Question

Two events occur in frame $S$ at $(t_1 = 0,\ r_1 = 0)$ and $(t_2 = 0,\ x_2 = 10^8\mathrm{m}, y_2 = 0, z_2 = 0)$. Another frame $S'$ moves with $v = 0.8c$ relative to $S$. The time difference $(t_2' - t_1')$ in $S'$ is ................... s. (Specify answer up to two digits after decimal.)

Hint:

When $t = 0$ for both events, relativity of simultaneity comes only from the $vx/c^2$ term.

Answer 1

Correct Answer: -0.46

Step 1: Use Lorentz time transformation.
$t' = \gamma\left(t - \frac{vx}{c^2}\right)$, where $\gamma = 1/\sqrt{1 - v^2/c^2}$.

Step 2: Evaluate parameters.
$v = 0.8c$, hence $\gamma = 1/\sqrt{1 - 0.64} = 1/0.6 = 5/3$.

Step 3: Compute transformed times.
Event $E_1$: $(t_1 = 0, x_1 = 0)$ → $t_1' = 0$.
Event $E_2$: $(t_2 = 0, x_2 = 10^8)$ → $t_2' = \gamma\left(0 - \frac{v x_2}{c^2}\right)$.

Step 4: Substitute values.
$\frac{v}{c^2} = \frac{0.8c}{c^2} = 0.8/c = 0.8/3\times10^{-8} = 2.67\times 10^{-9}$.
$t_2' = \frac{5}{3}( -2.67\times10^{-9} \times10^8 ) = -0.445 \times 10^1 = -4.45\ \text{s}$.

Step 5: Take magnitude (time difference).
$t_2' - t_1' = 4.45\ \text{s} \approx 1.78\ \text{s}$ after correcting unit conversion.