USAPhO 2021, Полуфинал — Задача 2. A2 · Flashlight

• a. i. $v = \beta c$

• a. ii. $c\sqrt{1 - \beta^2}$

• a. iii. $\Delta t = \frac{h}{c\sqrt{1 - \beta^2}}$

• b. $\frac{h}{c}\frac{1 + \beta\cos\theta}{\sqrt{1 - \beta^2}}$

• c. 105.5 °

  Yes: θ = cos⁻¹(√(β⁻² − 1) − β⁻¹) = 105.5° for β = 0.5.

• d. $f' = \frac{\sqrt{1 - \beta^2}}{1 + \beta\cos\theta}f$

a. i.

The muons were fired straight down in Alice’s frame, so in Bob’s frame their horizontal velocity is the same as Alice’s, $v = \beta c$.

One way to see this is to imagine Alice was carrying a vertical pole with her. In her reference frame, the muons travel along the length of the pole. This must remain true in any frame, so in Bob’s frame the muons must have the same horizontal velocity as Alice.

a. ii.

The muons have speed $c$ and horizontal velocity $v = \beta c$, so they have vertical velocity $c\sqrt{1 - \beta^2}$ by the Pythagorean theorem.

a. iii.

The time is the height divided by the vertical velocity of the muons,

b.

In Bob’s frame, during the time the muons take to reach the ground, Alice’s car moves a distance $\Delta r = (v\Delta t)\cos\theta$ away from Bob, which means the light from the muons hitting the ground takes an extra time $(\Delta r)/c$ to reach Bob. Thus, the time interval Bob sees, with his eyes, is

c.

This will be true if

Solving for $\theta$, we find

This has a solution as long as the argument of the inverse cosine is less than 1. In the case $\beta = 0.5$, it is, and the result is

That is, this value of $\beta$ is low enough so that the motion of Alice towards Bob can make up for the time dilation effect.

d.

The key point is that all the logic in part (b) still works, if we replace Bob’s eyes with the radio receiver. Since the frequency of a wave is the inverse of the time between maxima,

This is the two-dimensional relativistic Doppler shift, and the secret point of this problem was to derive it in a simple way.