3rd Feb 2021

# (a)

Let event 1 be lamp $$A$$ turning on and event 2 for lamp $$B$$. Let frame $$F'$$ be the moving along the rod, at velocity $$V$$ away from frame $$F$$ that is stationary with respect to $$O$$.

• To see that lamp $$A$$ turns on $$\frac{\gamma L_0V}{c^2}$$ seconds before lamp $$B$$ in frame $$F$$, we compute $t_2 - t_1 = \gamma\left[(t_2'-t_1') - (x_2'-x_1')\frac{-V}{c^2}\right] = \frac{\gamma L_0V}{c^2} > 0.$

• Light from lamp $$A$$ will reach the observer $$O$$ first as it comes on first and the light travels a shorter distance from lamp $$A$$.

• To find time delay $$\Delta t$$ between the two lamps, first consider an observer $$O_A$$ positioned at lamp $$A$$ on the moving rod. $$O_A$$ sees lamp $$B$$ come on $$\frac{L_0}c$$ seconds after lamp $$A$$. We treat it as a pulse with period $$T' = \frac{L_0}{c}$$ enamating from end $$A$$ of the rod. Applying Doppler effect, observer at $$O$$ observes that $\Delta t = \sqrt{\frac{c+V}{c-V}}T' = \sqrt{\frac{c+V}{c-V}}\frac{L_0}{c}.$

# (b)

Start from the frame $$F$$ where rod 1 is at rest and the other is moving at speed $$V$$. We want to find a frame $$F'$$ moving at velocity $$u$$ away from rod 1 such that $V' = \frac{V - u}{1 - \frac{u}{c^2}V}$ and $$V' = u$$, this will make both rods have the same length. Solving for $$u$$, we have $u = \frac{c^2}{v}\left(1 - 2\sqrt{1 - \frac{v^2}{c^2}}\right)$ reject the other quadratic solution since that will make $$u\geq c$$.

# (c)

Total energy twice as rest energy, this means $\gamma mc^2 = 2 mc^2 \implies \gamma = 2.$

Solving $\frac12 = \sqrt{1 - \beta^2}$ we have $u = \frac{\sqrt{3}}2 c.$