Identical or not?

Suppose, Supantho and Mohaimen are moving to each other at a relative velocity of 3 c 5 \frac{3c}{5} .

Now, consider the situation in Supantho's reference frame (first figure): The distance between Supantho and Mohaimen is 1.44 × 1 0 12 m 1.44 \times 10^{12} \text{ m} . Both Supantho and Mohaimen's clock show 4:00.

Again, consider the situation according to Mohaimen's reference frame (second figure): Supantho's clock shows 4:00. What will be the distance between them and what does Mohaimen's clock show?

Details and Assumptions

The speed of light is c = 3 × 1 0 8 ms 1 c = 3 \times 10^8 \text{ ms}^{-1} .

1.152 × 1 0 12 m , 5 : 00 1.152 \times 10^{12}\text { m},\ 5:00 1.44 × 1 0 12 m , 3 : 00 1.44 \times 10^{12}\text { m},\ 3:00 1.44 × 1 0 12 m , 4 : 00 1.44 \times 10^{12}\text { m},\ 4:00 1.8 × 1 0 12 m , 4 : 00 1.8 \times 10^{12} \text { m},\ 4:00 1.8 × 1 0 12 m , 3 : 00 1.8 \times 10^{12} \text { m},\ 3:00

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Arpon Paul
Oct 19, 2015

Let's format the problem in a structural way. Look, in this problem, there are 3 events in the spacetime: 1. Supantho in his rocket at 4:00 according to his clock(Fig. 1) 2. Mohaimen in his rocket at 4:00 according to his clock(Fig. 1) 3. Mohaimen in his rocket at time '?' according to his clock(Fig. 2) These 3 events are denoted by P 1 , P 2 , P 3 P_1, P_2, P_3 respectively in fig. 3: The primed coordinates are of Mohaimen's reference frame, and the unprimed ones of Supantho's (I mean a referance frame which is staionary w.r.t Supantho). I took P 2 P_2 on the origin of both reference frames, so, x 2 = x 2 = x 3 = 0 x_2 = x_2' = x_3'= 0 t 2 = t 2 = t 1 = 0 t_2' = t_2 =t_1 =0 t 3 = t 1 t_3' = t_1' x 1 = 1.44 × 1 0 12 m x_1 = 1.44 \times 10^{12}m We have to calculate x 1 x_1' & t 3 t_3' ; Using Lorentz's transformation, x 1 = x 1 v t 1 1 v 2 c 2 = 1.8 × 1 0 12 m x_1' = \frac{x_1 - vt_1}{\sqrt{1-\frac{v^2}{c^2}}} = 1.8 \times 10^{12} m t 3 = t 1 = t 1 v x 1 c 2 1 v 2 c 2 = 3600 s = 1 h r t_3' = t_1' = \frac{t_1 - \frac{vx_1}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}} = -3600s = -1hr So the time would be 1 hour less than 4:00, that is 3:00.

did the same!

jafar badour - 5 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...