Why Relativistic Momentum Is Greater Than mv at High Speeds

An electron enters a magnetic field at \(0.60c\). Another enters the same field at \(0.80c\). The second electron is travelling only one-third faster, so you might expect its path to have a radius one-third larger.

But the correct prediction is a radius about 78% larger. Why?

At high speeds, momentum is not simply \(mv\) because Newton’s momentum equation is only a low-speed approximation. Special relativity requires an extra factor that grows with speed:

\[ p=\gamma m_0v \]

Here, \(m_0\) is the particle’s constant rest mass, \(v\) is its speed relative to the observer, and \(\gamma\) is the Lorentz factor. The particle doesn’t acquire more rest mass. Instead, momentum increases more rapidly than speed because the classical relationship between motion, space, and time no longer describes how different observers measure and conserve momentum.

The interesting question is how physicists could tell. A fast electron doesn’t arrive with its momentum printed on it. They needed ways to measure its motion without assuming that \(p=mv\) was already correct.

01Why a magnetic field can reveal momentum

Picture a cyclist rounding a bend. The cyclist’s speed might stay almost constant, but their direction changes. That requires a sideways force.

An electron moving through a magnetic field behaves similarly, except that the magnetic force is perpendicular to its velocity and does no work. In a uniform field, it changes the electron’s direction without changing its speed.

The analogy has a limit. A bicycle turns through contact forces and may lose energy to friction. An ideal electron moving perpendicular to a uniform magnetic field follows circular motion without losing kinetic energy.

For a particle whose velocity is perpendicular to a magnetic field, the magnetic force has magnitude

\[ F_B=|q|vB \]

where:

• \(F_B\) is magnetic force in newtons (N).
• \(|q|\) is the magnitude of the particle’s electric charge in coulombs (C).
• \(v\) is speed in metres per second (\(\mathrm{m/s}\)).
• \(B\) is magnetic flux density in tesla (T).

For circular motion, the direction of the momentum vector continually changes. If the momentum magnitude is \(p\) and the radius is \(r\), the required inward force has magnitude

\[ F_\perp=\frac{pv}{r} \]

This follows from how quickly the momentum vector turns: a particle travelling a distance \(v\Delta t\) around a circle turns through an angle approximately \(v\Delta t/r\).

Equating the two forces gives

\[ |q|vB=\frac{pv}{r} \]

Cancelling the non-zero speed,

\[ p=|q|Br \]

This is a particularly useful experimental relationship.

A magnetic field can reveal a particle’s momentum from its path radius without first assuming \(p=m_0v\).

The result applies to circular motion in a uniform field perpendicular to the velocity. It also assumes other forces, including electric forces, are negligible.

What does a larger radius mean?

For particles with the same charge magnitude in the same magnetic field:

• Larger momentum produces a larger circular radius.
• Smaller momentum produces a smaller circular radius.
• Reversing the charge reverses the direction of curvature, not its radius at the same momentum magnitude.

There is a trap here.

You might say a fast electron travels in a wider circle because the magnetic force is weaker.

In fact, \(F_B=|q|vB\) says the magnetic force is stronger at greater speed when \(q\) and \(B\) are fixed.

The wider circle occurs because the electron’s momentum grows faster than the magnetic force does. The stronger force is still insufficient to turn that larger momentum as sharply.

The magnetic field tells us momentum. To test the relationship between momentum and speed, however, we also need to measure speed independently.

02The historical measurement problem: how fast were the electrons really moving?

Early experiments with fast electrons raised an uncomfortable question for classical mechanics: could the familiar relationships between speed, energy, and momentum still be trusted?

Investigators including Walter Kaufmann used electric and magnetic fields to study fast electrons in the early twentieth century. Those experiments were difficult to interpret because electron speeds, deflections, experimental geometry, and possible scattering had to be accounted for together.

A different experimental approach could test one part of the problem more directly: measure how long fast electrons took to travel a known distance.

Bertozzi’s 1964 experiment

In 1964, physicist William Bertozzi reported an experiment at MIT that compared electron speed with kinetic energy.

The apparatus had several important parts arranged along the electron beam’s path.

1. An electron gun inside a Van de Graaff electrostatic generator produced short bursts of electrons.
2. The electrons emerged from the generator and travelled towards a linear accelerator.
3. A short insulated metal tube near the beginning of the flight path produced an electrical timing signal when the electron burst passed.
4. The electrons travelled along an evacuated beamline with an 8.4 m flight path.
5. At the far end, an aluminium disc stopped the electrons and produced a second electrical signal.
6. Cables carried the two signals to an oscilloscope, where their time separation was measured.

The signal cables were checked for differences in transmission delay. Otherwise, a delay in the electronics could be mistaken for a delay in the electron’s flight.

What should the experiment have shown?

Classical physics predicts kinetic energy from

\[ K=\frac12m_0v^2 \]

If this relationship remained exact, adding enough kinetic energy would eventually predict an electron speed greater than the speed of light.

Special relativity instead predicts

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

Here, \(K\) is kinetic energy, measured in joules, and \(c\) is the speed of light in a vacuum, approximately \(3.00\times10^8\ \mathrm{m/s}\).

The relativistic prediction is that continued energy input increases \(\gamma\), while the electron’s speed approaches \(c\) without reaching it.

What was observed?

The following are selected measurements reported in Bertozzi’s 1964 paper, Speed and Kinetic Energy of Relativistic Electrons. They are historical experimental data, not simulated results.

A mega-electronvolt (MeV) is one million electronvolts, a unit of energy.

A nanosecond (ns) is \(10^{-9}\) seconds. The inferred speeds use rounded measurements.

There is an important experimental qualification. At 0.5 and 1.5 MeV, the linear accelerator was not supplying further acceleration along the flight path. At higher energies, electrons gained energy during part of their flight. Consequently, \(L/t\) represented an average speed, not necessarily the final speed.

The 28.0 ns measurement also does not establish that electrons reached exactly \(c\). That time is the light-travel time over 8.4 m to the reported precision, and the experiment could not resolve the extremely small remaining difference.

A useful scale estimate

How long would light take to cross 8.4 m?

\[ \begin{aligned} t&=\frac{L}{c}\\ &=\frac{8.4\ \mathrm{m}} {3.00\times10^8\ \mathrm{m/s}}\\ &=2.80\times10^{-8}\ \mathrm{s} \end{aligned} \]

That’s 28.0 ns.

An electron with non-zero rest mass cannot cross the same distance in less than this time in a vacuum, assuming both measurements use the same laboratory frame.

The experimental data showed that increasing the electron’s kinetic energy eventually produced only a very small further change in travel time.

This behaviour supported the relativistic relationship between energy and speed, rather than the classical prediction.

Why the energy measurement also needed checking

There was another possible weakness.

The accelerating voltage could be used to calculate the energy supplied to an electron. But that relies on the relationship between electric work and energy, and it leaves questions about possible energy losses inside the apparatus.

Bertozzi also used calorimetry to check the electron energy.

When electrons struck the aluminium disc, much of their kinetic energy was transferred to it as heat. A thermocouple measured the temperature response, while the collected electric charge helped determine how many electrons had arrived.

The calorimetric measurements provided an independent check on kinetic energy. They agreed with the expected accelerating-energy values within the reported experimental uncertainty.

The important historical conclusion is precise: Bertozzi’s experiment directly investigated electron speed and kinetic energy. It was not itself a direct measurement of the complete relativistic momentum equation.

To test momentum against speed, we can combine the two ideas: independently measure speed using timing, and momentum using magnetic curvature.

03Why the relativistic momentum equation has an extra factor

Before combining those measurements, we need the prediction they are testing.

The Lorentz factor is

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

It has no units.

At small speeds compared with \(c\), the fraction \(v^2/c^2\) is tiny, so \(\gamma\) is approximately 1.

At speeds approaching \(c\), the denominator becomes smaller and \(\gamma\) increases sharply.

The relativistic momentum magnitude is therefore

\[ p=\gamma m_0v \]

For motion in three dimensions, momentum is a vector:

\[ \mathbf p=\gamma m_0\mathbf v \]

The bold symbols indicate that momentum and velocity have directions. For positive rest mass, the momentum vector points in the same direction as velocity.

Why is this necessary?

In Newtonian mechanics, observers moving at different constant velocities use the same universal time. That assumption allows \(p=mv\) to work consistently.

Special relativity changes the relationship between space and time. Different inertial observers do not generally agree on elapsed time or distances between events.

A momentum law must still allow momentum and energy to transform consistently between those observers, with conservation laws remaining valid.

The classical expression \(m_0v\) cannot meet that requirement at relativistic speeds.

The corrected momentum is \(\gamma m_0v\).

In the deeper spacetime description, beyond the HSC calculation, the spatial part of a particle’s four-momentum is constructed using its own proper time rather than universal Newtonian time. The relationship between proper time and laboratory time introduces the Lorentz factor.

That is why \(\gamma\) is not an arbitrary correction invented to fit measurements.

What happens at different speeds?

The following are calculated predictions for a particle with constant rest mass \(m_0\), not experimental observations.

At \(0.10c\), the difference is small.

At \(0.80c\), the relativistic momentum is already about 67% larger than \(m_0v\).

Rest mass is not the extra momentum

The rest mass \(m_0\) is an invariant quantity. Every inertial observer agrees on its value for the same individual particle.

But the particle’s momentum and energy depend on which frame observes it.

For a massive particle, compare the following quantities:

The rest frame is the inertial reference frame in which the particle is stationary.

Rest energy is \(E_0=m_0c^2\), while total relativistic energy is \(E=\gamma m_0c^2\). Both are measured in joules.

An older approach sometimes calls \(\gamma m_0\) the “relativistic mass”.

That terminology can make the momentum equation look like \(p=mv\) again, but it hides the distinction between invariant mass and frame-dependent energy. For calculations, keeping \(m_0\) constant and using \(\gamma\) explicitly is clearer.

04How to test both predictions with one electron beam

Now imagine a teaching experiment designed specifically to compare the two momentum equations.

This is an illustrative experimental design, not a claim about Bertozzi’s apparatus.

The equipment is arranged from left to right:

• An electron gun and accelerator create a beam of known charged particles.
• Two timing detectors, separated by a measured distance \(L\), determine the electron’s speed through a field-free drift region.
• The electron enters a region containing a calibrated magnetic field perpendicular to its motion.
• A position-sensitive detector records its curved path, allowing the radius \(r\) to be determined.

The timing measurement gives

\[ v=\frac{L}{t} \]

The magnetic measurement gives

\[ p=|q|Br \]

Neither requires using the classical equation \(p=m_0v\) to determine both quantities.

That independence is the essential feature of the test.

05The decision that prevents most momentum mistakes

In an HSC question, first decide what the information actually measures.

Three ideas need to stay separate.

First, momentum is a vector. A magnetic field can change momentum by changing its direction, even when its magnitude stays constant.

Second, rest mass is invariant. The Lorentz factor doesn’t mean the electron’s rest mass has physically increased.

Third, energy and momentum are related but aren’t interchangeable. Increasing kinetic energy doesn’t imply an equally large percentage increase in speed.

For more on why a magnetic force changes direction without supplying kinetic energy, see Why Magnetic Fields Change a Charged Particle’s Direction, Not Speed.

06Questions and solutions

Assume all measurements are made in the laboratory frame unless another frame is specified.

Use these constants where needed:

• Speed of light: \(c=3.00\times10^8\ \mathrm{m/s}\)
• Electron rest mass: \(m_e=9.11\times10^{-31}\ \mathrm{kg}\)
• Proton rest mass: \(m_p=1.67\times10^{-27}\ \mathrm{kg}\)
• Elementary charge magnitude: \(e=1.60\times10^{-19}\ \mathrm{C}\)
• Electron rest energy: \(m_ec^2=0.511\ \mathrm{MeV}\)

Ignore gravitational effects, radiation losses, and interactions with matter unless otherwise stated.

Question 1

A proton moves east at \(0.60c\).

Calculate:

a. Its classical momentum.

b. Its relativistic momentum.

c. The percentage by which the relativistic momentum exceeds the classical result.

d. Its momentum as measured by an observer travelling alongside it at the same velocity.

Solution 1

The proton’s relativistic momentum is \(3.76\times10^{-19}\ \mathrm{kg\,m/s}\) east, which is 25% greater than its classical momentum. In the proton’s own rest frame, its momentum is zero.

Part a: Classical momentum

The proton’s speed is

\[ v=0.60c=1.80\times10^8\ \mathrm{m/s} \]

Therefore,

\[ \begin{aligned} p_{\mathrm{classical}}&=m_pv\\ &=(1.67\times10^{-27})\\ &\quad\times(1.80\times10^8)\\ &=3.01\times10^{-19}\ \mathrm{kg\,m/s} \end{aligned} \]

The direction is east.

Part b: Relativistic momentum

At \(0.60c\),

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

Therefore,

\[ \begin{aligned} p&=\gamma m_pv\\ &=1.25(3.01\times10^{-19})\\ &=3.76\times10^{-19}\ \mathrm{kg\,m/s} \end{aligned} \]

Part c: Percentage difference

Because \(p_{\mathrm{rel}}=\gamma p_{\mathrm{classical}}\),

\[ \begin{aligned} \%\text{ increase}&=(\gamma-1)\times100\\ &=25\% \end{aligned} \]

Part d: Rest-frame momentum

The observer travelling alongside the proton measures its velocity as zero, so its momentum is also zero.

The tempting mistake is to assume momentum must have one fixed value for all observers. Momentum is frame-dependent; invariant mass is not.

Question 2

An electron travelling at \(0.80c\) enters one of two magnetic deflectors.

Both deflectors contain uniform magnetic fields directed into the page, perpendicular to the electron’s initial motion. Each field region is shaped to turn the electron through exactly \(90^\circ\).

Deflector A has magnetic field strength \(0.0100\ \mathrm{T}\). Deflector B has twice that field strength.

The electron enters either deflector moving horizontally to the right.

a. Calculate the path radius in each deflector.

b. Calculate the magnitude and direction of the change in momentum during the \(90^\circ\) turn.

c. Which deflector produces the greater total impulse?

d. Explain whether either magnetic field increases the electron’s kinetic energy.

Solution 2

Both deflectors produce the same total impulse magnitude, \(5.15\times10^{-22}\ \mathrm{N\,s}\), even though deflector B has twice the magnetic field strength. Neither increases the electron’s kinetic energy.

The surprising point is that impulse depends on the total change in momentum, not simply on how strong the force is.

Part a: Calculate both radii

At \(0.80c\), the Lorentz factor is \(5/3\).

The electron’s momentum magnitude is

\[ \begin{aligned} p&=\gamma m_ev\\ &=\frac53(9.11\times10^{-31})\\ &\quad\times(2.40\times10^8)\\ &=3.64\times10^{-22}\ \mathrm{kg\,m/s} \end{aligned} \]

For deflector A,

\[ \begin{aligned} r_A&=\frac{p}{eB_A}\\ &=\frac{3.64\times10^{-22}} {(1.60\times10^{-19})(0.0100)}\\ &=0.228\ \mathrm{m} \end{aligned} \]

Deflector B has twice the magnetic field, so its radius is half as large:

\[ r_B=0.114\ \mathrm{m} \]

Part b: Calculate the momentum change

Take right as positive \(x\) and upwards as positive \(y\).

The initial momentum is

\[ \mathbf p_i=(p,0) \]

After the clockwise quarter-turn, it is

\[ \mathbf p_f=(0,-p) \]

Therefore,

\[ \Delta\mathbf p=\mathbf p_f-\mathbf p_i =(-p,-p) \]

Its magnitude is

\[ \begin{aligned} |\Delta\mathbf p|&=\sqrt{p^2+p^2}\\ &=\sqrt2\,p\\ &=5.15\times10^{-22}\ \mathrm{kg\,m/s} \end{aligned} \]

The momentum-change vector points downwards and to the left, \(45^\circ\) below the negative \(x\)-axis.

Part c: Compare the impulses

Impulse is the change in momentum:

\[ \mathbf J=\Delta\mathbf p \]

Both deflectors produce the same \(90^\circ\) change in direction at the same speed, so both produce the same impulse magnitude:

\[ J_A=J_B=5.15\times10^{-22}\ \mathrm{N\,s} \]

Deflector B applies twice the magnetic force, but the electron follows a path with half the radius.

For a quarter-circle, the transit time is

\[ t=\frac{\pi r}{2v} \]

So deflector B also acts for half as long.

The stronger force and shorter interaction time produce the same net impulse.

Part d: Kinetic energy

Both magnetic forces remain perpendicular to velocity. Neither does work.

The electron leaves either deflector with the same speed, kinetic energy, and momentum magnitude it had on entry.

The trap is assuming that a stronger force must always produce a larger total momentum change. The time and direction of the force matter too.

Question 3

An electron starts from rest and passes through two identical accelerating stages.

Each stage increases its kinetic energy by \(0.511\ \mathrm{MeV}\). The stages are separated by a region in which the electron travels at constant speed.

After each stage, the electron’s momentum could be measured using the same uniform magnetic analyser.

a. Determine the electron’s speed after the first stage.

b. Determine its speed after the second stage.

c. Calculate the ratio of the magnetic curvature radii after the second and first stages.

d. A student argues that doubling the total kinetic energy should double the magnetic radius. Explain precisely why this prediction fails.

Solution 3

After the first stage, the electron travels at approximately \(0.866c\). After the second, it travels at \(0.943c\). Its magnetic curvature radius increases by a factor of approximately 1.63, not 2.

Part a: Speed after stage 1

The relativistic kinetic-energy equation is

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

After the first stage,

\[ K_1=0.511\ \mathrm{MeV}=m_ec^2 \]

Therefore,

\[ \gamma_1-1=1 \]

So \(\gamma_1=2\).

Using the Lorentz factor,

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

and rearranging,

\[ \frac vc=\sqrt{1-\frac1{\gamma^2}} \]

Thus,

\[ \frac{v_1}{c}=\sqrt{1-\frac14}=0.866 \]

Therefore, \(v_1=0.866c\).

Part b: Speed after stage 2

The total kinetic energy is now

\[ K_2=1.022\ \mathrm{MeV}=2m_ec^2 \]

Hence,

\[ \gamma_2-1=2 \]

So \(\gamma_2=3\).

Therefore,

\[ \frac{v_2}{c} =\sqrt{1-\frac19} =0.943 \]

The second stage adds the same kinetic energy, but the electron’s speed increases only from \(0.866c\) to \(0.943c\).

Part c: Compare curvature radii

In the same magnetic field,

\[ r=\frac{p}{eB} \]

Therefore,

\[ \frac{r_2}{r_1}=\frac{p_2}{p_1} \]

The relativistic momentum can be expressed as \(p=\gamma m_ev\).

For stage 1,

\[ p_1=2m_ec\sqrt{1-\frac14} =\sqrt3\,m_ec \]

For stage 2,

\[ p_2=3m_ec\sqrt{1-\frac19} =\sqrt8\,m_ec \]

Hence,

\[ \begin{aligned} \frac{r_2}{r_1} &=\frac{\sqrt8\,m_ec}{\sqrt3\,m_ec}\\ &=\sqrt{\frac83}\\ &\approx1.63 \end{aligned} \]

Part d: Explain the failed prediction

Equal increases in kinetic energy do not produce equal increases in speed or momentum.

The classical relationship \(K=\frac12m_ev^2\) is no longer valid at these speeds.

More importantly, the magnetic radius follows momentum, not kinetic energy directly.

The electron receives twice as much total kinetic energy after two stages, but its relativistic momentum grows by a factor of approximately 1.63.

The analyser measures this momentum change through curvature, not through how much energy the accelerator has supplied.

Question 4

A beam passes through a velocity selector followed by a magnetic analyser.

In the selector:

• The electric field points vertically downwards.
• The magnetic field points into the page.
• Particles enter moving horizontally to the right.
• The electric field has magnitude \(4.80\times10^6\ \mathrm{N/C}\).
• The magnetic flux density is \(0.0200\ \mathrm{T}\).

The selector can be treated as uniform, and edge effects are negligible.

The downstream magnetic analyser has a uniform field of \(0.0100\ \mathrm{T}\) directed into the page, with no electric field.

Three electron beams enter the selector at \(0.70c\), \(0.80c\), and \(0.90c\). A positron beam also enters at \(0.80c\). A positron has the same rest mass as an electron but positive charge \(+e\).

a. Which beams pass straight through the selector?

b. Determine the direction of the initial deflection for the \(0.70c\) and \(0.90c\) electrons.

c. Determine the radius and direction of curvature of each undeflected particle once it enters the magnetic analyser.

d. Explain why the selector condition doesn’t need a Lorentz factor, even though the analyser does.

Solution 4

The \(0.80c\) electron and the \(0.80c\) positron pass straight through the selector. The slower electron initially deflects upwards, while the faster electron initially deflects downwards. In the analyser, the selected electron and positron follow equal-radius paths in opposite directions.

Part a: Find the selected speed

A particle passes undeflected when the electric and magnetic forces cancel.

The magnitudes must satisfy

\[ |q|E=|q|vB \]

Cancelling charge magnitude,

\[ v=\frac EB \]

Substitution gives

\[ \begin{aligned} v&=\frac{4.80\times10^6}{0.0200}\\ &=2.40\times10^8\ \mathrm{m/s}\\ &=0.80c \end{aligned} \]

Both the electron and positron travelling at \(0.80c\) pass straight through. Charge sign reverses both forces together, so it doesn’t change their cancellation.

Part b: Determine deflection directions

For motion to the right with the magnetic field into the page, the magnetic force on a positive charge points upwards. For an electron, it points downwards.

The downward electric field pushes an electron upwards.

At \(0.70c\), the electron’s magnetic force is too weak to balance the upward electric force. It initially deflects upwards.

At \(0.90c\), the downward magnetic force dominates. It initially deflects downwards.

Part c: Analyse the selected particles

Both selected particles travel at \(0.80c\).

Their common momentum magnitude is

\[ \begin{aligned} p&=\gamma m_ev\\ &=\frac53(9.11\times10^{-31})\\ &\quad\times(2.40\times10^8)\\ &=3.64\times10^{-22}\ \mathrm{kg\,m/s} \end{aligned} \]

Using the analyser’s magnetic field,

\[ \begin{aligned} r&=\frac{p}{eB}\\ &=\frac{3.64\times10^{-22}} {(1.60\times10^{-19})(0.0100)}\\ &=0.228\ \mathrm{m} \end{aligned} \]

The electron curves downwards and the positron curves upwards. Both have radius 0.228 m.

Part d: Why the Lorentz factor appears in only one calculation

The velocity selector compares two actual forces acting on the same particle.

Both forces contain the charge magnitude, and they cancel at \(E=vB\). There is no need to calculate momentum to find this balance.

The magnetic analyser is different. It measures how rapidly a moving momentum vector can be turned.

For that calculation, the momentum magnitude is \(\gamma m_0v\), not \(m_0v\).

The tempting mistake is to insert a Lorentz factor wherever relativistic particles appear. Relativity changes the relationship between speed and momentum; it does not automatically multiply every electromagnetic force formula by \(\gamma\).

Question 5

Two particle beams approach a collision chamber from opposite sides.

An electron enters from the left travelling right at \(0.80c\). A positron enters from the right travelling left at \(0.80c\).

They have equal rest masses \(m_e\).

Both particles annihilate in the chamber. Exactly two photons emerge, one travelling vertically upwards and the other vertically downwards.

Assume no other particles or radiation carry away energy or momentum.

For a photon, its momentum magnitude is related to its energy by

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

a. Calculate the momentum of each incoming particle and the total initial momentum.

b. Determine the energy and momentum magnitude of each outgoing photon.

c. Each photon has a larger momentum magnitude than either incoming particle. Explain how this can happen without violating momentum conservation.

d. The invariant mass \(M\) of a complete system is defined, in its zero-total-momentum frame, by \(E_{\mathrm{system}}=Mc^2\). Determine the initial two-particle system’s invariant mass in terms of \(m_e\). Explain why this doesn’t mean either particle acquired additional rest mass.

This final part extends beyond standard HSC exam difficulty but uses the same conservation principles.

Solution 5

The initial total momentum is zero, but the initial total energy is not. Each outgoing photon has energy approximately \(0.852\ \mathrm{MeV}\) and momentum magnitude \(4.56\times10^{-22}\ \mathrm{kg\,m/s}\), greater than either incoming particle’s momentum magnitude. Both momentum and energy are conserved.

The key is that conservation applies to the complete system. It does not require each outgoing particle to inherit the same momentum magnitude as one incoming particle.

Part a: Incoming momenta

At \(0.80c\),

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

The momentum magnitude of either particle is

\[ \begin{aligned} p&=\gamma m_ev\\ &=\frac53(9.11\times10^{-31})\\ &\quad\times(2.40\times10^8)\\ &=3.64\times10^{-22}\ \mathrm{kg\,m/s} \end{aligned} \]

Taking right as positive,

\[ p_{e^-}=+3.64\times10^{-22}\ \mathrm{kg\,m/s} \]

\[ p_{e^+}=-3.64\times10^{-22}\ \mathrm{kg\,m/s} \]

Therefore,

\[ p_{\mathrm{total}}=0 \]

The system has zero total momentum even though neither particle is stationary.

Part b: Outgoing photon energies and momenta

Each incoming particle has total relativistic energy

\[ E=\gamma m_ec^2 \]

The initial system energy is

\[ E_{\mathrm{initial}}=2\gamma m_ec^2 \]

Because the initial total momentum is zero and the two photons travel in opposite directions, the photons must have equal momentum magnitudes.

Since photon momentum is \(E_\gamma/c\), they must also have equal energies.

Energy conservation therefore gives

\[ E_\gamma=\gamma m_ec^2 \]

Using \(m_ec^2=0.511\ \mathrm{MeV}\),

\[ \begin{aligned} E_\gamma&=\frac53(0.511\ \mathrm{MeV})\\ &\approx0.852\ \mathrm{MeV} \end{aligned} \]

In joules, using \(1\ \mathrm{MeV}=1.60\times10^{-13}\ \mathrm{J}\),

\[ E_\gamma\approx1.37\times10^{-13}\ \mathrm{J} \]

The photon momentum magnitude is then

\[ \begin{aligned} p_\gamma&=\frac{E_\gamma}{c}\\ &=\frac{1.37\times10^{-13}}{3.00\times10^8}\\ &\approx4.56\times10^{-22}\ \mathrm{kg\,m/s} \end{aligned} \]

Part c: Explain the apparently larger photon momenta

The incoming particles each have momentum magnitude \(3.64\times10^{-22}\ \mathrm{kg\,m/s}\).

The outgoing photons each have momentum magnitude \(4.56\times10^{-22}\ \mathrm{kg\,m/s}\).

There is no contradiction.

Initially, the electron and positron momentum vectors cancel because they point in opposite directions.

Finally, the two photon momentum vectors also cancel because they point in opposite directions.

The total momentum is zero both before and after annihilation.

But the total energy includes the incoming particles’ rest energy and kinetic energy.

The outgoing photons carry that energy even though each photon has zero rest mass. A photon can have non-zero momentum without possessing invariant mass.

The false assumption is that momentum conservation preserves individual particle momentum magnitudes. It preserves the vector sum of all momenta in the system.

There is no requirement that the photons travel in the same directions as the incoming electron and positron, provided total momentum and energy are conserved.

Part d: Invariant mass of the complete initial system

The laboratory is also the zero-total-momentum frame of the electron-positron pair, because the incoming momenta cancel.

The system’s total energy is

\[ E_{\mathrm{system}}=2\gamma m_ec^2 \]

Using the supplied definition,

\[ Mc^2=2\gamma m_ec^2 \]

Therefore,

\[ M=2\gamma m_e \]

At \(0.80c\),

\[ M=\frac{10}{3}m_e\approx3.33m_e \]

The two-particle system has an invariant mass equivalent to approximately 3.33 electron rest masses.

Yet the sum of the individual particle rest masses is only \(2m_e\).

This is not a contradiction.

The invariant mass of a complete system depends on its total energy and total momentum. In its zero-momentum frame, the energy includes the particles’ kinetic energy as well as their individual rest energies.

Each electron or positron still has invariant mass \(m_e\).

After annihilation, the two photons have zero individual rest masses, but their combined system still has invariant mass \(M\), because their total energy is non-zero while their momenta cancel.

This is the deeper distinction behind relativistic momentum: the invariant mass of an individual particle, the invariant mass of a collection of particles, total energy, and momentum are related quantities, but they are not interchangeable.

06What the measurement teaches us next

The reason \(p=mv\) fails at high speed isn’t that physicists discovered a mysterious resistance to acceleration. The classical equation assumes a relationship between motion, mass, and time that is only approximately valid when speeds are small compared with \(c\).

Relativistic momentum preserves the correct relationship:

\[ p=\gamma m_0v \]

Magnetic curvature gives a way to measure momentum. Timing gives a way to measure speed. Energy measurements provide a further independent test. Together, they show why a particle can gain substantial momentum and energy while its speed changes very little.

The next useful distinction is between adding momentum and adding kinetic energy. A magnetic field can change a momentum vector without changing kinetic energy, while an electric field can transfer energy and change momentum magnitude.

For the broader relationship between relativistic momentum, applied forces, and the limiting speed \(c\), continue with Relativistic Momentum: How HSC Physics Explains the Speed Limit.