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Space, Time and Motion - Space-time Diagrams

Grade 12IBPhysics

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

🔑Concepts

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A Minkowski space-time diagram represents the motion of objects by plotting position xx on the horizontal axis and time (multiplied by the speed of light) ctct on the vertical axis. Both axes then have units of distance (meters).

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A worldline is the path an object takes through space-time. A vertical worldline represents an object at rest, a straight slanted line represents constant velocity, and a curved line represents acceleration.

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The worldline of a light pulse is always a straight line at an angle of 45∘45^\circ to the axes, because for light, x=ctx = ct.

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In a moving frame S′S' traveling at velocity vv relative to frame SS, the ct′ct' and x′x' axes are tilted towards the 45∘45^\circ light line. The angle of tilt θ\theta is given by tan⁡θ=β=vc\tan \theta = \beta = \frac{v}{c}.

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The lines of simultaneity for an observer in S′S' are lines parallel to the x′x' axis. Events that are simultaneous in SS (horizontal line) are generally not simultaneous in S′S'.

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The space-time interval (Δs)2(\Delta s)^2 is invariant, meaning it has the same value for all inertial observers: (Δs)2=(cΔt)2−(Δx)2(\Delta s)^2 = (c\Delta t)^2 - (\Delta x)^2.

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Proper time Δτ\Delta \tau is the time interval between two events occurring at the same spatial location in a specific frame. It is related to the interval by c2Δτ2=(cΔt)2−(Δx)2c^2 \Delta \tau^2 = (c\Delta t)^2 - (\Delta x)^2.

📐Formulae

β=vc\beta = \frac{v}{c}

tan⁡θ=β\tan \theta = \beta

γ=11−β2\gamma = \frac{1}{\sqrt{1 - \beta^2}}

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

ct′=γ(ct−βx)ct' = \gamma (ct - \beta x) Ratio of scales on axes

x′=γ(x−βct)x' = \gamma (x - \beta ct)

💡Examples

Problem 1:

A spacecraft moves past the Earth at a constant speed of v=0.8cv = 0.8c. Calculate the angle that the spacecraft's time axis ct′ct' makes with the Earth's time axis ctct on a space-time diagram.

Solution:

We use the relationship tan⁡θ=vc\tan \theta = \frac{v}{c}. Given v=0.8cv = 0.8c, then tan⁡θ=0.8\tan \theta = 0.8. θ=arctan⁡(0.8)≈38.7∘\theta = \arctan(0.8) \approx 38.7^\circ.

Explanation:

In a Minkowski diagram, the axis representing the worldline of a moving observer (ct′ct') is tilted relative to the stationary observer's axis (ctct) by an angle whose tangent is the fraction of the speed of light at which the observer is moving.

Problem 2:

Two lightning bolts strike points AA (at x=0x = 0) and BB (at x=2000 mx = 2000\text{ m}) simultaneously in the Earth's frame. If a rocket moves in the positive xx direction, which event happens first according to the rocket pilot?

Solution:

On a space-time diagram, the rocket's lines of simultaneity (the x′x' axis and lines parallel to it) slope upwards. As we move up the ctct axis, a line with positive slope will intersect event BB (x=2000x = 2000) at a lower ct′ct' coordinate than event AA (x=0x = 0). Therefore, event BB occurs first.

Explanation:

Relativity of simultaneity: Events that are simultaneous in the stationary frame occur at different times in the moving frame. Because the x′x' axis tilts 'up' in the direction of motion, events further along the direction of motion (+x+x) occur 'earlier' in the S′S' frame time.

Space-time Diagrams - Revision Notes & Key Formulas | IB Grade 12 Physics