Question

plot a, b and c, describe motions of a and c from b's perspective then estimate time measured by observer b when c meets b ln graph

BCC AaBbCD Aabbcc AaBbCD AalbCD AaBbCD ABCD AABBCCDC AABBCODE Aabba Subtitle Subtle Em... Emphasis intense E. Strong Quote intense (.. Subtle Ref. Intense Re. Book Tit Styles Observer A: On Earth Observer B: moving toward Star X, 1 light-years away from Earth, with the velocity, 0.50 Observer C: moving toward Earth from the position, 2 light-years away from Earth, with the velocity -0.5c. (Ctrl) (6) The graph below is the B's frame of referenceſt x'). Please do the following tasks. (6 pts) 1) On the graph below, plot the motions of A, B, and C. 2) Describe the motions of A and C from B's perspective. 3) Estimate time measured by observer B when C meets B on the graph (the unit of the x-axis is a light-year, the unit of the time-axis is a year.) We call this time, Ts (You can get more accurate time from the simulation) t(year) 2 x'tight-year)

Answer 1

1)

The motion of A in the space-time diagram is represented by the line marked A.

It is basically the world line. From the perspective of the earth, it is always stationary. Its position is 0 at all points in time, so it is along the t axis.

As for B, it starts from earth at time 0 (t,x)=0,0 and at time t=2 years it reaches star X (). You can see that the position of B remains x'=0 in the transformed coordinates for all t'.

= 2 x 0.50 = 1 lightyear

C's motion is represented by the line marked C. It starts from 2 lightyears. Its velocity is -0.5c hence the slope is negative. It takes 4 years to travel to earth.

= 4x(-0.5C) = -2 lightyears

2)

Motion of A wrt B:

With respect to the spaceship B, the earth is moving away from it with the velocity:

Velocity of earth (A) with respect to B .

VA = -0.50

You can see this in the ST diagram as, with respect to the (t',x') frame, the line A has a negative slope, which is exactly opposite to the velocity of B

Motion of C wrt B:

C is moving towards earth with a velocity of -0.5c and B is moving with a velocity of 0.5c. Therefore, the velocity with which it approaches B is the relative velocity between them. the relative velocity of B and C is (Using Einstein's velocity addition rule):

u +v= u+u 1+ 4y

-0.50 - 0.50 -C Vc = 1 + 0.250. 2 1 +0.25 = -0.8c

The velocity is negative since C is moving opposite to B.

2)

The two events in (t,x)

Launch of B:(0,0):E1(t1,x1)

Meeting of B and C:(2,1) E2(t2,x2)

We will apply the Lorentz transform to find the Events in (t',x') frame:

A= ( A1-877 79 -89

(%) = (****) () = 0

( 7 (-8 — 8y) 7 ) (27 — By -28y+y)

Time difference in the frame of B

At = t, - t = 27 - B9 - 0

Using v=0.5c

7= VI-0.26 = 1.155

$\beta=\frac{v}{c}=0.5$

At = 2 x 1.155 -0.5 x 1.155 = 1.7325 years