to Question #154264 in Mechanics | Relativity for VINIT MEENA

Question

Two rockets A and B depart from earth at a steady speed of 0.6c in opposite directions, having synchronized clocks with each other and with earth at departure. After one year as measured in Earth’s frame, rocket B emits a light signal. At what time, in the frames of earth and rockets A and B, does rocket A receive the signal? Work out the problem in all three frames and verify the answers are same. Use unit of distance as light year and that of time as year.

Answer 1

Solution:

In this problem we have 3 inertial frames to consider, so we will use coordinates x and t to denote quantities measured in the Earth`s frame, $x_A$, and $t_A$ to denote quantities measured in A`s frame, and $x_B$ and $t_B$ to denote quantities measured in B`s frame. We will assume that, according to an observer in the Earth`s frame, rocket A is travelling at speed v in the positive x-direction and rocket B in the negative x-direction. With these assumptions, we can write down the Lorentz transformations:

$x=\gamma(x_A+vt_A)$ , $x_A=\gamma(x-vt)$

$t=\gamma(t_A+\dfrac{vx_A}{c^2}),$ $t_A=\gamma(t+\dfrac{vx}{c^2}),$

$x=\gamma(x_B-vt_B),$ $x_B=\gamma(x+vt),$

$t=\gamma(t_B-\dfrac{vx_B}{c^2}), \; t_B=\gamma(t+\dfrac{vx}{c^2}),$

where

$v=0.6c$

and

$\gamma=(1-\dfrac{v^2}{c^2})^{-\frac{1}{2}}=\dfrac{5}{4}$

It will be convenient to use a system of units in which time is measured in years and distance in light-years, in which case c has a value of 1. Again, it is helpful to draw a space-time diagram in the Earth`s frame of reference, as shown in the following picture:

so at time $t\gt1$ the x-coordinate of the light signal is

$$x = -0.6+(t-1)=t-1.6$$

At time t, the x-coordinate of rocket A is

$$x = +0.6t$$

The x-coordinates must be equal at event 2, so for this event, we must have

$$t-1.6=0.6t$$

hence t=4. Thus according to an observer in the Earth frame, event 2 occurs after 4 years. The x-coordinate of this event us therefore

$$0.6\cdot 4=2.4$$

We can now substitute these coordinates into the Lorentz transformation formulae to find the time coordinates of event 2 in A`s and B`s frame of reference:

$$t_A=\gamma(t-vx)$$

(remember that c=1)

$$=\dfrac{5}{4}(4-0.6\cdot2.4)=3.2$$

so that, according to A, event 2 occurs after 3.2 years.

$$t_B=\gamma(t+vx)=\dfrac{5}{4}(4+0.6\cdot2.4)=6.8$$

so according to B, event 2 occurs after 6.8 years.