F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Special RelativityRelativistic Momentum and the Cosmic Speed Limit
F2.3 F2.1 F1.2About 70 minutes
Learning goals- I can explain why p = mv cannot be conserved in all inertial frames at high speeds.
- I can calculate relativistic momentum p = γmv and find a speed from a given momentum.
- I can explain, using momentum and force, why no object with mass can reach the speed of light.
- I can describe the 'relativistic mass' idea and the modern view that mass is invariant.
- I can apply relativistic momentum to charged particles in magnetic fields and accelerators.
Words to know
relativistic momentum
p = γmv, the quantity conserved in collisions in every inertial frame.
mass (rest mass, invariant mass)
The mass m measured in the object's rest frame; the same for every observer. Modern physics uses 'mass' to mean this.
relativistic mass
An older idea, m_rel = γm, describing a 'mass' that increases with speed. It reproduces p = m_rel v but is no longer used by most physicists.
particle accelerator
A machine that uses electric fields to speed up charged particles and magnetic fields to steer them.
cyclotron
An accelerator in which charged particles spiral outward in a magnetic field while an alternating electric field speeds them up.
When Newton's momentum breaks
Momentum is one of physics' great conservation laws: in any isolated collision, total momentum before equals total momentum after. In the unit Momentum and Collisions you used p = mv. Postulate 1 demands that momentum conservation hold in every inertial frame. But if you analyse a collision of fast particles in one frame using p = mv, and then transform all the velocities into another frame using the relativistic velocity addition rule (lesson 9), p = mv is no longer conserved. Either we give up momentum conservation, or we change the definition of momentum.
Physicists kept the conservation law and changed the definition. The quantity that is conserved in every frame is
p = γmv = mv / √(1 − v²/c²)
where m is the object's (rest) mass. At low speeds γ ≈ 1 and p ≈ mv, so everything you learned before still works. At high speeds, p grows much faster than mv.
Worked exampleExample 1: A fast electron
An electron (m = 9.11 × 10⁻³¹ kg) moves at 0.90c. Find its relativistic momentum and compare with mv.
- γ = 1/√(1 − 0.81) = 2.294.
- v = 0.90 × 3.00 × 10⁸ = 2.70 × 10⁸ m/s.
- p = γmv = 2.294 × 9.11 × 10⁻³¹ × 2.70 × 10⁸ = 5.64 × 10⁻²² kg·m/s.
- mv = 9.11 × 10⁻³¹ × 2.70 × 10⁸ = 2.46 × 10⁻²² kg·m/s.
p = 5.64 × 10⁻²² kg·m/s, 2.29 times the Newtonian value.
Worked exampleExample 2: Speed from momentum
A proton (m = 1.67 × 10⁻²⁷ kg) has momentum 1.00 × 10⁻¹⁸ kg·m/s. How fast is it moving?
- Start with p = mv/√(1 − v²/c²). Square: p²(1 − v²/c²) = m²v².
- Collect v²: p² = v²(m² + p²/c²), so v = p/√(m² + p²/c²).
- p/c = 1.00 × 10⁻¹⁸ ÷ 3.00 × 10⁸ = 3.33 × 10⁻²⁷ kg. m² + (p/c)² = (1.67 × 10⁻²⁷)² + (3.33 × 10⁻²⁷)² = 2.79 × 10⁻⁵⁴ + 1.11 × 10⁻⁵³ = 1.39 × 10⁻⁵³.
- √(1.39 × 10⁻⁵³) = 3.73 × 10⁻²⁷. v = 1.00 × 10⁻¹⁸ ÷ 3.73 × 10⁻²⁷ = 2.68 × 10⁸ m/s.
v ≈ 2.68 × 10⁸ m/s ≈ 0.894c. (Newton's v = p/m would give 6.0 × 10⁸ m/s, twice c, which is impossible.)
The cosmic speed limit
Newton's second law in its general form is F = Δp/Δt: a net force changes momentum. This form still works in relativity, with p = γmv. Apply a constant force to a particle and its momentum rises steadily, without limit. But because γ → ∞ as v → c, more and more momentum buys less and less extra speed.
For example, if a force adds 0.5mc of momentum every second to an object starting at rest, its speed after each second (using v = p/√(m² + p²/c²)) is: 0.447c, 0.707c, 0.832c, 0.894c, 0.928c, 0.949c, 0.962c, 0.970c… The speed keeps rising, but it never reaches c. To reach c, the momentum (and, as you will see in the next lesson, the energy) would have to be infinite.
Therefore no object with mass can be accelerated to the speed of light. Only massless particles, such as photons, travel at c, and they always do. This also protects causality (lesson 4).
'Relativistic mass' and modern language
Older textbooks, and some course materials, describe the γ in p = γmv as an increase in mass: m_rel = γm, so p = m_rel v and a fast object 'gets heavier'. This relativistic mass idea gives correct momenta, and you may see it in problems. But it can mislead: m_rel does not work in F = ma or in ½mv², and it depends on the observer.
Most physicists today use mass to mean the invariant (rest) mass m, the same for all observers, and put all the speed dependence in γ. In this course: when a question asks for 'relativistic mass', calculate γm; otherwise, m means the rest mass.
Worked exampleExample 3: Relativistic mass
A proton (m = 1.67 × 10⁻²⁷ kg) moves at 0.995c. Find γ and its 'relativistic mass'. Why is it harder to speed it up than a proton at rest?
- γ = 1/√(1 − 0.995²) = 1/√0.009975 = 10.0.
- m_rel = γm = 10.0 × 1.67 × 10⁻²⁷ = 1.67 × 10⁻²⁶ kg.
- A small extra speed requires a big increase in p = γmv, because γ rises steeply near c (the graph above).
γ ≈ 10.0 and m_rel ≈ 1.67 × 10⁻²⁶ kg. In modern terms: the mass is still 1.67 × 10⁻²⁷ kg, but its momentum is 10 times mv, and each extra bit of speed costs a lot more momentum.
Accelerators: relativistic momentum at work
In a magnetic field B, a charged particle moving perpendicular to the field follows a circle. In the unit Gravitational, Electric and Magnetic Fields you found r = mv/(qB). For fast particles the correct radius uses relativistic momentum:
r = p/(qB) = γmv/(qB)
Accelerator designers must use this formula; with mv, particles would hit the walls. Canada has world-class examples:
- TRIUMF in Vancouver is home to one of the largest cyclotrons in the world. It accelerates negative hydrogen ions to energies of up to 520 MeV, about three-quarters of the speed of light. Its beams are used for physics research, materials science and producing isotopes for medicine.
- The Canadian Light Source in Saskatoon is a synchrotron. Electrons circle its storage ring at 2.9 GeV, so fast that their speed differs from c by only about 1.6 parts in 100 million. As the electrons curve, they give off intense X-rays and other light ('synchrotron light') that scientists use to study everything from proteins to soil samples to battery materials.
You will use energies in electron volts (eV, MeV, GeV) in the next lesson, Mass–Energy Equivalence: E = mc², Total Energy and Kinetic Energy.
Worked exampleExample 4: Radius in a magnetic field
A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.60 × 10⁻¹⁹ C) moves at 0.60c perpendicular to a 1.5 T magnetic field. Find the radius of its path, relativistically and non-relativistically.
- γ = 1.25; v = 1.80 × 10⁸ m/s.
- p = γmv = 1.25 × 1.67 × 10⁻²⁷ × 1.80 × 10⁸ = 3.76 × 10⁻¹⁹ kg·m/s.
- r = p/(qB) = 3.76 × 10⁻¹⁹ ÷ (1.60 × 10⁻¹⁹ × 1.5) = 1.57 m.
- Newtonian: r = mv/(qB) = 3.01 × 10⁻¹⁹ ÷ 2.40 × 10⁻¹⁹ = 1.25 m.
r ≈ 1.57 m (the Newtonian formula would give 1.25 m, 20% too small).
Momentum toolkit
- p = γmv; at low speed p ≈ mv.
- To find v from p: v = p/√(m² + p²/c²). Check that v < c.
- F = Δp/Δt still works; F = ma does not (at high speed).
- 'Relativistic mass' γm: use it only when a question asks for it.
- Charged particle in B: r = γmv/(qB).
Just for funA proton walks into a bar going 0.999c. The bartender says, "Why the long face?" The proton says, "I'm not long, you're contracted. And my momentum is 22 times what you think it is."
Hands-on activityPushing toward the speed limit (calculator or spreadsheet)
You need: a calculator (or a free spreadsheet program), graph paper, a ruler and a pencil.
Imagine a spacecraft whose engine gives it the same push every month: each month its momentum increases by 0.25mc (where m is its mass).
- Make a table with columns: month, p/(mc), Newtonian speed v/c = p/(mc), relativistic speed v/c = (p/mc) ÷ √(1 + (p/mc)²).
- Fill it in for months 0 to 16.
- Plot both speeds against time on the same graph. Draw a dashed horizontal line at v/c = 1.
- When does the Newtonian speed pass c? What is the relativistic speed at that moment? At month 16?
What to look for: at month 4, p = 1.00mc: Newton says v = c, but relativity gives v = 0.707c. At month 16, p = 4.00mc: Newton says 4c (impossible), relativity says 0.970c. The relativistic curve flattens and creeps toward c without reaching it. Explain in a sentence why an engine that 'keeps pushing' cannot get the ship to c.
Your turn
1Calculate the momentum of an electron (m = 9.11 × 10⁻³¹ kg) moving at 0.50c.
γ = 1.155; p = 1.155 × 9.11 × 10⁻³¹ × 1.50 × 10⁸ = 1.58 × 10⁻²² kg·m/s.
2Calculate the momentum of a proton (m = 1.67 × 10⁻²⁷ kg) moving at 0.80c.
γ = 1.667; p = 1.667 × 1.67 × 10⁻²⁷ × 2.40 × 10⁸ = 6.68 × 10⁻¹⁹ kg·m/s.
3At what speed is relativistic momentum twice the Newtonian value mv?
4At what speed does p = mc?
p/(mc) = γβ = 1 → β²/(1 − β²) = 1 → β² = 0.5 → v = 0.707c.
5An electron has momentum 3.0 × 10⁻²² kg·m/s. Find its speed.
p/c = 1.0 × 10⁻³⁰ kg. v = p/√(m² + (p/c)²) = 3.0 × 10⁻²² ÷ √((9.11 × 10⁻³¹)² + (1.0 × 10⁻³⁰)²) = 3.0 × 10⁻²² ÷ 1.353 × 10⁻³⁰ = 2.22 × 10⁸ m/s ≈ 0.739c.
6Why can't a constant force accelerate a rocket to c?
The force increases momentum steadily (F = Δp/Δt), but p = γmv → ∞ as v → c. Reaching c would take infinite momentum (and energy), so v only approaches c.
7What is the 'relativistic mass' of an electron moving at 0.99c?
γ = 7.089; m_rel = 7.089 × 9.11 × 10⁻³¹ = 6.46 × 10⁻³⁰ kg.
More practice: warm-up
1Write the formula for relativistic momentum.
p = γmv = mv/√(1 − v²/c²).
2Which form of Newton's second law works in relativity?
3Does an object's (rest) mass change with speed, in the modern view?
No. The rest mass is invariant; the speed dependence is in γ.
4Name a Canadian particle accelerator.
TRIUMF (Vancouver) or the Canadian Light Source (Saskatoon).
5Which particles can travel at exactly c?
Massless particles, such as photons.
More practice: on your own
1A 2.0 kg object moves at 0.60c. Find its momentum and its 'relativistic mass'.
γ = 1.25. p = 1.25 × 2.0 × 1.80 × 10⁸ = 4.5 × 10⁸ kg·m/s. m_rel = 2.5 kg.
2By what percentage does p = mv underestimate the momentum of a car at 30 m/s? (Use γ − 1 ≈ v²/(2c²).)
γ − 1 ≈ 900 ÷ 1.80 × 10¹⁷ = 5.0 × 10⁻¹⁵, so the error is about 5 × 10⁻¹³ %, utterly negligible.
3An electron moves in a circle of radius 0.20 m in a 0.010 T field. Find its momentum and speed (q = 1.60 × 10⁻¹⁹ C).
p = qBr = 1.60 × 10⁻¹⁹ × 0.010 × 0.20 = 3.2 × 10⁻²² kg·m/s. p/c = 1.067 × 10⁻³⁰ kg. v = 3.2 × 10⁻²² ÷ √((9.11 × 10⁻³¹)² + (1.067 × 10⁻³⁰)²) = 3.2 × 10⁻²² ÷ 1.403 × 10⁻³⁰ = 2.28 × 10⁸ m/s (0.76c).
4Two protons have speeds 0.50c and 0.90c. Find the ratio of their momenta.
p₂/p₁ = (γ₂v₂)/(γ₁v₁) = (2.294 × 0.90)/(1.155 × 0.50) = 2.065/0.577 = 3.6.
5Explain why accelerator designers cannot use r = mv/(qB) for fast protons.
At high speed the momentum is γmv, so the real radius is γ times larger. Magnets designed with mv would be too weak to keep the particles on the planned path.
Challenge
1Derive v = p/√(m² + p²/c²) from p = γmv, and show that v < c for any finite p.
p² = m²v²/(1 − v²/c²) → p² − p²v²/c² = m²v² → p² = v²(m² + p²/c²) → v = p/√(m² + p²/c²). Since √(m² + p²/c²) > √(p²/c²) = p/c (because m > 0), v < p ÷ (p/c) = c.
2A constant net force of 1.0 × 10⁻¹³ N acts on a proton initially at rest. How long does it take to reach 0.90c? How long would Newtonian physics predict?
p needed = γmv = 2.294 × 1.67 × 10⁻²⁷ × 2.70 × 10⁸ = 1.034 × 10⁻¹⁸ kg·m/s. Δt = Δp/F = 1.0 × 10⁻⁵ s. Newtonian: mv/F = 4.51 × 10⁻¹⁹/1.0 × 10⁻¹³ = 4.5 × 10⁻⁶ s, less than half as long.
3Evaluate: 'Relativistic mass increase proves that fast objects contain more matter.' Agree or disagree, with reasons.
Disagree. The number of particles does not change with speed. What grows is the momentum (and energy) for a given speed, which older language described as 'mass increase'. The invariant (rest) mass, the same for every observer, is unchanged. Also, an observer moving with the object measures its normal mass, which would be impossible if extra matter had appeared.
Quiz
1. What is the relativistic momentum of a 1.0 kg object at 0.60c?
2. Why did physicists redefine momentum as p = γmv?
3. A constant force acts on a particle for a long time. What happens to its speed?
4. At what speed is a particle's momentum equal to mc?
5. In modern physics, what does 'mass' usually mean?
6. A proton moves at 0.60c in a magnetic field. Compared with the Newtonian prediction, its circular path has a radius that is:
Big idea (F2.3, F1.2): momentum is still conserved at high speeds if we define it as p = γmv. Because γ grows without limit near c, nothing with mass can be pushed to the speed of light. Accelerators such as TRIUMF and the Canadian Light Source are designed with these equations.
How to help: Have your teen explain the activity graph: why the Newtonian line shoots past c and the relativistic curve flattens. Ask: 'If the engine keeps pushing, where does the extra momentum go?' (Into γ, not into much more speed.)
Common difficulties:
- Forgetting γ, or using the speed in units of c inside p (p needs v in m/s).
- Solving for v from p: square carefully; the formula v = p/√(m² + p²/c²) avoids errors.
- Mixing up 'relativistic mass' (γm) and rest mass m. Read the question.
- Using F = ma at high speeds. Use F = Δp/Δt.