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Space, Time and Motion - Galilean and Special Relativity (HL)

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An Inertial Frame of Reference is a frame that is either at rest or moving with a constant velocity. Newton's laws of motion are valid in all inertial frames.

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Galilean Relativity assumes that time is absolute and the same for all observers. The transformations are given by x′=x−vtx' = x - vt and t′=tt' = t.

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The First Postulate of Special Relativity states that the laws of physics are the same in all inertial frames of reference.

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The Second Postulate of Special Relativity states that the speed of light in a vacuum, c=3.00×108 m s−1c = 3.00 \times 10^8 \text{ m s}^{-1}, is constant for all observers, regardless of the motion of the source or the observer.

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The Lorentz Factor γ\gamma determines the magnitude of relativistic effects and is defined as γ=11−v2c2\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}. Note that γ≥1\gamma \geq 1.

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Time Dilation: The time interval between two events is shortest when measured in the rest frame of the events (Proper Time, Δt0\Delta t_0). For a moving observer, the time interval is dilated: Δt=γΔt0\Delta t = \gamma \Delta t_0.

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Length Contraction: The length of an object is longest when measured in its rest frame (Proper Length, L0L_0). In a frame moving relative to the object, the length LL in the direction of motion is contracted: L=L0γL = \frac{L_0}{\gamma}.

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Simultaneity: Two events that are simultaneous in one inertial frame are not necessarily simultaneous in another frame moving relative to the first.

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Relativistic Velocity Addition: To ensure no object exceeds the speed of light, velocities are added using u=u′+v1+u′vc2u = \frac{u' + v}{1 + \frac{u'v}{c^2}}.

📐Formulae

γ=11−v2c2\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}

Δt=γΔt0\Delta t = \gamma \Delta t_0

L=L0γL = \frac{L_0}{\gamma}

u=u′+v1+u′vc2u = \frac{u' + v}{1 + \frac{u'v}{c^2}}

x′=γ(x−vt)x' = \gamma(x - vt)

t′=γ(t−vxc2)t' = \gamma(t - \frac{vx}{c^2})

(Δs)2=(cΔt)2−(Δx)2(\Delta s)^2 = (c\Delta t)^2 - (\Delta x)^2

💡Examples

Problem 1:

A muon is traveling at v=0.95cv = 0.95c relative to the laboratory. If the mean lifetime of a muon at rest is 2.2×10−6 s2.2 \times 10^{-6} \text{ s}, calculate the mean lifetime as measured by a technician in the laboratory.

Solution:

First, calculate the Lorentz factor γ\gamma: γ=11−(0.95)2\gamma = \frac{1}{\sqrt{1 - (0.95)^2}} γ≈3.203\gamma \approx 3.203 Now, apply the time dilation formula using the proper time Δt0=2.2×10−6 s\Delta t_0 = 2.2 \times 10^{-6} \text{ s}: Δt=γΔt0\Delta t = \gamma \Delta t_0 Δt=3.203×2.2×10−6\Delta t = 3.203 \times 2.2 \times 10^{-6} Δt≈7.05×10−6 s\Delta t \approx 7.05 \times 10^{-6} \text{ s}

Explanation:

Because the muon is moving at a high fraction of the speed of light, its internal 'clock' appears to run slower to the laboratory observer, resulting in a longer measured lifetime.

Problem 2:

A spaceship moving at 0.80c0.80c relative to Earth fires a missile forward at 0.50c0.50c relative to the spaceship. Calculate the velocity of the missile as measured by an observer on Earth.

Solution:

Identify the given velocities: v=0.80cv = 0.80c (velocity of the spaceship frame) and u′=0.50cu' = 0.50c (velocity of the missile in the spaceship frame). Use the relativistic velocity addition formula: u=u′+v1+u′vc2u = \frac{u' + v}{1 + \frac{u'v}{c^2}} u=0.50c+0.80c1+(0.50c)(0.80c)c2u = \frac{0.50c + 0.80c}{1 + \frac{(0.50c)(0.80c)}{c^2}} u=1.30c1+0.40u = \frac{1.30c}{1 + 0.40} u=1.30c1.40≈0.929cu = \frac{1.30c}{1.40} \approx 0.929c

Explanation:

Under Galilean relativity, the speed would be 1.30c1.30c, which is impossible. The relativistic formula ensures the resultant velocity remains below cc.

Problem 3:

A meter stick (L0=1.00 mL_0 = 1.00 \text{ m}) moves past an observer at a speed of 0.60c0.60c. What is the length of the meter stick as measured by the observer?

Solution:

First, calculate the Lorentz factor γ\gamma: γ=11−(0.60)2=11−0.36=10.64\gamma = \frac{1}{\sqrt{1 - (0.60)^2}} = \frac{1}{\sqrt{1 - 0.36}} = \frac{1}{\sqrt{0.64}} γ=10.8=1.25\gamma = \frac{1}{0.8} = 1.25 Apply the length contraction formula: L=L0γL = \frac{L_0}{\gamma} L=1.001.25L = \frac{1.00}{1.25} L=0.80 mL = 0.80 \text{ m}

Explanation:

The observer measures a shorter length (contraction) because the meter stick is moving relative to their frame of reference.