One Event, Two Descriptions

Imagine a spacecraft passing Earth at a constant velocity.

An event occurs—for example, a light flashes somewhere near the spacecraft.

An observer on Earth records:

where the flash occurred

and:

when the flash occurred.

An observer travelling with the spacecraft also records the position and time of exactly the same event.

Will they obtain the same coordinates?

Not necessarily.

Special Relativity tells us how to convert between their measurements using the:

Lorentz transformations.

---

What Is an Event?

In relativity, an event is something that occurs at a specific:

place and time.

Examples include:

• a light switching on
• two particles colliding
• a spacecraft passing a marker
• a detector recording a particle
• a signal being transmitted
• an explosion occurring

An event therefore needs both:

spatial coordinates

and:

a time coordinate.

In one dimension, we can describe an event as:

(x, t).

---

Two Reference Frames

Consider two inertial reference frames:

S

and:

S′

Suppose S′ moves in the positive x-direction relative to S with constant velocity:

v.

At:

t = t′ = 0

their origins coincide:

x = x′ = 0.

Observer S assigns an event the coordinates:

(x, t).

Observer S′ assigns the same event:

(x′, t′).

Our job is to calculate:

x′ and t′.

---

The Lorentz Transformations

For motion along the x-axis:

x′ = γ(x − vt)

and:

t′ = γ(t − vx/c²)

where:

γ = 1/√(1 − v²/c²)

The y- and z-coordinates do not change:

y′ = y

z′ = z

These equations allow us to transform an event from:

S → S′.

---

The Variables

Understanding the symbols is essential before attempting calculations.

The speed of light is approximately:

c = 3.00 × 10⁸ m/s

---

The Lorentz Factor

Before transforming coordinates, we usually calculate:

γ

using:

γ = 1/√(1 − v²/c²).

The Lorentz factor tells us how significant relativistic effects are.

At low speeds:

γ ≈ 1

At speeds approaching c:

γ becomes much larger.

---

Useful Lorentz Factors

Notice how slowly γ changes at first.

Near:

c

it increases very rapidly.

---

A Useful Problem-Solving Strategy

When transforming coordinates, use the same sequence every time:

Step 1: Identify the reference frames.

Step 2: Identify x, t, and v.

Step 3: Calculate γ.

Step 4: Calculate x′.

Step 5: Calculate t′.

Step 6: Include units.

Step 7: Interpret what the transformed coordinates mean.

Step 8: Verify the answer if possible.

This systematic approach prevents many common mistakes.

---

Worked Example 1: Calculating γ

A spacecraft travels at:

v = 0.60c

Calculate γ.

Use:

γ = 1/√(1 − v²/c²)

Substitute:

γ = 1/√(1 − 0.60²)

γ = 1/√(1 − 0.36)

γ = 1/√0.64

Therefore:

γ = 1.25

This value can now be used in both coordinate transformations.

---

Worked Example 2: Transforming Position

A spacecraft moves at:

0.60c

relative to Earth.

An event occurs in Earth's frame at:

x = 9.0 × 10⁸ m

and:

t = 4.0 s.

Find the position of the event in the spacecraft frame.

We already know:

γ = 1.25

Use:

x′ = γ(x − vt)

First calculate:

v = 0.60(3.0 × 10⁸)

v = 1.8 × 10⁸ m/s

Then:

vt = (1.8 × 10⁸)(4.0)

vt = 7.2 × 10⁸ m

Therefore:

x′ = 1.25[(9.0 × 10⁸) − (7.2 × 10⁸)]

x′ = 1.25(1.8 × 10⁸)

x′ = 2.25 × 10⁸ m

So the spacecraft observer assigns the event the position:

x′ = 2.25 × 10⁸ m.

---

What Does x′ Mean?

The result does not mean the event physically moved after it occurred.

Instead:

x′ = 2.25 × 10⁸ m

means that the observer in S′ assigns that spatial coordinate to:

the same event.

Different observers can assign different positions to an event because their coordinate systems are:

moving relative to one another.

---

Worked Example 3: Transforming Time

Use the same event:

x = 9.0 × 10⁸ m

t = 4.0 s

v = 0.60c

γ = 1.25

Use:

t′ = γ(t − vx/c²).

First calculate:

vx/c².

Since:

v = 1.8 × 10⁸ m/s

then:

vx = (1.8 × 10⁸)(9.0 × 10⁸)

and:

c² = (3.0 × 10⁸)².

Therefore:

vx/c² = 1.8 s

Now:

t′ = 1.25(4.0 − 1.8)

t′ = 1.25(2.2)

t′ = 2.75 s

So the spacecraft observer assigns the event the time:

t′ = 2.75 s.

---

The Complete Transformation

Earth frame:

x = 9.0 × 10⁸ m

t = 4.0 s

Spacecraft frame:

x′ = 2.25 × 10⁸ m

t′ = 2.75 s

We can write:

(9.0 × 10⁸ m, 4.0 s)

transforms to:

(2.25 × 10⁸ m, 2.75 s).

These are two coordinate descriptions of:

the same event.

---

Space and Time Both Change

This example reveals something important.

The observers disagree about:

where the event occurred

and:

when the event occurred.

The Lorentz transformations therefore transform both:

space and time.

This is very different from Galilean relativity, where:

t′ = t.

---

Why Does Position Affect Time?

Look carefully at:

t′ = γ(t − vx/c²).

The transformed time depends on:

x.

This means that the time assigned to an event depends partly on:

where the event occurs.

This connection between space and time is fundamental to:

Special Relativity.

---

Using ct Instead of t

Relativity calculations can sometimes be easier if we use:

ct

instead of t.

Because:

c × time

has units of distance.

The transformations can then be written:

x′ = γ(x − βct)

and:

ct′ = γ(ct − βx)

where:

β = v/c.

This form is especially useful when speeds are written as:

0.60c, 0.80c, 0.95c, etc.

---

What Is β?

The Greek letter beta is defined as:

β = v/c.

It represents speed as a fraction of:

the speed of light.

For example:

If:

v = 0.80c

then:

β = 0.80.

The Lorentz factor can then be written:

γ = 1/√(1 − β²).

This often makes calculations simpler.

---

Worked Example 4: Using β

Suppose:

v = 0.80c

Therefore:

β = 0.80

and:

γ = 1.667.

An event occurs at:

x = 6.0 × 10⁸ m

and:

t = 3.0 s.

Calculate x′.

Use:

x′ = γ(x − vt).

First:

v = 0.80(3.0 × 10⁸)

v = 2.4 × 10⁸ m/s.

Then:

vt = (2.4 × 10⁸)(3.0)

vt = 7.2 × 10⁸ m.

Therefore:

x′ = 1.667[(6.0 × 10⁸) − (7.2 × 10⁸)]

x′ = 1.667(−1.2 × 10⁸)

x′ ≈ −2.0 × 10⁸ m.

---

What Does a Negative x′ Mean?

We found:

x′ = −2.0 × 10⁸ m.

A negative coordinate does not mean something went wrong.

It means the event occurs on the:

negative side of the S′ origin.

If positive x′ is defined as the direction of the spacecraft's motion, then the event occurs:

behind the origin of S′.

The sign has:

physical meaning.

---

Worked Example 5: Transforming the Time

Continue the previous example.

Given:

x = 6.0 × 10⁸ m

t = 3.0 s

v = 0.80c

γ = 1.667

Use:

t′ = γ(t − vx/c²).

Since:

vx/c² = 1.6 s

we obtain:

t′ = 1.667(3.0 − 1.6)

t′ = 1.667(1.4)

t′ ≈ 2.33 s.

Therefore, the transformed coordinates are approximately:

x′ = −2.0 × 10⁸ m

t′ = 2.33 s.

---

Transforming an Event at the Origin

Suppose an event occurs at the origin of S:

x = 0.

Then:

x′ = γ(0 − vt)

so:

x′ = −γvt.

This makes sense.

From the perspective of S′, the origin of S is moving in the:

negative x′ direction.

---

Transforming an Event at t = 0

Suppose an event occurs at:

t = 0

but:

x ≠ 0.

Then:

t′ = γ(0 − vx/c²)

so:

t′ = −γvx/c².

This result is extremely important.

An event occurring at:

t = 0

in S does not necessarily occur at:

t′ = 0

in S′.

This is connected to the:

relativity of simultaneity.

---

Worked Example 6: Simultaneous Events

Suppose two flashes occur simultaneously in S at:

t = 0.

Flash A occurs at:

x = 0.

Flash B occurs at:

x = 3.0 × 10⁸ m.

Suppose S′ moves at:

0.60c.

For Flash A:

t′A = 0.

For Flash B:

t′B = γ(0 − vx/c²).

Using:

γ = 1.25

and:

v = 0.60c

we get:

vx/c² = 0.60 s.

Therefore:

t′B = 1.25(−0.60)

t′B = −0.75 s.

So S′ does not consider the events:

simultaneous.

This is a direct consequence of the Lorentz transformation.

---

Transforming a Light Event

A particularly useful check involves:

light.

Suppose a light pulse is emitted from the common origin at:

t = t′ = 0.

In S:

x = ct.

After:

2.0 s

the light has travelled:

x = 6.0 × 10⁸ m.

Suppose S′ moves at:

0.60c.

---

Worked Example 7: Light Pulse

Given:

x = 6.0 × 10⁸ m

t = 2.0 s

v = 0.60c

γ = 1.25

Position:

x′ = γ(x − vt)

x′ = 1.25[(6.0 × 10⁸) − (1.8 × 10⁸)(2.0)]

x′ = 1.25(2.4 × 10⁸)

x′ = 3.0 × 10⁸ m.

Time:

t′ = γ(t − vx/c²)

Here:

vx/c² = 1.2 s.

Therefore:

t′ = 1.25(2.0 − 1.2)

t′ = 1.0 s.

Now calculate the light speed in S′:

x′/t′ = (3.0 × 10⁸)/(1.0)

x′/t′ = 3.0 × 10⁸ m/s

Therefore:

x′/t′ = c.

Both observers measure:

the same speed of light.

---

Why This Is an Excellent Verification

If a light ray satisfies:

x = ct

then after a correct Lorentz transformation it must also satisfy:

x′ = ct′.

If it does not, you should check your:

• signs
• units
• value of γ
• substitution
• arithmetic

This provides a powerful way to verify:

transformed coordinates.

---

The Inverse Transformation

Sometimes we know:

x′ and t′

and want to find:

x and t.

We then use the inverse Lorentz transformations:

x = γ(x′ + vt′)

and:

t = γ(t′ + vx′/c²).

Notice that the minus signs become:

plus signs.

This corresponds to replacing:

v with −v.

---

Forward vs Inverse Transformations

S → S′

x′ = γ(x − vt)

t′ = γ(t − vx/c²)

S′ → S

x = γ(x′ + vt′)

t = γ(t′ + vx′/c²)

The value of γ does not change because γ contains:

v².

Therefore:

γ(v) = γ(−v).

---

Worked Example 8: Transform Back

Earlier we found:

x′ = 2.25 × 10⁸ m

t′ = 2.75 s

for a frame moving at:

0.60c.

Let's transform back.

Use:

x = γ(x′ + vt′).

With:

γ = 1.25

v = 1.8 × 10⁸ m/s

we obtain:

x = 1.25[(2.25 × 10⁸) + (1.8 × 10⁸)(2.75)]

x = 1.25(7.20 × 10⁸)

x = 9.0 × 10⁸ m.

This matches the original:

x.

---

Verify the Time

Now use:

t = γ(t′ + vx′/c²).

Calculate:

vx′/c² = 0.45 s.

Therefore:

t = 1.25(2.75 + 0.45)

t = 1.25(3.20)

t = 4.0 s.

We have recovered:

x = 9.0 × 10⁸ m

and:

t = 4.0 s.

This confirms that the original transformation was:

consistent.

---

Verification Method 1: Inverse Transformation

One of the best checks is:

transform forward

then:

transform backward.

You should recover the original coordinates.

Symbolically:

(x,t) → (x′,t′) → (x,t).

Small differences may appear because of:

rounding.

---

Verification Method 2: Check the Units

For:

x′ = γ(x − vt)

both:

x

and:

vt

must have units of:

metres.

For:

t′ = γ(t − vx/c²)

both:

t

and:

vx/c²

must have units of:

seconds.

If the units do not match, the calculation is:

incorrect.

---

Why Does vx/c² Have Units of Time?

Consider:

vx/c².

Units:

(m/s)(m)/(m²/s²)

The numerator is:

m²/s.

The denominator is:

m²/s².

Therefore:

(m²/s) ÷ (m²/s²) = s.

So:

vx/c²

has units of:

time.

This is an excellent dimensional check.

---

Verification Method 3: Check the Low-Speed Limit

If:

v ≪ c

then:

γ ≈ 1.

Therefore:

x′ ≈ x − vt

and:

t′ ≈ t.

If your calculation gives enormous relativistic effects for a bicycle travelling at:

5 m/s,

something has probably gone:

wrong.

---

Verification Method 4: Check Light Speed

For a light signal:

x = ct.

After transformation:

x′ = ct′.

Therefore:

x′/t′ = c.

This is one of the strongest conceptual checks available.

---

Verification Method 5: Check the Spacetime Interval

Lorentz transformations preserve the:

spacetime interval.

For an event measured relative to the common origin:

s² = c²t² − x²

in one dimension.

After transformation:

s′² = c²t′² − x′².

A correct Lorentz transformation gives:

s² = s′².

---

Worked Example 9: Checking the Interval

Return to:

x = 9.0 × 10⁸ m

t = 4.0 s.

Calculate:

ct = (3.0 × 10⁸)(4.0)

ct = 1.2 × 10⁹ m.

Then:

s² = (ct)² − x²

s² = (1.2 × 10⁹)² − (9.0 × 10⁸)²

s² = 6.3 × 10¹⁷ m².

Now use:

x′ = 2.25 × 10⁸ m

t′ = 2.75 s.

Then:

ct′ = 8.25 × 10⁸ m.

So:

s′² = (8.25 × 10⁸)² − (2.25 × 10⁸)²

s′² = 6.3 × 10¹⁷ m².

Therefore:

s² = s′².

The spacetime interval is:

invariant.

---

A Spacetime Diagram

Coordinate transformations can also be represented visually using:

spacetime diagrams.

Usually:

• horizontal axis = x
• vertical axis = ct

An event appears as a:

point.

Different inertial observers use different coordinate axes through the same spacetime.

The event does not change.

What changes is the:

coordinate system used to describe it.

---

Physical Interpretation of Coordinates

Suppose an event transforms from:

(x,t)

to:

(x′,t′).

This does not mean:

• the event occurred twice
• one observer is wrong
• the event jumped through space
• one clock is defective

It means two observers using different inertial frames assign different:

space and time coordinates

to the same physical occurrence.

---

Coordinates Are Frame-Dependent

Quantities such as:

x

and:

t

are generally:

frame-dependent.

They can change when we change reference frames.

However, certain quantities remain invariant.

Examples include:

• the spacetime interval
• the speed of light in vacuum
• rest mass

Understanding the distinction between:

frame-dependent quantities

and:

invariant quantities

is central to Special Relativity.

---

A Useful Analogy

Imagine two people describing the same city using different coordinate systems.

One map might place a building at:

(4, 7).

Another rotated coordinate system might assign:

(7.8, 1.4).

The building has not:

moved.

Only the coordinate description changed.

Relativity extends this idea to:

space and time together.

---

Transforming Differences Between Events

Often we compare two events rather than one event relative to the origin.

Then we use:

Δx′ = γ(Δx − vΔt)

and:

Δt′ = γ(Δt − vΔx/c²).

These equations are extremely useful for analyzing:

• time dilation
• length contraction
• simultaneity
• signal travel
• particle motion

---

Worked Example 10: Two Events

Suppose two events in S are separated by:

Δx = 6.0 × 10⁸ m

and:

Δt = 3.0 s.

S′ moves at:

0.80c.

Therefore:

γ = 1.667.

Calculate the spatial separation:

Δx′ = γ(Δx − vΔt).

Since:

vΔt = (2.4 × 10⁸)(3.0)

vΔt = 7.2 × 10⁸ m,

then:

Δx′ = 1.667[(6.0 × 10⁸) − (7.2 × 10⁸)]

Δx′ ≈ −2.0 × 10⁸ m.

Now calculate:

Δt′ = γ(Δt − vΔx/c²).

Here:

vΔx/c² = 1.6 s.

Therefore:

Δt′ = 1.667(3.0 − 1.6)

Δt′ ≈ 2.33 s.

So:

Δx′ ≈ −2.0 × 10⁸ m

and:

Δt′ ≈ 2.33 s.

---

A Very Useful Shortcut: Light-Seconds

Relativity problems often contain large numbers such as:

3.0 × 10⁸ m.

A useful distance unit is the:

light-second.

One light-second is the distance light travels in one second:

1 light-second = 3.00 × 10⁸ m.

Therefore:

6.0 × 10⁸ m = 2 light-seconds.

This can make coordinate calculations much easier to visualize.

---

Example Using Light-Seconds

Suppose:

x = 4 light-seconds

t = 5 s

and:

v = 0.60c.

Since:

γ = 1.25

and the spacecraft travels:

0.60 light-seconds per second,

then:

vt = 0.60 × 5

vt = 3 light-seconds.

Therefore:

x′ = 1.25(4 − 3)

x′ = 1.25 light-seconds.

This avoids repeatedly writing:

3.00 × 10⁸.

---

Be Careful with Signs

Suppose S′ moves in the:

+x direction.

Then the forward transformation is:

x′ = γ(x − vt).

If S′ instead moves in the:

−x direction,

then v is negative.

The equation itself does not need to be replaced.

Instead, use the correct:

signed velocity.

---

Example with Negative Velocity

Suppose:

v = −0.60c.

Then:

γ = 1.25

because γ depends on:

v².

But:

x′ = γ[x − (−0.60c)t]

becomes:

x′ = γ(x + 0.60ct).

Direction therefore matters through the:

sign of v.

---

Calculator Strategy

When using a calculator, calculate γ first and store it if possible.

For example, for:

v = 0.80c

enter:

1 ÷ √(1 − 0.80²)

to obtain:

1.6667...

Keep several digits during calculations.

Round only the:

final answer.

This reduces rounding error.

---

Common Error: Forgetting γ

Incorrect:

x′ = x − vt

This is the:

Galilean transformation.

Correct:

x′ = γ(x − vt).

At relativistic speeds, forgetting γ can produce a significantly incorrect answer.

---

Common Error: Using t′ = γt Every Time

The equation:

t′ = γt

is not the general coordinate transformation.

The general transformation is:

t′ = γ(t − vx/c²).

Simple time-dilation formulas apply only under specific:

conditions.

Always identify what the problem is asking before selecting an equation.

---

Common Error: Ignoring Position When Transforming Time

Because:

t′ = γ(t − vx/c²),

you need both:

t and x

to transform the time coordinate of a general event.

Knowing t alone is usually:

not enough.

---

Common Error: Mixing Units

Do not use:

x in kilometres

with:

c in metres per second

unless you convert the units first.

A safe standard is:

• distance → metres
• time → seconds
• velocity → metres per second

Consistency is essential.

---

Common Error: Treating γ as Having Units

The Lorentz factor is:

dimensionless.

It has:

no units.

This is because:

v²/c²

is a ratio of two squared velocities.

---

Common Error: Confusing Frames

Before calculating, clearly write:

S: x = ?, t = ?

S′: x′ = ?, t′ = ?

v = ?

Then determine whether you need:

forward

or:

inverse

transformations.

This simple step prevents many sign mistakes.

---

A Reliable Calculation Template

For any coordinate transformation problem, write:

Given:

x =

t =

v =

c = 3.00 × 10⁸ m/s

Step 1: Lorentz factor

γ = 1/√(1 − v²/c²)

Step 2: Position

x′ = γ(x − vt)

Step 3: Time

t′ = γ(t − vx/c²)

Step 4: Interpretation

The observer in S′ assigns the event:

(x′, t′).

Step 5: Verification

Use the inverse transformation, interval, units, or light-speed check.

---

Connecting the Mathematics to Physics

Lorentz transformations are more than an algebra exercise.

They show mathematically that:

space and time measurements depend on reference frame.

They also explain why:

• moving clocks behave differently
• moving lengths are measured differently
• distant simultaneity depends on reference frame
• all inertial observers still measure c

The transformation equations are the mathematical foundation connecting these:

relativistic effects.

---

Check Your Understanding

1. What is an event in Special Relativity?

2. What do x and t represent?

3. What do x′ and t′ represent?

4. What does v represent?

5. Write the equation for the Lorentz factor.

6. Write the Lorentz transformation for x′.

7. Write the Lorentz transformation for t′.

8. Calculate γ when v = 0.60c.

9. Calculate γ when v = 0.80c.

10. An event occurs at x = 6.0 × 10⁸ m and t = 3.0 s. Calculate x′ for an observer moving at 0.60c.

11. For Question 10, calculate t′.

12. Explain physically what x′ represents.

13. What does a negative value of x′ mean?

14. Why can two observers assign different times to the same event?

15. Write the inverse Lorentz transformations.

16. Explain how an inverse transformation can be used to check an answer.

17. Why must x and vt have the same units?

18. Show that vx/c² has units of time.

19. Explain how a light pulse can be used to verify a Lorentz transformation.

20. What quantity involving x and t remains invariant under Lorentz transformations?

---

Key Terms

• Event: Physical occurrence at a particular position and time.
• Coordinate: Number specifying an event's location or time within a reference frame.
• Inertial frame: Non-accelerating reference frame in which a free object moves at constant velocity.
• Lorentz transformation: Equations connecting space and time coordinates between inertial frames.
• Lorentz factor (γ): Dimensionless factor 1/√(1 − v²/c²).
• Beta (β): Dimensionless speed ratio v/c.
• Forward transformation: Transformation from S to S′.
• Inverse transformation: Transformation from S′ back to S.
• Spacetime: Unified description of three dimensions of space and one dimension of time.
• Spacetime interval: Frame-invariant combination of spatial and temporal separation between events.
• Invariant: Quantity having the same value for all inertial observers.
• Frame-dependent: Quantity whose measured value can differ between reference frames.
• Light-second: Distance travelled by light in one second, approximately 3.00 × 10⁸ m.
• Relativity of simultaneity: Principle that events simultaneous in one inertial frame need not be simultaneous in another.
• Coordinate difference: Separation between two events, represented using quantities such as Δx and Δt.

---

Key Takeaways

• An event has both a position and a time.
• Different inertial observers can assign different coordinates to the same event.
• Lorentz transformations convert coordinates between inertial reference frames.
• For motion along x, x′ = γ(x − vt).
• The time transformation is t′ = γ(t − vx/c²).
• The Lorentz factor is γ = 1/√(1 − v²/c²).
• The variables x and t describe an event in S, while x′ and t′ describe it in S′.
• The velocity v describes the motion of S′ relative to S.
• The transverse coordinates are unchanged: y′ = y and z′ = z.
• A negative transformed position indicates which side of the moving frame's origin the event occupies.
• A negative transformed time is mathematically valid and means the event occurs before the chosen t′ = 0 event in that frame.
• Space and time coordinates are frame-dependent.
• Transforming coordinates does not change the physical event; it changes its coordinate description.
• The inverse transformations are x = γ(x′ + vt′) and t = γ(t′ + vx′/c²).
• A powerful verification method is to transform coordinates forward and then transform them back again.
• Units provide another important check: x and vt must both have units of distance, while t and vx/c² must both have units of time.
• For a light signal, a correct transformation preserves x′ = ct′.
• Lorentz transformations preserve the spacetime interval.
• The equations can also be written using β = v/c, which simplifies many relativistic calculations.
• At low speeds, the transformations approach the familiar Galilean transformations.
• Coordinate transformations provide the mathematical foundation for understanding time dilation, length contraction, and relativity of simultaneity.
• Successful relativity problem solving requires more than obtaining a number: you should identify the frame, calculate carefully, interpret the result, and verify it.