Lessons 11–15 cover special relativity — Einstein's postulates, inertial frames, time dilation, length contraction, relativistic energy and momentum, and the resolution of famous paradoxes. This checkpoint assesses your ability to apply these concepts to solve problems and explain relativistic phenomena.
Principle of relativity; constancy of $c$; inertial reference frames
Proper time; $\\Delta t = \\gamma \\Delta \\tau$; muon decay; twin effect
Proper length; $L = L_0/\\gamma$; perpendicular dimensions unchanged
$E = \\gamma mc^2$; $E_k = (\\gamma-1)mc^2$; $E^2 = (pc)^2 + (mc^2)^2$
Relativity of simultaneity; barn-ladder; twin paradox; GPS evidence
15 questions — instant feedback
Q1. Einstein's second postulate states that:
Correct: B. The second postulate states that $c$ is invariant — all observers measure the same speed of light, regardless of their motion.
Q2. A clock on a spaceship moving at $0.6c$ ticks once per second in the ship's frame. Earth observers measure the interval between ticks as:
Correct: A. $\\gamma = 1/\\sqrt{1-0.36} = 1.25$. $\\Delta t = \\gamma \\Delta \\tau = 1.25 \\times 1 = 1.25$ s.
Q3. A rod moving at relativistic speed is measured to have:
Correct: C. Only lengths parallel to motion contract ($L = L_0/\\gamma$). Perpendicular dimensions are unchanged.
Q4. A spaceship of proper length 200 m is measured as 100 m by Earth observers. Its speed is:
Correct: D. $\\gamma = L_0/L = 200/100 = 2$. $v = c\\sqrt{1 - 1/4} = c\\sqrt{3}/2 = 0.866c$.
Q5. The rest energy of a particle is given by:
Correct: B. Rest energy $E_0 = mc^2$ is the energy of a particle at rest ($\\gamma = 1$).
Q6. An electron ($m_e c^2 = 0.511$ MeV) has kinetic energy 1.022 MeV. Its total energy is:
Correct: A. $E = E_k + mc^2 = 1.022 + 0.511 = 1.533$ MeV. $\\gamma = E/mc^2 = 1.533/0.511 = 3$.
Q7. As a particle's speed approaches $c$, its relativistic kinetic energy:
Correct: C. As $v \\to c$, $\\gamma \\to \\infty$, so $E_k = (\\gamma - 1)mc^2 \\to \\infty$. Infinite energy would be required to reach $c$.
Q8. In the twin paradox, the travelling twin is younger because:
Correct: B. The situations are not symmetric. The travelling twin must accelerate to turn around, changing inertial frames. The stay-at-home twin remains in one inertial frame.
Q9. Two events simultaneous in one inertial frame are also simultaneous in another if:
Correct: D. If $\\Delta x = 0$, then $\\Delta t' = \\gamma(\\Delta t - v\\Delta x/c^2) = \\gamma \\Delta t$. Simultaneity ($\\Delta t = 0$) is preserved when events occur at the same position.
Q10. A proton ($m_p c^2 = 938$ MeV) has total energy 1876 MeV. Its Lorentz factor is:
Correct: A. $\\gamma = E/mc^2 = 1876/938 = 2.00$. $E_k = (2-1)mc^2 = 938$ MeV.
Q11. Muons created at 15 km altitude reach Earth's surface because:
Correct: C. In Earth's frame, the muon's lifetime is extended by $\\gamma$, allowing it to travel farther. In the muon's frame, the atmosphere is length-contracted, so it reaches the surface in its short lifetime. Both explanations are correct.
Q12. GPS satellites require relativistic corrections because:
Correct: B. GPS satellites experience both special relativistic time dilation (due to orbital speed) and general relativistic gravitational time dilation (due to altitude). Both must be corrected.
Q13. A particle has momentum $p = 2mc$. Its total energy is:
Correct: D. $E^2 = (pc)^2 + (mc^2)^2 = (2mc \\cdot c)^2 + (mc^2)^2 = 4m^2c^4 + m^2c^4 = 5m^2c^4$. $E = \\sqrt{5}mc^2$.
Q14. In the barn-ladder paradox, the resolution depends on:
Correct: A. In the barn frame, both doors close simultaneously with the ladder inside. In the ladder frame, the door-closing events are not simultaneous. The relativity of simultaneity resolves the apparent contradiction.
Q15. A mass of 1.0 kg is completely converted to energy. The energy released is approximately:
Correct: C. $E = mc^2 = 1.0 \\times (3.0\\times10^8)^2 = 9.0\\times10^{16}$ J. This is roughly the annual output of a large power station.
5 questions — model answers revealed
SAQ 1. Distinguish between an inertial and a non-inertial reference frame. Give one example of each. Explain why Einstein's principle of relativity applies only to inertial frames in special relativity. (3 marks)
Model answer (3 marks):
An inertial frame is one that moves at constant velocity (zero acceleration); Newton's first law holds, and objects move in straight lines at constant speed unless acted upon by a force (1 mark). A non-inertial frame is accelerating or rotating; fictitious forces appear (e.g., you feel pushed back in an accelerating car) (1 mark). Einstein's principle of relativity in special relativity states that physical laws are identical in all inertial frames; accelerating frames require general relativity because the geometry of spacetime is curved (1 mark).
SAQ 2. A spacecraft of proper length 120 m flies past Earth at $0.8c$. (a) Calculate the length measured by Earth observers. (b) Calculate the time for the spacecraft to completely pass a fixed point on Earth, according to Earth clocks. (c) The spacecraft captain measures the same passing event. What time does she measure, and why is it different? (4 marks)
Model answer (4 marks):
(a) $\\gamma = 1/\\sqrt{1-0.64} = 1.667$ (0.5 mark). $L = L_0/\\gamma = 120/1.667 = 72$ m (1 mark).
(b) Time = distance/speed = $72/(0.8c) = 72/(0.8 \\times 3.0\\times10^8) = 3.0\\times10^{-7}$ s = 300 ns (1 mark).
(c) In the captain's frame, Earth moves past the 120 m ship at $0.8c$ (0.5 mark). Time = $120/(0.8c) = 5.0\\times10^{-7}$ s = 500 ns (0.5 mark). This is the proper time for the two events (front and back passing the point) because they occur at the same position in the captain's frame (0.5 mark). Earth measures a longer time due to time dilation: $\\Delta t = \\gamma \\Delta \\tau = 1.667 \\times 500 = 833$ ns... wait, that does not match 300 ns. Let me recalculate: actually, in Earth's frame the ship length is 72 m, so time = 72/(0.8c) = 300 ns. In the ship frame, the point on Earth travels 120 m at 0.8c, taking 500 ns. But 500 ns is not $\\Delta t/\\gamma$... Hmm, the two events (front passes point, back passes point) are at the same position in Earth's frame, so Earth measures proper time! Let me correct: In Earth's frame, the two events occur at the same point, so Earth measures proper time $\\Delta \\tau = 300$ ns. In the ship frame, they occur at different positions, so $\\Delta t' = \\gamma \\Delta \\tau = 1.667 \\times 300 = 500$ ns. Yes, that matches. So: The captain measures 500 ns because the events are separated in space in her frame; this is the dilated time (0.5 mark).
SAQ 3. A proton is accelerated from rest through 1.0 GV. ($m_p c^2 = 938$ MeV, $e = 1.60\\times10^{-19}$ C). (a) Calculate the proton's kinetic energy in GeV. (b) Calculate its Lorentz factor. (c) Calculate its speed as a fraction of $c$. (d) Calculate its momentum in GeV/$c$. (4 marks)
Model answer (4 marks):
(a) $E_k = qV = e \\times 1.0$ GV = 1.0 GeV (1 mark).
(b) $E_k = (\\gamma - 1)mc^2$; $1.0 = (\\gamma - 1)(0.938)$; $\\gamma - 1 = 1.066$; $\\gamma = 2.07$ (1 mark).
(c) $v/c = \\sqrt{1 - 1/\\gamma^2} = \\sqrt{1 - 1/2.07^2} = \\sqrt{1 - 0.233} = \\sqrt{0.767} = 0.876$ (1 mark).
(d) $E = E_k + mc^2 = 1.0 + 0.938 = 1.938$ GeV. $pc = \\sqrt{E^2 - (mc^2)^2} = \\sqrt{1.938^2 - 0.938^2} = \\sqrt{3.756 - 0.880} = \\sqrt{2.876} = 1.70$ GeV. So $p = 1.70$ GeV/$c$ (1 mark).
SAQ 4. (a) Explain the relativity of simultaneity with reference to a thought experiment. (b) A train of proper length 300 m moves at $0.6c$ past a platform. Lightning strikes the front and back of the train simultaneously in the platform frame. In the train frame, which strike occurs first, and by how much time? (c) Explain why there is no contradiction between the two frames. (5 marks)
Model answer (5 marks):
(a) Consider a train moving past a platform. Two lightning bolts strike the platform at points equidistant from a platform observer, simultaneously in the platform frame (1 mark). A passenger on the train moves toward one bolt and away from the other; since $c$ is constant, she sees the bolt ahead first and concludes the strikes were not simultaneous (1 mark). Neither observer is wrong; simultaneity is frame-dependent.
(b) In the platform frame, $\\Delta x = 300/1.25 = 240$ m (contracted train length) (0.5 mark). In the train frame, the platform moves at $0.6c$ and the strikes are separated by 300 m (proper length). $\\Delta t' = \\gamma v \\Delta x/c^2 = 1.25 \\times 0.6c \\times 300/c^2 = 225/c = 7.5\\times10^{-7}$ s = 750 ns (1 mark). The strike at the back of the train (which the train is moving away from in the platform frame) occurs first in the train frame (0.5 mark).
(c) There is no contradiction because the two frames disagree about whether the events are simultaneous, which is exactly what special relativity predicts (1 mark). Both frames correctly apply the laws of physics; they simply measure different time intervals for spatially separated events.
SAQ 5. (a) Distinguish between rest energy, total relativistic energy and kinetic energy. (b) Explain why the classical kinetic energy formula fails at relativistic speeds. (c) Calculate the speed at which a proton's kinetic energy equals its rest energy. (d) Describe one piece of experimental evidence that confirms special relativity. (5 marks)
Model answer (5 marks):
(a) Rest energy $E_0 = mc^2$ is the energy of a particle at rest (0.5 mark). Total relativistic energy $E = \\gamma mc^2$ includes both rest energy and kinetic energy (0.5 mark). Kinetic energy $E_k = E - mc^2 = (\\gamma - 1)mc^2$ is the energy due to motion (0.5 mark).
(b) Classical $E_k = \\frac{1}{2}mv^2$ assumes velocities add linearly and time is absolute (0.5 mark). At relativistic speeds, velocities do not add linearly, time dilates and space contracts, so $E_k$ grows faster than the classical prediction, diverging to infinity as $v \\to c$ (0.5 mark).
(c) $E_k = mc^2$ means $(\\gamma - 1)mc^2 = mc^2$, so $\\gamma = 2$ (0.5 mark). $v = c\\sqrt{1 - 1/4} = c\\sqrt{3}/2 = 0.866c$ (0.5 mark).
(d) Cosmic-ray muons are created at ~15 km altitude and reach Earth's surface despite their 2.2 $\\mu$s half-life (0.5 mark). In Earth's frame, time dilation extends their lifetime; in the muon frame, length contraction reduces the distance. Both predictions match observations (0.5 mark).