1. Derivation of Lorentz Transformations

 

 

 

5

How Do Different Observers Describe the Same Event?

Imagine two observers.

Observer S is standing on Earth.

Observer S′ is travelling in a spacecraft at constant velocity:

v

relative to S.

An event occurs somewhere in space.

Observer S describes it using coordinates:

(x, y, z, t)

Observer S′ describes the same event using:

(x′, y′, z′, t′)

The central question is:

How are these two sets of coordinates related?

In classical physics, we use:

Galilean transformations.

In Special Relativity, we must use:

Lorentz transformations.

Coordinate Transformations

A coordinate transformation is a mathematical rule that allows us to convert measurements from one reference frame into:

another reference frame.

For example, two observers may disagree about an object's:

• position
• velocity
• time
• distance

But their measurements must be connected by consistent:

mathematical relationships.

Setting Up Two Reference Frames

Consider two inertial reference frames:

S and S′

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

v.

At:

t = t′ = 0

the origins coincide:

x = x′ = 0

The axes are aligned.

Motion occurs along the:

x-axis.

 

 

 

5

This simple arrangement allows us to derive the transformation equations.

The Galilean Transformation

In classical mechanics, the relationship between positions is:

x′ = x − vt

The other spatial coordinates remain:

y′ = y

z′ = z

Classical physics also assumes:

t′ = t

This means:

time is absolute.

Every observer agrees on the same time interval.

Understanding x′ = x − vt

Suppose S′ moves to the right at:

10 m/s.

After:

5 s

its origin has travelled:

vt = 50 m

relative to S.

If an object is at:

x = 80 m

in S, then S′ assigns:

x′ = 80 − 50

x′ = 30 m

This works extremely well at:

ordinary speeds.

The Classical Assumption About Time

The most important Galilean assumption is:

t′ = t

If 10 seconds pass for S, then:

10 seconds

also pass for S′.

Time is treated as completely independent of:

motion.

This idea was central to classical Newtonian physics.

The Problem with Light

Now suppose S observes a pulse of light travelling in the positive x-direction.

Its speed is:

c.

Classical velocity transformation would predict:

u′ = u − v

Therefore:

c′ = c − v

If S′ moved toward the light, classical physics would predict a different measured:

light speed.

But this conflicts with the central postulate of Special Relativity.

Einstein's Second Postulate

Albert Einstein proposed that:

the speed of light in vacuum is the same for all inertial observers.

Therefore:

c′ = c

not:

c′ = c − v

This means the Galilean transformation cannot be the correct transformation at:

relativistic speeds.

What Must Change?

The Galilean transformation assumes:

space changes between frames

but:

time does not.

Special Relativity requires something fundamentally different.

To keep the speed of light invariant:

both space and time coordinates must transform.

This leads to the:

Lorentz transformations.

Assumptions Behind the Lorentz Transformations

The derivation begins with several important assumptions.

1. Principle of Relativity

The laws of physics are the same in all:

inertial reference frames.

There is no preferred inertial frame.

2. Constancy of the Speed of Light

All inertial observers measure the same vacuum light speed:

c.

3. Homogeneity of Space and Time

The laws of physics do not depend on:

where or when an experiment occurs.

4. Isotropy of Space

Physics does not fundamentally depend on:

direction.

5. Linearity

For inertial frames moving uniformly relative to one another, the coordinate transformation is taken to be:

linear.

This preserves uniform motion.

Why Assume a Linear Transformation?

Suppose an object moves at constant velocity in one inertial frame.

It should also move uniformly in another:

inertial frame.

If the transformation were strongly nonlinear, uniform motion in one frame could become:

accelerated motion

in another.

That would conflict with the equivalence of inertial frames.

Therefore, we seek transformations that are:

linear in x and t.

Start with a General Form

Because S′ moves along the x-axis, suppose:

x′ = A(x − vt)

where A is a factor that may depend on:

v.

Why use:

x − vt?

Because the origin of S′ satisfies:

x = vt.

At the S′ origin:

x′ = 0

so:

x − vt = 0.

This ensures the moving origin is correctly described.

The Transformation for Time

Unlike Galilean relativity, we cannot simply assume:

t′ = t.

Instead, time must also depend on position.

We eventually obtain:

t′ = A(t − vx/c²)

The same factor A appears in both transformations.

We now need to determine:

what A must be.

Use a Pulse of Light

Imagine that when the two origins coincide:

t = t′ = 0

a pulse of light is emitted from the common origin.

According to observer S:

x = ct

According to observer S′:

x′ = ct′

because both observers must measure the same:

speed of light c.

 

 

 

6

Substitute the Light Path

We proposed:

x′ = A(x − vt)

and:

t′ = A(t − vx/c²)

For light:

x = ct

Substitute into the position equation:

x′ = A(ct − vt)

Factor:

x′ = At(c − v)

Now substitute into the time equation:

t′ = A(t − vct/c²)

Simplify:

t′ = At(1 − v/c)

Multiplying by c:

ct′ = At(c − v)

Therefore:

x′ = ct′

as required.

The transformation preserves:

the speed of light.

Finding the Factor A

We still need to determine:

A.

The principle of relativity requires symmetry between:

S and S′.

If S′ moves at velocity v relative to S, then S moves at velocity:

−v

relative to S′.

Therefore, the inverse transformation must have the same form:

x = A(x′ + vt′)

Substitute the Forward Transformations

Start with:

x = A(x′ + vt′)

Substitute:

x′ = A(x − vt)

and:

t′ = A(t − vx/c²)

Then:

x = A[A(x − vt) + vA(t − vx/c²)]

Factor out A²:

x = A²[x − vt + vt − v²x/c²]

The middle terms cancel:

−vt + vt = 0

Therefore:

x = A²x(1 − v²/c²)

Divide by x:

1 = A²(1 − v²/c²)

So:

A² = 1/(1 − v²/c²)

Therefore:

A = 1/√(1 − v²/c²)

This factor is called:

the Lorentz factor.

The Lorentz Factor

The Lorentz factor is represented by the Greek letter:

γ

and is defined as:

γ = 1/√(1 − v²/c²)

Therefore:

A = γ

and our transformations become:

x′ = γ(x − vt)

and:

t′ = γ(t − vx/c²)

These are the central:

Lorentz transformation equations.

Complete Lorentz Transformations

For relative motion along the x-axis:

x′ = γ(x − vt)

y′ = y

z′ = z

t′ = γ(t − vx/c²)

where:

γ = 1/√(1 − v²/c²)

 

 

 

5

What Do These Equations Mean?

The equations show that measurements of:

space and time are interconnected.

Notice:

x′ depends on x and t

and:

t′ depends on t and x.

This means different inertial observers do not simply disagree about:

position.

They can also disagree about:

time.

This is one of the deepest differences between classical and relativistic physics.

The Inverse Lorentz Transformations

To transform from S′ back to S, replace:

v with −v.

Therefore:

x = γ(x′ + vt′)

and:

t = γ(t′ + vx′/c²)

The transverse coordinates remain:

y = y′

z = z′

The symmetry reflects the principle that neither inertial frame is:

preferred.

Why γ Matters

The Lorentz factor controls the size of:

relativistic effects.

Recall:

γ = 1/√(1 − v²/c²)

When:

v ≪ c

then:

v²/c² ≈ 0

so:

γ ≈ 1.

Therefore, Lorentz transformations become very close to:

Galilean transformations.

Values of the Lorentz Factor

 

 

 

5

At low speeds:

γ ≈ 1

Near the speed of light:

γ increases rapidly.

Recovering the Galilean Transformation

Consider:

x′ = γ(x − vt)

At low speeds:

γ ≈ 1

Therefore:

x′ ≈ x − vt

which is exactly the Galilean position transformation.

Now consider:

t′ = γ(t − vx/c²)

At low speeds:

γ ≈ 1

and:

vx/c²

is extremely small.

Therefore:

t′ ≈ t

which is the classical assumption of:

absolute time.

Classical Physics Is an Approximation

This is an important scientific idea.

Special Relativity does not mean that classical mechanics is:

useless or completely wrong.

Instead:

Galilean transformations are the low-speed approximation of Lorentz transformations.

When:

v ≪ c

the difference becomes negligible.

This is why classical mechanics works extremely well for:

• cars
• bicycles
• aircraft
• falling objects
• most engineering systems

Galilean vs Lorentz Transformations

Worked Example 1: Calculate γ

A spacecraft travels at:

0.60c

Calculate the Lorentz factor.

γ = 1/√(1 − v²/c²)

Since:

v = 0.60c

then:

v²/c² = 0.36

Therefore:

γ = 1/√(1 − 0.36)

γ = 1/√0.64

γ = 1.25

Worked Example 2: Transforming Position

Suppose:

v = 0.60c

and an event occurs at:

x = 9.0 × 10⁸ m

at:

t = 4.0 s

We know:

γ = 1.25

Use:

x′ = γ(x − vt)

Calculate:

vt = (0.60)(3.0 × 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

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

= (0.60c)(9.0 × 10⁸)/c²

Using:

c = 3.0 × 10⁸ m/s

this becomes:

1.8 s

Therefore:

t′ = 1.25(4.0 − 1.8)

t′ = 1.25(2.2)

t′ = 2.75 s

So S′ assigns the event:

x′ = 2.25 × 10⁸ m

t′ = 2.75 s

Same Event, Different Coordinates

Observer S describes the event as:

(9.0 × 10⁸ m, 4.0 s)

Observer S′ describes it as:

(2.25 × 10⁸ m, 2.75 s)

These are not different:

events.

They are different coordinate descriptions of:

the same event.

This is similar to two maps assigning different coordinates to the same location.

Worked Example 4: Transforming a Light Pulse

Suppose a light pulse travels according to:

x = ct

Take:

t = 2.0 s

Then:

x = 6.0 × 10⁸ m

Suppose S′ moves at:

0.60c.

Then:

γ = 1.25

Position:

x′ = γ(x − vt)

x′ = 1.25[(6.0 × 10⁸) − (0.60)(3.0 × 10⁸)(2.0)]

x′ = 3.0 × 10⁸ m

Time:

t′ = γ(t − vx/c²)

t′ = 1.0 s

Therefore:

x′/t′ = 3.0 × 10⁸ m/s

So S′ also measures:

c.

This is exactly what the Lorentz transformations must accomplish.

Galilean Transformation of the Same Light Pulse

Classically:

x′ = x − vt

Using the same example:

x′ = 6.0 × 10⁸ − 3.6 × 10⁸

x′ = 2.4 × 10⁸ m

Galilean physics also says:

t′ = 2.0 s

Therefore:

u′ = x′/t′

u′ = 1.2 × 10⁸ m/s

That is:

0.40c

But Special Relativity requires:

c.

This demonstrates why Galilean transformations fail for:

light and relativistic motion.

Space and Time Become Mixed

The Lorentz equations are:

x′ = γ(x − vt)

t′ = γ(t − vx/c²)

Notice that:

position transformation contains time

and:

time transformation contains position.

Space and time can no longer be treated as completely:

independent quantities.

They form a unified structure called:

spacetime.

 

 

 

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Lorentz Transformations and Time Dilation

Time dilation follows from the:

Lorentz transformations.

Consider a clock at rest in S′.

For that clock:

Δx′ = 0.

Applying the inverse transformation leads to:

Δt = γΔt′

If the clock measures proper time:

Δτ = Δt′

then:

Δt = γΔτ

This is the familiar:

time dilation equation.

Lorentz Transformations and Length Contraction

Length contraction also follows from the transformations.

Suppose an object has proper length:

L₀

in its own rest frame.

An observer who sees the object moving must measure the positions of both ends:

simultaneously in that observer's frame.

Applying the Lorentz transformation gives:

L = L₀/γ

This is the:

length contraction equation.

Lorentz Transformations and Simultaneity

Suppose two events are simultaneous in S:

Δt = 0

but occur at different positions:

Δx ≠ 0.

Then:

Δt′ = γ(Δt − vΔx/c²)

so:

Δt′ = −γvΔx/c²

Therefore:

Δt′ ≠ 0

in general.

This produces the:

relativity of simultaneity.

One Transformation, Several Effects

The Lorentz transformations explain:

time dilation

length contraction

relativity of simultaneity

relativistic velocity addition

These are not independent assumptions.

They are consequences of the same underlying:

transformation between inertial frames.

Relativistic Velocity Transformation

Galilean physics predicts:

u′ = u − v

But Special Relativity gives:

u′ = (u − v)/(1 − uv/c²)

This equation ensures that if:

u = c

then:

u′ = c.

Example: Light Speed

Suppose:

u = c

and:

v = 0.80c.

Then:

u′ = (c − 0.80c)/(1 − (c)(0.80c)/c²)

u′ = 0.20c/(1 − 0.80)

u′ = 0.20c/0.20

u′ = c

The moving observer still measures:

the speed of light as c.

Example: Two Spacecraft

Suppose spacecraft A moves at:

0.80c

relative to Earth.

Spacecraft B moves in the same direction at:

0.60c

relative to Earth.

Classically, A would measure B moving at:

0.20c

relative to it.

Relativistically:

u′ = (0.60c − 0.80c)/(1 − 0.60 × 0.80)

u′ = −0.20c/0.52

u′ ≈ −0.385c

The magnitude of the relative velocity is:

0.385c.

At relativistic speeds, velocities do not simply:

subtract classically.

Why Can't Massive Objects Reach c?

Consider:

γ = 1/√(1 − v²/c²)

As:

v → c

then:

1 − v²/c² → 0

Therefore:

γ → ∞

Many relativistic quantities involving γ grow without bound as a massive object's speed approaches:

c.

This is one mathematical indication that an object with nonzero rest mass cannot be accelerated to:

the speed of light.

The Spacetime Interval

Lorentz transformations change measurements of:

space and time.

But they preserve an important quantity called the:

spacetime interval.

For two events:

Δs² = c²Δt² − Δx² − Δy² − Δz²

All inertial observers calculate the same:

Δs².

This is called:

Lorentz invariance.

Compare with Distance in Ordinary Geometry

Imagine rotating coordinate axes on a sheet of paper.

The x and y coordinates of a point change.

But the distance:

r² = x² + y²

does not.

Similarly, Lorentz transformations change:

space and time coordinates

while preserving the:

spacetime interval.

This provides a useful geometric way to understand:

Special Relativity.

Minkowski Spacetime

Hermann Minkowski developed a geometric interpretation of Special Relativity.

Instead of treating space and time separately, events are represented in:

four-dimensional spacetime.

Coordinates can be written:

(ct, x, y, z).

Lorentz transformations describe how different inertial observers assign coordinates within this:

spacetime.

Galilean Spacetime vs Relativistic Spacetime

Galilean View

Space:

relative

Time:

absolute

Transformation:

x′ = x − vt

t′ = t

Relativistic View

Space:

frame-dependent

Time:

frame-dependent

Transformation:

x′ = γ(x − vt)

t′ = γ(t − vx/c²)

Invariant quantity:

spacetime interval

Why the Lorentz Factor Appears Everywhere

You have already encountered γ in:

time dilation

Δt = γΔτ

and:

length contraction

L = L₀/γ

Now we see where it comes from.

It is not an arbitrary correction added to equations.

It emerges from requiring the coordinate transformations to satisfy:

• the principle of relativity
• invariance of c
• symmetry between inertial frames

The Lorentz factor is therefore built into the:

geometry of spacetime.

At Everyday Speeds

Suppose a car travels at:

30 m/s.

Then:

v/c ≈ 10⁻⁷

so:

γ ≈ 1

and:

vx/c²

is extremely small for ordinary distances.

Therefore:

x′ ≈ x − vt

and:

t′ ≈ t.

Galilean transformations are therefore perfectly adequate for:

most everyday situations.

At Relativistic Speeds

Suppose:

v = 0.90c

Then:

γ ≈ 2.294

Now γ is far from:

1.

The differences between Lorentz and Galilean transformations become:

very large.

At these speeds, classical transformations cannot accurately describe:

space, time or velocity.

Worked Comparison

Suppose an event occurs at:

x = 3.0 × 10⁸ m

t = 2.0 s

and S′ moves at:

0.80c.

Galilean Transformation

x′ = x − vt

x′ = 3.0 × 10⁸ − (0.80)(3.0 × 10⁸)(2)

x′ = −1.8 × 10⁸ m

and:

t′ = 2.0 s

Lorentz Transformation

For:

v = 0.80c

γ = 1.667

Therefore:

x′ = 1.667(−1.8 × 10⁸)

x′ ≈ −3.0 × 10⁸ m

For time:

t′ = γ(t − vx/c²)

t′ = 1.667(2.0 − 0.80)

t′ ≈ 2.0 s

In this particular event, the transformed time happens to remain 2.0 s, while the transformed position differs substantially. Other event coordinates generally produce differences in both.

The important point is that the two theories make:

different quantitative predictions.

Choosing the Correct Transformation

Use Galilean transformations when:

• speeds are much smaller than c
• relativistic precision is unnecessary
• classical mechanics provides a sufficient approximation

Use Lorentz transformations when:

• speeds are a significant fraction of c
• light propagation is important
• relativistic precision is required
• studying particle physics or high-energy processes

Common Misconception: Lorentz Transformations Were Invented Just to Fix Time Dilation

No.

Lorentz transformations provide the fundamental coordinate relationship between:

inertial frames in Special Relativity.

Time dilation is one:

consequence.

So are:

• length contraction
• relativity of simultaneity
• relativistic velocity addition

Common Misconception: Galilean Transformations Are Wrong

They are not useless or meaningless.

They are an excellent:

low-speed approximation.

When:

v/c → 0

Lorentz transformations approach:

Galilean transformations.

This is called the:

correspondence principle.

A more general theory should reproduce the successful predictions of an older theory in the conditions where the older theory works.

Common Misconception: γ Changes the Speed of Light

No.

The Lorentz factor helps ensure that different inertial observers all measure:

the same c.

It changes how their space and time coordinates are:

related.

Common Misconception: Time Is Universal

Galilean transformations assume:

t′ = t.

Lorentz transformations show:

t′ = γ(t − vx/c²).

Therefore, the time assigned to a distant event depends on:

• the event's position
• relative velocity
• reference frame

Time is not universally:

absolute.

Common Misconception: Different Coordinates Mean Different Events

No.

Two observers can assign different:

x and t coordinates

to the same event.

The event itself is the same physical occurrence.

Coordinates depend on:

reference frame.

Check Your Understanding

1. What is a coordinate transformation?

2. State the Galilean transformation for position.

3. What assumption does Galilean relativity make about time?

4. Why does the Galilean transformation fail for light?

5. State Einstein's two postulates.

6. List three assumptions used when deriving Lorentz transformations.

7. Why is the transformation assumed to be linear?

8. State the Lorentz transformation for position.

9. State the Lorentz transformation for time.

10. Define the Lorentz factor.

11. Calculate γ for v = 0.60c.

12. Explain why γ approaches 1 at low speeds.

13. Explain how the Lorentz transformations reduce to Galilean transformations when v ≪ c.

14. Why does time transformation contain a position term?

15. Show that a light pulse travelling at c in S also travels at c in S′.

16. Explain how time dilation follows from Lorentz transformations.

17. Explain how length contraction follows from Lorentz transformations.

18. Explain how relativity of simultaneity follows from Lorentz transformations.

19. What quantity remains invariant under Lorentz transformations?

20. Why do physicists use Galilean transformations for everyday motion but Lorentz transformations for relativistic motion?

Key Terms

• Coordinate transformation: Mathematical relationship connecting coordinates assigned by different reference frames.
• Galilean transformation: Classical transformation assuming absolute time.
• Lorentz transformation: Relativistic transformation connecting space and time coordinates between inertial frames.
• Inertial reference frame: Non-accelerating frame in which a free object moves at constant velocity.
• Lorentz factor (γ): Factor 1/√(1 − v²/c²) controlling the magnitude of relativistic effects.
• Principle of relativity: Laws of physics have the same form in all inertial frames.
• Constancy of light speed: All inertial observers measure the same vacuum light speed c.
• Linearity: Property in which transformed coordinates depend linearly on the original coordinates.
• Homogeneity: Principle that the laws of physics do not depend on absolute position or time.
• Isotropy: Principle that physical laws do not depend on spatial direction.
• Spacetime: Unified four-dimensional description of space and time.
• Spacetime interval: Lorentz-invariant separation between events.
• Lorentz invariance: Property that fundamental physical laws and the spacetime interval have the appropriate invariant form under Lorentz transformations.
• Inverse transformation: Transformation converting coordinates back to the original frame.
• Correspondence principle: Requirement that a newer theory reproduce the successful predictions of an older theory in the older theory's valid limit.

Key Takeaways

• Coordinate transformations connect measurements made in different reference frames.
• Classical mechanics uses Galilean transformations.
• Galilean transformations assume t′ = t, meaning time is absolute.
• Classical velocity addition would predict different measured speeds of light for different observers.
• This conflicts with Einstein's postulate that all inertial observers measure the same vacuum speed of light, c.
• Special Relativity therefore requires Lorentz transformations.
• The derivation assumes the principle of relativity, invariance of c, homogeneity, isotropy and linearity.
• For relative motion along the x-axis, x′ = γ(x − vt).
• The time transformation is t′ = γ(t − vx/c²).
• The Lorentz factor is γ = 1/√(1 − v²/c²).
• The Lorentz factor arises from requiring symmetry between inertial frames while preserving c.
• Lorentz transformations mix space and time coordinates.
• This mixing is a fundamental feature of spacetime.
• Time dilation follows from Lorentz transformations.
• Length contraction follows from Lorentz transformations.
• Relativity of simultaneity follows from Lorentz transformations.
• Relativistic velocity addition also follows from the same framework.
• Lorentz transformations preserve the spacetime interval.
• At low speeds, γ ≈ 1 and vx/c² ≈ 0.
• Therefore, Lorentz transformations reduce approximately to Galilean transformations.
• Galilean physics remains an excellent approximation for ordinary speeds.
• At speeds approaching c, the differences become significant and Lorentz transformations are essential.
• The Lorentz factor is not an arbitrary correction—it is a fundamental consequence of the geometry and symmetry of Special Relativity.

2. Transforming Coordinates

 

 

 

5

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.

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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.

 

 

 

4

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′.

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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′.

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The Variables

Understanding the symbols is essential before attempting calculations.

Symbol	Meaning	SI Unit
x	Position of event measured in S	m
x′	Position of event measured in S′	m
t	Time of event measured in S	s
t′	Time of event measured in S′	s
v	Velocity of S′ relative to S	m/s
c	Speed of light	m/s
γ	Lorentz factor	no unit

The speed of light is approximately:

c = 3.00 × 10⁸ m/s

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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.

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Useful Lorentz Factors

v	γ
0.10c	1.005
0.20c	1.021
0.50c	1.155
0.60c	1.250
0.80c	1.667
0.90c	2.294
0.95c	3.203
0.99c	7.089

 

 

 

Notice how slowly γ changes at first.

Near:

c

it increases very rapidly.

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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.

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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.

 

 

 

5

---

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.

 

 

 

6

---

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.

 

 

 

5

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.

 

 

 

3. Relativistic Velocity Addition

 

 

 

5

Can Two Velocities Add to More Than the Speed of Light?

Imagine a spacecraft travelling away from Earth at:

0.80c

It launches a probe forward at:

0.70c

relative to the spacecraft.

Classical physics would suggest:

0.80c + 0.70c = 1.50c

But Special Relativity tells us that the probe cannot be measured travelling at:

1.50c.

Instead, velocities combine according to a different rule:

relativistic velocity addition.

Classical Velocity Addition

At ordinary speeds, velocities simply add or subtract.

Suppose a train moves at:

20 m/s

and a passenger walks forward inside the train at:

2 m/s.

An observer standing beside the tracks measures approximately:

20 + 2 = 22 m/s.

So:

u = u′ + v

where:

• u = object's velocity measured in S
• u′ = object's velocity measured in S′
• v = velocity of S′ relative to S

For everyday motion, this works extremely well.

Why Does Classical Addition Work?

For speeds much smaller than:

c

relativistic corrections are tiny.

A car travelling at 30 m/s and another object moving at 10 m/s relative to it do not require complicated relativistic calculations.

We can simply use:

30 + 10 = 40 m/s.

But this approximation breaks down when velocities become a significant fraction of:

the speed of light.

The Problem with Classical Addition

Suppose a spacecraft travels at:

0.80c

and fires a probe forward at:

0.70c.

Classically:

u = 0.80c + 0.70c

u = 1.50c

This creates a problem.

Special Relativity requires that massive objects cannot be accelerated through the light-speed limit, and all inertial observers measure light in vacuum travelling at:

c.

Therefore, ordinary velocity addition cannot apply at:

relativistic speeds.

The Relativistic Velocity Addition Equation

For motion along the same straight line:

u = (u′ + v)/(1 + u′v/c²)

This equation gives the velocity u measured in S when:

• S′ moves at velocity v relative to S
• an object moves at velocity u′ relative to S′

The denominator is the key difference from:

classical velocity addition.

Another Form of the Equation

Sometimes we know the object's velocity in S and want its velocity in S′.

Then:

u′ = (u − v)/(1 − uv/c²)

This is the relativistic equivalent of the classical equation:

u′ = u − v.

Which form you use depends on:

which frame contains the known velocity.

Understanding the Variables

Always define your:

reference frames

before substituting numbers.

This prevents many sign and direction errors.

Where Does the Equation Come From?

The relativistic velocity equation follows directly from the:

Lorentz transformations.

Recall:

x′ = γ(x − vt)

and:

t′ = γ(t − vx/c²).

Velocity is:

u = dx/dt

and:

u′ = dx′/dt′.

Therefore:

u′ = dx′/dt′

becomes:

u′ = [γ(dx − vdt)] / [γ(dt − vdx/c²)].

The γ factors cancel:

u′ = (dx − vdt)/(dt − vdx/c²).

Divide numerator and denominator by dt:

u′ = (dx/dt − v)/(1 − v(dx/dt)/c²).

Since:

dx/dt = u,

we obtain:

u′ = (u − v)/(1 − uv/c²).

So relativistic velocity addition is not an extra rule added separately to Special Relativity.

It follows from the:

Lorentz transformations.

Worked Example 1: Two Spacecraft

A spacecraft travels at:

0.80c

relative to Earth.

It launches a probe forward at:

0.70c

relative to the spacecraft.

What velocity does Earth measure?

Use:

u = (u′ + v)/(1 + u′v/c²).

Substitute:

u = (0.70c + 0.80c)/(1 + (0.70c)(0.80c)/c²)

Simplify:

u = 1.50c/(1 + 0.56)

u = 1.50c/1.56

Therefore:

u ≈ 0.962c.

Earth measures the probe travelling at about:

0.962c.

Not:

1.50c.

Classical vs Relativistic Result

For the same problem:

Classical

u = 0.70c + 0.80c

u = 1.50c

Relativistic

u = (0.70c + 0.80c)/(1 + 0.70 × 0.80)

u ≈ 0.962c

The difference is enormous because the speeds are:

relativistic.

Why the Denominator Matters

Consider:

u = (u′ + v)/(1 + u′v/c²).

The denominator:

1 + u′v/c²

reduces the result compared with simple addition.

At low speeds:

u′v ≪ c²

so:

u′v/c² ≈ 0.

Then:

u ≈ (u′ + v)/1

and therefore:

u ≈ u′ + v.

So classical velocity addition appears naturally as the:

low-speed approximation.

Worked Example 2: Moderate Speeds

Suppose:

v = 0.30c

and:

u′ = 0.40c.

Classically:

u = 0.30c + 0.40c

u = 0.70c.

Relativistically:

u = (0.40c + 0.30c)/(1 + 0.40 × 0.30)

u = 0.70c/1.12

u = 0.625c.

Even at these speeds, the difference is already noticeable.

Worked Example 3: Lower Speeds

Suppose:

v = 0.01c

and:

u′ = 0.02c.

Classically:

u = 0.03c.

Relativistically:

u = 0.03c/(1 + 0.0002)

u ≈ 0.029994c.

The difference is extremely small.

This explains why we normally use:

classical velocity addition

in everyday life.

A Visual Comparison

 

 

 

Classical velocity addition increases without a built-in limit.

Relativistic velocity addition approaches:

c

without allowing massive objects to cross it.

What Happens If the Object Is Light?

This is one of the most important tests of the equation.

Suppose S′ moves at velocity:

v

relative to S.

A light pulse moves forward in S′ at:

u′ = c.

Use:

u = (u′ + v)/(1 + u′v/c²).

Substitute:

u = (c + v)/(1 + cv/c²).

Simplify the denominator:

1 + v/c.

Therefore:

u = (c + v)/(1 + v/c).

Factor c from the numerator:

u = c(1 + v/c)/(1 + v/c).

So:

u = c.

Every inertial observer still measures the light travelling at:

c.

Example: Chasing a Beam of Light

Suppose a spacecraft travels at:

0.90c.

A light beam travels forward past the spacecraft.

Classical physics might suggest that the spacecraft measures the light travelling at:

c − 0.90c = 0.10c.

But use the relativistic transformation:

u′ = (u − v)/(1 − uv/c²).

Set:

u = c

and:

v = 0.90c.

Then:

u′ = (c − 0.90c)/(1 − 0.90)

u′ = 0.10c/0.10

Therefore:

u′ = c.

The spacecraft still measures:

the full speed of light.

 

 

 

Why This Is So Important

The invariance of the speed of light is one of the foundations of:

Special Relativity.

Relativistic velocity addition ensures that different inertial observers do not obtain:

different vacuum light speeds.

This is one reason classical velocity addition must be replaced at:

relativistic speeds.

Multiple Moving Observers

Now consider three observers:

Earth

Spacecraft A

Spacecraft B

Suppose A travels at:

0.70c

relative to Earth.

B travels forward at:

0.60c

relative to A.

What velocity does Earth measure for B?

Use:

u = (u′ + v)/(1 + u′v/c²).

Substitute:

u = (0.60c + 0.70c)/(1 + 0.60 × 0.70)

u = 1.30c/1.42

Therefore:

u ≈ 0.915c.

Earth measures B travelling at:

0.915c.

Three Frames, Three Measurements

This example demonstrates an important idea.

The velocity of an object is always measured:

relative to a reference frame.

Spacecraft B can have:

0.60c relative to A

while simultaneously having:

0.915c relative to Earth.

There is no contradiction.

Velocity is:

frame-dependent.

Opposite Directions

Signs become particularly important when objects move in:

opposite directions.

Define rightward as:

positive.

Then leftward velocities are:

negative.

The same equation still works:

u′ = (u − v)/(1 − uv/c²).

Do not automatically add magnitudes.

Use:

signed velocities.

Worked Example 4: Opposite Directions

Earth observes:

Spacecraft A moving right at:

+0.80c

Spacecraft B moving left at:

−0.70c.

What velocity does A measure for B?

Let:

u = −0.70c

and:

v = +0.80c.

Use:

u′ = (u − v)/(1 − uv/c²).

Substitute:

u′ = (−0.70c − 0.80c)/(1 − (−0.70)(0.80))

u′ = −1.50c/(1 + 0.56)

u′ = −1.50c/1.56

u′ ≈ −0.962c.

Therefore A measures B travelling at approximately:

0.962c in the opposite direction.

Not:

1.50c.

Relative Speed Between Two Spacecraft

This result is often surprising.

Earth can observe:

• A moving right at 0.80c
• B moving left at 0.70c

Classically their relative speed would be:

1.50c.

But the speed of B measured in A's inertial frame is:

0.962c.

This is an important distinction between:

coordinate-frame relative velocity

and simply adding two speed magnitudes measured by a third observer.

Worked Example 5: Finding Velocity in the Moving Frame

Earth observes a probe travelling at:

0.90c.

A spacecraft travels in the same direction at:

0.60c.

What velocity does the spacecraft measure for the probe?

Use:

u′ = (u − v)/(1 − uv/c²).

Substitute:

u′ = (0.90c − 0.60c)/(1 − 0.90 × 0.60)

u′ = 0.30c/(1 − 0.54)

u′ = 0.30c/0.46

Therefore:

u′ ≈ 0.652c.

Classically we would obtain:

0.30c.

At relativistic speeds, that classical answer is substantially incorrect.

Worked Example 6: Finding an Unknown Velocity

Earth measures a probe travelling at:

0.90c.

The probe moves at:

0.50c

relative to a spacecraft travelling in the same direction.

Find the spacecraft's speed relative to Earth.

Start with:

u = (u′ + v)/(1 + u′v/c²).

Substitute:

0.90c = (0.50c + v)/(1 + 0.50v/c).

Let:

β = v/c.

Then:

0.90 = (0.50 + β)/(1 + 0.50β).

Multiply:

0.90(1 + 0.50β) = 0.50 + β

0.90 + 0.45β = 0.50 + β

0.40 = 0.55β

β ≈ 0.727.

Therefore:

v ≈ 0.727c.

A Useful Dimensionless Form

When all speeds are given as fractions of c, calculations become easier.

Define:

βu = u/c

βu′ = u′/c

βv = v/c.

Then:

βu = (βu′ + βv)/(1 + βu′βv).

For example:

βu′ = 0.70

βv = 0.80

Then:

βu = (0.70 + 0.80)/(1 + 0.70 × 0.80)

βu = 0.962.

Therefore:

u = 0.962c.

Why Massive Objects Cannot Reach c

The velocity-addition equation ensures that combining sub-light velocities produces another velocity below:

c.

But there is a deeper reason massive objects cannot be accelerated to the speed of light.

Recall the Lorentz factor:

γ = 1/√(1 − v²/c²).

As:

v → c

then:

1 − v²/c² → 0

and therefore:

γ → ∞.

Relativistic Energy

The total energy of a particle with rest mass m is:

E = γmc².

As:

v → c

then:

γ → ∞.

Therefore, the energy required to continue accelerating a massive object toward c grows without bound.

Reaching exactly:

v = c

would require unbounded energy in this framework.

A massive object therefore cannot be accelerated from below c to:

c.

What About Light?

Light is different.

Photons have:

zero rest mass.

They travel in vacuum at:

c.

The equation:

E = γmc²

should not be applied to photons by simply setting m = 0 and v = c, because that produces an undefined limiting expression.

For photons, the appropriate energy relationship is:

E = pc.

Can Anything Travel Faster Than Light?

Within Special Relativity, ordinary matter and information cannot be accelerated through the invariant speed:

c.

The causal structure of spacetime prevents ordinary signals from being transmitted locally faster than:

light in vacuum.

This does not mean every speed-like quantity encountered in physics must be below c.

For example, apparent motion, certain wave velocities, and the increasing distance between sufficiently distant galaxies in cosmology require more careful interpretation.

They do not represent an ordinary local object overtaking a nearby beam of light.

The Speed Limit Is Local

This distinction is important.

Special Relativity states that locally, in an inertial frame:

c is the invariant speed of light in vacuum.

No massive object can locally accelerate through:

c.

In cosmology, however, General Relativity allows the distance between very distant objects to increase in ways that can correspond to recession rates greater than c because:

space itself is dynamically evolving.

That is not ordinary velocity addition.

What Happens as Speeds Approach c?

Suppose a spacecraft travels at:

0.99c.

It launches a probe forward at:

0.99c

relative to itself.

Classically:

u = 1.98c.

Relativistically:

u = (0.99c + 0.99c)/(1 + 0.99²)

u = 1.98c/1.9801

u ≈ 0.99995c.

Even two velocities extremely close to c combine to produce:

less than c.

What If One Velocity Equals c?

Suppose:

u′ = c.

Then:

u = (c + v)/(1 + v/c).

This always simplifies to:

u = c.

No matter how quickly the observer moves, the transformed speed of light remains:

c.

This is built directly into the mathematics of:

Lorentz transformations.

Mathematical Proof for Two Sub-Light Speeds

Suppose:

|u′| < c

and:

|v| < c.

Relativistic addition gives:

u = (u′ + v)/(1 + u′v/c²).

For same-direction positive velocities, define:

a = u′/c

and:

b = v/c

where:

0 ≤ a < 1

and:

0 ≤ b < 1.

Then:

u/c = (a + b)/(1 + ab).

To ask whether this is below 1, compare:

a + b

with:

1 + ab.

Their difference is:

1 + ab − a − b

which factors as:

(1 − a)(1 − b).

Because both factors are positive:

1 + ab > a + b.

Therefore:

u/c < 1

and hence:

u < c.

Classical and Relativistic Addition Compared

The two models agree closely at:

low speeds.

They diverge dramatically as velocities approach:

c.

Relativistic Addition Is Symmetric in One Dimension

For two velocities in the same direction:

u = (u′ + v)/(1 + u′v/c²).

Notice that swapping u′ and v gives:

u = (v + u′)/(1 + vu′/c²).

The result is unchanged.

This symmetry reflects the structure of:

one-dimensional relativistic velocity composition.

Direction Still Matters

Velocity is a:

vector quantity.

In one-dimensional problems, direction can be represented using:

positive and negative signs.

For example:

right:

+

left:

−

A negative final answer does not mean the speed is negative.

It means the velocity points in the:

negative direction.

Beyond One Dimension

So far, we have considered motion along one:

straight line.

If an object also has velocity components perpendicular to the motion of the reference frame, the transformation becomes more complicated.

For a frame moving along x:

u′x = (ux − v)/(1 − uxv/c²)

while the perpendicular components also involve:

γ.

This means relativistic velocity transformation is fundamentally:

three-dimensional.

For introductory problems, however, one-dimensional motion is usually the most important case.

Connection to Lorentz Transformations

 

 

 

Relativistic velocity addition is a direct consequence of transforming both:

space

and:

time.

Classically, only the position transformation matters because:

t′ = t.

Relativity instead gives:

t′ = γ(t − vx/c²).

Because time itself transforms, the ratio:

distance/time

also transforms differently.

That is why velocities cannot simply:

add and subtract classically.

Worldlines and Velocity

On a spacetime diagram, an object's motion is represented by a:

worldline.

The slope of that worldline is related to:

velocity.

 

 

 

Light forms the boundary:

x = ±ct.

Massive objects have worldlines that remain:

inside the light cone.

They cannot be continuously accelerated so that their worldlines cross the light-cone boundary.

A Practical Problem-Solving Method

When solving relativistic velocity problems:

Step 1: Identify the reference frames.

Step 2: Choose a positive direction.

Step 3: Assign signs to every velocity.

Step 4: Identify u, u′, and v.

Step 5: Choose the appropriate transformation.

Step 6: Substitute carefully.

Step 7: Check that the result makes physical sense.

For ordinary massive objects, you should expect:

|u| < c.

Example Problem-Solving Setup

Write:

S = Earth

S′ = spacecraft

v = velocity of spacecraft relative to Earth

u′ = velocity of probe relative to spacecraft

u = velocity of probe relative to Earth

Then use:

u = (u′ + v)/(1 + u′v/c²).

Writing the frames explicitly prevents one of the most common errors:

mixing velocities measured in different frames.

Common Misconception: 0.8c + 0.8c = 1.6c

Only under classical velocity addition.

Relativistically:

u = (0.8c + 0.8c)/(1 + 0.8²)

u = 1.6c/1.64

u ≈ 0.976c.

The combined velocity remains below:

c.

Common Misconception: A Fast Spacecraft Almost Catches Light

Suppose a spacecraft moves at:

0.999c.

It does not measure a forward-moving light beam travelling at:

0.001c.

It still measures:

c.

This is one of the most important departures from:

classical intuition.

Common Misconception: Light Gets an Extra c from a Moving Source

Suppose a spacecraft travels at:

0.70c

and switches on a laser pointing forward.

Earth does not measure the light at:

1.70c.

Both Earth and the spacecraft measure the light travelling locally at:

c.

The motion of the source does not add to the vacuum speed of:

light.

Common Misconception: Opposite Spacecraft Can Measure Each Other Above c

Suppose Earth sees:

A = +0.90c

and:

B = −0.90c.

Earth may note that their coordinate separation is increasing at:

1.80c.

But when A measures B's velocity in A's inertial frame:

u′ = (−0.90c − 0.90c)/(1 + 0.81)

u′ = −1.80c/1.81

u′ ≈ −0.9945c.

So neither spacecraft measures the other locally moving faster than:

c.

Common Misconception: Relativistic Addition Is Needed for Cars

Technically, relativity applies to:

all velocities.

But at everyday speeds:

u′v/c²

is extraordinarily small.

Therefore:

u ≈ u′ + v.

Classical velocity addition is usually more than accurate enough.

When Should You Use Relativistic Velocity Addition?

Use it when:

• velocities are significant fractions of c
• high-energy particles are involved
• spacecraft move at relativistic speeds
• comparing measurements between rapidly moving frames
• light or other relativistic signals are involved

For ordinary everyday speeds, classical addition remains an excellent:

approximation.

Real-World Applications

Relativistic velocity transformations are important in:

• particle accelerator physics
• cosmic-ray studies
• astrophysics
• relativistic jets
• high-energy particle collisions
• theoretical spacecraft problems
• fundamental tests of Special Relativity

 

 

 

6

Particles in accelerators routinely travel at velocities extremely close to:

c.

Their motion cannot be accurately analyzed using ordinary:

Galilean velocity addition.

A Final Example

Spacecraft A travels at:

0.95c

relative to Earth.

It launches a probe forward at:

0.80c

relative to itself.

Classically:

u = 1.75c.

Relativistically:

u = (0.80c + 0.95c)/(1 + 0.80 × 0.95)

u = 1.75c/1.76

u ≈ 0.9943c.

Despite both velocities being very large, the final result remains:

below c.

This is exactly what Special Relativity requires.

Check Your Understanding

1. State the classical velocity addition equation.

2. Why does classical velocity addition work well at everyday speeds?

3. Write the relativistic velocity addition equation for motion in the same direction.

4. Explain the meanings of u, u′, and v.

5. A spacecraft moves at 0.60c and launches a probe forward at 0.50c relative to itself. Calculate the probe's velocity relative to Earth.

6. Compare the classical and relativistic answers to Question 5.

7. A spacecraft travels at 0.80c and fires a probe forward at 0.80c. Calculate the probe's velocity relative to Earth.

8. Explain why the answer to Question 7 is not 1.60c.

9. Show mathematically that if u′ = c, then u = c.

10. A spacecraft travels at 0.90c while a light pulse travels in the same direction. What speed does the spacecraft measure for the light?

11. Earth sees spacecraft A travelling at +0.70c and spacecraft B at −0.60c. Calculate B's velocity in A's frame.

12. Why are signs important in velocity transformation problems?

13. Explain why relativistic velocity addition approaches classical addition at low speeds.

14. What happens to γ as a massive object's speed approaches c?

15. Explain why a massive object cannot be accelerated to c.

16. Two velocities of 0.99c are combined in the same direction. Calculate the resulting velocity.

17. Explain why the motion of a light source does not increase the measured vacuum speed of its light.

18. How does relativistic velocity addition follow from the Lorentz transformations?

19. Why can two observers measure different velocities for the same object without either being wrong?

20. Explain how relativistic velocity addition preserves the invariant speed c.

Key Terms

• Velocity addition: Rule used to determine an object's velocity when measurements are made from different moving reference frames.
• Classical velocity addition: Low-speed approximation u = u′ + v.
• Relativistic velocity addition: Special Relativity equation u = (u′ + v)/(1 + u′v/c²).
• Velocity transformation: Conversion of a measured velocity from one inertial frame to another.
• Reference frame: Coordinate system relative to which position, time and motion are measured.
• Inertial frame: Non-accelerating reference frame.
• Relative velocity: Velocity of one object measured from another reference frame.
• Speed of light (c): Invariant vacuum speed approximately 3.00 × 10⁸ m/s.
• Lorentz transformation: Transformation connecting space and time coordinates between inertial frames.
• Lorentz factor (γ): 1/√(1 − v²/c²).
• Beta (β): Dimensionless velocity ratio v/c.
• Light cone: Spacetime boundary formed by light travelling from an event.
• Worldline: Path of an object through spacetime.
• Invariant: Quantity that remains the same under the relevant transformation.
• Rest mass: Invariant mass of an object measured in its rest frame.

Key Takeaways

• Velocities do not simply add at relativistic speeds.
• Classical velocity addition is u = u′ + v.
• Relativistic velocity addition is u = (u′ + v)/(1 + u′v/c²).
• To transform the other way, use u′ = (u − v)/(1 − uv/c²).
• Relativistic velocity addition follows directly from the Lorentz transformations.
• At low speeds, u′v/c² ≈ 0, so the relativistic equation reduces to classical velocity addition.
• At speeds approaching c, the difference between classical and relativistic predictions becomes large.
• Two sub-light velocities do not combine to produce a massive object's speed greater than c.
• If one of the velocities is exactly c, the transformed velocity remains c.
• Every inertial observer measures light in vacuum travelling locally at c.
• A spacecraft cannot reduce the measured speed of a light beam simply by chasing it.
• Motion of a light source does not add its speed to the vacuum speed of the emitted light.
• Velocities are frame-dependent, so the same object can have different velocities in different inertial frames.
• Direction matters, so relativistic velocity problems should use signed velocities.
• Multiple moving-observer problems become easier when each reference frame is identified explicitly.
• The Lorentz factor grows without bound as a massive object's speed approaches c.
• Accelerating an object with nonzero rest mass to exactly c would require unbounded energy.
• Light follows a different energy-momentum relationship because photons have zero rest mass.
• Relativistic velocity addition is essential in particle physics, accelerator physics and astrophysics.
• The equation preserves the invariant speed c, making it a fundamental consequence of Special Relativity.

4. Transformation of Time Intervals

 

 

 

5

Do All Observers Measure the Same Time Interval?

Imagine an astronaut travelling past Earth at a very high speed.

Inside the spacecraft, a clock measures:

10 seconds

between two events.

Will an observer on Earth also measure exactly:

10 seconds?

At everyday speeds, the difference would be far too small to notice.

At relativistic speeds, however, the answer can be:

no.

Time intervals depend on the:

reference frame.

To understand how they transform, we need to distinguish carefully between:

proper time

and:

observed coordinate time.

Events and Time Intervals

A time interval is measured between:

two events.

Suppose Event 1 occurs at:

t₁

and Event 2 occurs at:

t₂.

The time interval is:

Δt = t₂ − t₁.

In another reference frame:

Δt′ = t′₂ − t′₁.

Special Relativity tells us that:

Δt and Δt′ are not generally equal.

The Lorentz Transformation for Time

For a single event:

t′ = γ(t − vx/c²)

where:

γ = 1/√(1 − v²/c²).

For two events, subtract the two time transformations.

This gives:

Δt′ = γ(Δt − vΔx/c²).

This is the general transformation equation for:

time intervals.

The Variables

The speed of light is:

c ≈ 3.00 × 10⁸ m/s.

Why Does Position Appear in a Time Equation?

Notice:

Δt′ = γ(Δt − vΔx/c²).

The transformed time interval depends not only on:

Δt

but also on:

Δx.

This is a major difference from classical physics.

In Special Relativity:

space and time are interconnected.

Two observers can disagree about the time between events partly because they also disagree about the:

spatial relationship between those events.

Proper Time

The proper time between two events is the time measured by a clock that is present at:

both events.

Equivalently, the two events occur at the same spatial location in the clock's:

rest frame.

Proper time is usually represented by:

Δτ.

If the two events occur at the same location in S′:

Δx′ = 0,

then S′ measures the:

proper time.

So:

Δt′ = Δτ.

A Simple Example of Proper Time

Imagine an astronaut starts a stopwatch while sitting in a spacecraft.

Ten seconds later, according to the same stopwatch, the astronaut stops it.

In the spacecraft frame:

• Event 1: stopwatch starts
• Event 2: stopwatch stops

Both events occur at the same place relative to the astronaut and stopwatch.

Therefore, the spacecraft clock measures:

proper time.

 

 

 

5

Coordinate Time

An observer in another frame sees the spacecraft:

moving.

The two stopwatch events therefore occur at:

different positions

in that observer's frame.

The time interval measured using synchronized clocks in that frame is a:

coordinate time interval.

For the standard time-dilation situation:

Δt = γΔτ.

The coordinate interval is larger than the:

proper time.

Time Dilation

The time-dilation equation is:

Δt = γΔτ

where:

• Δτ = proper time
• Δt = time interval measured in a frame where the clock is moving
• γ = Lorentz factor

Because:

γ ≥ 1

we have:

Δt ≥ Δτ.

The proper time is the shortest time interval between the same pair of timelike-separated events.

Why Does Time Dilation Follow from the Lorentz Transformation?

Suppose a clock is at rest in S′.

The two events occur at the same position in S′:

Δx′ = 0.

Use the inverse time transformation:

Δt = γ(Δt′ + vΔx′/c²).

Since:

Δx′ = 0,

we obtain:

Δt = γΔt′.

Since S′ measures proper time:

Δt′ = Δτ.

Therefore:

Δt = γΔτ.

So time dilation is not a separate rule.

It follows directly from the:

Lorentz transformations.

The Lorentz Factor

Recall:

γ = 1/√(1 − v²/c²).

Some useful values are:

 

 

 

4

At low velocities:

γ ≈ 1.

Near the speed of light:

γ increases rapidly.

Worked Example 1: Basic Time Dilation

An astronaut measures:

Δτ = 8.0 s

between two events on the spacecraft.

The spacecraft moves at:

0.60c

relative to Earth.

Calculate the time interval measured by Earth.

First:

γ = 1.25.

Use:

Δt = γΔτ.

Therefore:

Δt = 1.25(8.0)

Δt = 10.0 s.

The astronaut measures:

8.0 s.

Earth measures:

10.0 s.

Interpreting the Result

The result does not mean the astronaut's clock is:

malfunctioning.

Nor does it mean Earth's clocks are:

incorrect.

Each observer measures time normally within their own inertial frame.

The difference arises from the:

geometry of spacetime.

Worked Example 2: Faster Spacecraft

A process takes:

20.0 s

according to a clock travelling with a spacecraft moving at:

0.80c.

Find the time interval measured by Earth.

At:

0.80c

the Lorentz factor is:

γ ≈ 1.667.

Therefore:

Δt = γΔτ

Δt = 1.667(20.0)

Δt ≈ 33.3 s.

Earth measures approximately:

33.3 seconds.

Worked Example 3: Finding Proper Time

Earth measures a moving process lasting:

50 s.

The object carrying the clock moves at:

0.60c.

Find the proper time.

Use:

Δt = γΔτ.

Rearrange:

Δτ = Δt/γ.

Therefore:

Δτ = 50/1.25

Δτ = 40 s.

The clock travelling with the process measures:

40 seconds.

Which Time Is Proper Time?

This is one of the most important questions to ask.

Do not decide that the smaller number is proper time simply because it is smaller.

Instead ask:

Which frame has both events occurring at the same place?

That frame measures:

proper time.

For a single clock that records both events, the clock's own rest frame measures:

Δτ.

Example: A Spacecraft Clock

Suppose:

Event A: spacecraft clock reads 0 s.

Event B: spacecraft clock reads 12 s.

Both events occur at:

the spacecraft clock.

Therefore:

12 s

is the proper time interval.

An Earth observer sees the spacecraft move between Event A and Event B.

Earth therefore measures a:

dilated coordinate time interval.

Example: Events at an Earth Laboratory

Now suppose:

Event A: a machine in an Earth laboratory switches on.

Event B: the same machine switches off.

Both events occur at the same location in the:

Earth frame.

Therefore, Earth measures:

proper time.

A passing spacecraft does not.

Proper time is not automatically associated with:

spacecraft.

It belongs to whichever inertial frame places both events at:

the same spatial coordinate.

General Time Transformation

Time dilation applies to a particular arrangement of events.

For the general case, use:

Δt′ = γ(Δt − vΔx/c²).

If:

Δx ≠ 0

then you cannot usually use the simple time-dilation equation:

Δt′ = γΔt.

The spatial separation must also be considered.

Worked Example 4: General Time Transformation

Suppose two events in S have:

Δt = 5.0 s

and:

Δx = 6.0 × 10⁸ m.

Frame S′ moves at:

0.60c.

Find:

Δt′.

We know:

γ = 1.25.

Use:

Δt′ = γ(Δt − vΔx/c²).

Calculate:

vΔx/c²

= (0.60c)(6.0 × 10⁸)/c²

Since:

c = 3.0 × 10⁸ m/s,

this gives:

vΔx/c² = 1.2 s.

Therefore:

Δt′ = 1.25(5.0 − 1.2)

Δt′ = 1.25(3.8)

Δt′ = 4.75 s.

So S′ measures:

Δt′ = 4.75 s.

Why Isn't the Answer 6.25 s?

Someone might incorrectly calculate:

1.25 × 5.0 = 6.25 s.

But that assumes the simple time-dilation situation.

Here:

Δx ≠ 0.

Therefore, we must use:

Δt′ = γ(Δt − vΔx/c²).

This distinction is extremely important.

Time Dilation vs General Time Transformation

Time dilation

Use:

Δt = γΔτ

when one of the frames measures the two events at the:

same location.

General time transformation

Use:

Δt′ = γ(Δt − vΔx/c²)

when transforming arbitrary events between:

reference frames.

Time dilation is therefore a:

special case

of the Lorentz time transformation.

Worked Example 5: Events at the Same Position in S

Suppose two events occur at the same position in S.

Therefore:

Δx = 0.

Let:

Δt = 12 s

and:

v = 0.80c.

Then:

γ = 1.667.

Use:

Δt′ = γ(Δt − vΔx/c²).

Because:

Δx = 0,

we get:

Δt′ = γΔt

Δt′ = 1.667(12)

Δt′ ≈ 20.0 s.

Here, S measures the:

proper time.

Proper Time Is Frame-Specific

Notice what happened.

Earlier, S′ measured proper time.

In this example:

S measures proper time.

There is no universal frame that always measures proper time.

The defining condition is:

the two events occur at the same location in that frame.

A Spacetime View of Proper Time

 

 

 

5

A clock traces a path through spacetime called its:

worldline.

Two readings of the same clock correspond to two events on that:

worldline.

The time recorded directly by that clock between the events is:

proper time.

The Spacetime Interval

For two events separated along one spatial dimension:

c²Δτ² = c²Δt² − Δx²

when the separation is timelike.

Divide by c²:

Δτ² = Δt² − Δx²/c².

Therefore:

Δτ = √(Δt² − Δx²/c²).

This provides another way to calculate the:

proper time.

Worked Example 6: Finding Proper Time from Two Events

Suppose two events are separated by:

Δt = 5.0 s

and:

Δx = 9.0 × 10⁸ m.

Since:

c = 3.0 × 10⁸ m/s,

we have:

Δx/c = 3.0 s.

Therefore:

Δτ = √(5.0² − 3.0²)

Δτ = √(25 − 9)

Δτ = √16

Δτ = 4.0 s.

So the proper time between the events is:

4.0 seconds.

Why Proper Time Is Invariant

Different inertial observers may disagree about:

Δt

and:

Δx.

However, they agree on the combination:

c²Δt² − Δx².

Therefore they agree on:

Δτ.

Proper time is related to the invariant:

spacetime interval.

This is why proper time has such an important role in relativity.

Worked Example 7: Particle Lifetime

A particle has a proper lifetime of:

2.2 μs.

It travels at:

0.90c.

How long does its lifetime appear in the laboratory frame?

At:

0.90c

γ ≈ 2.294.

Use:

Δt = γΔτ.

Therefore:

Δt = 2.294(2.2 μs)

Δt ≈ 5.05 μs.

The laboratory measures approximately:

5.05 μs.

Particle Lifetimes and Evidence for Time Dilation

 

 

 

5

High-speed unstable particles provide important experimental tests of:

time dilation.

For example, muons produced in Earth's atmosphere can reach detectors at Earth's surface in greater numbers than a simple nonrelativistic lifetime argument would suggest.

In the Earth frame, their moving decay clocks are:

time-dilated.

In the muon's frame, the atmospheric travel distance is:

length-contracted.

Both descriptions are consistent.

Worked Example 8: A Longer Journey

A spacecraft travels at:

0.80c.

According to astronauts aboard the spacecraft, a certain phase of the journey lasts:

3.0 years.

How much time passes in Earth's frame?

Use:

γ = 1.667.

Then:

Δt = γΔτ

Δt = 1.667(3.0)

Δt ≈ 5.0 years.

The astronauts measure:

3.0 years.

Earth measures:

5.0 years.

What Does "Moving Clocks Run Slow" Mean?

You may often hear:

moving clocks run slow.

This is a useful shorthand, but it must be interpreted carefully.

It means that when comparing appropriate clock readings between inertial frames, a clock moving relative to an inertial observer accumulates less proper time between the relevant events.

It does not mean that someone looking at their own clock notices it:

physically ticking strangely.

Every observer sees their own local clock behaving normally.

The Symmetry Question

If Earth says the spacecraft clock is moving, couldn't the spacecraft say:

Earth is moving?

Yes.

For observers in uniform relative motion, each can describe the other's moving clocks as:

time-dilated.

This is not a contradiction because comparisons of distant clocks depend on:

simultaneity.

The simple symmetry changes if one observer turns around or accelerates, as in the:

twin paradox.

Time Dilation and the Twin Paradox

 

 

 

7

Suppose one twin remains on Earth while another travels away at high speed and later returns.

The travelling twin changes inertial frames during the journey.

Their paths through spacetime are:

different.

When reunited, the twins can directly compare clocks at the same location.

The elapsed proper times along their worldlines can therefore:

differ.

Transforming Simultaneous Events

Suppose two events are simultaneous in S.

Then:

Δt = 0.

But if they occur at different positions:

Δx ≠ 0.

The transformation becomes:

Δt′ = γ(0 − vΔx/c²)

or:

Δt′ = −γvΔx/c².

Therefore:

Δt′ ≠ 0.

The events are not simultaneous in S′.

This is the:

relativity of simultaneity.

Worked Example 9: Simultaneous in One Frame

Two flashes occur simultaneously in S:

Δt = 0.

They are separated by:

Δx = 3.0 × 10⁸ m.

Frame S′ moves at:

0.60c.

Since:

γ = 1.25,

use:

Δt′ = −γvΔx/c².

Now:

vΔx/c² = 0.60 s.

Therefore:

Δt′ = −1.25(0.60)

Δt′ = −0.75 s.

The two events are separated by:

0.75 s

in S′.

The negative sign tells us that their:

time order in the chosen coordinate labeling

is opposite to the order implied by positive Δt′.

Can Event Order Change?

For some pairs of events:

yes.

If two events are separated enough in space that light cannot travel between them during their time separation, they are:

spacelike separated.

Different observers may disagree about which occurred:

first.

For events that can be causally connected:

timelike or lightlike separated events,

all inertial observers preserve the causal ordering.

Cause cannot become:

effect after its own consequence.

Transforming Back

The inverse time-interval transformation is:

Δt = γ(Δt′ + vΔx′/c²).

This can be used to:

verify a calculation.

If you transform from S to S′ and then back again, you should recover the original:

time interval.

Worked Example 10: Verification

Suppose:

Δt = 5.0 s

Δx = 6.0 × 10⁸ m

v = 0.60c

Earlier we found:

Δt′ = 4.75 s.

The corresponding position transformation is:

Δx′ = γ(Δx − vΔt).

Calculate:

vΔt = (1.8 × 10⁸)(5.0)

vΔt = 9.0 × 10⁸ m.

Therefore:

Δx′ = 1.25(6.0 × 10⁸ − 9.0 × 10⁸)

Δx′ = −3.75 × 10⁸ m.

Now transform the time back:

Δt = γ(Δt′ + vΔx′/c²).

Here:

vΔx′/c² = −0.75 s.

Therefore:

Δt = 1.25(4.75 − 0.75)

Δt = 1.25(4.00)

Δt = 5.00 s.

The original value is recovered.

Verification Using the Spacetime Interval

Another check is:

c²Δt² − Δx² = c²Δt′² − Δx′².

If both sides agree, the transformed measurements are:

consistent.

This is often a powerful check in more advanced relativity problems.

Dimensional Check

Consider:

vΔx/c².

Its units are:

(m/s)(m)/(m²/s²).

This simplifies to:

seconds.

Therefore:

Δt − vΔx/c²

is dimensionally valid because both terms are:

time intervals.

Always checking units can catch:

calculation errors.

Choosing the Correct Equation

A common challenge is deciding which equation to use.

If you know proper time:

Use:

Δt = γΔτ.

If you know the dilated interval:

Use:

Δτ = Δt/γ.

If the two events occur at different positions in the known frame:

Use:

Δt′ = γ(Δt − vΔx/c²).

If transforming back:

Use:

Δt = γ(Δt′ + vΔx′/c²).

A Reliable Problem-Solving Method

For each problem:

Step 1: Identify the two events.

Step 2: Identify the reference frames.

Step 3: Determine whether either frame sees both events at the same position.

Step 4: If yes, identify the proper time.

Step 5: Calculate γ.

Step 6: Choose the appropriate equation.

Step 7: Substitute with consistent units.

Step 8: Interpret the result physically.

Step 9: Check whether the answer is reasonable.

Common Error: Assuming Δt Is Always Proper Time

The symbol:

Δt

does not automatically mean proper time.

Proper time is normally written:

Δτ.

But what really determines proper time is the physical condition:

both events occur at the same position in that frame.

Always examine the:

events and frame.

Common Error: Multiplying by γ in Every Problem

You cannot automatically use:

Δt′ = γΔt.

The general transformation is:

Δt′ = γ(Δt − vΔx/c²).

The simple time-dilation equation applies only when the events satisfy the appropriate:

same-location condition.

Common Error: Thinking Proper Time Belongs to Earth

Proper time is not automatically:

Earth time.

If both events occur at the same position on a spacecraft, then the spacecraft measures proper time.

If both occur at the same Earth laboratory, Earth measures proper time.

Proper time depends on:

the pair of events.

Common Error: Thinking Proper Time Is "True Time"

Proper time is not a universal time that is more correct than all others.

It is the time measured along a particular:

worldline.

Other inertial frames can measure different coordinate time intervals without being:

incorrect.

Common Error: Ignoring Relativity of Simultaneity

Comparing distant clocks requires deciding which distant events are:

simultaneous.

Different inertial observers generally disagree about this.

This is essential for understanding why time dilation does not create a contradiction between:

moving observers.

Common Error: Confusing Time Dilation with Signal Delay

Suppose you look through a telescope at a distant spacecraft.

What you literally see is affected by the:

travel time of light.

Time dilation is a different physical effect.

Relativistic measurements are defined after accounting for:

signal propagation.

So time dilation is not simply caused by light taking longer to:

reach an observer.

Everyday Speeds

Suppose an aircraft travels at:

250 m/s.

Compared with:

c = 3.00 × 10⁸ m/s,

this is extremely slow.

Therefore:

γ ≈ 1.

The time-dilation effect is extremely small.

For ordinary activities, we can safely treat:

Δt ≈ Δτ.

Relativistic Speeds

At:

0.99c

the Lorentz factor is approximately:

7.09.

Suppose a process lasts:

1.0 hour

in its own rest frame.

Another inertial frame in which the process moves at 0.99c measures:

Δt = γΔτ

Δt ≈ 7.09 hours.

At very high speeds, the difference becomes:

dramatic.

Time Intervals in Particle Physics

Relativistic time transformations are essential when studying:

• muons
• unstable particles
• particle accelerators
• cosmic rays
• high-energy collisions

Particles may have extremely short:

proper lifetimes.

But because they travel near c, laboratory observers can measure significantly longer:

coordinate lifetimes.

Time Intervals in Space Travel

Relativistic time dilation would also matter for hypothetical spacecraft travelling at a substantial fraction of:

c.

Travellers could experience less elapsed proper time than observers remaining in another frame between appropriately defined departure and reunion events.

This is not because their biological processes somehow escape physics.

Their clocks, chemical reactions and biological processes all evolve according to their own:

proper time.

Connecting Time Transformations to Previous Topics

The time transformation:

Δt′ = γ(Δt − vΔx/c²)

connects several major ideas in Special Relativity.

If:

Δx′ = 0

we obtain:

time dilation.

If:

Δt = 0

we obtain:

relativity of simultaneity.

Combined with the spatial transformation:

Δx′ = γ(Δx − vΔt),

we obtain the complete transformation of:

spacetime intervals.

These effects are therefore not separate phenomena.

They are different consequences of the same:

Lorentz transformations.

Check Your Understanding

1. What is a time interval?

2. Write the general Lorentz transformation for a time interval.

3. Define proper time.

4. What condition identifies the frame that measures proper time?

5. What symbol is commonly used for proper time?

6. State the time-dilation equation.

7. A spacecraft moves at 0.60c. Calculate γ.

8. A spacecraft clock measures 12 s while travelling at 0.60c. What interval does Earth measure in the standard time-dilation setup?

9. Earth measures 40 s for a moving process at 0.60c. Determine the proper time.

10. Explain why the proper time is not automatically the time measured on Earth.

11. Two events occur at different positions in S. Why can't you automatically use Δt′ = γΔt?

12. Two events have Δt = 4.0 s, Δx = 3.0 × 10⁸ m, and v = 0.60c. Calculate Δt′.

13. Explain physically what a transformed time interval represents.

14. How does the Lorentz transformation produce the time-dilation equation?

15. What happens to time dilation as v approaches c?

16. Why is time dilation not simply caused by the travel time of light?

17. If two events are simultaneous in S but occur at different positions, must they be simultaneous in S′? Explain.

18. Write the equation relating proper time to the spacetime interval.

19. Describe one method for verifying a transformed time interval.

20. Explain the relationship between time transformation, time dilation and relativity of simultaneity.

Key Terms

• Event: Physical occurrence at a particular place and time.
• Time interval: Difference between the time coordinates of two events.
• Proper time (Δτ): Time measured by a clock present at both events; equivalently, the interval measured in the frame where the events occur at the same spatial position.
• Coordinate time: Time interval between events measured using the clocks of a particular reference frame.
• Time dilation: Relationship in which a moving clock accumulates less proper time between appropriate events than the coordinate time measured in another inertial frame.
• Lorentz transformation: Equations relating space and time coordinates between inertial frames.
• Lorentz factor (γ): Factor 1/√(1 − v²/c²).
• Reference frame: Coordinate system used to measure positions, times and motion.
• Inertial frame: Non-accelerating reference frame.
• Worldline: Path of an object through spacetime.
• Spacetime interval: Invariant combination of spatial and temporal separation between events.
• Invariant: Quantity that has the same value for all inertial observers.
• Relativity of simultaneity: Principle that events simultaneous in one inertial frame need not be simultaneous in another.
• Timelike separation: Separation between events that permits a slower-than-light causal connection.
• Spacelike separation: Separation between events for which no causal signal travelling at or below c can connect them.

Key Takeaways

• Time intervals are measured between two events.
• Different inertial frames can measure different time intervals between the same events.
• The general transformation is Δt′ = γ(Δt − vΔx/c²).
• Time transformation therefore depends on both temporal and spatial separation.
• Proper time is written Δτ.
• Proper time is measured by a clock that is present at both events.
• Equivalently, proper time is measured in the frame where the two events occur at the same position.
• Proper time does not automatically belong to Earth, a spacecraft, or any preferred frame.
• For the standard time-dilation situation, Δt = γΔτ.
• Because γ ≥ 1, the coordinate time interval in that setup is at least as large as the proper time.
• Time dilation is a special case of the Lorentz time transformation.
• You should not automatically multiply every time interval by γ.
• When Δx ≠ 0, the full Lorentz time transformation may be required.
• If Δt = 0 but Δx ≠ 0, another inertial observer generally measures Δt′ ≠ 0.
• This produces the relativity of simultaneity.
• Proper time is related to the invariant spacetime interval by Δτ² = Δt² − Δx²/c² for timelike-separated events.
• Moving-particle lifetimes provide important experimental evidence for relativistic time dilation.
• Time dilation is not simply a visual effect caused by light-travel delay.
• At everyday speeds, γ ≈ 1, so relativistic time differences are usually extremely small.
• At speeds close to c, transformed time intervals can differ substantially.
• A strong solution should identify the events, identify the frames, determine which frame—if any—measures proper time, calculate the transformation, and interpret the result physically.

5. Implications of Lorentz Transformations

 

 

 

5

What Do Lorentz Transformations Really Tell Us?

The Lorentz transformations are more than equations for changing coordinates.

They reveal something fundamental about the universe:

space and time are not independent absolute quantities.

Observers moving relative to one another can disagree about:

• the position of an event
• the time of an event
• the distance between events
• the time between events
• whether distant events are simultaneous

Yet their measurements are connected by precise mathematical rules.

Those rules are the:

Lorentz transformations.

The Lorentz Transformations

For two inertial frames S and S′, with S′ moving at velocity v along the x-axis relative to S:

x′ = γ(x − vt)

t′ = γ(t − vx/c²)

where:

γ = 1/√(1 − v²/c²).

For the perpendicular coordinates:

y′ = y

z′ = z.

These equations transform the coordinates of the:

same physical event

between different inertial frames.

What Is Being Transformed?

Suppose an event occurs at:

(x, t)

according to observer S.

Another observer S′ assigns the same event:

(x′, t′).

The event itself has not changed.

What changes is its:

coordinate description.

This distinction is central to understanding relativity.

Space and Time Become Connected

Look carefully at:

x′ = γ(x − vt).

The transformed position depends on:

time.

Now examine:

t′ = γ(t − vx/c²).

The transformed time depends on:

position.

Therefore:

space affects transformed time

and:

time affects transformed space.

This mixing of space and time is one of the deepest implications of the Lorentz transformations.

 

 

 

6

From Space and Time to Spacetime

Classical physics treats space and time as largely separate.

We might imagine:

3 dimensions of space + an independent universal time.

Special Relativity instead leads naturally to:

spacetime.

An event is described using four coordinates:

(x, y, z, t).

Different inertial observers divide spacetime into space and time differently, but they remain describing:

the same spacetime events.

No Universal Time

In Newtonian physics:

t′ = t.

Time is assumed to pass identically for everyone.

Lorentz transformations instead give:

t′ = γ(t − vx/c²).

Therefore:

t′ ≠ t

in general.

There is no single universal clock shared by all inertial observers.

Time measurements depend on:

reference frame.

No Universal Length

Spatial measurements are also frame-dependent.

The Lorentz transformations lead to:

length contraction.

If an object has proper length:

L₀

then an observer who sees the object moving at velocity v measures:

L = L₀/γ.

Since:

γ ≥ 1,

we have:

L ≤ L₀.

A moving object's length parallel to the direction of relative motion is measured to be:

shorter.

Time Dilation

Lorentz transformations also lead directly to:

time dilation.

If:

Δτ

is the proper time between two events, then another inertial frame in which that clock moves measures:

Δt = γΔτ.

Therefore:

Δt ≥ Δτ.

The coordinate time interval is larger than the proper time.

 

 

 

6

One Transformation, Many Effects

Time dilation and length contraction can sometimes appear to be separate rules.

They are not.

Both follow from:

the Lorentz transformations.

The same equations also explain:

• relativity of simultaneity
• relativistic velocity addition
• invariance of the speed of light
• transformation of energy and momentum
• preservation of causal structure

These are interconnected consequences of:

the same spacetime geometry.

How Time Dilation Emerges

Consider a clock at rest in S′.

Two ticks of the clock occur at the same location in S′:

Δx′ = 0.

The clock measures the proper time:

Δt′ = Δτ.

Using the inverse Lorentz transformation:

Δt = γ(Δt′ + vΔx′/c²).

Because:

Δx′ = 0,

we obtain:

Δt = γΔτ.

This is exactly the:

time-dilation equation.

Worked Example: Time Dilation

A spacecraft moves at:

0.80c.

A clock aboard the spacecraft measures:

6.0 s.

At:

0.80c,

γ ≈ 1.667.

Therefore:

Δt = γΔτ

Δt = 1.667(6.0)

Δt ≈ 10.0 s.

The spacecraft measures:

6.0 s.

Earth measures:

10.0 s.

Both measurements are valid in their respective:

reference frames.

How Length Contraction Emerges

Length measurement requires determining the positions of both ends of an object:

at the same time in the observer's frame.

This condition is essential.

Suppose a rod is at rest in S′.

Its proper length is:

L₀ = Δx′.

An observer in S measures both ends simultaneously:

Δt = 0.

The spatial Lorentz transformation gives:

Δx′ = γ(Δx − vΔt).

Since:

Δt = 0,

Δx′ = γΔx.

Therefore:

L₀ = γL.

So:

L = L₀/γ.

This is:

length contraction.

Worked Example: Length Contraction

A spacecraft has a proper length of:

100 m.

It travels past Earth at:

0.80c.

Since:

γ ≈ 1.667,

Earth measures:

L = L₀/γ

L = 100/1.667

L ≈ 60 m.

The astronauts still measure their spacecraft as:

100 m long.

Earth measures:

60 m.

Neither measurement is incorrect.

They are made in different:

reference frames.

Why Simultaneity Matters for Length

To measure the length of a moving object, an observer must record:

where the front is

and:

where the back is

at the same time.

But simultaneity is:

frame-dependent.

Events simultaneous in S may not be simultaneous in S′.

Therefore, length contraction is deeply connected to:

relativity of simultaneity.

Relativity of Simultaneity

Suppose two events are simultaneous in S:

Δt = 0.

The time transformation is:

Δt′ = γ(Δt − vΔx/c²).

Therefore:

Δt′ = −γvΔx/c².

If:

Δx ≠ 0

then:

Δt′ ≠ 0.

So two spatially separated events that occur simultaneously in one frame generally do not occur simultaneously in:

another moving frame.

 

 

 

4

Einstein's Train Example

Imagine lightning strikes the front and back of a train.

An observer standing midway along the platform might determine that the strikes occurred:

simultaneously.

An observer at the middle of the moving train can assign the two strike events:

different time coordinates.

The disagreement is not merely due to eyesight or signal delay.

After correcting for signal travel, the observers can still disagree about:

distant simultaneity.

That is a fundamental consequence of the:

Lorentz transformations.

Why This Matters

Without absolute simultaneity, there cannot be one universal definition of:

"right now everywhere."

Observers moving relative to one another divide spacetime into sets of simultaneous events differently.

This is one of the most significant conceptual changes from:

Newtonian physics.

The Speed of Light Remains c

One of Einstein's postulates states that all inertial observers measure the same vacuum speed of light:

c.

Lorentz transformations preserve this property.

Suppose a light pulse satisfies:

x = ct.

Transform its coordinates:

x′ = γ(x − vt)

and:

t′ = γ(t − vx/c²).

Substituting:

x = ct

leads to:

x′ = ct′.

Therefore:

x′/t′ = c.

The second observer also measures:

c.

Why Galilean Transformations Fail

Classical mechanics uses:

x′ = x − vt

and:

t′ = t.

If light travels at c in S, Galilean transformation would predict:

c − v

in S′.

Experiments do not support such a classical transformation of vacuum light speed.

Lorentz transformations instead preserve:

c.

The Spacetime Interval

Although observers disagree about distances and times separately, they agree on an important combination:

s² = c²Δt² − Δx²

for one-dimensional motion.

More generally:

s² = c²Δt² − Δx² − Δy² − Δz².

This quantity is called the:

spacetime interval.

Lorentz transformations preserve it.

Therefore:

s² = s′².

What Does Invariant Mean?

An invariant is a quantity that remains the same when changing between the relevant reference frames.

For Lorentz transformations:

c²Δt² − Δx² = c²Δt′² − Δx′².

Observers can disagree about:

Δt

and:

Δx,

while agreeing on the:

spacetime interval.

This is similar to how rotations in ordinary geometry change x- and y-coordinates while preserving:

distance.

Lorentz Transformations as Spacetime Rotations

There is a useful mathematical analogy.

In ordinary geometry, rotating coordinate axes changes:

x and y

while preserving:

x² + y².

In spacetime, Lorentz transformations mix:

space and time

while preserving:

c²t² − x².

They can therefore be thought of, with important mathematical differences from ordinary rotations, as:

rotations in spacetime.

 

 

 

5

Light Cones

Consider an event at the origin.

Light travelling outward satisfies:

x = ±ct.

On a spacetime diagram, these paths form the boundaries of a:

light cone.

 

 

 

6

The light cone divides spacetime into important regions:

• causal future
• causal past
• spacelike-separated region

This structure helps us understand:

causality.

Timelike Separation

Two events are timelike separated when:

c²Δt² > Δx².

There is enough time for an object travelling slower than light to travel between the events.

One event can potentially:

cause the other.

For timelike-separated events, all inertial observers agree on their:

temporal order.

Lightlike Separation

Two events are lightlike separated when:

c²Δt² = Δx².

Only a signal travelling at:

c

can connect the events.

Examples include:

emission and later detection of the same light pulse.

All inertial observers agree that the separation is:

lightlike.

Spacelike Separation

Two events are spacelike separated when:

c²Δt² < Δx².

Light cannot travel between the events quickly enough for one to cause the other.

Different inertial observers may disagree about:

which event occurred first.

This does not violate causality because the events cannot be:

causally connected.

Lorentz Transformations Preserve Causality

This is one of their most important implications.

Lorentz transformations preserve whether an interval is:

• timelike
• lightlike
• spacelike

Therefore, if Event A can causally influence Event B, all inertial observers preserve the relevant:

causal ordering.

A cause cannot become an effect that happens:

after its own consequence.

Worked Example: Causal Events

Suppose Event A occurs at:

x = 0

t = 0.

Event B occurs at:

x = 3.0 × 10⁸ m

t = 2.0 s.

Light could travel:

6.0 × 10⁸ m

during 2.0 s.

Since the spatial separation is only:

3.0 × 10⁸ m,

the events are:

timelike separated.

A slower-than-light signal could travel from A to B.

Therefore, their temporal order cannot be reversed by a Lorentz transformation.

Worked Example: Spacelike Events

Suppose Event B instead occurs:

0.50 s

after Event A and:

3.0 × 10⁸ m

away.

During 0.50 s, light travels only:

1.5 × 10⁸ m.

The events are farther apart than light could travel during that interval.

Therefore they are:

spacelike separated.

Different observers can assign different temporal orderings without violating:

causality.

Why Faster-Than-Light Signalling Is a Problem

If usable information could propagate faster than c, then some Lorentz-transformed frames could describe the reception of the signal as occurring:

before its transmission.

Combined with appropriate return signalling, this could create causal paradoxes.

The invariant causal structure associated with c therefore plays a fundamental role in preserving:

cause and effect.

Relativistic Velocity Addition

Lorentz transformations also lead to:

relativistic velocity addition.

For motion along one dimension:

u = (u′ + v)/(1 + u′v/c²).

This replaces the classical rule:

u = u′ + v.

The relativistic equation ensures that combining ordinary sub-light velocities does not accelerate a massive object beyond:

c.

Example

A spacecraft travels at:

0.80c.

It launches a probe forward at:

0.70c

relative to itself.

Classically:

0.80c + 0.70c = 1.50c.

Relativistically:

u = (0.80c + 0.70c)/(1 + 0.80 × 0.70)

u = 1.50c/1.56

u ≈ 0.962c.

The resulting velocity remains:

below c.

Many Relativistic Effects Have One Origin

This is an important organizational idea.

You do not need to think of Special Relativity as a collection of unrelated strange effects.

Instead:

Einstein's postulates

lead to:

Lorentz transformations

which lead to:

time dilation

length contraction

relativity of simultaneity

relativistic velocity addition

invariant spacetime intervals

and:

preserved causal structure.

That is why Lorentz transformations form the mathematical foundation of:

Special Relativity.

Low-Speed Limit

A successful theory should reproduce older theories where those theories are known to work.

When:

v ≪ c,

the Lorentz factor becomes:

γ ≈ 1.

Also:

vx/c²

becomes extremely small.

Therefore:

x′ ≈ x − vt

and:

t′ ≈ t.

These are approximately the:

Galilean transformations.

So Newtonian mechanics appears naturally as the:

low-speed limit of Special Relativity.

Why We Don't Notice These Effects Every Day

A car might travel at:

30 m/s.

But:

c ≈ 300,000,000 m/s.

Therefore:

v/c ≈ 0.0000001.

At such speeds:

γ is extraordinarily close to 1.

Time dilation, length contraction and simultaneity differences are therefore far too small to notice in:

ordinary life.

At Relativistic Speeds

When:

v → c,

the Lorentz factor increases dramatically.

For example:

Relativistic effects become increasingly important as:

v approaches c.

Experimental Significance

Lorentz transformations are not merely mathematical speculation.

Relativistic predictions have been tested through many phenomena and technologies, including:

• high-speed particle experiments
• particle lifetimes
• accelerator physics
• precision atomic clocks
• satellite navigation systems
• electromagnetic phenomena

 

 

 

These applications require relativistic effects to be taken into account at the appropriate precision.

Particle Accelerators

Modern particle accelerators routinely accelerate particles to speeds extremely close to:

c.

At these speeds, classical equations cannot accurately describe:

• energy
• momentum
• particle lifetimes
• collisions
• trajectories

Lorentz-compatible relativistic physics is essential.

Muons and Time Dilation

Muons produced high in Earth's atmosphere have very short proper lifetimes.

Yet many reach Earth's surface.

In Earth's frame, the muons' decay times are:

dilated.

In the muon's frame, the atmosphere is:

length-contracted.

These are not competing explanations.

They are two frame-dependent descriptions of the:

same physical events.

GPS and Relativity

Satellite navigation provides an important real-world example of relativistic clock effects.

Satellite clocks move relative to receivers on Earth, producing a:

Special Relativistic timing correction.

Gravity also affects satellite clocks, requiring:

General Relativity.

Accurate satellite navigation therefore depends on accounting for relativistic timing effects.

Electromagnetism and Relativity

Lorentz transformations also reveal a deep connection between:

electric and magnetic fields.

What one observer describes as a particular combination of electric and magnetic fields can be described differently by:

another moving observer.

Electricity and magnetism are therefore closely connected through:

relativistic transformations.

This helped establish Special Relativity as a natural framework for:

electromagnetism.

Energy and Momentum

Classical momentum is:

p = mv.

Relativistically:

p = γmv.

Total relativistic energy is:

E = γmc².

These quantities transform consistently between inertial frames.

They satisfy the invariant relationship:

E² = p²c² + m²c⁴.

So the implications of Lorentz symmetry extend far beyond:

space and time coordinates.

Mass-Energy Equivalence

For an object at rest:

p = 0.

Therefore:

E² = m²c⁴,

giving:

E₀ = mc².

This is the object's:

rest energy.

The famous relationship between mass and energy fits naturally within the relativistic structure based on:

Lorentz invariance.

There Is No Preferred Inertial Frame

Lorentz transformations work between:

any inertial reference frames.

There is no experimentally privileged inertial frame in Special Relativity that represents:

absolute rest.

Each inertial observer can apply the same laws of physics.

This is Einstein's:

principle of relativity.

What Observers Can Disagree About

Different inertial observers may disagree about:

• position
• elapsed coordinate time
• length
• simultaneity
• velocity
• energy
• momentum

These quantities can be:

frame-dependent.

What Observers Agree About

Observers agree on important invariant structures and quantities, including:

• the vacuum speed of light
• the spacetime interval
• rest mass
• whether an interval is timelike, spacelike or lightlike
• causal relationships between causally connected events

This distinction between:

frame-dependent quantities

and:

invariants

is central to modern physics.

A Geometrical View

 

 

 

6

Special Relativity can be understood as a theory of the:

geometry of spacetime.

Different observers use different coordinate systems.

But the underlying spacetime structure remains:

consistent.

Lorentz transformations tell us how to move mathematically between those coordinate descriptions while preserving:

the physical laws and invariant spacetime structure.

Why Lorentz Transformations Are So Significant

Before Einstein, space and time were usually treated as:

absolute backgrounds.

Lorentz transformations reveal that measurements of space and time depend on:

relative motion.

Yet physics does not become arbitrary.

Instead, deeper quantities remain:

invariant.

This represents an important shift:

the coordinates change, but the underlying physical relationships remain consistent.

From Newton to Einstein

Newtonian picture

Space:

absolute

Time:

absolute

Simultaneity:

universal

Velocity addition:

u = u′ + v

Transformations:

Galilean

Relativistic picture

Space:

frame-dependent

Time:

frame-dependent

Simultaneity:

frame-dependent

Vacuum speed of light:

invariant

Transformations:

Lorentz

A Useful Concept Map

The structure of Special Relativity can be summarized as:

Einstein's postulates

↓

Lorentz transformations

↓

Space and time mix

↓

Time dilation

Length contraction

Relativity of simultaneity

Relativistic velocity addition

↓

Invariant spacetime interval

↓

Light-cone structure

↓

Preserved causality

This is why Lorentz transformations are not simply another equation in the unit.

They connect nearly:

every major idea in Special Relativity.

Common Misconception: Relativity Means Everything Is Relative

Special Relativity does not mean:

everything is relative.

Some quantities are frame-dependent.

Others are:

invariant.

For example, observers can disagree about:

time intervals and spatial distances,

while agreeing on:

the spacetime interval.

Relativity therefore identifies both what changes and:

what remains unchanged.

Common Misconception: Time Dilation Is an Optical Illusion

Time dilation is not simply caused by:

seeing a distant clock through delayed light.

After signal-travel effects are properly accounted for, different inertial observers still obtain the relativistic relationship predicted by:

Lorentz transformations.

It is a property of spacetime measurements.

Common Misconception: Length Contraction Means Objects Are Damaged

An object does not experience itself being:

crushed.

In its own rest frame, its length remains:

its proper length.

Length contraction describes how another inertial frame measures the distance between the object's endpoints simultaneously in:

that observer's frame.

Common Misconception: Different Time Orders Violate Causality

Only sufficiently separated:

spacelike events

can have their time ordering reversed between inertial frames.

Such events cannot causally influence one another without faster-than-light signalling.

Causally connected events retain their causal ordering.

Therefore Lorentz transformations preserve:

causality.

Common Misconception: Newtonian Physics Is Wrong Everywhere

Newtonian physics remains an extremely accurate approximation when:

v ≪ c.

Special Relativity does not simply discard classical physics.

It explains:

when and why classical physics works.

The Galilean transformations emerge as the low-speed approximation of:

Lorentz transformations.

Evaluating the Significance

Lorentz transformations are significant because they provide a single mathematical framework that:

• preserves the laws of physics between inertial frames
• preserves the measured vacuum speed of light
• connects space and time measurements
• predicts time dilation
• predicts length contraction
• explains relativity of simultaneity
• produces relativistic velocity transformations
• preserves spacetime intervals
• preserves causal structure
• reduces to classical physics at low speeds

Few equations reorganized our understanding of physical measurement as profoundly as:

the Lorentz transformations.

Check Your Understanding

1. Write the Lorentz transformations for x′ and t′.

2. Explain what it means to transform the coordinates of an event.

3. How does the equation for x′ show that space and time are connected?

4. How does the equation for t′ show the same connection?

5. Explain why Special Relativity does not contain a universal time.

6. State the time-dilation equation.

7. Explain how time dilation follows from the Lorentz transformations.

8. State the length-contraction equation.

9. Why is simultaneity important when measuring length?

10. Two events are simultaneous in S but occur at different locations. Are they necessarily simultaneous in S′? Explain.

11. What is the spacetime interval?

12. What does it mean for a quantity to be invariant?

13. Distinguish between timelike, lightlike and spacelike separations.

14. Explain why spacelike-separated events may have different temporal orderings in different frames without violating causality.

15. Explain why causally connected events cannot have their causal order reversed.

16. How do Lorentz transformations preserve the speed of light?

17. Why do Lorentz transformations approach Galilean transformations at low speeds?

18. Describe one experimental or technological situation where relativistic effects are important.

19. Explain why time dilation and length contraction should not be viewed as unrelated effects.

20. Why can Lorentz transformations be described as the mathematical foundation of Special Relativity?

Key Terms

• Lorentz transformation: Equations relating space and time coordinates between inertial reference frames.
• Lorentz factor (γ): Factor 1/√(1 − v²/c²) governing many relativistic effects.
• Spacetime: Unified four-dimensional description of space and time.
• Event: Physical occurrence at a specific position and time.
• Reference frame: Coordinate system used to describe events and motion.
• Inertial frame: Non-accelerating reference frame.
• Time dilation: Difference in measured time intervals between relatively moving frames under the appropriate conditions.
• Proper time: Time interval measured by a clock present at both events.
• Length contraction: Reduced length measured parallel to relative motion for an object moving relative to an observer.
• Proper length: Length measured in the object's rest frame.
• Relativity of simultaneity: Principle that distant events simultaneous in one frame need not be simultaneous in another.
• Spacetime interval: Invariant combination of temporal and spatial separation between events.
• Invariant: Quantity unchanged by a Lorentz transformation.
• Light cone: Boundary separating regions of spacetime according to possible causal connections.
• Timelike interval: Separation permitting a slower-than-light causal connection.
• Lightlike interval: Separation connected by light travelling at c.
• Spacelike interval: Separation for which no signal travelling at or below c can connect the events.
• Causality: Principle that causes precede their effects within causal relationships.
• Worldline: Path followed by an object through spacetime.
• Lorentz invariance: Property that physical laws retain their appropriate form under Lorentz transformations.

Key Takeaways

• Lorentz transformations describe how space and time coordinates change between inertial frames.
• The transformed position depends on time, while the transformed time depends on position.
• This reveals that space and time are components of a unified spacetime.
• There is no universal time shared by all inertial observers.
• There is no universal measurement of spatial length independent of reference frame.
• Time dilation follows directly from the Lorentz transformations.
• Length contraction also follows directly from the same transformations.
• Measuring the length of a moving object requires simultaneous endpoint measurements, connecting length contraction to the relativity of simultaneity.
• Events simultaneous in one inertial frame need not be simultaneous in another.
• Lorentz transformations preserve the vacuum speed of light c.
• They replace Galilean transformations when relativistic speeds are important.
• At low speeds, Lorentz transformations reduce approximately to Galilean transformations.
• Observers can disagree about space and time separately while agreeing on the spacetime interval.
• The spacetime interval allows event separations to be classified as timelike, lightlike, or spacelike.
• Lorentz transformations preserve these classifications.
• Causally connected events retain their causal ordering.
• Spacelike-separated events may have different temporal orderings because neither can causally influence the other without faster-than-light signalling.
• The resulting light-cone structure provides the causal organization of Special Relativity.
• Relativistic velocity addition also follows from Lorentz transformations and preserves c as the invariant limiting speed.
• Special Relativity does not mean "everything is relative"; important quantities and structures remain invariant.
• Time dilation, length contraction and relativity of simultaneity are not separate coincidences. They are interconnected consequences of the same transformations.
• Relativistic effects have been confirmed in particle physics, precision timing and other experiments and technologies.
• Lorentz transformations provide the mathematical connection between Einstein's postulates and the observable consequences of Special Relativity.
• Their importance extends beyond kinematics to relativistic momentum, energy and electromagnetism.
• They form the foundation of Special Relativity because they specify how physical measurements made by different inertial observers remain mathematically and physically consistent.