How fast do you have to go for γ = 2?

Rearrange γ = 1/√(1 − β²) to β = √(1 − 1/γ²). For γ = 2, β = √(1 − 1/4) = √0.75 = 0.866, so v = 0.866c = 2.60 × 10⁸ m/s. At that speed a moving clock runs at half rate. Pick β from γ under Solve for to see the steps for any γ.

What is γ at 0.6c, 0.8c and 0.99c?

γ = 1/√(1 − β²) gives 1.25 at 0.6c, 1.667 at 0.8c and 7.09 at 0.99c. A moving clock runs slow by that factor and a moving rod is shortened by it.

Why are the moving frame's axes tilted on a spacetime diagram?

The ct' axis is the worldline of the moving observer, x = βct, so it leans towards the light line. Their x' axis is the set of events they call simultaneous with the origin, ct = βx, which leans the other way by the same angle. As β approaches 1 both axes close in on the 45° light line.

What is the spacetime interval?

s² = (ct)² − x² between two events. Every inertial observer gets the same value, even though they disagree about the time and the distance separately. If s² is positive the events can be causally linked and √s² is the proper time a clock travelling between them would show.

Who is really younger in the twin paradox?

The travelling twin. The situation is not symmetric: the traveller changes frame at the turnaround, while the stay-at-home twin stays in one inertial frame. At 0.8c over 10 Earth years the traveller ages 6 years.

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Spacetime diagram: special relativity

Draw time up the page and space across it, and light travels at 45°. A moving observer slices the same diagram differently: their time and space axes tilt towards the light line.

Lorentz factor γ against speed β

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What's happening

On a Minkowski diagram each event is a point (x, ct), and anything moving at speed v draws a straight worldline at slope c/v; light draws the 45° lines of the light cone. An observer moving at β = v/c has their own axes: the ct' axis is their worldline, and the x' axis, their line of 'now', tilts up by the same angle. The Lorentz transformation, ct' = γ(ct − βx) and x' = γ(x − βct) with γ = 1/√(1 − β²), converts between the two sets of coordinates. The unit ticks on the tilted axes are not the same length as on the upright ones; the hyperbolae (ct)² − x² = constant mark them, because that interval is the same in every frame. Events that are simultaneous for one observer are not for the other. A clock moving with S' ticks once per γ units of S time, which is time dilation, and a rod at rest in S' is measured in S as L₀/γ, which is length contraction. In the twin paradox the traveller turns round, switching frames, so the two twins' worldlines have different lengths of proper time and the traveller comes home younger.

γ = 1 / √(1 − β²)ct' = γ(ct − βx)x' = γ(x − βct)s² = (ct)² − x²u = (u' + v) / (1 + u'v/c²)

First-year university physics: special relativity, Minkowski diagrams, the Lorentz transformation, the invariant interval and the twin paradox. A-Level Physics (AQA option, OCR): time dilation and length contraction.

Work through the numbers with Physics Formulas and Scientific Calculator.

Challenge

Predict first: a ship leaves Earth at 0.8c and comes straight back, and 10 years pass on Earth. How many years pass on the ship? Turn on the twin paradox in Advanced with a 10-year trip at β = 0.8 and check.

FAQ

How fast do you have to go for γ = 2?: Rearrange γ = 1/√(1 − β²) to β = √(1 − 1/γ²). For γ = 2, β = √(1 − 1/4) = √0.75 = 0.866, so v = 0.866c = 2.60 × 10⁸ m/s. At that speed a moving clock runs at half rate. Pick β from γ under Solve for to see the steps for any γ.
What is γ at 0.6c, 0.8c and 0.99c?: γ = 1/√(1 − β²) gives 1.25 at 0.6c, 1.667 at 0.8c and 7.09 at 0.99c. A moving clock runs slow by that factor and a moving rod is shortened by it.
Why are the moving frame's axes tilted on a spacetime diagram?: The ct' axis is the worldline of the moving observer, x = βct, so it leans towards the light line. Their x' axis is the set of events they call simultaneous with the origin, ct = βx, which leans the other way by the same angle. As β approaches 1 both axes close in on the 45° light line.
What is the spacetime interval?: s² = (ct)² − x² between two events. Every inertial observer gets the same value, even though they disagree about the time and the distance separately. If s² is positive the events can be causally linked and √s² is the proper time a clock travelling between them would show.
Who is really younger in the twin paradox?: The travelling twin. The situation is not symmetric: the traveller changes frame at the turnaround, while the stay-at-home twin stays in one inertial frame. At 0.8c over 10 Earth years the traveller ages 6 years.