Question

An alien spaceship streaks past Spanos football stadium along the direction of play at 0.8c. A football field is 120 yards long by 55 yards wide. Eager to learn about earth games, an alien on the spaceship makes some measurements. Conveniently the alien has already studied Earth units of measurement. (i) What does the alien measure for the length and width of the stadium? (ii) The game begins at 1:30 PM Pacific Standard Time (PST) and ends at 4:30 PM PST. How much time has elapsed according to the alien? (iii) Who sees the proper length of the stadium? Why? (iv) Who sees the proper time for the game? Why?

Answer 1

Answer:

(i) Length of stadium, L₀ = 72 yards; width of stadium W = 55 yards (ii) 5 hours (iii) The people on earth. This is because they are not moving at relativistic speed relative to the game. (iv) The people on earth. This is because they are not moving at relativistic speed relative to the game.

Explanation:

(i) Given Length of stadium, L₀ = 120 yards, width of stadium, W₀ = 55 yards, speed of alien spaceship in direction of play, v = 0.8c

We use the equation for length contraction to determine the length of stadium L measured by the alien spaceship, since it is moving in the direction of game play and length contraction only occurs in the direction of motion. So, L = $\sqrt{1-\beta^{2} }$L₀ where β = $\frac{v}{c}$  v= speed of alien spaceship and c= speed of light. So, β = $\frac{v}{c}$ = $\frac{0.8c}{c}$ = 0.8. Therefore, L = $\sqrt{1-0.8^{2} }$L₀= $\sqrt{1-0.64}$L₀= $\sqrt{0.36}$L₀ = 0.6L₀ = 0.6 × 120 = 72 yards. So, the alien spaceship measures a length of 72 yards. The alien spaceship measures a width,W of 55 yards because there is no length contraction in the direction perpendicular to that of its motion. So, W = W₀ = 55 yards.

(ii) Since the game begins at 1:30 PM Pacific Standard Time (PST) and ends 4:30 PM Pacific Standard Time (PST), the proper time t₀, which is the duration of the event is 4:30 - 1:30 = 3 hours. The time measured by the alien spaceship t (since the time dilates)is given by t = t₀/$\sqrt{1 - \beta ^{2} }$. From (i) above β = 0.8 and t₀ = 3 hours. So, t = 3/$\sqrt{1 - 0.8^{2} }$ = 3/0.6 = 5 hours.

(iii) The people in Spanos football stadium. This is because they are not moving at relativistic speed relative to the game. Since $\frac{v}{c}$ ≈ 0 then β ≈ 0 So, L = $\sqrt{1-\beta^{2} }$L₀= So, L = $\sqrt{1-0}$L₀thus L =  L₀ for the spectators in spanos football field (iv) The people in Spanos football stadium. This is because they are not moving at relativistic speed relative to the game. Since $\frac{v}{c}$ ≈ 0 then β ≈ 0. So, t = t₀/$\sqrt{1 - \beta ^{2} }$ = t₀/$\sqrt{1-0}$ = t₀.Thus t =  t₀ for the spectators in spanos football field.