Special Relativity Practice: Reference Frames, Time, and Length

Two lamps flash at opposite ends of a lab. Their light reaches the same detector at the same instant, even though one lamp is four times as far from the detector. Which lamp fired first? Start with what the detector clock actually recorded, then work backward.

Use the same approach for time dilation and length contraction. Write down the two events and the frame that supplies the readings before choosing an equation. The four worked examples and five practice problems use inertial frames, motion along one line, and basic algebra.

Write the frame next to the reading

An event happens at one place and time, like a flash leaving a lamp. An inertial frame gives you a coordinate grid and synchronized clocks at rest in that grid. A lab and a pod coasting past it are two such frames. Each can assign its own position and time to the same event.

If a clock is next to an event, use its local reading. If you only know when a distant detector received a flash, subtract the light’s travel time in the same frame. One light-microsecond is the distance light travels in vacuum in one microsecond.

To find	Identify
Proper time between two events	The frame where both events happen at the same position, so one clock at rest there can time both
Proper length of an object	The frame where the object is at rest
Length of a moving object	The positions of both ends at the same time in the measuring frame
Whether distant events are simultaneous	Whether synchronized local clocks in the chosen frame give the events the same time

The Lorentz factor is γ = 1 / √(1 − (v/c)²). Here v is the relative speed and c is the speed of light in vacuum. At 0.60c, γ is 5/4. At 0.80c, it’s 5/3.

1. Work back from the detector

The two lamps and the detector are at rest in the lab. The left lamp is two light-microseconds from the detector and the right lamp is eight light-microseconds away. Both flashes reach the detector at lab time 12 μs.

Lamp	Travel time	Emission time in the lab
Left	2 μs	12 μs − 2 μs = 10 μs
Right	8 μs	12 μs − 8 μs = 4 μs

The right lamp fired 6 μs earlier. Clocks synchronized in the lab and placed beside the lamps would record 10 and 4 μs directly. Both signals reach the same detector at 12 μs, so their arrivals coincide at one place and time.

2. Time the same two events in both frames

A pod coasts to the right at 0.60c. A lamp at its midpoint sends a pulse through vacuum toward each end. The events to compare are the two local detections, one at the rear and one at the front.

In the pod frame, the end detectors are stationary and equally far from the lamp. Both pulses travel at speed c, so the synchronized pod clocks record simultaneous arrivals.

In the lab frame, suppose each detector is four light-microseconds from the lamp at the instant it fires. The rear detector moves toward the left-going pulse. The front detector moves away from the right-going pulse. Both pulses still travel at speed c in the lab.

• The gap to the rear closes at c + 0.60c = 1.60c. The pulse reaches the rear after 4 light-μs / (1.60c) = 2.5 μs in the lab.
• The gap to the front closes at c − 0.60c = 0.40c. The pulse reaches the front after 4 light-μs / (0.40c) = 10 μs in the lab.

The rear detection happens 7.5 μs earlier in the lab. Those times describe the detections themselves. Synchronized lab clocks at the detection locations would record that 7.5 μs difference, regardless of when anyone elsewhere receives news of the detections. OpenStax on the relativity of simultaneity

3. Find the clock that’s there both times

The nose of a pod moving at 0.60c passes the lab’s Gate A, then Gate B. The clock on the nose reads 3 μs at A and 11 μs at B. The crossings happen at the same position in the pod frame, so the nose clock measures the proper time, Δτ = 11 μs − 3 μs = 8 μs.

In the lab frame, the crossings happen at different positions. Read the synchronized lab clock at A for the first crossing and the one at B for the second. The difference in those readings is Δt = γ × Δτ = (5/4) × 8 μs = 10 μs.

The comparison works in the other direction, too. Pick two ticks of the lab clock at Gate A. That one lab clock measures the proper time between its ticks, while two synchronized pod clocks time them at different pod positions. The pod assigns a longer interval to those ticks, so it measures that lab clock as running slow. The event pair has changed from two gate crossings to two ticks of a single gate clock. OpenStax on time dilation

4. Measure both ends at one lab time

A calibration bar is 54 m long at rest. It moves to the right at 0.80c along its length. The 54 m is the bar’s proper length, L₀. To measure it in the lab, record the positions of the front and rear at the same lab time. Subtract the rear position from the front position to get L.

L = L₀ / γ = 54 m / (5/3) = 32.4 m

Suppose the lab records the front a little later than the rear. The front moves right during the delay, so subtracting the two recorded positions gives more than 32.4 m. The two positions have to come from a single lab time. OpenStax on length contraction

Five special relativity practice problems

For each problem, name the two events and the frame before looking at the answers.

1. Two lamps and a detector are at rest in a lab. The first lamp is three light-microseconds away and its flash reaches the detector at 14 μs on the lab clock. The second lamp is one light-microsecond away and its flash arrives at 12 μs. Were the emissions simultaneous in the lab?
2. A pulse leaves a laser at the front of a coasting pod, reflects off a mirror at the rear, and returns to the laser. Which frame gives the proper time between emission and return? Could the clock at the laser directly time the interval from emission to reflection?
3. Two ticks of a clock moving at 0.60c are 25 s apart in the lab, as measured by synchronized lab clocks. What interval does the moving clock itself record?
4. A pole moves along its length at 0.80c. Its ends are 24 m apart when measured at the same lab time. What is its proper length?
5. Repeat the midpoint-pulse experiment with the pod moving to the left at 0.60c. The front is now the left end. Which end receives its pulse first in the lab? What do the pod clocks record?

Worked answers

1. Yes. The first emission happened at 14 μs − 3 μs = 11 μs in the lab. The second happened at 12 μs − 1 μs = 11 μs. The signals arrived at different times because they traveled different distances.
2. The pod frame gives the proper time from emission to return. Both events happen at the laser, so one clock there records both. The reflection happens at the mirror at the rear of the pod. Timing that event in the pod frame uses a separate synchronized clock at the mirror.
3. 20 s. The 25 s comes from the lab frame. The interval on the moving clock is the proper time because that clock is present at both ticks. Divide the lab interval by γ to get Δτ = 25 s / 1.25 = 20 s.
4. 40 m. The lab gives the contracted length L = 24 m. Multiply by γ to get the rest length, L₀ = 24 m × (5/3) = 40 m.
5. The rear, now the right end, receives its pulse first in the lab. The rear moves left toward the right-going pulse. The front moves left away from the left-going pulse. In the pod frame, both ends are still at rest at equal distances from the lamp, so the detections are simultaneous.

If you miss a problem, save the decision you want to recall next time. Try “Which clock is present at both gate crossings?” or “In which frame must the two endpoint readings share a time?” Answer from memory and revisit the prompt in a later session. The guide to making physics flashcards shows how to turn setup mistakes into short prompts.

AI assistance was used to draft and edit this worksheet. The numerical examples were checked by calculation against the relationships cited above.