Special Relativity for HSC Physics: A Clear, Worked Guide
Understand special relativity without the mystique. This HSC Physics guide explains reference frames, relativity of simultaneity, time dilation, length contraction, and mass-energy through clear explanations and worked examples.
A spacecraft passes Earth at \(0.80c\). Earth assigns a longer interval between two ticks of the spacecraft’s clock than the astronaut does. The astronaut, meanwhile, says that Earth clocks run slow. Before reading on, predict: is one observer simply wrong? No. Each uses a different inertial frame, and the disagreement comes from how time and space are measured at high speeds.
The same two ticks, different elapsed times
| Astronaut’s rest frame | Δτ = 3.0 s Both ticks occur at the same cabin location. The cabin clock measures proper time. |
|---|---|
| Earth’s frame | Δt = γΔτ = 5.0 s Synchronised Earth clocks assign the interval. Same two events; neither clock is faulty. |
01Frames: what is each observer actually measuring?
A frame of reference is a coordinate system for assigning positions and times to events. It can be represented by rulers and a network of synchronised clocks.
An inertial frame is one that is not accelerating or rotating. For this guide, assume that gravity is negligible and that the spacecraft and Earth move at constant relative velocity whenever we compare them.
In everyday life, we can usually treat time as if it were universal. If two events happen at the same time for you, you expect them to happen at the same time for everyone else.
Special relativity challenges that assumption. It starts with two rules:
- The laws of physics are the same in every inertial frame.
- Every inertial observer measures the same speed of light in a vacuum, regardless of the motion of the light source or observer.
The speed of light is
\[ c=3.00\times10^8\text{ m s}^{-1}. \]
Suppose you move towards a beam of light. You might expect to measure its speed as \(c+v\), where \(v\) is your speed towards the beam. You don’t. You still measure \(c\).
For this to work, observers moving relative to one another cannot always agree on time intervals, lengths, or which distant events happened simultaneously.
This is not about what their eyes happen to see. Relativistic measurements use synchronised clocks and account for the time light takes to travel. The disagreement remains even after that correction.
Same events, different coordinates
| Platform frame S | A (0 s, 0 m); B (0 s, 600 m) Own synchronised clocks and position scale. Events A and B are simultaneous. |
|---|---|
| Rocket frame S′ | A (0 s, 0 m); B (−2.67 μs, 1000 m) Own synchronised clocks and position scale. Event B occurs earlier than A. |
02Simultaneity: two events can happen at different times in different frames
Imagine two lamps, A and B, at opposite ends of a long platform. Clocks at rest with the platform are synchronised, and both lamps flash at platform time \(t=0\).
Before reading further, predict: must a rocket passing the platform agree that the flashes happened simultaneously?
No. The rocket can assign different times to the two flashes.
Simultaneous means that two events have the same time coordinate in a specified frame. It does not automatically mean that every inertial frame assigns them the same time.
To make the situation precise, suppose:
- Lamp A is at \(x=0\).
- Lamp B is at \(x=600\text{ m}\).
- The platform is at rest in frame \(S\).
- The rocket’s frame \(S'\) moves in the positive \(x\)-direction at speed \(v\) relative to \(S\).
In the platform frame, the flashes are simultaneous. In the rocket frame, the flash at B happens earlier than the flash at A.
Why that direction? The time-coordinate transformation for these two events is
\[ \Delta t'=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right), \]
where
\[ \gamma=\frac{1}{\sqrt{1-v^2/c^2}}. \]
Here, \(\Delta t=t_B-t_A\) and \(\Delta x=x_B-x_A\) are the time and position differences measured in frame \(S\). The primed time difference is measured in frame \(S'\), which moves in the positive \(x\)-direction relative to \(S\).
For the simultaneous platform flashes,
\[ \Delta t=0,\qquad \Delta x>0. \]
Therefore,
\[ \Delta t'=-\gamma\frac{v\Delta x}{c^2}<0. \]
The negative sign tells us that B’s flash occurs earlier in the rocket frame.
Relativity of simultaneity: required timeline structure
| Platform frame S | Δt = 0 Events A and B share the same time coordinate. |
|---|---|
| Rocket frame S′ | Δt′ < 0 Event B occurs earlier than event A. |
One important qualification: the order in which an observer receives two flashes is not, by itself, proof of their event times. Light has to travel to the observer, and the distances can differ. Relativity of simultaneity concerns the times assigned by synchronised clocks after accounting for light travel.
This distinction becomes especially useful when a question involves two distant events rather than two ticks of one clock.
03Time dilation: a moving clock has a longer measured tick interval
Now consider two ticks of a single clock attached to a spacecraft.
The two ticks occur at the same location in the spacecraft’s rest frame. The elapsed time measured by that clock is called the proper time, written \(\Delta\tau\).
An observer in another inertial frame, where the spacecraft is moving at speed \(v\), assigns a longer time interval to those same two events. This is time dilation.
The equation is
\[ \boxed{\Delta t=\gamma\Delta\tau} \]
where:
- \(\Delta t\) is the time interval measured in the frame where the clock is moving.
- \(\Delta\tau\) is the proper time measured by the clock in its rest frame.
- \(\gamma\) is the Lorentz factor.
- \(v\) is the relative speed between the frames.
- \(c\) is the speed of light.
Since \(\gamma\geq1\), the moving clock’s tick interval is never shorter than its proper-time interval.
For example, at \(v=0.80c\),
\[ \gamma=\frac{1}{\sqrt{1-0.80^2}} =\frac{1}{0.60} =1.67. \]
The other frame measures an interval \(1.67\) times as long as the clock’s own interval.
One clock, two measured time intervals
| Spacecraft frame | Δτ = 10.0 min Both ticks occur at the same cabin location. The cabin clock measures proper time. |
|---|---|
| Laboratory frame | Δt = γΔτ = 16.7 min The cabin clock is moving at 0.80c. Synchronised laboratory clocks assign a longer interval. |
Worked example: A clock on a fast spacecraft
A spacecraft moves at \(0.80c\) relative to Earth. Its onboard clock measures \(10.0\text{ min}\) between two ticks. What time interval does Earth assign to those same ticks?
Step 1
The two ticks occur at the same location in the spacecraft’s rest frame, so the onboard reading is the proper time:
\[ \Delta\tau=10.0\text{ min}. \]
Step 2
\[ \begin{aligned} \gamma &=\frac{1}{\sqrt{1-(0.80)^2}}\\ &=\frac{1}{0.60}\\ &=1.67. \end{aligned} \]
Step 3
\[ \begin{aligned} \Delta t&=\gamma\Delta\tau\\ &=(1.67)(10.0\text{ min})\\ &=16.7\text{ min}. \end{aligned} \]
Earth assigns an interval of \(16.7\text{ min}\), compared with the spacecraft clock’s \(10.0\text{ min}\). The astronaut’s clock behaves normally in its own rest frame; the difference is in the time intervals assigned by the two frames.
Why can both observers say the other’s clock runs slow?
This sounds contradictory, but the observers are not making a direct comparison of the same pair of ticks on both clocks.
Earth uses its own synchronised clock network to assign times to the spacecraft’s ticks. The astronaut uses a different synchronised clock network to assign times to Earth’s distant clock ticks. Because simultaneity is frame-dependent, the two comparisons do not use the same assignments of distant events.
If the astronaut turns around and later reunites with Earth, the situation changes again. The journey includes a change of inertial frame, so the two observers’ complete journeys are no longer equivalent. They can accumulate different amounts of proper time.
The key rule is simple: time dilation applies to the same two events, with the proper-time interval measured in the frame where those events occur at one location.
04Length contraction: the measured length of a moving object changes
Time dilation concerns the interval between events. Length contraction concerns the distance between the ends of an object.
Imagine a spacecraft at rest. You measure its length using a ruler and find \(150\text{ m}\). This is its proper length, \(L_0\), because the measurement is made in the spacecraft’s rest frame.
Now suppose the spacecraft moves past Earth at a high speed. Earth measures its length by recording the positions of its front and rear at the same Earth-frame time.
That last condition matters. To measure a length in a particular frame, you must record both endpoint positions simultaneously in that frame.
The moving spacecraft’s measured length is
\[ \boxed{L=\frac{L_0}{\gamma}} \]
or, equivalently,
\[ L=L_0\sqrt{1-\frac{v^2}{c^2}}, \]
where:
- \(L\) is the length measured in the frame where the object moves.
- \(L_0\) is the object’s proper length, measured in its rest frame.
- \(v\) is the object’s speed relative to the measuring frame.
- \(\gamma\) is the Lorentz factor.
Only the dimension parallel to the motion contracts. The object’s perpendicular dimensions are unchanged by this length-contraction relationship.
Length contraction affects only the direction of motion
| Spacecraft at rest | L = L₀ L₀ is the proper length. Perpendicular dimension: h₀. |
|---|---|
| Spacecraft moving at speed v | L = L₀/γ Length contracts parallel to motion. Perpendicular dimension remains h₀. |
A common mistake is to imagine that the spacecraft feels itself being physically squeezed. It doesn’t measure a shorter length in its own rest frame. The difference arises because length measurements in different frames use different sets of simultaneous endpoint positions.
Worked example: A spacecraft passing a detector
A probe has a proper length of \(150\text{ m}\) and passes a stationary detector at \(0.80c\).
Calculate:
- the length Earth measures,
- the time between the probe’s nose and tail passing the detector in Earth’s frame, and
- the time between those same events in the probe’s frame.
Step 1
\[ \gamma=\frac{1}{\sqrt{1-0.80^2}}=1.67. \]
Step 2
\[ \begin{aligned} L&=\frac{L_0}{\gamma}\\ &=\frac{150\text{ m}}{1.67}\\ &=90.0\text{ m}. \end{aligned} \]
Earth measures the moving probe as \(90.0\text{ m}\) long.
Step 3
The probe moves at
\[ v=0.80(3.00\times10^8) =2.40\times10^8\text{ m s}^{-1}. \]
The nose and tail are separated by \(90.0\text{ m}\) in Earth’s frame. Therefore, the time between their passing the stationary detector is
\[ \begin{aligned} \Delta t_E&=\frac{L}{v}\\ &=\frac{90.0\text{ m}} {2.40\times10^8\text{ m s}^{-1}}\\ &=3.75\times10^{-7}\text{ s}\\ &=0.375\,\mu\text{s}. \end{aligned} \]
Step 4
In the probe’s frame, the detector moves past the probe’s full \(150\text{ m}\) proper length at the same relative speed. Thus,
\[ \begin{aligned} \Delta t_P&=\frac{L_0}{v}\\ &=\frac{150\text{ m}} {2.40\times10^8\text{ m s}^{-1}}\\ &=6.25\times10^{-7}\text{ s}\\ &=0.625\,\mu\text{s}. \end{aligned} \]
The two crossing events occur at the same location in Earth’s frame, so the detector’s \(0.375\,\mu\text{s}\) interval is a proper-time interval. In the probe’s frame, the detector is moving, and the interval is longer:
\[ \gamma\Delta t_E =(1.67)(0.375\,\mu\text{s}) =0.625\,\mu\text{s}. \]
The calculations agree. The same pair of events has different time intervals in the two frames, while the probe’s measured length depends on the frame in which both endpoints are recorded simultaneously.
Same crossing events, different time intervals
| Earth frame | A → B: 0.375 μs Probe length: 90 m; detector is stationary. |
|---|---|
| Probe frame | A → B: 0.625 μs Probe length: 150 m; detector moves past it. |
05Mass-energy: what does \(E=mc^2\) actually mean?
Consider a nucleus forming from separate protons and neutrons. After the nucleus forms, its rest mass is slightly less than the combined rest mass of the separated particles.
Has some matter simply disappeared?
No. The system has released energy, and the energy carried away must be included in the accounting. Mass and energy are related.
The rest energy of an object is
\[ \boxed{E_0=mc^2} \]
where:
- \(E_0\) is the rest energy.
- \(m\) is the invariant, or rest, mass.
- \(c\) is the speed of light.
For a change in rest mass, the corresponding change in rest energy is
\[ \boxed{\Delta E_0=\Delta m c^2}. \]
The factor \(c^2\) is enormous. A mass difference of just \(1.0\text{ g}\) corresponds to a rest-energy difference of
\[ \begin{aligned} E_0&=(1.0\times10^{-3}\text{ kg}) (3.00\times10^8\text{ m s}^{-1})^2\\ &=9.0\times10^{13}\text{ J}. \end{aligned} \]
That is a huge amount of energy for a tiny mass difference.
Mass decreases; total energy is conserved
| Before energy escapes | Rest mass = m Initial rest energy: E₀ = mc². |
|---|---|
| After radiation escapes | Remaining mass = m − Δm Radiation carries ΔE = Δmc². Remaining rest energy + radiation energy = initial energy. |
What changes when energy enters or leaves a system?
If energy escapes from a system, the remaining system can have a lower rest mass. If energy is added to a system and retained internally, its rest mass increases, provided the system is compared at rest before and after.
This does not mean that energy is destroyed when mass decreases. For an isolated system, total energy is conserved. The energy may have left the remaining object as radiation or kinetic energy, but it still exists.
A bound nucleus provides a useful example. The difference between the combined rest mass of its separated nucleons and the rest mass of the bound nucleus is called the mass defect. The corresponding energy difference is the binding energy.
A necessary qualification: rest mass is not the same as total energy
The equation \(E_0=mc^2\) gives rest energy. For an object moving at speed \(v\), its total energy is
\[ E=\gamma mc^2, \]
and its kinetic energy is
\[ K=(\gamma-1)mc^2. \]
Here, \(m\) remains the object’s invariant rest mass. The factor \(\gamma\) changes the object’s total energy, not its invariant mass.
Some older treatments use the term relativistic mass for \(\gamma m\). The modern convention generally keeps mass invariant and describes the increase in terms of energy and momentum instead.
Photons, for example, carry energy and momentum despite having zero rest mass. So \(E=mc^2\) should not be interpreted as meaning that every form of energy must belong to a particle with non-zero rest mass.
Worked example: Finding energy from a mass difference
A stationary source releases radiation, and its remaining rest mass decreases by \(0.50\text{ mg}\). Assuming the entire mass difference corresponds to energy carried away by the radiation, estimate the energy released.
Step 1
\[ \begin{aligned} \Delta m&=0.50\text{ mg}\\ &=5.0\times10^{-7}\text{ kg}. \end{aligned} \]
Step 2
\[ \begin{aligned} E&=\Delta m c^2\\ &=(5.0\times10^{-7}\text{ kg}) (3.00\times10^8\text{ m s}^{-1})^2\\ &=4.5\times10^{10}\text{ J}. \end{aligned} \]
Step 3
The released energy is \(4.5\times10^{10}\text{ J}\). A mass difference of only \(0.50\text{ mg}\) corresponds to an enormous amount of energy.
The source’s remaining rest mass has decreased, but the energy has not vanished. When the emitted radiation is included, the total energy accounting remains consistent.
06Choosing the right relationship
The most reliable way to solve a special relativity question is to identify what is being measured before choosing a formula.
| What the question describes | Relevant relationship | Check before substituting |
|---|---|---|
| Time between two ticks of a moving clock | \(\Delta t=\gamma\Delta\tau\) | Are the two events at the same location in the clock’s rest frame? |
| Length of an object moving relative to the observer | \(L=L_0/\gamma\) | Is \(L_0\) the proper length, and are both endpoints measured simultaneously in the observer’s frame? |
| Two events at different positions | Relativity of simultaneity | Do the events occur at different locations in the frame where their times are being compared? |
| Change in a system’s rest energy | \(\Delta E_0=\Delta m c^2\) | Which system is being considered, and where has the energy gone? |
Four checks catch many common errors:
- Check the Lorentz factor. For \(0\leq v<c\), \(\gamma\geq1\).
- Check the events. Time dilation is not a rule for every pair of events. The proper-time condition matters.
- Check the measurement. Length contraction applies along the direction of motion, and a length measurement requires simultaneous endpoint positions in the measuring frame.
- Check the energy accounting. A decrease in the rest mass of one object does not mean that the total energy of an isolated system has disappeared.
07Questions and solutions
Question 1
A clock fixed to a probe measures \(3.0\text{ s}\) between two ticks. The probe moves at \(0.60c\) relative to a laboratory.
Calculate the time interval the laboratory assigns to the same two ticks.
Solution 1
The laboratory assigns an interval of \(3.75\text{ s}\), or \(3.8\text{ s}\) to two significant figures.
The onboard clock measures the proper time because both ticks occur at the same location in the probe’s rest frame.
Step 1: Calculate the Lorentz factor.
\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.60^2}}\\ &=\frac{1}{0.80}\\ &=1.25. \end{aligned} \]
Step 2: Apply time dilation.
\[ \begin{aligned} \Delta t&=\gamma\Delta\tau\\ &=(1.25)(3.0\text{ s})\\ &=3.75\text{ s}. \end{aligned} \]
The laboratory measures a longer interval than the probe clock does. The result is consistent with time dilation.
Question 2
A probe has a proper length of \(80\text{ m}\). A hangar is \(60\text{ m}\) long in its own rest frame. The probe passes the hangar at \(0.80c\).
An engineer argues that the probe can fit entirely inside the hangar at one instant because its contracted length is shorter than the hangar. The pilot argues that this is impossible because the hangar is shorter than the probe in the pilot’s frame.
Assume ideal doors that can open and close instantaneously, and ignore material damage.
Are both observers’ length calculations correct? Can the probe be entirely inside the hangar at one instant? Explain how the two descriptions are consistent.
Solution 2
Both length calculations are correct. The disagreement is about whether the probe is entirely inside the hangar at one instant, because simultaneity depends on the frame.
Step 1: Calculate the Lorentz factor.
\[ \gamma=\frac{1}{\sqrt{1-0.80^2}}=1.67. \]
Step 2: Calculate the probe’s length in the hangar frame.
The probe’s proper length is \(L_0=80\text{ m}\).
\[ \begin{aligned} L_{\text{probe}}&=\frac{L_0}{\gamma}\\ &=\frac{80\text{ m}}{1.67}\\ &=48\text{ m}. \end{aligned} \]
Since \(48\text{ m}<60\text{ m}\), the engineer is correct that the probe can fit entirely inside the hangar at one hangar-frame instant, provided it is positioned appropriately.
Step 3: Calculate the hangar’s length in the probe frame.
The hangar’s proper length is \(60\text{ m}\). In the probe frame, the hangar moves, so its measured length is
\[ \begin{aligned} L_{\text{hangar}}&=\frac{60\text{ m}}{1.67}\\ &=36\text{ m}. \end{aligned} \]
The pilot is also correct: the moving hangar is only \(36\text{ m}\) long in the probe frame, while the probe’s proper length remains \(80\text{ m}\).
Step 4: Resolve the apparent contradiction.
The phrase “entirely inside at one instant” means that the positions of both ends of the probe are inside the hangar at the same time in a specified frame.
In the hangar frame, there is an instant when both ends are inside. In the probe frame, there is no instant when the entire \(80\text{ m}\) probe fits inside the \(36\text{ m}\) hangar.
The key is that the two door-closing events, simultaneous in the hangar frame, are not simultaneous in the probe frame. The observers are not describing the same pair of simultaneous events. Neither description requires the probe to be physically destroyed or compressed in its own rest frame.
Question 3
A bound nucleus has a rest mass \(2.0\times10^{-28}\text{ kg}\) lower than the combined rest mass of its separated nucleons.
Assume that both states are compared at rest and that the energy released during binding escapes the nucleus.
Calculate the binding energy released, in joules, and explain why the lower rest mass of the final nucleus does not violate conservation of energy.
Solution 3
The binding energy equivalent of the mass difference is \(1.8\times10^{-11}\text{ J}\). The lower rest mass is consistent with energy conservation because the released energy leaves the system.
Step 1: Identify the mass difference.
The magnitude of the mass defect is
\[ \Delta m=2.0\times10^{-28}\text{ kg}. \]
Step 2: Apply mass-energy equivalence.
\[ \begin{aligned} E&=\Delta m c^2\\ &=(2.0\times10^{-28}\text{ kg}) (3.00\times10^8\text{ m s}^{-1})^2\\ &=(2.0\times10^{-28})(9.00\times10^{16})\\ &=1.8\times10^{-11}\text{ J}. \end{aligned} \]
Step 3: Explain the energy accounting.
The bound nucleus has less rest energy than the separated nucleons. The difference appears as energy released during binding, such as radiation or kinetic energy carried away by other products.
The final nucleus has a lower rest mass because energy has left the nucleus. When the released energy is included, the total energy remains conserved.
Question 4
A spacecraft has an invariant rest mass of \(1.0\text{ kg}\) and moves at \(0.80c\) relative to Earth.
A student calculates \(\gamma=1.67\) and claims that the spacecraft’s mass has increased to \(1.67\text{ kg}\).
Using
\[ E_0=mc^2,\qquad E=\gamma mc^2,\qquad K=(\gamma-1)mc^2, \]
calculate the spacecraft’s rest energy, total energy, and kinetic energy. Assess the student’s claim.
Solution 4
The rest energy is \(9.0\times10^{16}\text{ J}\), the total energy is \(1.5\times10^{17}\text{ J}\), and the kinetic energy is \(6.0\times10^{16}\text{ J}\). Under the invariant-mass convention, the spacecraft’s rest mass remains \(1.0\text{ kg}\).
Step 1: Calculate the rest energy.
The rest mass is \(m=1.0\text{ kg}\).
\[ \begin{aligned} E_0&=mc^2\\ &=(1.0\text{ kg})(3.00\times10^8\text{ m s}^{-1})^2\\ &=9.0\times10^{16}\text{ J}. \end{aligned} \]
Step 2: Calculate the total energy.
\[ \begin{aligned} E&=\gamma mc^2\\ &=(1.67)(1.0\text{ kg})(3.00\times10^8\text{ m s}^{-1})^2\\ &\approx1.5\times10^{17}\text{ J}. \end{aligned} \]
Step 3: Calculate the kinetic energy.
\[ \begin{aligned} K&=(\gamma-1)mc^2\\ &=(1.67-1)(9.0\times10^{16}\text{ J})\\ &\approx6.0\times10^{16}\text{ J}. \end{aligned} \]
The results are consistent because the total energy is the sum of rest energy and kinetic energy:
\[ E=E_0+K. \]
Step 4: Assess the student’s claim.
The student has correctly identified that \(\gamma\) multiplies the rest energy to give the total energy. The mistake is interpreting this factor as an increase in invariant rest mass.
The spacecraft’s kinetic energy increases with speed, but its invariant rest mass remains \(1.0\text{ kg}\). This distinction matters whenever a question combines mass-energy equivalence with the energy of a moving object.
Question 5
A platform is \(600\text{ m}\) long in its rest frame \(S\). Two lamps, A and B, are mounted at its ends, with A at \(x=0\) and B at \(x=+600\text{ m}\). They flash simultaneously at \(t=0\) in \(S\).
Each flash launches a light pulse towards the platform’s midpoint. In frame \(S\), the two pulses meet at the midpoint.
A rocket’s rest frame \(S'\) moves in the positive \(x\)-direction at \(0.80c\) relative to \(S\).
For this question, use the time-coordinate relation
\[ \Delta t'=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right), \]
where \(\Delta t=t_B-t_A\) and \(\Delta x=x_B-x_A\) are measured in \(S\).
(a) Which flash occurs first in \(S'\), and by how much?
(b) The platform’s length in \(S'\) is \(360\text{ m}\). A student claims that the time gap between the flashes must therefore be \(360\text{ m}/c=1.20\,\mu\text{s}\). Explain the exact mistake.
(c) Must the two light pulses still meet at the same event in \(S'\)? Explain how that can be true even though the flashes are no longer simultaneous.
Solution 5
Flash B occurs first in \(S'\), by \(2.67\,\mu\text{s}\). The pulses still meet at the same event, because all inertial frames agree on the intersection of the two light pulses, even when they assign different times to the emission events.
Step 1: Calculate the Lorentz factor and identify the event differences.
The relative speed is \(v=0.80c\), so
\[ \gamma=\frac{1}{\sqrt{1-0.80^2}} =\frac{5}{3}. \]
The flashes are simultaneous in \(S\), and B is at the larger position coordinate. Therefore,
\[ \Delta t=0,\qquad \Delta x=+600\text{ m}. \]
Step 2: Calculate the time difference in \(S'\).
Substitute into the supplied time-coordinate relation:
\[ \begin{aligned} \Delta t' &=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right)\\ &=\frac{5}{3}\left(0-\frac{(0.80c)(600\text{ m})}{c^2}\right)\\ &=-\frac{800\text{ m}}{c}\\ &=-2.67\times10^{-6}\text{ s}. \end{aligned} \]
The negative result means \(t'_B<t'_A\). B’s flash occurs \(2.67\,\mu\text{s}\) before A’s flash in the rocket frame.
Step 3: Explain why the \(360\text{ m}/c\) calculation fails.
The student’s calculation gives
\[ \frac{360\text{ m}}{3.00\times10^8\text{ m s}^{-1}} =1.20\,\mu\text{s}. \]
The arithmetic is correct, but it answers the wrong question.
The \(360\text{ m}\) is the platform’s length measured in \(S'\), obtained by measuring both endpoints at the same rocket-frame time. The two flash events are not simultaneous in \(S'\), so their event positions cannot be treated as the endpoints of that simultaneous length measurement.
The time difference between two events is not generally the measured length divided by \(c\). The student has mixed a length measured using simultaneous endpoint positions with two events that occur at different times in the same frame.
Step 4: Explain why the pulses still meet.
In \(S\), each pulse travels \(300\text{ m}\) to the midpoint. The meeting event therefore occurs at
\[ t=\frac{300\text{ m}}{c} =1.00\,\mu\text{s}, \qquad x=300\text{ m}. \]
This is one event in spacetime. Every inertial frame agrees that both pulses meet at that event, even though the coordinates assigned to it differ.
We can check the rocket-frame timing. Applying the supplied relation between flash A and the meeting event gives
\[ \begin{aligned} t'_{\text{meet}} &=\gamma\left(1.00\,\mu\text{s} -\frac{(0.80c)(300\text{ m})}{c^2}\right)\\ &=\frac{5}{3}(1.00-0.80)\,\mu\text{s}\\ &=0.333\,\mu\text{s}. \end{aligned} \]
Flash A occurs at \(t'_A=0\), so its pulse takes \(0.333\,\mu\text{s}\) of rocket-frame coordinate time to reach the meeting event.
Flash B occurs at \(t'_B=-2.67\,\mu\text{s}\), so its pulse takes
\[ 0.333-(-2.67)=3.00\,\mu\text{s} \]
to reach that same event.
The different emission times and travel intervals are consistent with the same meeting event. The mistake would be to assume that simultaneous emission in one frame must remain simultaneous in every frame.
The next useful step is to learn how to represent events on a spacetime diagram. For HSC problems, start by identifying the frame and asking whether the events occur at the same location in that frame. That decision helps you choose between time dilation, length contraction, and relativity of simultaneity before you begin calculating.