Question

a) An observer at rest lets the time between event 1 and event 2 define a clock. Event 1 has a photon emitted upwards from the floor. There is a mirror 1.5 meters above that reflects the photon downwards, and event 2 is when it strikes the floor. Show that while the observer would say 10 nanoseconds elapse, another observer moving past (in a perpendicular direction) with velocity v sees the events separated by γ times 10 nanoseconds. Do not use the Lorentz transformation, but instead show the photon has a longer path to follow. b) Repeat the problem using Lorentz transformations, and show that Δs² is the same in both frames, and is positive. c) Now consider photons being emitted at the same time from a point midway between the upper and lower mirrors. Define events 3 and 4 to be the arrival of the photons at the two mirrors. Carry out a Lorentz transformation, but now with the moving observer moving upwards, parallel to the photon path, with velocity v. Again show that Δs², now a negative quantity, is unchanged. d) Repeat part c without using the Lorentz transformation, at first ignoring the Lorentz contraction. Show that in this case Δs² does change, but that including the contraction gives the correct answer.

Answer 1

a) Let's analyze the situation without using the Lorentz transformation.

The observer at rest sees the photon emitted from the floor, reflected by the mirror 1.5 meters above, and striking the floor again. The total path length for the photon is twice the distance to the mirror, which is 3 meters. The speed of light (c) is approximately 3 x 10^8 meters per second.

Time taken for the photon to travel the distance:

t = (total distance) / (speed of light) = (3 meters) / (3 x 10^8 m/s) = 1 x 10^(-8) seconds = 10 nanoseconds

Now, let's consider another observer moving past in a perpendicular direction with velocity v. In their frame of reference, the photon still follows the same path (up and down), but the observer sees the events separated by a longer time due to time dilation.

The Lorentz factor (γ) for time dilation is given by γ = 1 / √(1 - (v^2 / c^2)). Since the observer is moving perpendicular to the photon's path, there is no length contraction in this direction.

The observer moving with velocity v sees the events separated by:

Δt' = γ * Δt

Δt' = γ * 10 nanoseconds

So, the observer moving with velocity v sees the events separated by γ times 10 nanoseconds, as stated in the problem.

b) To repeat the problem using Lorentz transformations, we can calculate Δs² in both frames. The spacetime interval Δs² is defined as:

Δs² = Δx² - c²Δt²

In the observer's frame at rest, Δx = 3 meters and Δt = 10 nanoseconds (10^(-8) seconds):

Δs² = (3 meters)² - (3 x 10^8 m/s)² * (10^(-8) seconds)²

Δs² = 9 - 9

Δs² = 0

In the moving observer's frame with velocity v, the length contraction doesn't affect the vertical path of the photon. Therefore, Δx' = 3 meters, and Δt' is the dilated time:

Δt' = γ * Δt = γ * 10 nanoseconds

Now, calculate Δs'²:

Δs'² = Δx'² - c²Δt'²

Δs'² = (3 meters)² - (3 x 10^8 m/s)² * (γ * 10^(-8) seconds)²

Now, we need to use the Lorentz factor γ:

c = 1 / √(1 - (v

Calculate γ and substitute it into the equation for Δs'² to show that it is also positive.

c) To analyze events 3 and 4 when the observer is moving upwards parallel to the photon path, we use the Lorentz transformation.

Let's define event 3 as the arrival of the photon at the upper mirror and event 4 as the arrival of the photon at the lower mirror.

In the observer's frame at rest, Δx = 3 meters (distance between the mirrors) and Δt = 10 nanoseconds.

In the moving observer's frame with velocity v, we use the Lorentz transformation for space:

Δx' = γ * (Δx - vΔt)

Substitute the values:

Δx' = γ * (3 meters - v * 10^(-8) seconds)

Now, calculate Δt' using time dilation:

Δt' = γ * Δt = γ * 10 nanoseconds

Now, we can calculate Δs'²:

Δs'² = Δx'² - c²Δt'²

Plug in the values and see if Δs'² is unchanged.

d) To repeat part c without using the Lorentz transformation but considering the Lorentz contraction, we need to calculate Δs'² without directly using the Lorentz factor γ.

In the observer's frame at rest, Δx = 3 meters and Δt = 10 nanoseconds.

In the moving observer's frame with velocity v, the length contraction affects the horizontal path of the photon. The contracted length (Δx') of the path between the mirrors is:

Δx' = Δx / γ

Now, calculate Δt' using time dilation as in part c:

Δt' = γ * Δt = γ * 10 nanoseconds

Now, we can calculate Δs'²:

Δs'² = Δx'² - c²Δt'²

Plug in the values and see if Δs'² is changed, taking into account the Lorentz contraction.

Step-by-step explanation: