How to Find Rest Length from a Contracted Length

A laboratory laser beam stays blocked for \(0.500\ \mu\mathrm{s}\) as a probe travels past at \(0.60c\). The timing equipment is accurate, but does that measurement tell you how long the probe is when it’s at rest?

Not quite. The beam measures how long the moving probe takes to pass a fixed point. That gives us its contracted length, not its rest length.

To find the rest length, also called the proper length, we need to reverse length contraction:

\[
L_0=\frac{L}{\sqrt{1-v^2/c^2}}
\]

Here:

• \(L_0\) is the rest length, measured in metres (\(\mathrm{m}\)).
• \(L\) is the contracted length, measured in metres (\(\mathrm{m}\)).
• \(v\) is the relative speed between the probe and the observer, measured in \(\mathrm{m\,s^{-1}}\).
• \(c\) is the speed of light in a vacuum, \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).

At \(0.60c\), the contraction factor is \(\sqrt{1-0.60^2}=0.80\). Our detector’s blocked time corresponds to a moving length of \(90.0\ \mathrm{m}\), so the rest length is \(90.0/0.80=112.5\ \mathrm{m}\).

The probe is longer in its own rest frame than the laboratory measures it to be.

01Why the rest length is the larger measurement

Imagine measuring a train carriage while you’re sitting inside it.

The front and back of the carriage are stationary relative to you. You can measure their separation with a ruler, and that is the carriage’s rest length.

Now imagine the train travelling at a substantial fraction of the speed of light. Someone standing beside the track measures the positions of the front and back simultaneously in the track frame.

Special relativity predicts that the track observer measures a shorter length along the direction of motion.

The train analogy helps you picture the two frames, but ordinary trains are nowhere near fast enough for this shortening to be noticeable. More importantly, the difference isn’t caused by the carriage physically compressing under a force. It arises because simultaneous position measurements are defined differently in relatively moving reference frames.

For the same pair of endpoints, the rest length \(L_0\) is measured in the frame where both endpoints are stationary. The contracted length \(L\) is measured in a frame where those endpoints move parallel to the length being measured.

The equation is often written:

\[
L=\frac{L_0}{\gamma}
\]

where \(\gamma\), the Lorentz factor, is

\[
\gamma=\frac{1}{\sqrt{1-v^2/c^2}}
\]

Since \(\gamma\geq1\), rearranging gives:

\[
L_0=\gamma L
\]

This is a useful mental check. If you’ve measured a moving object’s length along its motion and you’re recovering its rest length, you multiply by \(\gamma\), not divide by it.

For a fuller explanation of the underlying relationship, see Length Contraction in HSC Physics: Proper Length and Calculations.

02How could a laboratory measure the contracted length?

Knowing the equation is one thing. Obtaining a trustworthy value for \(L\) is another.

Consider an idealised experiment with a long, opaque probe travelling in a straight line at constant velocity. We’ll take motion to the right as positive.

This is a thought experiment with illustrative measurements, not a report of an experiment that has been performed.

The first design: two photographs at different times

Suppose a researcher wants to determine the probe’s laboratory-frame length by recording its front and back positions.

The first attempt uses two cameras along the track. One records the front at time \(t_1\), while the other records the back at time \(t_2\).

The researcher subtracts the recorded positions.

The prediction is straightforward: the difference should give the probe’s length.

But there’s a problem. What if \(t_1\) and \(t_2\) aren’t the same?

At \(0.60c\), the probe travels:

\[
\begin{aligned}
\Delta x&=v\Delta t\\
&=(0.60)(3.00\times10^8)\\
&\quad\times(100\times10^{-9})\\
&=18.0\ \mathrm{m}
\end{aligned}
\]

Here \(\Delta x\) is the distance travelled, and \(\Delta t=100\ \mathrm{ns}\) is the difference between the photographs, with \(1\ \mathrm{ns}=10^{-9}\ \mathrm{s}\).

If the front is recorded first and the back \(100\ \mathrm{ns}\) later, the back has moved another \(18.0\ \mathrm{m}\) forwards. Simply subtracting those positions gives a length \(18.0\ \mathrm{m}\) too small.

That error isn’t length contraction. It’s a timing mistake.

The researcher needs to measure both endpoint positions at the same laboratory time or use a method that reconstructs that simultaneous separation.

The revised design: one beam and one clock

There’s a simpler way.

Place a laser beam across the probe’s path. Connect its detector to a laboratory clock.

As the probe travels right:

1. Its nose reaches the beam, blocking the light. The clock records the start.
2. The probe continues moving, keeping the beam blocked.
3. Its tail passes the beam. Light reaches the detector again, and the clock records the end.

Both events occur at the same fixed laboratory position.

If the probe travels at constant velocity and maintains a fixed length in the laboratory frame, the time between these events tells us how far it moves while passing the detector.

That distance is exactly its contracted length:

\[
L=v\Delta t
\]

where \(\Delta t\) is the laboratory time for which the beam is blocked.

You might notice something odd. Didn’t we just say that length measurements require simultaneous positions? Why are we now using two events that happen at different times?

Because we’re doing a different calculation. We aren’t subtracting endpoint positions recorded at different times. We’re using the known constant velocity to reconstruct the separation between the endpoints at any one laboratory time.

At the instant the nose blocks the beam, the tail is a distance \(L\) behind it. By the time the tail clears the same beam, the tail has travelled exactly that distance.

The method works because the speed and length remain constant during the measurement.

Measuring the speed as well

A single beam gives us the blockage time, but we still need the probe’s speed.

We can add a second laser detector farther along the track.

Call the detectors A and B, separated by a known laboratory distance \(d\).

The nose reaches A first and B later. If the time between those nose detections is \(\Delta t_n\), then

\[
v=\frac{d}{\Delta t_n}
\]

The blockage time at A independently determines the moving length:

\[
L=v\Delta t_A
\]

Here \(\Delta t_A\) is the time between the nose blocking A and the tail clearing A.

Combining the two results gives:

\[
L_0=\frac{v\Delta t_A}{\sqrt{1-v^2/c^2}}
\]

In a realistic timing system, the detector signals must be referenced to the same laboratory time system, with signal-transmission delays calibrated. A long cable can make a perfectly accurate detector appear to respond late.

05When you should not multiply a length by \(\gamma\)

The equation \(L_0=\gamma L\) is powerful, but it has conditions.

Length contraction only affects the dimension parallel to motion

Imagine a spacecraft travelling horizontally with a long boom pointing forwards and a vertical mast pointing upwards.

The boom’s horizontal length contracts in the laboratory frame. The mast’s vertical length does not.

Would you multiply the laboratory measurement of the mast by \(\gamma\) to find its rest length?

No. Its vertical dimension is perpendicular to the relative motion, so it is unchanged by Lorentz contraction.

A diagonal length is more complicated because it has components in both directions.

You must recover the rest-frame parallel component while leaving the perpendicular component unchanged, then combine them geometrically.

Simply multiplying an entire diagonal measurement by \(\gamma\) generally gives the wrong answer.

Rest length, invariant mass, and energy are not interchangeable

Another trap is treating everything labelled “rest” as though it changes in the same way.

Consider a probe with invariant mass \(m_0\).

Its invariant mass is the same in every inertial reference frame. The rest energy is \(E_0=m_0c^2\), where energy is measured in joules (\(\mathrm{J}\)) and mass in kilograms (\(\mathrm{kg}\)).

But its total energy in a frame where it moves is:

\[
E=\gamma m_0c^2
\]

The extra energy relative to \(E_0\) is kinetic energy.

So a moving probe has a shorter measured longitudinal length and greater total energy than in its rest frame, but its invariant mass does not increase merely because of its motion.

Older descriptions sometimes use the term “relativistic mass”. For this calculation, keeping invariant mass and total energy separate is clearer.

A quick decision table

These methods assume inertial frames, subluminal relative speed, and a probe with a stable rest-frame shape during uniform motion. Optical signal delays and detector timing errors must also be accounted for.

A useful final check is that \(L_0\geq L\) whenever \(L\) really is the contracted longitudinal length. Equality occurs when \(v=0\).

06Questions and solutions

Question 1

A spacecraft travels at \(0.60c\) relative to Earth. An Earth observer measures its length parallel to its motion as \(48.0\ \mathrm{m}\).

Calculate its rest length and identify the reference frame in which that length is measured.

Solution 1

The spacecraft’s rest length is \(60.0\ \mathrm{m}\), measured in the spacecraft’s own reference frame.

The moving length is \(L=48.0\ \mathrm{m}\), and the speed is \(v=0.60c\).

Use:

\[
L_0=\frac{L}{\sqrt{1-v^2/c^2}}
\]

Substitute:

\[
\begin{aligned}
L_0&=\frac{48.0\ \mathrm{m}}{\sqrt{1-0.60^2}}\\
&=\frac{48.0\ \mathrm{m}}{0.80}\\
&=60.0\ \mathrm{m}
\end{aligned}
\]

An observer moving with the spacecraft measures \(60.0\ \mathrm{m}\), because its nose and tail are stationary in that frame.

The tempting mistake is to multiply \(48.0\ \mathrm{m}\) by \(0.80\), obtaining an even smaller length. That calculates further contraction rather than recovering the rest length.

Question 2

A probe moves to the right past two stationary laser gates.

Gate A is at \(x=0\), and gate B is at \(x=24.0\ \mathrm{m}\).

Both gates connect to a central laboratory recorder, which displays:

A calibration test reveals that gate B’s electronics delay every recorded B event by \(50\ \mathrm{ns}\) relative to the A channel. Both the blocking and clearing events at A experience the same electronic delay.

Assume the probe travels at constant velocity and has a stable length.

Determine its speed and rest length. Explain why the electronic correction affects the speed measurement but not the blockage duration at A.

Solution 2

The probe travels at \(0.800c\) and has a rest length of \(204\ \mathrm{m}\). The \(50\ \mathrm{ns}\) delay must be removed from the A-to-B time difference, but it cancels when calculating the duration between two events recorded by A.

First, correct the nose’s travel time between gates.

The recorder reports \(150\ \mathrm{ns}\), but B responds \(50\ \mathrm{ns}\) later than A’s channel.

Therefore:

\[
\Delta t_n=150-50=100\ \mathrm{ns}
\]

The speed is:

\[
\begin{aligned}
v&=\frac{24.0\ \mathrm{m}}{100\times10^{-9}\ \mathrm{s}}\\
&=2.40\times10^8\ \mathrm{m\,s^{-1}}\\
&=0.800c
\end{aligned}
\]

Next, use the two A events.

Because both are delayed by the same amount, their time difference remains \(510\ \mathrm{ns}\).

The contracted length is:

\[
\begin{aligned}
L&=v\Delta t_A\\
&=(2.40\times10^8)\\
&\quad\times(510\times10^{-9})\\
&=122.4\ \mathrm{m}
\end{aligned}
\]

For \(0.800c\):

\[
\gamma=\frac{1}{\sqrt{1-0.800^2}}=\frac53
\]

Therefore:

\[
\begin{aligned}
L_0&=\gamma L\\
&=\frac53(122.4\ \mathrm{m})\\
&=204\ \mathrm{m}
\end{aligned}
\]

The misleading approach is to trust every timestamp equally just because the recorder is accurate.

An accurate recorder can still receive signals with unequal electronic delays. The A-to-B comparison involves two channels, so their relative delay matters. The blockage duration involves one channel, so its constant delay cancels.

Question 3

A spacecraft carries a rigid rectangular panel. Two sides of the panel are parallel to the spacecraft’s motion, and the other two are perpendicular to it.

The spacecraft moves at \(0.80c\) relative to a laboratory.

At one instant in the laboratory frame, the panel measures:

• \(9.00\ \mathrm{m}\) along the direction of motion.
• \(12.0\ \mathrm{m}\) perpendicular to the direction of motion.

A straight line is painted diagonally between opposite corners of the panel.

Calculate the rest length of this diagonal.

A student suggests calculating the diagonal’s laboratory length using Pythagoras’ theorem and then multiplying the result by \(\gamma\). Determine whether this method is valid and explain your reasoning.

Solution 3

The painted diagonal has a rest length of approximately \(19.2\ \mathrm{m}\). Multiplying the entire laboratory diagonal by \(\gamma\) is incorrect because only its component parallel to motion contracts.

Start with the horizontal side.

The laboratory measures \(9.00\ \mathrm{m}\), and:

\[
\gamma=\frac{1}{\sqrt{1-0.80^2}}=\frac53
\]

Its rest-frame horizontal length is:

\[
\begin{aligned}
L_{0x}&=\gamma L_x\\
&=\frac53(9.00\ \mathrm{m})\\
&=15.0\ \mathrm{m}
\end{aligned}
\]

Here \(L_x\) is the laboratory horizontal component, and \(L_{0x}\) is its rest-frame value.

The vertical length is perpendicular to the relative motion, so it remains:

\[
L_{0y}=12.0\ \mathrm{m}
\]

Using Pythagoras’ theorem in the panel’s rest frame:

\[
\begin{aligned}
L_{0d}&=\sqrt{L_{0x}^2+L_{0y}^2}\\
&=\sqrt{15.0^2+12.0^2}\\
&=\sqrt{369}\ \mathrm{m}\\
&=19.2\ \mathrm{m}
\end{aligned}
\]

Here \(L_{0d}\) is the rest length of the diagonal.

Now examine the student’s proposed method.

The laboratory diagonal is:

\[
L_d=\sqrt{9.00^2+12.0^2}
=15.0\ \mathrm{m}
\]

Multiplying by \(\gamma\) would give:

\[
L_{0d}=15.0\left(\frac53\right)
=25.0\ \mathrm{m}
\]

That is wrong.

The error is applying longitudinal contraction to the entire diagonal, including its perpendicular component.

The diagonal also has a different orientation relative to the motion in the two frames. Its horizontal dimension changes while its vertical dimension remains fixed.

The correct calculation must therefore restore the components separately before combining them.

This is a useful boundary of the simple equation \(L_0=\gamma L\): it applies directly to a length entirely parallel to the relative motion, not an arbitrary diagonal.

Question 4

An instrument package accelerates along a straight laboratory track.

During its acceleration, its nose reaches a fixed laser gate while travelling at \(0.50c\). The tail clears the gate \(400\ \mathrm{ns}\) later, when the package is travelling at \(0.70c\).

Later, after acceleration has ended, the package travels uniformly past a second stationary laser gate. This gate remains blocked for \(450\ \mathrm{ns}\).

During the second passage, an independent laboratory measurement determines that the package’s total energy is:

\[
E=\frac53 E_0
\]

where \(E_0\) is the package’s rest energy.

Assume there is no material exchange or change in the package’s internal energy, and that its rest-frame shape is stable during the second passage.

Determine:

(a) The package’s speed during the second passage.

(b) Its rest length during the second passage.

(c) Whether the first gate’s \(400\ \mathrm{ns}\) measurement is enough to determine its rest length during acceleration.

(d) Whether the difference between \(E\) and \(E_0\) means that the package’s invariant mass has increased.

Solution 4

During its second passage, the package travels at \(0.800c\) and has a rest length of \(180\ \mathrm{m}\). Its accelerating passage does not provide enough information to recover a rest length, and the increase in laboratory-frame total energy does not imply an increase in invariant mass.

Part (a): Speed during the second passage

For a package with unchanged invariant mass:

\[
E=\gamma m_0c^2
\]

Its rest energy is:

\[
E_0=m_0c^2
\]

Dividing gives:

\[
\frac{E}{E_0}=\gamma
\]

Therefore:

\[
\gamma=\frac53
\]

Use the definition of the Lorentz factor:

\[
\frac53=\frac{1}{\sqrt{1-v^2/c^2}}
\]

Rearrange:

\[
\begin{aligned}
\sqrt{1-v^2/c^2}&=\frac35\\
1-\frac{v^2}{c^2}&=\frac9{25}\\
\frac{v^2}{c^2}&=\frac{16}{25}\\
\frac vc&=\frac45
\end{aligned}
\]

Taking the positive root because \(v\) is the speed:

\[
v=0.800c
\]

Part (b): Rest length during steady motion

During uniform motion, the beam-blockage method is valid.

\[
\begin{aligned}
L&=v\Delta t\\
&=(0.800)(3.00\times10^8)\\
&\quad\times(450\times10^{-9})\\
&=108\ \mathrm{m}
\end{aligned}
\]

This is the laboratory-frame length.

Recover the rest length:

\[
\begin{aligned}
L_0&=\gamma L\\
&=\frac53(108\ \mathrm{m})\\
&=180\ \mathrm{m}
\end{aligned}
\]

Part (c): Why the accelerating measurement is insufficient

The first measurement cannot be converted into a rest length using one Lorentz factor.

The nose crosses the gate at \(0.50c\), while the tail clears it later at \(0.70c\).

A tempting approach is to average these speeds and use \(L=v\Delta t\), followed by the length-contraction formula.

But the two supplied speeds occur at different times, and knowing them does not determine the complete motion between those times.

The tail’s displacement depends on its actual velocity throughout the interval, not just the two stated speeds. The package’s length may also change during acceleration.

Without further information about the motion and deformation, there is no justified single constant-speed length calculation for the first passage.

The second passage is different: the package moves uniformly with a stable rest-frame shape, so the method applies.

Part (d): Invariant mass versus energy

The package’s invariant mass remains \(m_0\) under the stated assumptions.

The ratio \(E/E_0=5/3\) tells us its total energy in the laboratory frame is greater than its rest energy because it has kinetic energy.

It does not mean its invariant mass has become \(5m_0/3\).

The key is to distinguish invariant mass, which is frame-independent, from total energy, which depends on the reference frame.

Question 5

An idealised research facility contains a straight test bay with two thin shutters.

The entrance shutter is at \(x=0\), and the exit shutter is at \(x=75.0\ \mathrm{m}\). Both are stationary in the laboratory frame.

A long probe travels to the right at constant speed \(0.80c\).

A laser gate at the entrance detects the probe’s nose at laboratory time \(t=0\). The tail clears that gate at \(t=250\ \mathrm{ns}\).

The facility’s controller is programmed to close both shutters at exactly \(t=280\ \mathrm{ns}\) in the laboratory frame, then immediately reopen them.

Assume the shutters are idealised, have negligible thickness, and complete these instantaneous operations without disturbing the probe. The probe retains a constant rest length.

Answer the following.

(a) Determine the probe’s rest length.

(b) Calculate the laboratory time interval during which the entire probe is strictly inside the bay. Is the probe fully inside when both shutters close at \(280\ \mathrm{ns}\)?

(c) An observer travelling with the probe measures the bay to be shorter than the probe. Explain why this does not contradict the result in part (b).

(d) The simultaneous shutter closures occur at different times in the probe’s rest frame.

For this extension calculation, use the supplied Lorentz time-coordinate relation:

\[
\Delta t’=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right)
\]

Here:

• \(\Delta t’\) is the time difference between two events in the probe frame.
• \(\Delta t\) is their time difference in the laboratory frame.
• \(\Delta x\) is the laboratory position difference between the events, taken as exit minus entrance.
• \(v\) is positive because the probe moves towards increasing \(x\).

Determine which shutter closes first in the probe frame and by how much time.

Explain why two shutters closing simultaneously in one frame does not require them to close simultaneously in another.

Solution 5

The probe’s rest length is \(100\ \mathrm{m}\). It is completely inside the \(75.0\ \mathrm{m}\) laboratory bay at \(t=280\ \mathrm{ns}\), even though the probe-frame observer measures the bay as only \(45.0\ \mathrm{m}\) long. The resolution is that the exit shutter closes approximately \(333\ \mathrm{ns}\) before the entrance shutter in the probe frame.

The challenge is that a length comparison and a statement about two events occurring simultaneously must use the same reference frame.

Part (a): Recover the probe’s rest length

The laboratory gate is blocked for:

\[
\Delta t=250\times10^{-9}\ \mathrm{s}
\]

The speed is:

\[
v=0.80c=2.40\times10^8\ \mathrm{m\,s^{-1}}
\]

Therefore, the contracted length is:

\[
\begin{aligned}
L&=v\Delta t\\
&=(2.40\times10^8)\\
&\quad\times(250\times10^{-9})\\
&=60.0\ \mathrm{m}
\end{aligned}
\]

The Lorentz factor is:

\[
\gamma=\frac{1}{\sqrt{1-0.80^2}}=\frac53
\]

So:

\[
\begin{aligned}
L_0&=\gamma L\\
&=\frac53(60.0\ \mathrm{m})\\
&=100\ \mathrm{m}
\end{aligned}
\]

The probe has a rest length of \(100\ \mathrm{m}\), but it occupies only \(60.0\ \mathrm{m}\) along the track at any one laboratory time.

Part (b): Determine when the probe fits

At \(t=0\), the nose is at the entrance, \(x=0\), while the tail is \(60.0\ \mathrm{m}\) behind it.

The entire probe first enters the bay when its tail clears \(x=0\), at:

\[
t_{\mathrm{in}}=250\ \mathrm{ns}
\]

The nose reaches the exit at \(x=75.0\ \mathrm{m}\) after:

\[
\begin{aligned}
t_{\mathrm{out}}&=\frac{75.0\ \mathrm{m}}{v}\\
&=\frac{75.0}{2.40\times10^8}\\
&=312.5\times10^{-9}\ \mathrm{s}\\
&=312.5\ \mathrm{ns}
\end{aligned}
\]

The entire probe is strictly inside the bay for laboratory times satisfying:

\[
250\ \mathrm{ns}<t<312.5\ \mathrm{ns}
\]

That is a \(62.5\ \mathrm{ns}\) window.

The shutters close at \(280\ \mathrm{ns}\), which lies inside that interval.

We can verify the result from the positions.

At \(t=280\ \mathrm{ns}\), the nose is at:

\[
\begin{aligned}
x_n&=vt\\
&=(2.40\times10^8)\\
&\quad\times(280\times10^{-9})\\
&=67.2\ \mathrm{m}
\end{aligned}
\]

The tail is one contracted length behind the nose:

\[
\begin{aligned}
x_t&=67.2-60.0\\
&=7.20\ \mathrm{m}
\end{aligned}
\]

Both endpoints are between \(x=0\) and \(x=75.0\ \mathrm{m}\).

Therefore, the entire probe fits inside the bay at that instant in the laboratory frame, with clearance at both ends.

Part (c): Why the probe-frame observer disagrees about the fit

The bay is stationary in the laboratory frame, so its proper length is \(75.0\ \mathrm{m}\).

In the probe frame, the bay moves backwards at speed \(0.80c\).

Its contracted length is:

\[
\begin{aligned}
L_{\mathrm{bay}}&=\frac{75.0\ \mathrm{m}}{\gamma}\\
&=\frac{75.0}{5/3}\\
&=45.0\ \mathrm{m}
\end{aligned}
\]

The probe-frame observer therefore measures:

• Probe length: \(100\ \mathrm{m}\).
• Bay length: \(45.0\ \mathrm{m}\).

It seems impossible for the entire probe to be inside the shorter bay.

But the laboratory statement was more specific: both shutters close at the same laboratory time while the probe is fully inside.

The probe-frame observer does not agree that those two shutter closures occur at the same time.

That difference in simultaneity removes the apparent contradiction. The observer must consider the shutters’ individual events, rather than imagining both ends of the moving bay closing together in the probe frame.

Part (d): Find the order of the shutter closures

The closures are simultaneous in the laboratory frame, so:

\[
\Delta t=0
\]

The exit is \(75.0\ \mathrm{m}\) to the right of the entrance:

\[
\Delta x=+75.0\ \mathrm{m}
\]

Use the supplied time-coordinate relation:

\[
\Delta t’=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right)
\]

Substitute \(v=0.80c\) and \(\gamma=5/3\):

\[
\begin{aligned}
\Delta t’&=\frac53
\left(0-\frac{0.80c(75.0)}{c^2}\right)\\
&=-\frac53\left(\frac{60.0}{c}\right)\\
&=-3.33\times10^{-7}\ \mathrm{s}\\
&=-333\ \mathrm{ns}
\end{aligned}
\]

The negative sign means that the exit shutter’s closure occurs earlier than the entrance shutter’s closure in the probe frame.

The two closures are separated by approximately \(333\ \mathrm{ns}\) in that frame.

This is the conceptual hinge.

In the laboratory frame, the probe is shorter, and both shutters close simultaneously around it.

In the probe frame, the bay is shorter, but the shutters close at different times. There is no instant in the probe frame when both closures occur together around the stationary probe.

The instantaneous reopening assumption also matters. It means the shutters are not left closed to obstruct the probe as it continues moving. Real shutters have finite closing times, and those details would need to be included in a physical experiment.

There is no collision in the stated idealised setup. Whether a collision occurs is a physical fact on which all observers must agree, even when they disagree about lengths and the timing of separated events.

The tempting mistake is to take the laboratory’s simultaneous shutter closures and insert them unchanged into the probe frame. Special relativity does not allow that.

07What to carry into the next problem

When you need to recover a rest length, the calculation itself is short:

\[
L_0=\gamma L
\]

The real skill is knowing where the measured length came from.

If it’s a simultaneous laboratory separation between moving endpoints, or a length reconstructed correctly from uniform motion and gate timing, you can recover the longitudinal rest length using the Lorentz factor.

If the measurement mixes times, involves acceleration, or describes an oblique dimension, you need to inspect the measurement before applying the equation.

The same Lorentz factor also connects speed to energy and momentum. Once you can distinguish a quantity measured in an object’s rest frame from one measured in the laboratory, you’re ready to apply that reasoning to calculating relativistic momentum from invariant mass and speed.