Question #154264

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.


1
Expert's answer
2021-01-07T10:54:59-0500

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, xAx_A, and tAt_A to denote quantities measured in A`s frame, and xBx_B and tBt_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=γ(xA+vtA)x=\gamma(x_A+vt_A) , xA=γ(xvt)x_A=\gamma(x-vt)

t=γ(tA+vxAc2),t=\gamma(t_A+\dfrac{vx_A}{c^2}), tA=γ(t+vxc2),t_A=\gamma(t+\dfrac{vx}{c^2}),

x=γ(xBvtB),x=\gamma(x_B-vt_B), xB=γ(x+vt),x_B=\gamma(x+vt),

t=γ(tBvxBc2),  tB=γ(t+vxc2),t=\gamma(t_B-\dfrac{vx_B}{c^2}), \; t_B=\gamma(t+\dfrac{vx}{c^2}),


where

v=0.6cv=0.6c


and

γ=(1v2c2)12=54\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>1t\gt1 the x-coordinate of the light signal is



x=0.6+(t1)=t1.6x = -0.6+(t-1)=t-1.6

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



x=+0.6tx = +0.6t


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



t1.6=0.6tt-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



0.64=2.40.6\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:



tA=γ(tvx)t_A=\gamma(t-vx)


(remember that c=1)


=54(40.62.4)=3.2=\dfrac{5}{4}(4-0.6\cdot2.4)=3.2


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



tB=γ(t+vx)=54(4+0.62.4)=6.8t_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.


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