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6th Year HL · Physics
Relativistic, quantum, nuclear, and particle physics · Relativity

Length contraction, time dilation and relativistic energy

Before you start, make sure you know: Kinetic energy and potential energy (5th Year, Leaving Certificate)
Before you start, make sure you know: Power, energy consumption and efficiency (5th Year, Leaving Certificate)

Every second, muons created by cosmic rays several kilometres up in the atmosphere pass through your school, whether it is in Dublin, Cork or Galway. On their own clock they live for barely $2.2\ \mu\text{s}$: at the speed they travel, they should cover only a few hundred metres before decaying, a long way short of the ground. And yet detectors on the surface register them by the million. The explanation is not that muons live longer: it is that time itself passes differently for someone watching them rush by.

The Lorentz factor

When the lifetime of unstable particles is measured in accelerators at different speeds (for a particle with a lifetime of $2.2\ \mu\text{s}$ at rest: $0.1\cdot c \to 2.2\ \mu\text{s}$; $0.8\cdot c \to 3.7\ \mu\text{s}$; $0.99\cdot c \to 15.6\ \mu\text{s}$), the time observed in the laboratory shoots up compared with the time at rest: what mathematical factor determines the speed above which classical physics stops being valid?

Graph of the Lorentz factor γ against speed v/c, showing how it stays close to 1 at low speeds and diverges towards infinity as it approaches the speed of light. v/c γ v = c 0.1c → 1.005 0.5c → 1.15 0.9c → 2.29

This table of values gives a feel for how quickly $\gamma$ grows close to $c$: at $0.1c$ it is barely $1.005$ (effects of roughly half a per cent), but at $0.99c$ it is already $7.09$, and it keeps growing ever faster.

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Length contraction is given by $L = \frac{L_0}{\gamma}$. A spacecraft measures $L_0 = 100\ \text{m}$ at rest and travels at $\gamma = 2$. What length does an observer on…

Length contraction is given by $L = \frac{L_0}{\gamma}$. A spacecraft measures $L_0 = 100\ \text{m}$ at rest and travels at $\gamma = 2$. What length does an observer on Earth measure?

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The Lorentz factor is $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$, from which $v = c \times \sqrt{1 - \frac{1}{\gamma^2}}$. a) Calculate the speed of a spacecraft…

The Lorentz factor is $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$, from which $v = c \times \sqrt{1 - \frac{1}{\gamma^2}}$.

  1. a) Calculate the speed of a spacecraft travelling with $\gamma = 2$, expressed as a fraction of $c$.
  2. b) Express that speed in $\text{m/s}$ (given: $c = 3 \times 10^{8}\ \text{m/s}$).
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A space agency is considering a spacecraft with a proper length of $50\ \text{m}$ measuring only $10\ \text{m}$ from Earth. a) What factor $\gamma$ is needed for this…

A space agency is considering a spacecraft with a proper length of $50\ \text{m}$ measuring only $10\ \text{m}$ from Earth.

  1. a) What factor $\gamma$ is needed for this ($L = \frac{L_0}{\gamma}$)?
  2. b) Calculate the corresponding speed using $v = c \times \sqrt{1 - \frac{1}{\gamma^2}}$ and express it as a fraction of $c$.
  3. c) With this factor $\gamma$, how many years would pass on Earth for each year the crew ages ($\Delta t = \gamma \times \Delta t_0$)?
  4. d) In light of the results, evaluate the advantages and problems that interstellar travel at these speeds would present for human exploration.
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Time dilation

Light clock: at rest the light goes straight up and down; seen moving it travels a longer zigzag, so each tick lasts longer. at rest: tick = Δt₀ (proper time) v moving: the light travels a longer path → the tick lasts Δt = γ·Δt₀ > Δt₀
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A phenomenon lasts $\Delta t_0 = 10\ \text{s}$ as measured in its proper reference frame. For an observer seeing it pass at high speed, the Lorentz factor is…

A phenomenon lasts $\Delta t_0 = 10\ \text{s}$ as measured in its proper reference frame. For an observer seeing it pass at high speed, the Lorentz factor is $\gamma = 2$. Given that time dilation is $\Delta t = \gamma \times \Delta t_0$, how long does the phenomenon last for that observer?

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An astronaut travels in a spacecraft at $\gamma = 1.25$. From Earth, the trip is measured to last $8$ years. a) Using $\Delta t = \gamma \times \Delta t_0$, by how many…

An astronaut travels in a spacecraft at $\gamma = 1.25$. From Earth, the trip is measured to last $8$ years.

  1. a) Using $\Delta t = \gamma \times \Delta t_0$, by how many years does the astronaut age?
  2. b) If he left at $30$ years of age, how old is he upon return, and how old is his twin brother, who stayed on Earth?
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A muon is created in the upper atmosphere travelling at $v = 0.998c$ with $\gamma = 15$; its proper mean lifetime is $2.2\ \mu\text{s}$ (given:…

A muon is created in the upper atmosphere travelling at $v = 0.998c$ with $\gamma = 15$; its proper mean lifetime is $2.2\ \mu\text{s}$ (given: $c = 3 \times 10^{8}\ \text{m/s}$).

  1. a) Calculate its mean lifetime measured from Earth ($\Delta t = \gamma \times \Delta t_0$).
  2. b) Calculate the average distance it travels according to an observer on Earth.
  3. c) If the muon had been created at an altitude of $30\ \text{km}$, would it reach the surface on average? Justify your answer using the previous result.
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Length contraction and the twin paradox

Comparison of the length of a spacecraft at rest (L₀) and of the same spacecraft moving at relativistic speed, shorter in the direction of travel (L = L₀/γ), with the same height. at rest: L₀ in motion: L = L₀/γ v the same spacecraft, seen at rest and in motion
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Length contraction only affects the dimension parallel to motion. A spacecraft with a proper length of $50\ \text{m}$ and a proper height of $20\ \text{m}$ travels…

Length contraction only affects the dimension parallel to motion. A spacecraft with a proper length of $50\ \text{m}$ and a proper height of $20\ \text{m}$ travels horizontally at $\gamma = 1.25$. The diagram shows its proper dimensions.

v proper length: 50 m proper height: 20 m from Earth: L = ? h = ?
  1. a) What length is measured from Earth ($L = \frac{L_0}{\gamma}$)?
  2. b) What height is measured from Earth? Justify your answer.
Solve
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A spacecraft of proper length $L_0 = 120\ \text{m}$ travels at $\gamma = 1.5$. a) What length does an observer on Earth measure ($L = \frac{L_0}{\gamma}$)? b) If the…

A spacecraft of proper length $L_0 = 120\ \text{m}$ travels at $\gamma = 1.5$.

  1. a) What length does an observer on Earth measure ($L = \frac{L_0}{\gamma}$)?
  2. b) If the trip lasts $10$ years for Earth, how long does it last for the crew ($\Delta t = \gamma \times \Delta t_0$)?
  3. c) Reason which of the two observers measures the proper time of the journey and why.
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A spacecraft with a proper length of $100\ \text{m}$ passes through a space station at $\gamma = 7.09$. The station has a hangar of $15\ \text{m}$. The diagram…

A spacecraft with a proper length of $100\ \text{m}$ passes through a space station at $\gamma = 7.09$. The station has a hangar of $15\ \text{m}$. The diagram represents the situation.

hangar 15 m spacecraft: L₀ = 100 m v (γ = 7.09) L according to station = ?
  1. a) What length does the station staff measure for the spacecraft ($L = \frac{L_0}{\gamma}$)? According to them, does the entire spacecraft fit inside the hangar, even if only for an instant?
  2. b) For the pilot, it is the station that moves: what length does the pilot measure for the hangar? Does the spacecraft fit according to him?
  3. c) The apparent contradiction is resolved because "the spacecraft is entirely inside at the same time" does not mean the same thing to both observers. Name and explain the relativistic concept that resolves the paradox.
Solve

Mass-energy equivalence: $E = m\cdot c^2$

When the mass of the fragments after a nuclear fission reaction is measured with extreme precision, the final mass is $0.1\%$ less than the initial mass, while an immense amount of thermal energy is released: where has the missing mass gone?

In fission the mass of the fragments is slightly less than the initial mass: the missing mass is released as energy, E = Δm·c². m initial mass fragments: 0.999·m energy the missing mass, Δm, comes out as energy: E = Δm·c² c² is huge: 1 g of mass is equivalent to 9·10¹³ J

Fission, fusion and annihilation

In nuclear fission, a heavy nucleus splits into smaller ones. The fragments have slightly less total mass than the original nucleus, and the mass defect is released as energy. Fission also releases neutrons, which can split other nuclei. In an uncontrolled chain reaction, this grows very quickly, as in a nuclear weapon. In a controlled chain reaction, as in a nuclear power station, the number of reactions is kept steady.

In nuclear fusion, light nuclei join to form a heavier one, as hydrogen does in the Sun. The product has less mass than the nuclei that joined, and again the mass defect is released as $\Delta m\, c^2$. In annihilation, a particle and its antiparticle are completely converted into energy. A key example is positron–electron annihilation, where a positron meets an electron and their combined mass becomes two gamma photons, with energy $E = mc^2$ (used in PET scans). The proton–antiproton example above works in exactly the same way.

The mass defect is also linked to binding energy: the mass of a nucleus is less than the total mass of its separate nucleons, and the missing mass, $\Delta m$, corresponds to the binding energy $E = \Delta m\, c^2$ that holds the nucleus together.

Evaluating nuclear energy for electricity generation

Ireland has no nuclear power stations, but the ESB and EirGrid still import electricity through the National Grid, and the debate about nuclear power is a real one. When you evaluate fission and fusion for generating electricity, use secondary sources (such as the IAEA, EirGrid or the SEAI) and compare them on the same criteria.

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The energy associated with a rest mass is $E = m \times c^2$, with $c = 3 \times 10^{8}\ \text{m/s}$. Calculate the energy corresponding to $1\ \text{kg}$ of mass and…

The energy associated with a rest mass is $E = m \times c^2$, with $c = 3 \times 10^{8}\ \text{m/s}$. Calculate the energy corresponding to $1\ \text{kg}$ of mass and express it in standard form.

Solve
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The Sun radiates a power of $3.9 \times 10^{26}\ \text{W}$ by converting part of its mass into energy. a) Calculate the energy it emits each second. b) Using…

The Sun radiates a power of $3.9 \times 10^{26}\ \text{W}$ by converting part of its mass into energy.

  1. a) Calculate the energy it emits each second.
  2. b) Using $E = m \times c^2$ with $c = 3 \times 10^{8}\ \text{m/s}$, calculate the mass that the Sun converts into energy each second.
  3. c) Express the result in tonnes ($1\ \text{t} = 10^{3}\ \text{kg}$).
Solve
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In a positron emission tomography (PET) scan, a positron emitted by a radioactive tracer annihilates with an electron in the tissue, and all its mass is converted into…

In a positron emission tomography (PET) scan, a positron emitted by a radioactive tracer annihilates with an electron in the tissue, and all its mass is converted into energy, emitted as two gamma photons. The mass of each particle is $9.11 \times 10^{-31}\ \text{kg}$ (given: $c = 3 \times 10^{8}\ \text{m/s}$). The diagram represents the process.

electron positron annihilation γ γ E of each photon = ?
  1. a) Calculate the total mass annihilated in the event.
  2. b) Calculate the energy released ($E = m \times c^2$).
  3. c) What energy does each photon carry if the two share the energy equally?
  4. d) Explain why matter-antimatter annihilation is the most efficient possible process for converting mass into energy, and compare it with solar fusion, which converts less than $1\ \%$ of the mass.
Solve

Back to the beginning: at $v = 0.998\cdot c$ (where $\gamma \approx 15.8$), the lifetime measured from Earth is dilated to $\Delta t = \gamma\cdot\Delta t_0 \approx 15.8 \cdot 2.2\ \mu\text{s} \approx 35\ \mu\text{s}$, long enough to cover $d = v\cdot\Delta t \approx 10.4\ \text{km}$, so muons created at a height of $10\ \text{km}$ reach the detectors at the surface.

Summary

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