# 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.

Apr 12, 2018

(a) __according to observers in the ground frame;

$\Delta {t}_{G} = \frac{7000}{c} \text{ s}$

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

$\Delta {t}_{A} = \frac{5000}{c} \text{ s}$

#### Explanation:

(a) __according to observers in the ground frame;

A and B both have proper lengths ${L}_{p} = 1 \text{km}$, 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 $0.2 c$, 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:

$\Delta {t}_{G} = \frac{{L}_{A} + {L}_{P}}{0.8 c - 0.6 c}$

$= 1000 \frac{\sqrt{1 - {0.6}^{2}} + \sqrt{1 - {0.8}^{2}}}{0.2 c} = \frac{7000}{c} \text{ s}$

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

Use inverse Lorentz transform between G and A:

$\Delta {t}_{A} = {\gamma}_{A} \left(\Delta {t}_{G} - \frac{{v}_{A} \Delta {x}_{G}}{c} ^ 2\right)$

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:

$\Delta {x}_{G} = {v}_{B} \Delta {t}_{G} - {L}_{B} = 0.8 c \frac{7000}{c} - 600 = 5 \text{ km}$

Or:

$\Delta {x}_{G} = {v}_{A} \Delta {t}_{G} + {L}_{A} = 0.6 c \frac{7000}{c} + 800 = 5 \text{ km}$

Hence:

$\Delta {t}_{A} = \frac{1}{\sqrt{1 - {0.6}^{2}}} \left(\frac{7000}{c} - \frac{0.6 c \cdot 5000}{c} ^ 2\right) = \frac{5000}{c} \text{ s}$