Question

In Fig. $$37-9,$$ observer $$S$$ detects two flashes of light. A big flash occurs at $$x_{1}=1200 \mathrm{m}$$ and, $$5.00 \mu \mathrm{s}$$ later, a small flash occurs at $$x_{2}=480 \mathrm{m} .$$ As detected by observer $$S^{\prime},$$ the two flashes occur at a single coordinate $$x^{\prime} .$$ (a) What is the speed parameter of $$S^{\prime},$$ and(b) is $$S^{\prime}$$ moving in the positive or negative direction of the $$x$$ axis? To $$S^{\prime},(\mathrm{c})$$ which flash occurs first and(d) what is the time interval between the flashes?

Answer 1

(a) In frame $$S$$, our coordinates are such that $$x_{1}=+1200 \mathrm{m}$$ for the big flash, and $$X_{2}=$$ $$1200-720=480 \mathrm{m}$$ for the small flash (which occurred later).

Thus,

$$\Delta X=X_{2}-X_{1}=-720 \mathrm{m}\\$$

If we set $$\Delta x^{\prime}=0$$ in Eq. $$37-25,$$ we find

$$0=\gamma(\Delta x-v \Delta t)=\gamma\left(-720 \mathrm{m}-v\left(5.00 \times 10^{-6} \mathrm{s}\right)\right)\\$$

which yields $$V=-1.44 \times 10^{8} \mathrm{m} / \mathrm{s},$$ or $$\beta=v / c=0.480\\$$

(b) The negative sign in part (a) implies that frame $$S^{\prime}$$ must be moving in the $$-x$$ direction.

(c) Equation $$37-28$$ leads to

$$\Delta t^{\prime}=\gamma\left(\Delta t-\dfrac{V \Delta X}{c^{2}}\right)=\gamma\left(5.00 \times 10^{-6} \mathrm{s}-\dfrac{\left(-1.44 \times 10^{8} \mathrm{m} / \mathrm{s}\right)(-720 \mathrm{m})}{\left(2.998 \times 10^{8} \mathrm{m} / \mathrm{s}\right)^{2}}\right)\\$$

which turns out to be positive (regardless of the specific value of $$\gamma$$ ).

Thus, the order of the flashes is the same in the $$S^{\prime}$$ frame as it is in the $$S$$ frame (where $$\Delta t$$ is also positive).

Thus, the big flash occurs first, and the small flash occurs later.

(d) Finishing the computation begun in part (c), we obtain

$$\Delta t^{\prime}=\dfrac{5.00 \times 10^{-6} \mathrm{s}-\left(-1.44 \times 10^{8} \mathrm{m} / \mathrm{s}\right)(-720 \mathrm{m}) /\left(2.998 \times 10^{8} \mathrm{m} / \mathrm{s}\right)^{2}}{\sqrt{1-0.480^{2}}}=4.39 \times 10^{-6} \mathrm{s}\\$$