Relativity

Spacetime Diagrams

Start

Events, worldlines, and a geometric interpretation of space & time.

Lorentz Transformations

Here

Derive the coordinate transformations that preserve the speed of light for every inertial observer.

Relativistic Momentum and Energy

See how momentum and energy transform near the speed of light.

General Relativity

Finish

Connect accelerating frames to gravity through the equivalence principle, and move beyond flat spacetime.

Lorentz Transformations

Spacetime diagrams show that different inertial observers tilt their notions of space and time rather than sharing one universal clock. The Lorentz transformations turn that geometry into algebra. They tell us exactly how the coordinates $(x, t)$ used in one frame relate to $(x', t')$ used in another when the second frame moves at constant velocity $v$ along the shared $x$ -axis.

Why Galilean Time Fails

The concept of universal, absolute time would suggest that moving frames differ only by a shifted position coordinate:

x' = x - vt, \qquad t' = t,

but this cannot be correct if all inertial observers measure the same light speed. The takeaway is that time cannot remain unchanged: the corrected transformation has to mix $x$ and $t$ .

Show the contradiction with a light ray

A light pulse emitted from the origin satisfies

x = ct

in the unprimed frame. Under the Galilean assumption,

x' = x - vt = ct - vt = (c-v)t, \qquad t' = t.

Therefore the moving observer would infer

\frac{x'}{t'} = c-v,

not $c$ .

Linear Transformations

Inertial frames should map straight worldlines to straight worldlines, so the coordinate change is linear. If the two frames are aligned at the origin ( $x = x' = 0$ at $t = t' = 0$ ), then we can show

x' = A(x-vt).

Reduce the general linear form

Use the worldline of the moving origin to constrain the coefficients.

Start from the most general linear transformation:

x' = Ax + Bt, \qquad t' = Dx + Et.

The primed origin moves with speed $v$ in the unprimed frame, so its worldline is

x = vt.

Every event on that worldline must satisfy $x' = 0$ , so

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

Because this must hold for every $t$ ,

B = -Av.

Constancy of $c$

If we assume the speed of light is constant in all frames, a flash emitted from the shared origin. Both right-moving and left-moving light rays still satisfy speed $c$ after the transformation.

That requirement forces the time transformation to be

t' = A\left(t - \frac{vx}{c^2}\right).

This is one of the central results: the transformed time depends on position. That is why simultaneity is not universal.

Solve for the time-mixing term

Keep the time transformation in the form

t' = Dx + Et

for the moment. For a right-moving light ray,

x = ct, \qquad x' = ct'.

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

A(c-v)t = c(Dct + Et),

A(c-v) = c^2D + cE.

For a left-moving light ray,

x = -ct, \qquad x' = -ct'.

Then

A(-c-v)t = -c\bigl(D(-c)t + Et\bigr),

which simplifies to

-A(c+v) = c^2D - cE.

Add the two equations:

-2Av = 2c^2D \quad \Longrightarrow \quad D = -\frac{Av}{c^2}.

Subtract the second from the first:

2Ac = 2cE \quad \Longrightarrow \quad E = A.

No Inertial Frame Is Preferred

If we assume there is nothing special about any particular frame of reference, the inverse transformation must have the same form as the forward one, except that the relative velocity changes from $v$ to $-v$ .

That symmetry fixes the remaining scale factor:

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

Determine the overall scale factor

Apply the inverse transformation and require it to undo the forward one for every event.

At this stage we know

x' = A(x-vt), \qquad t' = A\left(t - \frac{vx}{c^2}\right).

The inverse must therefore be

x = A(x'+vt'), \qquad t = A\left(t' + \frac{vx'}{c^2}\right).

Substitute the forward expressions into the inverse formula for $x$ :

x = A\left[A(x-vt) + vA\left(t - \frac{vx}{c^2}\right)\right].

The middle terms cancel, leaving

x = A^2 x\left(1 - \frac{v^2}{c^2}\right).

Because this must hold for every $x$ ,

A^2\left(1 - \frac{v^2}{c^2}\right) = 1.

Choose the positive root so the transformation becomes the identity at $v = 0$ .

The Lorentz Transformations

Altogether, we have the standard Lorentz transformations:

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

with

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

The inverse relations are

x = \gamma(x'+vt'), \qquad t = \gamma\left(t' + \frac{vx'}{c^2}\right).

For motion entirely along the $x$ -direction, the transverse coordinates do not change:

y' = y, \qquad z' = z.

Problem Solving

Event Coordinates

Problem

Frame $S'$ moves to the right with $\beta = 0.60$ relative to frame $S$ . An event occurs at

x = 600\,\text{m}, \qquad t = 4.0\,\mu\text{s}

in $S$ . Find $x'$ and $t'$ .

Given

$\beta = 0.60$
$x = 600\,\text{m}$
$t = 4.0\,\mu\text{s}$

Use

\gamma = \frac{1}{\sqrt{1-\beta^2}} = \frac{1}{\sqrt{1-0.60^2}} = 1.25,

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

Solve

Since $v = 0.60c = 1.8 \times 10^8\,\text{m/s}$ ,

vt = (1.8 \times 10^8)(4.0 \times 10^{-6}) = 720\,\text{m}.

Therefore

x' = 1.25(600 - 720)\,\text{m} = -150\,\text{m}.

For the time coordinate,

\frac{vx}{c^2} = \frac{(0.60c)(600\,\text{m})}{c^2} = 0.60\frac{600}{3.0 \times 10^8}\,\text{s} = 1.2\,\mu\text{s}.

t' = 1.25(4.0 - 1.2)\,\mu\text{s} = 3.5\,\mu\text{s}.

Check

The negative value of $x'$ means the event lies slightly behind the primed origin when the primed observer assigns the time $3.5\,\mu\text{s}$ . The transformed time is not equal to $t$ , which is exactly what the Lorentz transformation predicts.

Time Dilation

Problem

A clock is fixed inside a spacecraft moving with $\beta = 0.90$ relative to Earth. Between two ticks, the spacecraft frame measures

\Delta x' = 0, \qquad \Delta t' = 12\,\mu\text{s}.

How much time $\Delta t$ passes in the Earth frame?

Given

$\beta = 0.90$
$\Delta x' = 0$
$\Delta t' = 12\,\mu\text{s}$

Use

For event differences, the inverse Lorentz transformation gives

\Delta t = \gamma\left(\Delta t' + \frac{v \Delta x'}{c^2}\right).

Since the clock is at rest in $S'$ ,

\Delta x' = 0 \quad \Longrightarrow \quad \Delta t = \gamma \Delta t'.

Also,

\gamma = \frac{1}{\sqrt{1-\beta^2}} = \frac{1}{\sqrt{1-0.90^2}} \approx 2.29.

Solve

Therefore

\Delta t = (2.29)(12\,\mu\text{s}) \approx 27.5\,\mu\text{s}.

Check

Earth records more elapsed time than the clock riding with the spacecraft. This effect is known as time dilation: the moving clock accumulates less proper time between the same two events.

Length Contraction

Problem

A cargo pod is at rest in frame $S'$ , where its proper length is

L_0 = \Delta x' = 18.0\,\text{m}.

The pod moves past a station with $\beta = 0.80$ . What length $L = \Delta x$ does the station measure?

Given

$\beta = 0.80$
$\Delta x' = 18.0\,\text{m}$
$\Delta t = 0$ in the station frame

Use

The station must record both ends of the pod at the same time in its frame (this is the definition of length), so use the spatial transformation for event differences:

\Delta x' = \gamma(\Delta x - v\Delta t).

Because the station measures the endpoints simultaneously,

\Delta t = 0 \quad \Longrightarrow \quad \Delta x' = \gamma \Delta x.

Thus

\Delta x = \frac{\Delta x'}{\gamma} = \frac{L_0}{\gamma}.

With

\gamma = \frac{1}{\sqrt{1-0.80^2}} = \frac{5}{3},

we can solve for the contracted length.

Solve

L = \Delta x = \frac{18.0\,\text{m}}{5/3} = 10.8\,\text{m}.

Check

The moving length is shorter than the proper length; this effect is known as length contraction.

Testing Simultaneity

Problem

In frame $S$ , two flashes occur simultaneously at $t = 0$ . Flash $A$ happens at $x_A = 0$ and flash $B$ happens at $x_B = 300\,\text{m}$ . Frame $S'$ moves to the right with $\beta = 0.80$ . Are the flashes simultaneous in $S'$ ?

Given

$\Delta t = 0$
$\Delta x = x_B - x_A = 300\,\text{m}$
$\beta = 0.80$

Use

For event differences,

\Delta t' = \gamma\left(\Delta t - \frac{v \Delta x}{c^2}\right).

Since

\gamma = \frac{1}{\sqrt{1-0.80^2}} = \frac{5}{3},

we have

\Delta t' = -\gamma \frac{v \Delta x}{c^2}.

Solve

First note that

\frac{\Delta x}{c} = \frac{300\,\text{m}}{3.0 \times 10^8\,\text{m/s}} = 1.0\,\mu\text{s}.

Therefore

\Delta t' = -\left(\frac{5}{3}\right)(0.80)(1.0\,\mu\text{s}) = -1.33\,\mu\text{s}.

Check

The flashes are not simultaneous in $S'$ . Because $\Delta t' < 0$ , the flash at the larger $x$ -coordinate occurs earlier in the moving frame. This is the algebraic form of the tilted simultaneity lines seen on a spacetime diagram.

Lorentz Transformations Checkpoint

Question 1 of 3

Why does the Lorentz derivation force the transformed time $t'$ to depend on position $x$?

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Spacetime Diagrams