A proton traveling through water at 0.9c emits a cone of blue light. A student claims this violates special relativity because 'nothing can travel faster than light.' What is the correct response?
AThe student is right — this situation is physically impossible
BThe proton is violating special relativity, but Cherenkov radiation is allowed as an exception
CSpecial relativity forbids exceeding c (vacuum speed), not c/n (phase velocity in the medium). The proton moves at 0.9c, well below c, while light in water moves at c/1.33 ≈ 0.75c — so the proton exceeds the phase velocity in water without violating relativity
DSpecial relativity only applies in vacuum, so there is no violation regardless of speed in matter
The key distinction is between c (vacuum speed of light) and c/n (phase velocity in a medium). Special relativity prohibits exceeding c, the vacuum speed. But light slows down in matter to c/n, and a particle can exceed this lower threshold while remaining below c. The proton at 0.9c is slower than c but faster than c/1.33 ≈ 0.75c — it is 'superluminal' relative to the medium, not relative to the vacuum. No physical law is violated.
Question 2 Multiple Choice
As a charged particle's speed increases from just above the Cherenkov threshold (v slightly above c/n) toward the relativistic limit (β → 1), the Cherenkov cone angle θ_c:
ADecreases from 90° toward 0° as the particle accelerates
BRemains constant — the angle depends only on the medium, not the particle speed
CIncreases from 0° at threshold toward a maximum of arccos(1/n)
DOscillates, because higher speeds produce more destructive interference
At threshold (v = c/n, β = 1/n), cos θ_c = 1/(βn) = 1, so θ_c = 0 — the cone is infinitely narrow and no radiation is emitted. As v increases (β increases), 1/(βn) decreases, so θ_c = arccos(1/(βn)) increases. At the relativistic limit β → 1, θ_c approaches arccos(1/n), which for water (n = 1.33) is about 41°. The angle opens as speed increases — exactly the opposite of sonic boom behavior, which many students expect to go the other way.
Question 3 True / False
Cherenkov radiation can primarily be emitted by particles moving faster than c, the vacuum speed of light.
TTrue
FFalse
Answer: False
False — Cherenkov radiation requires exceeding c/n (the phase velocity of light in the medium), not c (the vacuum speed). Since c/n < c for any medium with n > 1, a particle can emit Cherenkov radiation while remaining below c. This is the fundamental point: special relativity is not violated because the speed limit c remains inviolable; the medium has simply slowed light's phase velocity below c, creating a threshold that a slower-than-c particle can still cross.
Question 4 True / False
Measuring the Cherenkov angle θ_c from emitted radiation allows experimentalists to determine the velocity of the charged particle that produced it.
TTrue
FFalse
Answer: True
True — the relation cos θ_c = 1/(βn) directly encodes β = v/c. Knowing n (the refractive index of the medium) and measuring θ_c from the cone of emitted light gives v immediately. Combined with an independent momentum measurement (from a magnetic field, for instance), this yields the particle mass m, making Cherenkov detectors powerful tools for particle identification. This is how experiments like those at CERN distinguish pions from kaons from protons at the same momentum.
Question 5 Short Answer
Explain using the wave-interference picture why Cherenkov radiation forms a cone when v > c/n, but not when v < c/n.
Think about your answer, then reveal below.
Model answer: When v < c/n, the particle moves slower than the electromagnetic disturbances it generates, so those disturbances spread outward spherically faster than the particle travels. The waves from different points along the path interfere destructively in almost all directions, producing no coherent radiation. When v > c/n, the particle outruns its own electromagnetic wake — the spherical wavefronts pile up behind the particle and constructively interfere along a cone whose half-angle satisfies cos θ_c = c/(nv). The geometry is identical to a sonic boom: coherent radiation only forms along the envelope where wavefronts add constructively.
The cone is defined by the condition that the wavefronts emitted at different times all arrive simultaneously along its surface — the Mach cone condition. This is constructive interference by definition. Outside the cone, the phase relationships are wrong and interference is destructive. The angle shrinks to zero at threshold (v = c/n) because the wavefronts just barely pile up with an infinitesimally narrow cone.