Questions: Relativistic Dynamics and Acceleration

Question 1

A constant force F is applied continuously to a particle initially at rest. As the particle's speed approaches c, what happens to its acceleration?

• A. Acceleration remains constant — Newton's second law F = ma still holds, so constant F means constant a.
• B. Acceleration increases — as the particle gains energy, it becomes easier to add further velocity increments.
• C. Acceleration approaches zero — the γ³ factor in F = γ³ma means the same force produces less and less acceleration near c.
• D. Acceleration oscillates — relativistic corrections produce oscillatory dynamics at high speeds.

Answer: (C)

F = dp/dt with relativistic momentum p = γmv. For a force applied parallel to the velocity, this gives F = γ³ma. As v → c, γ → ∞, so γ³ → ∞, meaning the same force F requires γ³ times as much 'classical acceleration response.' The particle's acceleration diminishes to zero even though the force remains constant. This is why particle accelerators cannot push particles to c regardless of energy input — the acceleration they can produce shrinks faster than the velocity deficit.

Question 2

A force of equal magnitude is applied to a particle moving at 0.99c: in one case parallel to its velocity, in another case perpendicular to its velocity. How do the resulting accelerations compare?

• A. The accelerations are equal — the magnitude of acceleration depends only on force magnitude and rest mass.
• B. The parallel acceleration is larger — forces aligned with motion are more effective relativistically.
• C. The perpendicular acceleration is larger — the parallel case has a γ³ factor suppressing acceleration, while the perpendicular case has only γ.
• D. The perpendicular acceleration is zero — perpendicular forces cannot accelerate a relativistic particle.

Answer: (C)

This is a key asymmetry in relativistic dynamics. For force parallel to velocity: F = γ³ma. For force perpendicular to velocity: F = γma. At 0.99c, γ ≈ 7.1 and γ³ ≈ 357, so the perpendicular acceleration is about 50 times larger than the parallel acceleration for the same force. There is no classical analogue of this distinction. The γ³ vs γ asymmetry arises from the geometry of spacetime — the relativistic momentum formulation accounts differently for how force affects the γ factor when aligned vs perpendicular to velocity.

Question 3

A constant force applied to a relativistic particle produces a continuously decreasing acceleration as the particle's speed approaches c.

• T. True
• F. False

Answer: True

Directly from F = γ³ma (for parallel force): since γ grows without bound as v → c, the acceleration a = F/(γ³m) shrinks to zero even as the force remains constant. The particle never stops accelerating — it approaches c asymptotically — but the rate of velocity increase diminishes. This is why accelerating a particle from 0.99c to 0.999c requires enormously more energy than accelerating it from 0 to 0.99c, even though the velocity gain is smaller.

Question 4

The reason a massive particle cannot be accelerated to the speed of light is that its rest mass increases without bound as v → c, requiring infinite force.

• T. True
• F. False

Answer: False

This is the 'relativistic mass' misconception that the topic explicitly warns against. The suppression of acceleration is better understood through F = dp/dt with relativistic momentum p = γmv: as v → c, γ → ∞, so p → ∞ even for tiny additional velocity increments, requiring infinite energy — not infinite force. Rest mass m is an invariant property of the particle that does not change with velocity. The concept of 'relativistic mass' γm, while mathematically valid, obscures the physics by importing a Newtonian intuition that doesn't apply. The γ³ factor in the force-acceleration relation is a consequence of spacetime geometry, not mass growth.

Question 5

Explain why pushing a particle from 0.99c to 0.999c requires far more energy than pushing it from 0 to 0.99c, even though the velocity increment (0.009c) is much smaller than the initial push (0.99c).

Model answer: Relativistic kinetic energy is K = (γ - 1)mc². At 0.99c, γ ≈ 7.1, so K ≈ 6.1mc². At 0.999c, γ ≈ 22.4, so K ≈ 21.4mc². The energy required for the second push (from 0.99c to 0.999c) is about 15.3mc² — more than twice the energy that got the particle from 0 to 0.99c. This is because γ grows rapidly as v → c: the same small velocity increment corresponds to an enormous change in γ near c. Practically, the energy is going into increasing γ (and thus relativistic momentum), not into velocity. The speed limit c is enforced by an ever-increasing energy cost, not by any explicit barrier.

The key insight is that relativistic kinetic energy is nonlinear in velocity — it is linear in (γ - 1), and γ diverges at c. Work done by the force all goes into relativistic kinetic energy (γ - 1)mc², not into velocity linearly. Near c, tiny velocity gains correspond to large γ increments and therefore large energy requirements. This is why particle accelerators need enormous energy to push particles from 0.99c to 0.999c even though they're 'almost there' in velocity terms.