In the context of the theory of relativity, state what is meant by an event.

(i) Calculate the time interval t between event 1 and event 2 according to an observer in the box.

(ii) According to an observer on the ground the time interval between event 1 and event 2 is T. One student claims that \(T = \frac{t}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\) and another that \(T = t\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} \).

Explain why both students are wrong.

Relative to an observer on the ground,

(i) calculate the distance between S and D.

(ii) state the speed of the photon leaving S.

(iii) state an expression for the distance travelled by detector D in the time interval T (T is the interval in (b)(ii)).

(iv) determine T, using your answers to (c)(i), (ii) and (iii).

a point in spacetime / something happening at a particular time and a particular point in space;

(i) \(t = \frac{{6.0}}{{3.0 \times {{10}^8}}} = 2.0 \times {10^{ - 8}}{\rm{s}}\);

(ii) for either formula to be used one of the time intervals must be a proper time interval;
the two events occur at different points in space and so neither observer measures a proper time interval;
the proper time interval is that of the photons;

(i) \(\gamma  = \frac{1}{{\sqrt {1 - {{0.80}^2}} }} = \frac{5}{3} = 1.67\);
\(l = \frac{L}{\gamma } = \frac{{6.0}}{{1.67}} = 3.6{\rm{m}}\);
Award [2] for a bald correct answer.

(ii) c;

(iii) vT or 0.80cT;

(iv) cT=0.80cT+3.6;
\(T = \frac{{3.6}}{{0.20 \times 3.0 \times {{10}^8}}} = 6.0 \times {10^{ - 8}}{\rm{s}}\);
Award [2] for a bald correct answer.