Rest Mass vs Relativistic Energy: HSC Physics Explained

An electron accelerator keeps adding energy to a beam, but the electrons barely get any faster. If the energy is increasing, does that mean each electron is becoming more massive?

No. The electron’s rest mass, also called its invariant mass, stays the same regardless of its speed. What increases is its total energy as measured in the laboratory.

The distinction comes down to two equations:

\[ E_0=m_0c^2 \]

\[ E=\gamma m_0c^2 \]

The first gives the electron’s energy in its own rest frame. The second gives its total energy in a frame where it is moving. The extra energy is kinetic energy.

You’ll sometimes see \(\gamma m_0\) described as relativistic mass. It’s an older way of expressing \(E/c^2\), but it isn’t the electron’s invariant mass.

For HSC Physics, the reliable approach is to keep rest mass fixed, identify the observer’s reference frame, and calculate energy separately.

01Rest mass and energy: what actually changes?

Imagine an electron travelling through an evacuated tube.

You stand beside the tube. The electron moves past you at \(0.60c\), where \(c\) is the speed of light in a vacuum.

Now imagine an observer moving alongside the electron at exactly the same velocity.

From that observer’s perspective, the electron is stationary. From yours, it is moving at 60% of the speed of light.

Who measures the correct mass?

Both do. They agree on the electron’s rest mass.

They don’t agree on its kinetic energy, because kinetic energy depends on the reference frame.

The electron’s rest frame

An object’s rest frame is an inertial reference frame in which that object is stationary.

For a single massive particle in its rest frame:

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

Notice that zero kinetic energy does not mean zero total energy.

Even a stationary electron possesses rest energy.

The laboratory frame

When the electron moves through the laboratory, it has both rest energy and kinetic energy.

Its total energy is:

\[ E=E_0+K \]

Here:

• \(E\) is total relativistic energy, measured in joules (J).
• \(E_0\) is rest energy, measured in joules.
• \(K\) is kinetic energy, measured in joules.
• \(m_0\) is invariant mass, measured in kilograms (kg).
• \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\).

The two observers describe different amounts of kinetic energy, but neither observes a change in the electron’s invariant mass.

Each frame has a valid description. Energy conservation still applies within either frame; it does not require different observers to assign the same numerical energy to a moving particle.

02Why increasing energy doesn’t mean increasing rest mass

Think about the speedometer and energy meter in a car.

They tell you different things. Speed describes the car’s motion relative to the road, while energy tells you something about its capacity to do work.

At ordinary speeds, adding kinetic energy normally produces an easily noticeable increase in speed.

At relativistic speeds, the relationship becomes much less straightforward.

The comparison has a limit: an electron isn’t held below the speed of light by engine resistance or friction. The energy-speed relationship itself is different.

The Lorentz factor explains the difference

The Lorentz factor, represented by the Greek letter \(\gamma\), is:

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

Here \(v\) is the particle’s speed relative to the observer, in \(\mathrm{m\,s^{-1}}\).

For a massive particle:

\[ E=\gamma m_0c^2 \]

Because rest energy is \(E_0=m_0c^2\), kinetic energy must be:

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

The three energy quantities therefore fit together:

\[ \boxed{\text{Total energy = Rest energy + Kinetic energy}} \]

If an electron travels at \(0.80c\):

\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-0.80^2}}\\ &=\frac{1}{0.60}\\ &=1.667 \end{aligned} \]

Its total energy is approximately \(1.667\) times its rest energy.

Its rest mass is still \(m_0\), not \(1.667m_0\).

As speed approaches \(c\), \(\gamma\) grows without bound. This is why an object with non-zero invariant mass cannot be accelerated to the speed of light using a finite amount of energy.

At low speeds, the relativistic kinetic energy is approximately \(\tfrac12m_0v^2\), the familiar Newtonian expression. That approximation becomes unreliable when \(v\) is a substantial fraction of \(c\).

03How an electron experiment tested the energy-speed relationship

The distinction between mass and energy becomes especially important when interpreting accelerator experiments.

Before special relativity was established, Newtonian mechanics provided an excellent model of kinetic energy:

\[ K=\frac12m_0v^2 \]

Rearranging:

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

Imagine continually increasing the energy supplied to an electron.

Under this equation, its predicted speed would keep increasing. For sufficiently large \(K\), the calculation would eventually produce a speed greater than \(c\).

But special relativity predicts that the electron approaches \(c\) without reaching it, even while its kinetic energy continues increasing.

Which prediction matches measurements?

Bertozzi’s 1964 experiment

In 1964, William Bertozzi reported an experiment designed to test the relationship between electron speed and kinetic energy.

He used an electron accelerator at the Massachusetts Institute of Technology.

The important feature was that the apparatus could measure the electrons’ motion and check their kinetic energy independently.

Follow the electron beam from left to right.

1. An electron gun within a Van de Graaff electrostatic generator produced short bursts of electrons.
2. The bursts travelled towards a linear accelerator, which could add further kinetic energy.
3. A short insulated metal tube near the start of the flight path collected some electrons, producing an electrical timing signal.
4. The remaining electrons travelled along an evacuated beam path extending 8.4 metres to an aluminium disc.
5. The electrons struck the disc, producing an arrival signal. Wires carried the starting and arrival signals to an oscilloscope.

The oscilloscope displayed the separation between the two signals in time.

That gave a way to determine how quickly the electron bursts travelled.

What the measured times revealed

Bertozzi reported measurements at several electron energies.

Here are two of his historical results, rounded as reported:

A megaelectronvolt (MeV) is one million electronvolts, a unit of energy. A nanosecond (ns) is \(10^{-9}\) seconds.

The supplied kinetic energy increased by a factor of 30, but the measured travel time decreased only slightly.

Why didn’t the second burst cross the apparatus dramatically faster?

Consider the time light would take to cover the same distance in a vacuum:

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

Here \(d\) is the path length and \(t_c\) is the light travel time.

The high-energy electron measurement had approached that limiting travel time.

It had not demonstrated that electrons could reach or exceed the speed of light.

There’s an important experimental limitation here. In Bertozzi’s higher-energy runs, the electrons were still accelerating along part or much of the path. Dividing 8.4 m by the measured time therefore gives an average speed, not necessarily the final speed. The reported times also have finite precision.

A rounded result of \(28.0\ \mathrm{ns}\) is not evidence that a massive electron travelled at exactly \(c\).

The missing experimental check: where did the energy go?

Timing measurements alone established how the electron speeds behaved.

They didn’t independently establish how much kinetic energy the electrons had gained.

There was another issue. Calculating electron energy solely from accelerating voltages assumes that the electrical energy supplied is correctly accounted for, including any energy lost along the beamline.

How could an experimenter check that assumption?

Bertozzi also used calorimetry: measuring energy transferred as heat.

The aluminium stopping disc had a thermocouple attached to it, allowing its temperature rise to be measured. A calibrated heating resistor provided a comparison, while the charge collected by the disc revealed how many electrons had arrived.

This allowed the energy transferred per electron to be estimated independently of the accelerating voltage.

For a nominal electron kinetic energy of \(1.5\ \mathrm{MeV}\), the calorimetric estimate was approximately \(1.6\ \mathrm{MeV}\), agreeing within the experiment’s roughly 10% uncertainty.

The calorimeter measures the kinetic energy transferred when electrons are stopped, apart from experimental losses.

It does not measure their entire rest energy as heat. The electrons remain present in the absorber; they have not been annihilated.

That distinction is central to this topic.

The experiment supports the relativistic relationship between increasing kinetic energy and a speed approaching \(c\). It doesn’t directly weigh a moving electron or independently prove every aspect of invariant mass. The invariant-mass interpretation belongs to the broader theory that consistently describes those measurements.

04A decision method for exam questions

When a question mentions mass, energy, or an electron accelerator, don’t select an equation immediately.

First identify exactly what quantity is being asked for.

A useful quick calculation is:

\[ \frac{E}{c^2}=\gamma m_0 \]

Both sides have units of kilograms.

But a quantity having mass units does not automatically make it the particle’s invariant mass.

At \(0.80c\), \(E/c^2=1.667m_0\). That is the laboratory energy divided by \(c^2\), not a new intrinsic mass for the electron.

One important exception: the mass of a complete system

So far, we’ve discussed an individual particle that remains unchanged while accelerating.

A composite system needs more care.

Imagine a sealed, insulated metal box initially at rest. You add energy by heating its contents and allow the system to settle, without letting energy escape.

Its total energy in its rest frame has increased. Its invariant mass has therefore increased slightly.

This does not mean every particle inside has acquired a larger individual rest mass. The box’s total invariant mass includes its internal energy.

This becomes important in collisions, where moving particles can form a stationary composite object.

It also explains why a system can lose invariant mass when it releases radiation, even though individual stable particles do not lose rest mass simply by changing speed.

For calculations involving nuclear mass defects and energy release, the related guide Mass-Energy Equivalence: Using E = mc² in HSC Physics develops that separate application.

05Questions and solutions

The following problems use constructed scenarios and illustrative data, not historical experimental measurements.

Use \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\), and assume special relativity applies in inertial reference frames.

The questions progress from individual particle energy to experiments involving multiple measurements, and finally to the invariant mass of complete systems.

Question 1

A hypothetical particle has an invariant mass of \(4.00\times10^{-27}\ \mathrm{kg}\) and moves at \(0.80c\) relative to a laboratory observer.

Calculate:

a. Its rest energy.

b. Its total relativistic energy in the laboratory.

c. Its kinetic energy.

d. Its invariant mass as measured by an observer travelling alongside it.

Solution 1

The particle’s rest energy is \(3.60\times10^{-10}\ \mathrm{J}\), its laboratory total energy is \(6.00\times10^{-10}\ \mathrm{J}\), and its kinetic energy is \(2.40\times10^{-10}\ \mathrm{J}\). Its invariant mass remains \(4.00\times10^{-27}\ \mathrm{kg}\).

Part a: Rest energy

\[ \begin{aligned} E_0&=m_0c^2\\ &=(4.00\times10^{-27})\\ &\quad\times(9.00\times10^{16})\\ &=3.60\times10^{-10}\ \mathrm{J} \end{aligned} \]

Part b: Laboratory total energy

At \(0.80c\):

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

Therefore:

\[ \begin{aligned} E&=\gamma E_0\\ &=(1.667)(3.60\times10^{-10})\\ &=6.00\times10^{-10}\ \mathrm{J} \end{aligned} \]

Part c: Kinetic energy

\[ \begin{aligned} K&=E-E_0\\ &=(6.00-3.60)\times10^{-10}\\ &=2.40\times10^{-10}\ \mathrm{J} \end{aligned} \]

Part d: Invariant mass

The observer moving alongside the particle measures zero kinetic energy, but the same invariant mass:

\[ m_0=4.00\times10^{-27}\ \mathrm{kg} \]

The tempting mistake is to use \(E/c^2=6.67\times10^{-27}\ \mathrm{kg}\) as the particle’s new rest mass.

That number represents laboratory total energy divided by \(c^2\). It is not invariant mass.

Question 2

Two electron bursts, A and B, pass through the same field-free evacuated tube of length 5.00 m. Each burst takes \(20.0\ \mathrm{ns}\) to cross the tube.

After leaving the tube, the electrons are stopped in a calorimeter. The magnitude of the total electron charge and the measured thermal energy are recorded.

Each electron has rest mass \(9.11\times10^{-31}\ \mathrm{kg}\) and charge magnitude \(1.60\times10^{-19}\ \mathrm{C}\).

Assume identical electron speeds within each burst, no significant energy escapes the calorimeter, and all kinetic energy becomes heat.

An experimenter claims that electrons in burst B have twice the relativistic mass because the burst deposits approximately twice as much energy.

Determine the number of electrons and the kinetic energy per electron in each burst. Check the results against the relativistic prediction from the flight time, and assess the experimenter’s claim.

Solution 2

Burst B contains twice as many electrons as burst A. The kinetic energy per electron is approximately the same in both, about \(6.64\times10^{-14}\ \mathrm{J}\), consistent with their identical measured speeds. Neither the invariant mass nor the historical quantity called relativistic mass has doubled.

The experimenter has confused total burst energy with energy per electron.

First, calculate electron numbers.

Let \(q\) be the measured charge magnitude, \(e\) the charge magnitude of one electron, and \(N\) the number of electrons.

\[ N=\frac{q}{e} \]

For burst A:

\[ \begin{aligned} N_A&=\frac{16.0\times10^{-9}} {1.60\times10^{-19}}\\ &=1.00\times10^{11} \end{aligned} \]

For burst B:

\[ \begin{aligned} N_B&=\frac{32.0\times10^{-9}} {1.60\times10^{-19}}\\ &=2.00\times10^{11} \end{aligned} \]

Burst B has twice the number of electrons.

Next, calculate kinetic energy per electron.

Let \(Q\) represent the total measured thermal energy in joules.

\[ K=\frac{Q}{N} \]

For burst A:

\[ \begin{aligned} K_A&=\frac{6.64\times10^{-3}} {1.00\times10^{11}}\\ &=6.64\times10^{-14}\ \mathrm{J} \end{aligned} \]

For burst B:

\[ \begin{aligned} K_B&=\frac{13.3\times10^{-3}} {2.00\times10^{11}}\\ &=6.65\times10^{-14}\ \mathrm{J} \end{aligned} \]

The small difference comes from the rounded measurements.

Now check the flight-time prediction.

The speed in either run is:

\[ \begin{aligned} v&=\frac{d}{t}\\ &=\frac{5.00}{20.0\times10^{-9}}\\ &=2.50\times10^8\ \mathrm{m\,s^{-1}}\\ &=0.833c \end{aligned} \]

Therefore:

\[ \begin{aligned} \gamma&=\frac{1}{\sqrt{1-(5/6)^2}}\\ &\approx1.809 \end{aligned} \]

The relativistic kinetic energy prediction is:

\[ \begin{aligned} K&=(\gamma-1)m_0c^2\\ &=(0.809)(8.20\times10^{-14})\\ &=6.63\times10^{-14}\ \mathrm{J} \end{aligned} \]

This agrees with both calorimeter estimates to the precision of the supplied data.

The crucial experimental control is the collected charge. Without it, the total heat measurement alone cannot tell us whether there were more electrons or more kinetic energy per electron.

Both bursts have approximately the same speed and the same \(\gamma\). Their individual rest masses are identical, and even \(\gamma m_0\) is the same for electrons in both bursts.

Question 3

A moving electron and a photon enter separate stationary absorbers in the same laboratory.

The electron has an invariant mass of \(9.11\times10^{-31}\ \mathrm{kg}\) and travels at \(0.60c\).

The photon has exactly the same total laboratory energy as the moving electron.

Assume both are completely absorbed. The electron remains present in its absorber without annihilating, and any binding-energy changes, escaping radiation, and absorber recoil energy are negligible.

a. Calculate the total laboratory energy of each incoming particle.

b. Determine the thermal energy transferred to each absorber.

c. Compare their invariant masses and explain why only one of the particles has a rest frame.

A student argues that because both incoming particles have equal total energy, they must have equal rest mass and must deposit equal heat.

Assess both parts of that argument.

Solution 3

Both particles have total laboratory energy \(1.025\times10^{-13}\ \mathrm{J}\), but the electron transfers approximately \(2.05\times10^{-14}\ \mathrm{J}\) as heat, while the photon transfers approximately \(1.025\times10^{-13}\ \mathrm{J}\). Their invariant masses are different.

Equal total energy does not imply equal invariant mass or equal kinetic-energy release on stopping.

Part a: Total laboratory energy

First calculate the electron’s rest energy:

\[ \begin{aligned} E_0&=m_0c^2\\ &=(9.11\times10^{-31})\\ &\quad\times(9.00\times10^{16})\\ &=8.20\times10^{-14}\ \mathrm{J} \end{aligned} \]

At \(0.60c\), \(\gamma=1.25\).

\[ \begin{aligned} E_e&=\gamma E_0\\ &=1.25(8.20\times10^{-14})\\ &=1.025\times10^{-13}\ \mathrm{J} \end{aligned} \]

The question states that the photon has the same energy:

\[ E_\gamma=1.025\times10^{-13}\ \mathrm{J} \]

Here \(E_e\) is the electron’s total laboratory energy and \(E_\gamma\) is the photon’s energy.

Part b: Heat transferred

The electron’s kinetic energy is:

\[ \begin{aligned} K_e&=E_e-E_0\\ &=(1.025-0.820)\times10^{-13}\\ &=2.05\times10^{-14}\ \mathrm{J} \end{aligned} \]

When the electron stops without annihilating, its rest energy remains associated with the electron now in the absorber.

Under the stated assumptions, only its initial kinetic energy becomes measured heat:

\[ Q_e=2.05\times10^{-14}\ \mathrm{J} \]

A completely absorbed photon, however, transfers its entire energy to the absorber under the same assumptions:

\[ Q_\gamma=1.025\times10^{-13}\ \mathrm{J} \]

So the photon deposits five times as much heat as the electron in this idealised comparison.

Part c: Invariant mass and reference frames

The electron has a non-zero invariant mass:

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

A photon has zero invariant mass:

\[ m_\gamma=0 \]

Yet its energy is non-zero.

There is no inertial reference frame in which a photon is stationary. It travels at \(c\) in every inertial frame.

Therefore, \(E_\gamma/c^2\) must not be interpreted as a photon’s non-zero rest mass.

The student’s argument fails for two connected reasons: total energy is not necessarily rest energy, and stopping a massive particle without destroying it does not release its rest energy as heat.

Question 4

Two identical hypothetical particles, each with invariant mass \(1.00\times10^{-27}\ \mathrm{kg}\), travel towards each other at equal speeds of \(0.60c\) in a laboratory.

Take right as positive.

Particle A moves right at \(+0.60c\), while particle B moves left at \(-0.60c\).

They collide and form a single bound composite object inside an isolated chamber.

a. If no energy escapes and the composite object remains at rest, calculate its invariant mass.

b. In a second trial, exactly 40% of the particles’ original total kinetic energy escapes as two equal-energy photons travelling in opposite directions. All remaining energy stays in the composite object. Calculate its final invariant mass.

c. Explain why the result does not contradict the claim that invariant mass is independent of speed.

Assume no external forces act on the system, the composite remains intact, and radiation in part b has zero net momentum.

Solution 4

The composite object has invariant mass \(2.50\times10^{-27}\ \mathrm{kg}\) when no energy escapes. If 40% of the original kinetic energy escapes as balanced radiation, its invariant mass is \(2.30\times10^{-27}\ \mathrm{kg}\).

The important principle is that the invariant mass of a complete system includes its internal energy. It need not equal the sum of the original individual particle masses.

Part a: No energy escapes

For either particle:

\[ \gamma=\frac{1}{\sqrt{1-0.60^2}}=1.25 \]

The total initial energy is:

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

Substituting:

\[ \begin{aligned} E_i&=2(1.25)(1.00\times10^{-27})\\ &\quad\times(9.00\times10^{16})\\ &=2.25\times10^{-10}\ \mathrm{J} \end{aligned} \]

Initially, the particles have equal and opposite momenta, so the system’s total momentum is zero.

No energy escapes. After the collision, the stationary composite contains the original total energy.

Its invariant mass \(M\) is therefore:

\[ \begin{aligned} M&=\frac{E_i}{c^2}\\ &=\frac{2.25\times10^{-10}} {9.00\times10^{16}}\\ &=2.50\times10^{-27}\ \mathrm{kg} \end{aligned} \]

This exceeds the sum of the initial individual rest masses, \(2.00\times10^{-27}\ \mathrm{kg}\).

The extra contribution comes from the original kinetic energy, now stored as internal energy of the composite.

Part b: Some energy escapes as radiation

Calculate the original combined rest energy:

\[ \begin{aligned} E_{0,\mathrm{sum}}&=2m_0c^2\\ &=1.80\times10^{-10}\ \mathrm{J} \end{aligned} \]

The original combined kinetic energy is:

\[ \begin{aligned} K_i&=E_i-E_{0,\mathrm{sum}}\\ &=(2.25-1.80)\times10^{-10}\\ &=4.50\times10^{-11}\ \mathrm{J} \end{aligned} \]

The emitted radiation carries 40% of this energy:

\[ \begin{aligned} E_\gamma&=0.40K_i\\ &=1.80\times10^{-11}\ \mathrm{J} \end{aligned} \]

Therefore, the energy remaining in the composite is:

\[ \begin{aligned} E_f&=E_i-E_\gamma\\ &=2.25\times10^{-10}\\ &\quad-1.80\times10^{-11}\\ &=2.07\times10^{-10}\ \mathrm{J} \end{aligned} \]

The two photons have equal and opposite momenta. Consequently, the composite remains stationary.

Its invariant mass is:

\[ \begin{aligned} M_f&=\frac{E_f}{c^2}\\ &=\frac{2.07\times10^{-10}} {9.00\times10^{16}}\\ &=2.30\times10^{-27}\ \mathrm{kg} \end{aligned} \]

Part c: Why this is not a contradiction

Neither original particle acquired a larger rest mass simply by moving.

Instead, the particles combined to form a new system whose total rest-frame energy includes internal energy.

In the first trial, all the initial kinetic energy remains in the composite.

In the second trial, some energy leaves as radiation, reducing the composite’s invariant mass.

The tempting mistake is to add the original particle masses and ignore energy stored inside the new object. The system boundary and any energy escaping across it determine the result.

Question 5

Extension challenge – beyond typical HSC examination difficulty.

Two identical isolated spacecraft, A and B, are initially stationary in the same laboratory reference frame. Each spacecraft has initial invariant mass \(M\).

Both spacecraft release light carrying the same total energy in the original laboratory frame:

\[ E_{\mathrm{light}}=0.20Mc^2 \]

However, they emit the light differently.

• Spacecraft A emits two simultaneous pulses in opposite horizontal directions. Each pulse carries energy \(0.10Mc^2\).
• Spacecraft B emits one pulse towards the right carrying energy \(0.20Mc^2\).

Both spacecraft remain intact. They are free to recoil, and no external forces or other energy losses occur.

Use the fact that a photon with energy \(E_\gamma\) carries momentum of magnitude:

\[ p_\gamma=\frac{E_\gamma}{c} \]

For a massive object, total energy and momentum in a chosen inertial frame satisfy:

\[ E=\gamma mc^2,\qquad p=\gamma mv \]

Here \(m\) is its invariant mass, and \(p\) is signed momentum along the horizontal axis.

Determine:

a. The final direction of motion of each spacecraft.

b. Each spacecraft’s final invariant mass as a fraction of \(M\).

c. The recoil kinetic energy of spacecraft B.

d. Why the spacecraft lose different amounts of invariant mass despite emitting equal total light energy.

Take right as positive. Give numerical mass fractions to four significant figures.

Solution 5

Spacecraft A remains stationary and finishes with invariant mass \(0.8000M\). Spacecraft B recoils left at \(0.25c\) and finishes with invariant mass \(0.7746M\). Its recoil kinetic energy is \(0.0254Mc^2\).

The surprising result is that B loses more invariant mass than A, even though both emit the same laboratory light energy.

The reason is momentum conservation. Spacecraft B must also acquire recoil kinetic energy.

Part a: Determine motion using momentum conservation

Initially both spacecraft are stationary, so their momenta are zero.

For A, the two photons travel in opposite directions and have equal energies.

Their momenta cancel:

\[ \begin{aligned} p_{\mathrm{light},A} &=\frac{0.10Mc^2}{c}\\ &\quad-\frac{0.10Mc^2}{c}\\ &=0 \end{aligned} \]

Spacecraft A therefore remains stationary.

For B, the single photon carries momentum to the right:

\[ p_{\mathrm{light},B}=+0.20Mc \]

To conserve momentum, spacecraft B must recoil left:

\[ p_B=-0.20Mc \]

Part b: Calculate the final invariant masses

For A, there is no recoil kinetic energy.

Its initial rest energy is \(Mc^2\), and the photons remove \(0.20Mc^2\).

Thus:

\[ \begin{aligned} M_Ac^2&=Mc^2-0.20Mc^2\\ M_A&=0.8000M \end{aligned} \]

For B, subtracting the emitted energy gives the spacecraft’s final total laboratory energy:

\[ E_B=0.80Mc^2 \]

But B is moving, so we cannot immediately write \(M_B=0.80M\).

Instead, use:

\[ \frac{p_B}{E_B}=\frac{v_B}{c^2} \]

This follows by dividing \(p=\gamma mv\) by \(E=\gamma mc^2\).

Rearranging and substituting:

\[ \begin{aligned} v_B&=\frac{p_Bc^2}{E_B}\\ &=\frac{(-0.20Mc)c^2} {0.80Mc^2}\\ &=-0.25c \end{aligned} \]

The negative sign means recoil to the left.

Now calculate B’s Lorentz factor:

\[ \begin{aligned} \gamma_B&=\frac{1}{\sqrt{1-0.25^2}}\\ &=1.0328 \end{aligned} \]

Since \(E_B=\gamma_BM_Bc^2\):

\[ \begin{aligned} M_B&=\frac{E_B}{\gamma_Bc^2}\\ &=\frac{0.80M}{1.0328}\\ &=0.7746M \end{aligned} \]

Thus B’s final invariant mass is smaller than A’s.

Part c: Calculate B’s recoil kinetic energy

The spacecraft’s total energy is the sum of its rest and kinetic energies:

\[ E_B=M_Bc^2+K_B \]

Therefore:

\[ \begin{aligned} K_B&=E_B-M_Bc^2\\ &=(0.8000-0.7746)Mc^2\\ &=0.0254Mc^2 \end{aligned} \]

This is the kinetic energy spacecraft B gains by recoiling.

Part d: Explain the different mass losses

For A:

\[ \Delta M_A=M-0.8000M=0.2000M \]

For B:

\[ \Delta M_B=M-0.7746M=0.2254M \]

Spacecraft B loses more invariant mass because its initial rest energy supplies both the outgoing light energy and the spacecraft’s recoil kinetic energy.

The energy accounting for B, in the original laboratory frame, is:

\[ \begin{aligned} Mc^2 &=0.7746Mc^2\\ &\quad+0.0254Mc^2\\ &\quad+0.2000Mc^2 \end{aligned} \]

The terms on the right are the final spacecraft rest energy, spacecraft recoil kinetic energy, and emitted light energy.

Their sum equals the original energy.

The tempting route is to use \(\Delta M=E_{\mathrm{light}}/c^2\) for both spacecraft.

That works for A because its total momentum remains zero and its final total laboratory energy is also its rest energy.

It fails for B because its final laboratory energy includes recoil kinetic energy. Momentum conservation forces that recoil.

The deeper lesson is that equal energy transfers do not automatically produce equal changes in invariant mass. You must consider the momentum of the complete system and identify its rest frame.

06What to carry into the next problem

When a question asks you to distinguish rest mass from relativistic energy, make three decisions.

1. Identify the object or system. An unchanged individual particle has fixed invariant mass. A composite system can gain or lose invariant mass when its internal energy changes.
2. Identify the reference frame. A massive particle has total energy \(m_0c^2\) in its rest frame and \(\gamma m_0c^2\) when moving relative to the observer.
3. Identify what the measurement represents. A flight-time measurement gives speed, a calorimeter generally measures energy deposited, and neither by itself establishes that a particle’s rest mass has changed.

The link between energy and momentum is the natural next step. A moving electron’s total energy changes with the observer, and so does its momentum, while its invariant mass remains unchanged.

To calculate that momentum without confusing it with rest mass or energy, continue with How to Calculate Relativistic Momentum from Mass and Speed.