Question

You are on a high-speed train. traveling at a decent clip: \(0.8 c\). On the ground are two signal stations 5 km per. each with a status-reporting sign, which always give simultaneous reports. Ar precisely noon on the train's clocks. the conductor at the front of the train passes one station and sees a sign reading "All Clear," and another employee at the back passes the other station and sees a sign reading "Severe Electrical Storms Reported! Slow to \(0.1 c ! "\) (a) How long is the train? (b) Should it slow down? (c) Suppose that both reporting signs display the time very precisely, updated every microsecond. By how much would the two observed time readings differ, if at all?

Short answer

a) The length of the train is approximately 3 km. b) Yes, the train should slow down. c) The time difference on the clocks observed on the train and on signs would diverge as the train keeps moving.

Step-by-step solution

Find the Length of train

Using the Lorentz contraction formula, the length of the train (\(L')) is given by the stationary length (\(L\)) and train's velocity (\(v\)) as follows: \(L' = L \sqrt{1 - v^2/c^2}\). \nSwap the sides and solve for \(L\): \[L = L'/ \sqrt{1 - v^2/c^2}\]. \nSince the train just covers the two stations as it passes by, the stationary length \(L\) can be considered as the distance between the two stations or 5 km. The velocity \(v\) of the train is given as 0.8c. Solving for \(L'\) results in the length of the train.

Decide whether to slow down or not

In the train’s reference frame, the events '_front train passing the first station' and 'back train passing the second station' are simultaneous, but they are not simultaneous in the ground’s reference frame. To the person on the ground, the 'Electrical Storms Reported' sign turns on first and then the 'All Clear' sign turns on. So the train should slow down.

Difference in Time Readings

The difference in time readings in the train's frame is \( \Delta t' = L'/v \). To find the time difference in ground frame we use time dilation formula: \[ \Delta t = \Delta t' /( \sqrt{1 - v^2/c^2}) \]. The time difference would be the difference between \( \Delta t \) and \( \Delta t'\).

Key concepts

Lorentz contraction

Lorentz contraction is a fascinating concept from Einstein's theory of relativity. It explains how objects moving at high speeds appear shorter along their direction of motion to a stationary observer. In simpler terms, if you're zooming past at a significant fraction of the speed of light, you'll look a bit squashed to someone standing still.

For example, if a train is speeding down the tracks at 0.8 times the speed of light, its length seems shorter to an observer on the ground. The contraction can be calculated using the formula: \[ L' = L \sqrt{1 - \frac{v^2}{c^2}} \]where:

• \( L' \) is the contracted length seen by the observer,
• \( L \) is the train's actual length at rest,
• \( v \) is the velocity of the train,
• \( c \) is the speed of light.

This formula shows that the faster you move, the more pronounced the contraction. However, in everyday experiences at normal speeds, Lorentz contraction isn't noticeable because we move so much slower than light.

Reference frames

Reference frames are the viewpoints from which measurements and observations are made. In relativity, an observer's frame of reference significantly influences how events are perceived.

Consider our high-speed train scenario. For someone onboard, events may happen simultaneously, like seeing two signals at the same time. But for someone on the ground, these events might seem different. Why? Because each reference frame has its own clock, speed, and perspective.

Here's the key idea:

• In the train's reference frame, the front and back passing the signals might seem simultaneous.
• In the ground’s reference frame, one signal happens before the other.

This leads to different observations based on where you're "standing" in the universe. Understanding reference frames helps explain why two observers might disagree on the sequence or simultaneity of events, especially at high speeds.

Time dilation

Time dilation is another intriguing aspect of Einstein's relativity theory. It refers to the way time stretches or dilates for objects in motion, relative to a stationary observer. In essence, a fast-moving clock ticks slower than a stationary one.

This effect becomes significant at speeds close to the speed of light. Using our train example, passengers onboard might notice their watches ticking normally. However, to an observer standing still on the ground, those onboard clocks run slower. The formula to calculate this difference is:\[ \Delta t = \frac{\Delta t'}{\sqrt{1 - \frac{v^2}{c^2}}} \]where:

• \( \Delta t' \) is the time interval measured in the moving frame (train),
• \( \Delta t \) is the time interval measured in the stationary frame (ground),
• \( v \) is the velocity,
• \( c \) is the speed of light.

Time dilation helps explain how passengers on our rapid train perceive time differently than those watching from the sidelines. This fascinating concept answers why fast-moving observers experience time at a slower rate.