4. Lorentz Transformation for Space

Definitions

Reference Frames

• $S$ is the stationary reference frame with coordinates $(x, t)$

• $S'$ is the moving reference frame with coordinates $(x', t')$ moving at velocity $v$ relative to $S$

Spatial Coordinate

• $x$ is the position in the stationary frame

• $x'$ is the position in the moving frame

 

Derivations

consider two inertial frames $S$ with coordinates $(x, t)$ and $S'$ with coordinates $(x', t')$

assume $S'$ moves with constant velocity $v$ in the $+x$ direction relative to $S$

the origins coincide at $t = t' = 0$

uniform motion in one frame must remain uniform in another frame

if transformations were nonlinear, straight-line worldlines could curve

so we assume linear transformations

$$x' = Ax + Bt, \quad t' = Dx + Et$$

where $A, B, D, E$ depend only on $v$

the origin of $S'$ is always at $x' = 0$

in frame $S$, the origin of $S'$ moves as $x = vt$

plug into the transformation

$$0 = A(vt) + Bt = t(Av + B)$$

so $B = -Av$, thus

$$x' = A(x - vt)$$

this already gives the form of the spatial Lorentz transformation

Einstein's key postulate states that light travels at speed $c$ in all inertial frames

a light pulse emitted from the origin satisfies $x = ct$ and $x = -ct$

in $S'$, it must satisfy $x' = ct'$ and $x' = -ct'$

substitute $x = ct$ into the linear transformations

$$A(ct - vt) = c(Dct + Et)$$

factor out $t$

$$A(c - v) = c(Dc + E)$$

do the same for $x = -ct$

$$A(-c - v) = c(-Dc + E)$$

adding the equations gives

$$A(-2v) = 2cE \quad \Rightarrow \quad E = -\frac{Av}{c}$$

subtracting them gives

$$A(2c) = 2c^2D \quad \Rightarrow \quad D = \frac{A}{c}$$

so the time transformation is

$$t' = \frac{A}{c}x - \frac{Av}{c^2}t$$

use symmetry: transforming back from $S'$ to $S$ must have the same form with $v \rightarrow -v$

this implies

$$A = \frac{1}{\sqrt{1 - v^2/c^2}} \equiv \gamma$$

therefore the Lorentz transformations are

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

the $x - vt$ term comes from relative motion

the $\gamma$ factor comes from enforcing light-speed invariance

together, they lead directly to length contraction and time dilation

 

Useful Equations

$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}, \quad x' = \gamma(x - vt), \quad x = \gamma(x' + vt')$$

 

Example 1

A rocket is moving at $0.8c$ relative to Earth. An event occurs at $x = 1000$ m and $t = 2 \times 10^{-6}$ s in Earth's frame. What is the position of this event in the rocket's frame?

Answer

calculate the Lorentz factor

$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{1 - \frac{(0.8c)^2}{c^2}}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} \approx 1.67$$

apply the Lorentz transformation

$$x' = \gamma(x - vt) = 1.67(1000 - 0.8 \times 3 \times 10^8 \times 2 \times 10^{-6})$$ $$x' = 1.67(1000 - 480) = 1.67 \times 520 \approx 868 \text{ m}$$

the event occurs at approximately $868$ m in the rocket's frame

 

Example 2

Two events occur in a spaceship's frame at $x_1' = 0$ m and $x_2' = 500$ m, both at $t' = 0$ s. The spaceship moves at $0.6c$ relative to Earth. What is the spatial separation between these events in Earth's frame at $t = 0$ s?

Answer

calculate the Lorentz factor

$$\gamma = \frac{1}{\sqrt{1 - 0.36}} = \frac{1}{0.8} = 1.25$$

apply the inverse Lorentz transformation to both events

$$x_1 = \gamma(x_1' + vt') = 1.25(0 + 0) = 0 \text{ m}$$ $$x_2 = \gamma(x_2' + vt') = 1.25(500 + 0) = 625 \text{ m}$$

the spatial separation in Earth's frame is

$$\Delta x = x_2 - x_1 = 625 - 0 = 625 \text{ m}$$

note that simultaneity in the spaceship's frame does not guarantee simultaneity in Earth's frame for these spatially separated events

 

Example 3

A particle is created at position $x' = 0$ m in a moving frame and travels to $x' = 100$ m in time $t' = 5 \times 10^{-7}$ s. If the frame moves at $0.5c$ relative to the lab, where does the particle decay in the lab frame?

Answer

calculate the Lorentz factor

$$\gamma = \frac{1}{\sqrt{1 - 0.25}} = \frac{1}{\sqrt{0.75}} \approx 1.15$$

the particle decays at $x' = 100$ m and $t' = 5 \times 10^{-7}$ s in the moving frame

apply the inverse transformation

$$x = \gamma(x' + vt') = 1.15(100 + 0.5 \times 3 \times 10^8 \times 5 \times 10^{-7})$$ $$x = 1.15(100 + 75) = 1.15 \times 175 \approx 201 \text{ m}$$

the particle decays at approximately $201$ m in the lab frame