Why More Energy Cannot Push a Massive Particle Past Light Speed

In a 1964 electron-accelerator experiment, a burst of electrons took about 28.0 nanoseconds to travel 8.4 metres. Light takes 28.0 nanoseconds to cover the same distance in a vacuum. Had the electrons finally reached light speed?

No. The measurement was rounded, and the electrons were still gaining energy along much of the apparatus.

More importantly, a particle with non-zero invariant mass cannot reach the speed of light, no matter how much finite energy is supplied. As its speed approaches \(c\), more energy produces progressively smaller increases in speed. Its kinetic energy and momentum can keep increasing without limit, but its speed remains below \(c\).

The question is why the relationship between energy and speed behaves this way, and how an experiment can actually test it.

01What changes when an electron gains energy?

Imagine an electron travelling through a straight, evacuated tube.

An electric field pushes it forwards, increasing its speed. At first, this resembles a cyclist pedalling harder along a straight road: energy is supplied, and the cyclist moves faster.

But the analogy has a limit. A cyclist eventually encounters increasing air resistance and mechanical limitations. A particle’s light-speed limit is not caused by friction or an opposing force. It follows from the relationship between energy, momentum, and motion in special relativity.

To understand that relationship, first separate the particle’s mass from its energy.

Rest mass is not the same as total energy

An electron has an invariant mass, also called its rest mass, of approximately

\[
m_0=9.11\times10^{-31}\ \mathrm{kg}
\]

The symbol \(m_0\) represents the mass measured in the electron’s rest frame.

A rest frame is an inertial reference frame in which the particle is stationary. For an electron travelling freely at constant velocity, we can imagine an observer moving alongside it.

In that electron’s rest frame:

• Its velocity is zero.
• Its momentum is zero.
• Its kinetic energy is zero.
• Its invariant mass is \(m_0\).
• Its total energy equals its rest energy, \(E_0=m_0c^2\).

Here, \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\) is the speed of light in a vacuum.

Now consider an observer stationary beside the laboratory accelerator.

The electron moves past this observer, so it has kinetic energy and non-zero momentum. Its total energy is greater than its rest energy.

However, both observers agree on its invariant mass.

For every straight-line accelerator example in this article, take the electron beam’s direction of travel, from source to target, as the positive \(x\)-direction.

An electron moving towards the target has positive velocity \(v_x\). Its speed \(v\) is the positive magnitude of that velocity.

Because an electron has negative charge, an electric force accelerating it towards positive \(x\) requires an electric field pointing towards negative \(x\). Force direction and electric-field direction are opposite for an electron.

One further detail: while an electron is accelerating, there is no single inertial rest frame that follows it throughout the acceleration. We can instead consider its instantaneous rest frame at each moment.

Energy in the two reference frames

The electron’s rest energy is

\[
E_0=m_0c^2
\]

In a laboratory frame where the electron moves at speed \(v\), its total relativistic energy is

\[
E=\gamma m_0c^2
\]

The Lorentz factor, \(\gamma\), is

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

Therefore, its kinetic energy is

\[
K=(\gamma-1)m_0c^2
\]

Here:

• \(E_0\) is rest energy.
• \(E\) is total energy in the chosen reference frame.
• \(K\) is kinetic energy in that frame.
• \(m_0\) is invariant mass.
• \(v\) is speed relative to the observer.
• \(\gamma\) is dimensionless.

Energy can be measured in joules (J) or electronvolts (eV). For electron accelerators, megaelectronvolts (MeV) are convenient, where \(1\ \mathrm{MeV}=1.602\times10^{-13}\ \mathrm{J}\).

The electron’s rest energy is approximately \(0.511\ \mathrm{MeV}\).

The key relationship is

\[
E=E_0+K
\]

Adding energy to an individual stable particle increases its kinetic and total energy in the laboratory frame. It does not increase the particle’s invariant mass.

Older physics texts sometimes describe \(\gamma m_0\) as relativistic mass. That language can be misleading because it makes a changing energy quantity sound like a change in the particle’s intrinsic mass.

For a closer comparison of those quantities, see Rest Mass vs Relativistic Energy: HSC Physics Explained.

02Why the classical energy equation fails near light speed

At ordinary speeds, kinetic energy is calculated using

\[
K=\frac12m_0v^2
\]

This Newtonian equation works extremely well for objects travelling much slower than light.

Rearranging gives

\[
v=\sqrt{\frac{2K}{m_0}}
\]

The equation predicts that increasing kinetic energy always increases speed. It contains no limiting speed.

Suppose an electron receives kinetic energy equal to its own rest energy.

Using the classical equation:

\[
\begin{aligned}
K&=m_0c^2\\
v^2&=\frac{2m_0c^2}{m_0}\\
v&=\sqrt{2}\,c\\
&=1.41c
\end{aligned}
\]

The classical model predicts a speed 41% greater than the speed of light.

That isn’t evidence that electrons can travel faster than light. It tells us the classical equation is being applied outside the conditions where it is a good approximation.

The Lorentz factor changes the prediction

Recall that

\[
K=(\gamma-1)m_0c^2
\]

Because \(E_0=m_0c^2\), we can write

\[
K=(\gamma-1)E_0
\]

Rearranging gives

\[
\gamma=1+\frac{K}{E_0}
\]

This equation reveals something important: adding kinetic energy increases the Lorentz factor.

Now connect the Lorentz factor to speed:

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

As more energy is supplied, \(\gamma\) becomes larger. The fraction \(1/\gamma^2\) becomes smaller, so \(v/c\) approaches 1.

But for any finite kinetic energy, \(\gamma\) remains finite. Consequently, \(1/\gamma^2\) stays positive, and the square root remains less than 1.

The electron approaches light speed without reaching it.

Here is the difference numerically.

Illustrative model predictions, not experimental measurements

Notice the last two rows.

Increasing kinetic energy from \(E_0\) to \(9E_0\) raises the relativistic speed from about \(0.866c\) to \(0.995c\).

The electron has gained eight times its rest energy in additional kinetic energy, yet its speed has increased by only about \(0.129c\).

This does not mean the extra energy has disappeared.

It contributes to the electron’s total energy and relativistic momentum. The energy-speed relationship simply becomes much less sensitive to additional energy near \(c\).

Why not just accelerate it a little harder?

There is no final amount of energy that gets an electron across the light-speed boundary.

As \(v\) approaches \(c\),

\[
\gamma\longrightarrow\infty
\]

Therefore,

\[
K=(\gamma-1)m_0c^2
\]

also increases without limit for a particle with non-zero invariant mass.

Reaching exactly \(c\) would require infinite energy in this idealised model.

This doesn’t mean a sufficiently large accelerator would make the electron’s speed stop changing completely. Every additional positive energy gain still increases its speed slightly, provided the energy goes into the electron’s kinetic energy.

A photon is a useful contrast. A photon has zero invariant mass, carries energy and momentum, and travels at \(c\) in a vacuum. It has no rest frame. A photon is not an ordinary massive particle that has been accelerated until it reaches \(c\).

03How an electron accelerator tested the prediction

Special relativity was introduced in 1905. By the 1960s, particle accelerators already relied on relativistic physics.

However, physicist William Bertozzi wanted an experiment that would demonstrate the relationship between kinetic energy and speed through relatively direct measurements.

His experiment, reported in 1964, used accelerated electrons at the Massachusetts Institute of Technology.

The central question was straightforward:

If electrons receive increasingly large amounts of kinetic energy, will their measured speeds follow the unrestricted classical prediction, or approach a limiting value as relativity predicts?

The apparatus: following the electron from source to target

The experiment used a Van de Graaff electrostatic generator, an electron linear accelerator, a timing system, and an aluminium stopping target.

Follow the beam from left to right.

First: electron production and initial acceleration.

An electron gun inside the Van de Graaff system produced short bursts of electrons.

The Van de Graaff generator accelerated these electrons using a large electric potential difference.

Second: the starting timing signal.

The electrons emerged towards the linear accelerator.

A short insulated metal tube collected some electrons from each burst. This produced an electrical signal indicating the burst’s arrival near the beginning of the measured flight path.

The remaining electrons continued forwards.

Third: travel through the evacuated beamline.

The beam travelled along an approximately 8.4 m path through the linear accelerator.

For the lower-energy measurements, the linear accelerator was switched off. It effectively acted as an evacuated flight tube.

For higher-energy measurements, sections of the linear accelerator supplied additional energy while the electrons were travelling along the path.

Fourth: the stopping target and arrival signal.

At the far end, the electrons struck an aluminium disc.

Charge collected on the disc produced a second electrical signal indicating their arrival.

Fifth: measuring the interval.

Cables carried the starting and arrival signals to an oscilloscope, an instrument that displays electrical signals against time.

The separation between those signals gave the electron burst’s travel time, after accounting for signal delays.

The experiment’s timing method resembles measuring a cyclist between two timing gates.

If the cyclist moves at constant speed, distance divided by time gives that speed.

If the cyclist accelerates between the gates, distance divided by time gives the average speed, not necessarily the speed at the finish.

That distinction becomes important for Bertozzi’s highest-energy runs.

What should the timing measurements reveal?

The first useful comparison is with light.

For the 8.4 m path, the time light takes in a vacuum is

\[
\begin{aligned}
t_c&=\frac{d}{c}\\
&=\frac{8.4\ \mathrm m}{3.00\times10^8\ \mathrm{m\,s^{-1}}}\\
&=2.80\times10^{-8}\ \mathrm s\\
&=28.0\ \mathrm{ns}
\end{aligned}
\]

Here, \(d\) is the laboratory path length, and \(t_c\) is the vacuum light-travel time.

One nanosecond (ns) is \(10^{-9}\) seconds.

If electrons continued accelerating according to the Newtonian kinetic-energy equation, sufficiently energetic electrons would eventually be predicted to cross the path in less than 28.0 ns.

Relativity predicts that an electron with non-zero invariant mass cannot cross the same stationary path in less than the light-travel time.

The important comparison is between independently measured energy and travel time.

What Bertozzi actually observed

The experiment reported the following timing measurements.

Historical experimental measurements, Bertozzi (1964)

At the lower energies, the powered acceleration occurred before the measured flight through the otherwise field-free tube.

Increasing the kinetic energy from 0.50 to 1.50 MeV reduced the flight time from 32.3 ns to 29.2 ns.

The electron was moving faster, but the change was becoming relatively small.

At 15 MeV, the reported travel time was 28.0 ns.

This is the same as the light-travel time when both are rounded to the reported precision.

It does not establish that the electrons reached exactly \(c\).

Two experimental limitations matter.

First, no timing measurement has unlimited precision. A result reported as 28.0 ns does not establish an exactly measured interval of 28.000000… ns.

Second, in the 15 MeV run, the electrons continued accelerating along much of the measured flight path.

Dividing the full 8.4 m by the travel time therefore gives an average speed. It does not directly give the electron’s final speed at the aluminium target.

This is a particularly important point when interpreting the experiment: the final kinetic energy and the average speed refer to different aspects of the particle’s journey.

The second measurement: proving the electrons really gained energy

Timing alone wasn’t enough.

An accelerator’s voltage or electric-field settings provide a way to calculate the expected kinetic-energy gain. But the experiment was intended to test the relationship between energy and speed.

Relying only on calculated accelerator energy could introduce assumptions about how much energy the electrons actually received and how much they lost before reaching the target.

Bertozzi therefore included an independent energy measurement using the aluminium disc.

When electrons struck the disc and stopped, most of their kinetic energy became thermal energy in the disc.

A thermocouple attached to the disc measured its temperature rise. A calibrated electrical heating resistor established how the temperature response related to energy input.

The charge collected by the disc provided an estimate of the number of arriving electrons.

Combining these measurements gave the energy deposited per electron.

For an accelerator setting corresponding to approximately \(1.5\ \mathrm{MeV}\) per electron, the thermal measurement gave approximately \(1.6\ \mathrm{MeV}\).

The difference was consistent with the experiment’s uncertainty of roughly 10%.

This supported the accelerator-based energy estimate. The electrons were continuing to gain substantial kinetic energy while their measured speed approached the limiting speed predicted by special relativity.

The calorimeter did not measure the entire rest energy of each electron as heat. The electrons were stopped and absorbed, not annihilated.

Their kinetic energy was transferred to the target.

That experimental distinction supports the theoretical distinction between invariant mass and kinetic energy.

The experiment did not independently prove every aspect of special relativity or directly weigh a moving electron. It supplied direct evidence that the energy-speed relationship agreed with relativistic predictions rather than the unrestricted Newtonian model.

Historical source: William Bertozzi, “Speed and Kinetic Energy of Relativistic Electrons”, American Journal of Physics, volume 32, pages 551-555 (1964).

04Calculating what more energy actually does

For a massive particle with known invariant mass, the useful calculation sequence is:

1. Identify the reference frame in which the energy and speed are measured.
2. Find the rest energy \(E_0=m_0c^2\).
3. Determine whether the supplied energy is kinetic energy \(K\) or total energy \(E\).
4. Calculate \(\gamma=1+K/E_0\).
5. Find \(v/c=\sqrt{1-\gamma^{-2}}\).
6. If calculating travel time using \(t=d/v\), check that the particle’s speed is constant over the stated distance.

The last check prevents a common mistake when applying equations to an accelerating beam.

05Questions and solutions

Question 1

An electron has kinetic energy equal to its rest energy.

Its rest energy is \(0.511\ \mathrm{MeV}\).

Calculate:

a) Its Lorentz factor.

b) Its speed as a fraction of \(c\).

c) Whether increasing the kinetic energy further could make it reach exactly \(c\).

Solution 1

The electron has Lorentz factor \(2.00\) and speed \(0.866c\). Increasing its kinetic energy further will increase its speed, but no finite energy gain can make it reach exactly \(c\).

Part a: Lorentz factor

\[
\begin{aligned}
\gamma&=1+\frac{K}{E_0}\\
&=1+\frac{0.511}{0.511}\\
&=2.00
\end{aligned}
\]

Part b: Speed

\[
\begin{aligned}
\frac{v}{c}&=\sqrt{1-\frac{1}{2^2}}\\
&=\sqrt{0.75}\\
&=0.866
\end{aligned}
\]

Thus,

\[
v=0.866c
=2.60\times10^8\ \mathrm{m\,s^{-1}}
\]

Part c: The limiting speed

For the electron to reach \(c\), the Lorentz factor would have to increase without limit.

Since

\[
K=(\gamma-1)E_0
\]

the required kinetic energy would also increase without limit.

The tempting mistake is to use the classical equation and obtain \(v=\sqrt{2}c\). That result shows the classical approximation has failed, not that an electron can exceed light speed.

Question 2

A laboratory uses two stationary timing detectors separated by 6.00 m along a straight, field-free electron beamline.

The oscilloscope reports a time difference of \(19.76\ \mathrm{ns}\).

A calibration test then reveals that the cable carrying the start signal takes \(1.50\ \mathrm{ns}\) longer to reach the oscilloscope than the cable carrying the arrival signal.

An independent calorimeter measures the electrons’ kinetic energy as \(1.00\pm0.02\ \mathrm{MeV}\) per electron.

Assume the electrons have constant speed between the timing detectors and that all timing delays have been identified.

a) Calculate the corrected electron flight time.

b) Calculate the speed inferred from the corrected time.

c) Use the corrected speed to estimate the kinetic energy and determine whether it agrees with the calorimeter.

d) Explain why the original timing result appeared to indicate faster-than-light travel.

Use \(E_0=0.511\ \mathrm{MeV}\).

Solution 2

The corrected flight time is \(21.26\ \mathrm{ns}\), corresponding to approximately \(0.941c\). The inferred kinetic energy is \(0.996\ \mathrm{MeV}\), consistent with the calorimeter. There is no evidence of faster-than-light travel.

Part a: Correct the timing measurement

The oscilloscope measures the difference between the arrival times of the two electrical signals, not directly between the physical detector events.

Because the start signal has an additional \(1.50\ \mathrm{ns}\) cable delay, the observed interval is shorter than the electron’s actual flight time.

Therefore,

\[
\begin{aligned}
t_{\mathrm{flight}}
&=19.76+1.50\\
&=21.26\ \mathrm{ns}
\end{aligned}
\]

Part b: Calculate the speed

The laboratory distance is \(d=6.00\ \mathrm m\).

\[
\begin{aligned}
v&=\frac{d}{t}\\
&=\frac{6.00}{21.26\times10^{-9}}\\
&=2.82\times10^8\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Hence,

\[
\frac{v}{c}=0.9407
\]

The electron moves at approximately 94.1% of light speed.

Part c: Check against the calorimeter

First calculate the Lorentz factor.

\[
\begin{aligned}
\gamma&=\frac{1}{\sqrt{1-0.9407^2}}\\
&\approx2.95
\end{aligned}
\]

Using the unrounded speed in the calculation gives \(\gamma\approx2.949\).

Now calculate kinetic energy.

\[
\begin{aligned}
K&=(\gamma-1)E_0\\
&=(2.949-1)(0.511)\\
&=0.996\ \mathrm{MeV}
\end{aligned}
\]

This agrees with the calorimeter’s \(1.00\pm0.02\ \mathrm{MeV}\) result.

The two independent measurements support the same physical interpretation.

Part d: Explain the apparent faster-than-light result

If the original oscilloscope time were used without correcting for cable delays,

\[
\begin{aligned}
v_{\mathrm{apparent}}
&=\frac{6.00}{19.76\times10^{-9}}\\
&=3.04\times10^8\ \mathrm{m\,s^{-1}}\\
&\approx1.012c
\end{aligned}
\]

The apparent violation comes from treating the delay between electrical signals as though it were the actual particle flight time.

The exact problem is that the two signal paths have different propagation delays.

This is why an accelerator experiment must calibrate its timing system rather than immediately interpreting an unexpectedly short interval as a new physical effect.

Question 3

Two identical electrons enter separate 8.00 m straight beamlines with the same initial kinetic energy, \(K=E_0\).

Each receives an additional \(3E_0\) of kinetic energy through an idealised, negligibly short accelerating gap.

The gaps are positioned differently:

• Run A: The energy is added immediately after the starting timing detector at \(x=0\).
• Run B: The energy is added halfway along the beamline at \(x=4.00\ \mathrm m\).

Outside the gaps, there are no electric fields or energy losses.

Both electrons are stopped in identical calorimeters at \(x=8.00\ \mathrm m\).

a) Calculate the electron speeds before and after the acceleration.

b) Calculate the total flight time for each run.

c) Explain why the calorimeters measure equal energy deposition even though the flight times differ.

d) Identify the incorrect assumption in calculating both flight times using only the final speed.

Solution 3

Run A arrives approximately \(1.79\ \mathrm{ns}\) earlier than Run B, even though both electrons leave the accelerator with the same kinetic energy \(4E_0\). The difference comes entirely from how long each electron spends travelling at the higher speed.

Part a: Calculate the two speeds

Initially,

\[
\begin{aligned}
\gamma_i&=1+\frac{E_0}{E_0}\\
&=2
\end{aligned}
\]

Therefore,

\[
\begin{aligned}
\frac{v_i}{c}
&=\sqrt{1-\frac{1}{2^2}}\\
&=0.8660
\end{aligned}
\]

After gaining \(3E_0\), the kinetic energy is

\[
K_f=E_0+3E_0=4E_0
\]

The final Lorentz factor is

\[
\gamma_f=1+4=5
\]

Therefore,

\[
\begin{aligned}
\frac{v_f}{c}
&=\sqrt{1-\frac{1}{5^2}}\\
&=0.9798
\end{aligned}
\]

Part b: Calculate the flight times

Run A receives its energy immediately after the start detector. Under the ideal instantaneous-gap assumption, it travels essentially all 8.00 m at \(0.9798c\).

\[
\begin{aligned}
t_A&=\frac{8.00}{0.9798(0.300)}\\
&=27.22\ \mathrm{ns}
\end{aligned}
\]

Run B travels the first 4.00 m at \(0.8660c\), then the remaining 4.00 m at \(0.9798c\).

\[
\begin{aligned}
t_B&=\frac{4.00}{0.8660(0.300)}\\
&\quad+\frac{4.00}{0.9798(0.300)}\\
&=15.40+13.61\\
&=29.00\ \mathrm{ns}
\end{aligned}
\]

Hence,

\[
\begin{aligned}
\Delta t&=t_B-t_A\\
&=29.00-27.22\\
&=1.79\ \mathrm{ns}
\end{aligned}
\]

Part c: Explain the equal calorimeter readings

Both electrons begin with kinetic energy \(E_0\) and receive an additional \(3E_0\).

Both reach the target with kinetic energy \(4E_0\).

Assuming all their kinetic energy becomes measured heat, each deposits \(4E_0\) in its calorimeter.

The total energy gained does not determine how quickly that energy was supplied.

Part d: Identify the incorrect assumption

Using

\[
t=\frac{8.00}{v_f}
\]

for both runs assumes both electrons travel the full 8.00 m at their final speed.

That is true to a good approximation for Run A, but false for Run B.

Run B travels half the distance before receiving the additional energy.

The important distinction is between final speed and speed throughout the journey. Equal final kinetic energies do not guarantee equal travel times.

Question 4

An insulated box of empty rest mass \(M\) is stationary in a laboratory.

Inside it, two identical neutral particles, each of invariant mass \(m_0\), travel in exactly opposite directions at \(0.60c\).

The particles strike identical absorbers and come to rest relative to the box. Their kinetic energy becomes heat retained inside the box.

Assume the complete system of box and particles is isolated, no energy escapes, and interactions before absorption can be neglected. The opposite momenta cancel, so the box remains stationary.

a) Determine the total energy and invariant mass of the complete system before absorption.

b) Calculate the thermal energy produced.

c) Determine whether the invariant mass of the complete system changes after absorption.

d) Explain why adding energy to a massive particle and adding internal energy to a composite system must not be treated as identical statements about mass.

Solution 4

The invariant mass of the complete isolated system remains \(M+2.50m_0\) before and after the particles stop. Their kinetic energy becomes internal thermal energy, rather than disappearing or changing either particle’s invariant mass.

This is an extension of the usual single-particle HSC calculation: the invariant mass of a whole system must include its internal energy.

Part a: Initial total energy and system mass

Each particle has Lorentz factor

\[
\begin{aligned}
\gamma&=\frac{1}{\sqrt{1-0.60^2}}\\
&=1.25
\end{aligned}
\]

The box is initially at rest, so its energy is \(Mc^2\).

The two moving particles contribute

\[
2\gamma m_0c^2=2.50m_0c^2
\]

Therefore,

\[
E_i=(M+2.50m_0)c^2
\]

The particles have equal and opposite momentum. The box is also at rest.

Consequently, the complete system’s total momentum is zero.

In this centre-of-momentum frame, the system’s total energy equals its invariant mass multiplied by \(c^2\).

Thus,

\[
M_{\mathrm{system}}=M+2.50m_0
\]

Part b: Thermal energy produced

Each particle initially has kinetic energy

\[
\begin{aligned}
K&=(\gamma-1)m_0c^2\\
&=0.25m_0c^2
\end{aligned}
\]

Two particles produce

\[
Q=2K=0.50m_0c^2
\]

of thermal energy.

If \(m_0\) is expressed in kilograms and \(c\) in metres per second, \(Q\) is in joules.

Part c: System mass after absorption

After absorption, the two particles are stationary relative to the box.

Their combined rest energy is \(2m_0c^2\).

However, the box also contains the thermal energy produced when they stopped.

The total final energy is

\[
\begin{aligned}
E_f&=Mc^2+2m_0c^2+Q\\
&=Mc^2+2m_0c^2\\
&\quad+0.50m_0c^2\\
&=(M+2.50m_0)c^2
\end{aligned}
\]

Therefore,

\[
M_{\mathrm{system,final}}
=M+2.50m_0
\]

The invariant mass of the complete isolated system is unchanged.

Part d: Explain the distinction

The tempting argument is that once the particles stop moving, the system must have mass \(M+2m_0\).

That overlooks the thermal energy stored in the box.

Each individual particle still has invariant mass \(m_0\).

The complete system has additional energy stored internally, and that energy contributes to the system’s invariant mass.

If energy were supplied from outside to an initially stationary system without transferring net momentum, its invariant mass could increase by the supplied energy divided by \(c^2\).

Here, however, the system is isolated. Energy is only redistributed internally, so its total invariant mass remains constant.

This does not contradict the light-speed limit. The invariant mass of a single stable massive particle and the invariant mass of a composite system are related concepts, but they are not interchangeable.

Question 5

An electron beamline is 9.00 m long. Timing detectors are positioned at \(x=0\), \(x=6.00\ \mathrm m\), and \(x=9.00\ \mathrm m\).

Every electron enters the beamline travelling at \(0.60c\) and leaves at \(0.80c\).

Two different acceleration programmes are proposed.

Programme A

The electron travels the first 3.00 m at \(0.60c\), receives all its additional kinetic energy in a single idealised accelerating gap at \(x=3.00\ \mathrm m\), and travels the remaining 6.00 m at \(0.80c\).

Programme B

Immediately after passing the \(x=0\) timing detector, the electron receives an initial positive energy gain, increasing its speed to an unknown value \(u\).

It travels at speed \(u\) until \(x=6.00\ \mathrm m\), where a second idealised accelerating gap increases its speed to \(0.80c\).

Both programmes use only acceleration, never deceleration.

Assume the gaps have negligible length, energy losses are negligible, and the timing system is perfectly calibrated.

An experimenter claims that measurements from the existing three timing detectors, combined with a calorimeter at the exit, must uniquely reveal which programme was used.

Investigate this claim.

a) Calculate Programme A’s arrival times at \(x=6.00\ \mathrm m\) and \(x=9.00\ \mathrm m\).

b) Find the intermediate speed \(u\) that makes Programme B produce exactly the same arrival times at both detectors.

c) Prove that Programme B is physically consistent with the requirement that both gaps only add kinetic energy.

d) Determine one additional detector position that would distinguish the programmes, and calculate the difference in arrival times there.

e) Explain what this reveals about using time-of-flight and final-energy measurements to reconstruct an electron’s acceleration history.

Solution 5

The experimenter’s claim is false. Both programmes can give exactly the same arrival times at \(x=6.00\ \mathrm m\) and \(x=9.00\ \mathrm m\), as well as identical final kinetic energy, despite having different acceleration histories.

A further detector at \(x=3.00\ \mathrm m\) would distinguish them, with an arrival-time difference of approximately \(2.08\ \mathrm{ns}\).

The crucial insight is that a measured travel time constrains the total time spent crossing a distance, not the unique sequence of instantaneous speeds along that distance.

Part a: Programme A

For convenient calculation, use

\[
c=0.300\ \mathrm{m\,ns^{-1}}
\]

The electron travels the first 3.00 m at \(0.60c\).

\[
\begin{aligned}
t_1&=\frac{3.00}{0.60(0.300)}\\
&=16.667\ \mathrm{ns}
\end{aligned}
\]

It then travels at \(0.80c\).

The time from \(x=3.00\ \mathrm m\) to \(x=6.00\ \mathrm m\) is

\[
\begin{aligned}
t_2&=\frac{3.00}{0.80(0.300)}\\
&=12.500\ \mathrm{ns}
\end{aligned}
\]

So its arrival time at 6.00 m is

\[
\begin{aligned}
t_A(6)&=16.667+12.500\\
&=29.167\ \mathrm{ns}
\end{aligned}
\]

The final 3.00 m also takes 12.500 ns.

Therefore,

\[
\begin{aligned}
t_A(9)&=29.167+12.500\\
&=41.667\ \mathrm{ns}
\end{aligned}
\]

Part b: Find Programme B’s intermediate speed

Programme B must reach \(x=6.00\ \mathrm m\) at the same time as Programme A.

It travels that entire distance at constant speed \(u\), after the first gap.

Therefore,

\[
\begin{aligned}
u&=\frac{6.00}{29.167}\\
&=0.205714\ \mathrm{m\,ns^{-1}}
\end{aligned}
\]

As a fraction of light speed,

\[
\begin{aligned}
\frac{u}{c}&=\frac{0.205714}{0.300}\\
&=0.685714
\end{aligned}
\]

In exact fractional form,

\[
u=\frac{24}{35}c
\]

This is approximately \(0.686c\).

After reaching the 6.00 m detector, Programme B increases the electron’s speed to \(0.80c\).

The final 3.00 m therefore takes the same 12.500 ns as Programme A.

So,

\[
\begin{aligned}
t_B(9)&=29.167+12.500\\
&=41.667\ \mathrm{ns}
\end{aligned}
\]

Both detector arrival times match exactly.

Part c: Check whether the energy changes are physically possible

The condition that both gaps only add energy is essential.

We must show that the required intermediate speed is greater than the initial speed but less than the final speed.

Numerically,

\[
0.60c<0.685714c<0.80c
\]

This already establishes that both changes increase speed.

However, the stronger check is to calculate the associated kinetic energies.

Initially, at \(0.60c\),

\[
\begin{aligned}
\gamma_i&=\frac{1}{\sqrt{1-0.60^2}}\\
&=1.25
\end{aligned}
\]

So the initial kinetic energy is

\[
K_i=(1.25-1)E_0=0.250E_0
\]

At \(u=0.685714c\),

\[
\begin{aligned}
\gamma_u
&=\frac{1}{\sqrt{1-(24/35)^2}}\\
&\approx1.374
\end{aligned}
\]

Thus,

\[
K_u\approx0.374E_0
\]

Finally, at \(0.80c\),

\[
\begin{aligned}
\gamma_f&=\frac{1}{\sqrt{1-0.80^2}}\\
&=1.667
\end{aligned}
\]

So,

\[
K_f\approx0.667E_0
\]

The kinetic energies satisfy

\[
0.250E_0<0.374E_0<0.667E_0
\]

Both energy gains in Programme B are positive.

Programme B is therefore physically consistent with the stated conditions.

This matters because an apparent timing match would not be valid if it required an unmentioned braking process. No such process is needed here.

Part d: Add a detector at 3.00 m

Programme A takes 16.667 ns to reach \(x=3.00\ \mathrm m\).

Programme B travels the same distance at \(u=0.685714c\).

Therefore,

\[
\begin{aligned}
t_B(3)&=\frac{3.00}{0.685714(0.300)}\\
&=14.583\ \mathrm{ns}
\end{aligned}
\]

The difference is

\[
\begin{aligned}
\Delta t&=16.667-14.583\\
&=2.083\ \mathrm{ns}
\end{aligned}
\]

Programme B reaches the new detector approximately \(2.08\ \mathrm{ns}\) earlier than Programme A.

The extra detector reveals the difference in their early motion.

Part e: What the experiment can and cannot reveal

The final calorimeter cannot distinguish the two programmes because both electrons arrive with the same speed, \(0.80c\), and therefore the same final kinetic energy.

The original timing detectors cannot distinguish them either because the total time across each measured section is the same.

But their speed histories are different.

Programme A spends the first 3.00 m moving relatively slowly before receiving a single larger energy gain.

Programme B receives some of its energy earlier, moves at an intermediate speed for 6.00 m, and receives the remaining energy later.

Both produce the same measured outcomes at the original detector positions.

The tempting mistake is to assume that knowing average speed and final kinetic energy uniquely determines how the electron was accelerated.

It doesn’t.

An average over an interval does not specify what happened at every position inside that interval.

The additional timing detector provides new information that the original measurements did not contain.

This is a wider lesson about experimental reasoning: even an accurate measurement can leave several physical explanations possible. The experiment must measure something that distinguishes those explanations.

06What to check when an accelerator result looks impossible

When you see a question involving a massive particle receiving more energy near light speed, separate three ideas.

First, its invariant mass stays fixed if it remains the same stable particle. Its rest energy therefore stays fixed too.

Second, additional kinetic energy raises its total energy and relativistic momentum, while its speed approaches \(c\) without reaching it.

Third, a reported speed depends on how it was measured. Dividing distance by time gives a constant speed only when the particle actually travels at constant speed over that distance. Cable delays, acceleration within the flight path, and measurement uncertainty can all change the interpretation.

The next useful question is where that additional energy appears mathematically when speed changes only slightly. The answer leads to How to Calculate Relativistic Momentum from Mass and Speed, where increasing momentum provides another way to describe the continuing effect of energy supplied to a massive particle.