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Space, Time and Motion - Galilean and Lorentz Transformations

Grade 12IBPhysics

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

🔑Concepts

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Inertial Frame of Reference: A coordinate system in which Newton's First Law holds true, meaning the frame is either at rest or moving at a constant velocity vv.

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Postulates of Special Relativity: 1. The laws of physics are the same in all inertial frames. 2. The speed of light in a vacuum, c≈3×108 m s−1c \approx 3 \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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Galilean Transformations: Used in classical mechanics for v≪cv \ll c, assuming absolute space and time. It defines the relationship x′=x−vtx' = x - vt and t′=tt' = t.

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Lorentz Transformations: Relativistic transformations used when velocities approach cc. They account for the fact that time and space are not absolute but relative to the observer.

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Lorentz Factor (γ\gamma): A dimensionless factor that determines the magnitude of relativistic effects such as time dilation and length contraction.

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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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Proper Time (Δt0\Delta t_0) and Proper Length (L0L_0): Proper time is the time interval measured by an observer at rest relative to the events. Proper length is the length of an object measured by an observer at rest relative to the object.

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Time Dilation: A moving clock is observed to run slower than a clock at rest in the observer's frame.

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Length Contraction: The length of an object is measured to be shorter in the direction of motion when moving at a velocity vv relative to an observer.

📐Formulae

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

x′=γ(x−vt)x' = \gamma(x - vt)

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

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}}

u′=u−v1−uvc2u' = \frac{u - v}{1 - \frac{uv}{c^2}}

💡Examples

Problem 1:

A muon travels at a speed of v=0.95cv = 0.95c relative to the laboratory. If the muon's half-life in its own rest frame is 1.5×10−6 s1.5 \times 10^{-6} \text{ s}, calculate the half-life as measured by a technician in the laboratory.

Solution:

First, calculate the Lorentz factor: γ=11−0.952≈3.20\gamma = \frac{1}{\sqrt{1 - 0.95^2}} \approx 3.20. Then use the time dilation formula: Δt=γΔt0=3.20×(1.5×10−6 s)=4.8imes10−6 s\Delta t = \gamma \Delta t_0 = 3.20 \times (1.5 \times 10^{-6} \text{ s}) = 4.8 imes 10^{-6} \text{ s}.

Explanation:

Because the muon is moving at a relativistic speed relative to the laboratory, its internal 'clock' appears to slow down. The laboratory observer measures a dilated time Δt\Delta t, which is longer than the proper time Δt0\Delta t_0 measured in the muon's frame.

Problem 2:

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

Solution:

Use the relativistic velocity addition formula where v=0.80cv = 0.80c and u′=0.50cu' = 0.50c: u=0.50c+0.80c1+(0.50c)(0.80c)c2=1.30c1+0.40=1.30c1.40≈0.93cu = \frac{0.50c + 0.80c}{1 + \frac{(0.50c)(0.80c)}{c^2}} = \frac{1.30c}{1 + 0.40} = \frac{1.30c}{1.40} \approx 0.93c.

Explanation:

According to Galilean relativity, the speed would be 1.30c1.30c, which is impossible as it exceeds the speed of light. The Lorentz velocity transformation ensures that the resulting velocity uu remains below cc.

Galilean and Lorentz Transformations - Revision Notes & Key Formulas | IB Grade 12 Physics