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Topic 2 of 10

Lorentz Transformations

The Lorentz transformations replace the Galilean transformations of Newtonian mechanics. They mix space and time coordinates between inertial frames and reduce to the Galilean transform in the limit $v\ll c$. Any physical law that is Lorentz-covariant automatically satisfies the postulates of special relativity.

Key Concepts

Lorentz Boost

For frame

S'

moving at velocity

v

in the

x

-direction relative to

S

t'=\gamma(t-vx/c^2)

x'=\gamma(x-vt)

y'=y

z'=z

. The inverse replaces

v\to-v

Lorentz Factor

\gamma=1/\sqrt{1-\beta^2}

where

\beta=v/c

. At low speeds

\gamma\to1

. At

v=0.866c

\gamma=2

. As

v\to c

\gamma\to\infty

Simultaneity Shift

The time difference between two spatially separated events in

S'

\Delta t'=\gamma(\Delta t-v\Delta x/c^2)

. Even if

\Delta t=0

(simultaneous in

S

\Delta t'\ne0

unless

\Delta x=0

Group Structure

Successive Lorentz boosts compose as:

\beta_{\rm total}=(\beta_1+\beta_2)/(1+\beta_1\beta_2)

. The Lorentz group

O(3,1)

is the symmetry group of Minkowski spacetime.

Key Equations

Lorentz boost (x-direction)

t' = \gamma\left(t - \frac{vx}{c^2}\right), \quad x' = \gamma(x - vt)

S' moves at v relative to S along x. Inverse: replace v → -v.

Lorentz factor

\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-\beta^2}}

Always ≥ 1; diverges as v → c.

Velocity addition

u' = \frac{u - v}{1 - uv/c^2}

Speed of object moving at u in S, as measured in S' (moving at v relative to S).

Worked Example

Transforming Coordinates Between Frames

Problem

Frame $S'$ moves at $v=0.60c$ relative to $S$ . An event occurs at $t=5.0$ ns, $x=3.0$ m in $S$ . Find $t'$ and $x'$ . ( $c=3\times10^8$ m/s.)

Solution

$\gamma=1/\sqrt{1-0.36}=1.25$ . Note: $vt=0.60\times3\times10^8\times5\times10^{-9}=0.90$ m.

x' = \gamma(x-vt) = 1.25(3.0 - 0.90) = 1.25\times2.10 = 2.625\text{ m}

$vx/c^2 = 0.60\times3\times10^8\times3.0/(9\times10^{16})=0.60\times10^{-7}/10^8\times3=6\times10^{-9}$ s $=6.0$ ns.

t' = \gamma(t - vx/c^2) = 1.25(5.0 - 6.0)\text{ ns} = 1.25\times(-1.0) = -1.25\text{ ns}

Answer

x'=2.63

t'=-1.25

ns.

Practice

Exercises

7 problems

1 of 7

Explore the γ(β) curve — drag the slider to see how γ grows as a rocket approaches the speed of light. Then calculate γ for β = 0.80 using γ = 1/√(1−β²).

γ curve

Explore β γ(0.00) = 1.000

γ at β = 0.8

γ =

2 of 7

Watch the S′ axes (purple) tilt toward the light cone as β → 1 — at β ≈ 0.99 they nearly merge with it. Calculate γ for β = 0.990.

Lorentz boost

Boost speed β γ = 1.000

Watch the S′ axes (purple) tilt toward the light cone as β → 1. At β ≈ 0.99, γ grows dramatically.

γ at β = 0.99

γ =

3 of 7

Frame $S'$ moves at $v=0.60c$ ( $\gamma=1.25$ ). An event at $x=600$ m, $t=0$ in $S$ . Find $x'$ (in m).

Unlock Exercise 3