Question

Relativity problems?

Two trains (A and B), each of proper length 1 km run on parallel tracks. Train A has a velocity of 0.6 c while train B has a velocity of 0.8 c relative to the ground. How long does it take the faster train to fully pass the slower one (from the time when the front of B coincides with the rear of A to the time when the rear of B coincides with the front of A)? The answer is
(a) __according to observers in the ground frame;
(b) __according to observers in the frame of the slower train.

Answer 1

(a) __according to observers in the ground frame;

`Delta t_G = 7000/c " s"`

(b) __according to observers in the frame of the slower train.

`Delta t_A = 5000/c " s"`

Explanation:

(a) __according to observers in the ground frame;

A and B both have proper lengths `L_p = 1 "km"`, but in ground frame G, they are length contracted (`L_(A,B) = L_(p : A,B)/gamma_(A,B)`) ... and by different amounts as they are travelling at different velocities.

In ground frame G, the velocity of B rel to A is `0.2c`, and in order to transition from:

• (a) the front of B coinciding with the rear of A to

• (b) the rear of B coinciding with the front of A,

....B has to travel a distance `L_A + L_B` more than A. Draw it and see.

The time required for that is:

`Delta t_G = (L_A + L_P)/(0.8c - 0.6 c)`

`= 1000(sqrt(1-0.6^2) + sqrt(1-0.8^2))/(0.2c) = 7000/c " s"`

(b) __according to observers in the frame of the slower train.

Use inverse Lorentz transform between G and A:

`Delta t_A = gamma_A(Delta t_G - (v_A Delta x_G)/c^2)`

Everything is known apart from `Delta x_G`, which is either of:

• (a) the total distance travelled by B in `Delta t_G` less B's contracted length, or

• (b) the total distance travelled by A in `Delta t_G` plus A's contracted length.

So:

`Deltax_G = v_B Delta t_G - L_B = 0.8c 7000/c - 600 = 5 " km"`

Or:

`Deltax_G = v_A Delta t_G + L_A = 0.6c 7000/c + 800 = 5 " km"`

Hence:

`Delta t_A = 1/sqrt(1 - 0.6^2)(7000/c - (0.6c * 5000)/c^2) = 5000/c " s"`