Physics • Year 12 • Module 7 • Lesson 11

Special Relativity — Inertial Frames and Postulates

Apply the Lorentz factor, Einstein’s postulates, and the relativistic energy equations to a range of scenarios and data.

Apply · Data & Reasoning

1. Interpret data — Lorentz factor and relativistic energy at the Australian Synchrotron

Electrons at the Australian Synchrotron (Clayton, VIC) are accelerated to relativistic speeds. The table below gives several operating speeds and derived quantities. 8 marks

$v/c$ $\gamma$ (calculated) Total energy $E = \gamma m_e c^2$ (keV) Relativistic $E_k = (\gamma-1)m_e c^2$ (keV)
0.101.0055142.6
0.501.15559079
0.902.291170659
0.997.0936233112
0.9997calculatecalculatecalculate

Electron rest mass energy $m_e c^2 = 511$ keV. Note: $v = 0.9997c$ is close to the injection energy at the Australian Synchrotron.

1.1 For $v = 0.9997c$, calculate $\gamma$ (show working) and complete the last row of the table. 3 marks

1.2 Describe the trend in $\gamma$ as $v/c$ increases from 0.10 to 0.9997. What does this imply about the energy required to accelerate an electron to higher and higher speeds? 2 marks

1.3 At $v = 0.9997c$, compare the total energy to the rest mass energy (511 keV). What fraction of the total energy is kinetic energy? Is the classical formula $E_k = \tfrac{1}{2}m_e v^2$ a reasonable approximation at this speed? Justify qualitatively. 3 marks

Stuck? For Q1.1: $\gamma = 1/\sqrt{1 - (0.9997)^2}$. Then $E = \gamma \times 511$ keV; $E_k = (\gamma - 1) \times 511$ keV.

2. Interpret a graph — Lorentz factor vs speed

The graph below shows how the Lorentz factor $\gamma$ varies with speed $v/c$ for any massive particle. 7 marks

1 2 3 4 5 6 7 8 9 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Speed ($v/c$) Lorentz factor $\gamma$ $v=0.6c,\;\gamma=1.25$ $v=0.9c,\;\gamma=2.29$

Figure 2. Lorentz factor $\gamma$ as a function of speed $v/c$. Note the steep rise as $v \to c$.

2.1 Read the graph to estimate $\gamma$ at $v = 0.5c$ and $v = 0.8c$. Use the formula $\gamma = 1/\sqrt{1-v^2/c^2}$ to verify your readings. 2 marks

2.2 Describe what the graph tells you about the energy required to accelerate a massive particle as $v$ approaches $c$. Why can no massive particle ever reach exactly $v = c$? 2 marks

2.3 An electron ($m_e c^2 = 511$ keV) is accelerated to $v = 0.98c$. Calculate $\gamma$ and then find its total energy $E = \gamma m_e c^2$ in keV. Show full working. 3 marks

Stuck? For Q2.3: $\gamma = 1/\sqrt{1 - 0.98^2}$. Then $E = \gamma \times 511$ keV.

3. Compare Galilean and Einsteinian relativity

Complete the two-column table below. For each feature, state what Galilean (classical) relativity says and what Einstein’s special relativity says. 10 marks (1 per cell)

FeatureGalilean relativity predicts…Einstein’s relativity predicts…
Speed of light measured by moving observer
Velocity addition (two objects)
Time measurements in different frames
Length of moving object
Energy needed to reach $v = c$
Stuck? Revisit Cards 1 and 2 and the formula panel in the lesson.

4. Predict and justify — laser on a spaceship

A spaceship passes Earth at 0.80c. The captain shines a laser beam forward (in the direction of motion) and one backward (opposite to motion). 5 marks

4.1 (a) What does Einstein’s second postulate predict for the speed of the forward laser beam as measured by Earth observers? (b) What does Galilean velocity addition predict? (c) Which prediction matches experiment, and what does this tell us about space and time? 3 marks

4.2 The captain’s on-board clock records the journey to a star 4.0 light-years away taking 3.0 years ship time. Calculate $\gamma$ for the journey and determine the speed of the spaceship as a fraction of $c$. Use the time dilation formula $\Delta t = \gamma \Delta t_0$, where $\Delta t$ is the Earth-frame time and $\Delta t_0$ is the ship-frame time. (Hint: Earth-frame time = distance ÷ speed.) 2 marks

Stuck? For Q4.2: let $v$ be the speed. Earth-frame distance = 4.0 ly; Earth-frame time = 4.0/v years; ship time $\Delta t_0 = 3.0$ yr; $\Delta t = 4.0/v$; $\gamma = \Delta t/\Delta t_0$. Also $\gamma = 1/\sqrt{1-v^2/c^2}$; solve for $v$.

5. Spot the errors — a student’s summary of special relativity

A Year 12 student wrote the following summary. It contains three errors. Identify each error and write the correction. 6 marks (2 per error: 1 identify, 1 correct)

“Einstein’s first postulate says that the speed of light is the same in all frames. His second postulate says that the laws of physics are different for moving observers. The Michelson-Morley experiment confirmed these postulates by detecting the motion of Earth through the aether. The Lorentz factor $\gamma$ approaches zero as the speed of an object approaches the speed of light.”

5.1 Error 1: What is wrong?

Correction:

5.2 Error 2: What is wrong?

Correction:

5.3 Error 3: What is wrong?

Correction:

Stuck? Revisit the two postulates and the description of the Michelson-Morley experiment in Card 1 of the lesson.
Answers — Do not peek before attempting

Q1.1 — $\gamma$ at $v = 0.9997c$ (3 marks)

$\gamma = 1/\sqrt{1 - (0.9997)^2} = 1/\sqrt{1 - 0.99940009} = 1/\sqrt{0.000600} \approx 1/0.02449 \approx \mathbf{40.8}$ [1].

Total energy: $40.8 \times 511 = 20\,849$ keV $\approx$ 20.8 MeV [1]. Relativistic KE: $(40.8 - 1) \times 511 = 39.8 \times 511 = 20\,338$ keV $\approx$ 20.3 MeV [1].

Q1.2 — Trend in $\gamma$ (2 marks)

$\gamma$ increases slowly at low speeds but grows rapidly and non-linearly above $v \approx 0.8c$, diverging toward infinity as $v \to c$ [1]. This means the energy required ($E = \gamma m_e c^2$) also diverges — an infinite amount of energy would be needed to reach exactly $c$, so massive particles can never reach the speed of light [1].

Q1.3 — Fraction that is kinetic (3 marks)

At $v = 0.9997c$, $\gamma \approx 40.8$. Total energy $\approx 20\,849$ keV; rest energy = 511 keV [1]. KE fraction $= 20\,338/20\,849 \approx 97.5\%$ — almost all the energy is kinetic [1]. Classical formula $E_k = \tfrac{1}{2}m_e v^2 \approx \tfrac{1}{2}m_e c^2 = 256$ keV, far less than 20\,338 keV; the classical formula vastly underestimates KE at this speed and is not a reasonable approximation [1].

Q2.1 — Read $\gamma$ from graph (2 marks)

From graph: $v = 0.5c \Rightarrow \gamma \approx 1.15$; $v = 0.8c \Rightarrow \gamma \approx 1.67$ [1 for reading both correctly within reasonable tolerance]. Verification: $\gamma(0.5c) = 1/\sqrt{1-0.25} = 1/\sqrt{0.75} = 1.155$ ✓; $\gamma(0.8c) = 1/\sqrt{1-0.64} = 1/0.6 = 1.667$ ✓ [1].

Q2.2 — Why massive particles cannot reach $c$ (2 marks)

As $v \to c$, $\gamma \to \infty$, so $E = \gamma mc^2 \to \infty$ [1]. An infinite amount of energy would be required to accelerate a massive particle to exactly $c$, which is physically impossible. Only massless particles (photons) travel at $c$ [1].

Q2.3 — Electron at $v = 0.98c$ (3 marks)

$\gamma = 1/\sqrt{1 - 0.98^2} = 1/\sqrt{1 - 0.9604} = 1/\sqrt{0.0396} = 1/0.199 \approx \mathbf{5.03}$ [1].

$E = \gamma m_e c^2 = 5.03 \times 511 \approx \mathbf{2570}$ keV [1]. This is the total relativistic energy including rest mass [1].

Q3 — Compare Galilean and Einstein relativity

Speed of light: Galilean: observer moving toward source measures $c + v$. Einstein: all observers measure exactly $c$.

Velocity addition: Galilean: $u' = u + v$ (simple sum). Einstein: relativistic addition formula; result never exceeds $c$.

Time: Galilean: time is absolute, same for all frames. Einstein: time is relative; moving clocks run slow ($\Delta t = \gamma \Delta t_0$).

Length: Galilean: lengths are absolute, same for all frames. Einstein: moving objects contract in direction of motion ($L = L_0/\gamma$).

Energy to reach $c$: Galilean: finite; classical $KE = \tfrac{1}{2}mv^2$ is finite at $v = c$. Einstein: infinite; $E = \gamma mc^2 \to \infty$ as $v \to c$.

Q4.1 — Laser on spaceship (3 marks)

(a) Einstein’s second postulate: Earth observers measure the forward laser beam at speed $c$ [1]. (b) Galilean addition: Earth would measure $c + 0.80c = 1.80c$ [1]. (c) Einstein’s prediction ($c$) matches all experiments. This tells us that time and space are not absolute — they adjust so that all observers measure the same $c$. This gives rise to time dilation, length contraction, and the relativity of simultaneity [1].

Q4.2 — Speed from time dilation (2 marks)

Let $v$ be the speed. Earth-frame time $\Delta t = 4.0/v$ (where $v$ is in light-years/year, i.e. as a fraction of $c$). Ship frame $\Delta t_0 = 3.0$ yr. So $\Delta t = \gamma \Delta t_0 \Rightarrow 4.0/v = 3.0\gamma$. Also $\gamma = 1/\sqrt{1-v^2}$. Try $v = 0.8c$: $\Delta t = 4.0/0.8 = 5.0$ yr; $\gamma = 1/0.6 = 1.667$; $3.0 \times 1.667 = 5.0$ ✓ [1]. Speed = $\mathbf{0.8c}$, $\gamma = \mathbf{1.667}$ [1].

Q5 — Errors in student’s summary (6 marks)

5.1 Error 1 — first and second postulates swapped: The student assigned “speed of light is same in all frames” to the first postulate and “laws of physics are different” to the second. Both are wrong [1]. Correction: Einstein’s first postulate states that the laws of physics are identical in all inertial frames. His second postulate states that the speed of light in vacuum is constant for all observers [1].

5.2 Error 2 — Michelson-Morley “confirmed” the postulates by detecting Earth’s motion: The experiment produced a null result — no aether wind was detected [1]. Correction: The Michelson-Morley experiment failed to detect any motion of Earth through the aether. This null result undermined the aether hypothesis and was consistent with (though it did not directly prove) Einstein’s postulates [1].

5.3 Error 3 — $\gamma$ approaches zero as $v \to c$: This is the opposite of what happens [1]. Correction: The Lorentz factor $\gamma = 1/\sqrt{1-v^2/c^2}$ approaches infinity as $v \to c$. It equals 1 at $v = 0$ and grows without bound [1].