Lorentz Transformation

In classical mechanics, we assume that space and time are absolute. However, experiments in the late 19th and early 20th centuries, particularly those involving the speed of light, showed that this assumption is incorrect. The theory of Special Relativity, proposed by Albert Einstein in 1905, revolutionized our understanding of space and time. A cornerstone of this theory is the Lorentz transformation, which describes how measurements of space and time differ between two observers moving at constant velocities relative to each other.

Imagine two observers, Alice and Bob. Alice is stationary in her laboratory frame of reference (let's call it S), and Bob is moving with a constant velocity 'v' relative to Alice in the x-direction. Bob also has his own frame of reference (S'). The Lorentz transformation equations relate the space and time coordinates (x, y, z, t) measured by Alice to the coordinates (x', y', z', t') measured by Bob.

In classical mechanics (Galilean transformation), the transformation is simple: x' = x - vt, y' = y, z' = z, t' = t. This implies that time is absolute and the same for both observers. However, the Lorentz transformation acknowledges that time is relative and depends on the observer's motion.

The Lorentz Transformation Equations

For motion along the x-axis with relative velocity 'v', the Lorentz transformation equations are:

• x' = γ(x - vt)
• y' = y
• z' = z
• t' = γ(t - vx/c²)

Here, 'c' is the speed of light in a vacuum, a universal constant. The term 'γ' (gamma) is the Lorentz factor, defined as:

γ = 1 / √(1 - v²/c²)

The inverse Lorentz transformation equations, which give Alice's coordinates in terms of Bob's, are obtained by replacing 'v' with '-v' and swapping the primed and unprimed coordinates:

• x = γ(x' + vt')
• y = y'
• z = z'
• t = γ(t' + vx'/c²)

The Lorentz factor γ is always greater than or equal to 1. When the relative velocity 'v' is much smaller than the speed of light 'c' (v << c), the term v²/c² becomes very small, and γ approaches 1. In this low-velocity limit, the Lorentz transformations reduce to the Galilean transformations, which is why classical mechanics works so well for everyday phenomena.

Time Dilation

One of the most striking consequences of the Lorentz transformation is time dilation. It means that time passes more slowly for an observer who is moving relative to another observer.

Let's consider a clock at rest in Bob's frame S'. Alice, in frame S, observes Bob's clock moving with velocity 'v'. Suppose Bob measures a time interval Δt' for an event happening at the same spatial location in his frame (e.g., the tick-tock of his own clock). From Bob's perspective, this is the proper time interval, often denoted as Δτ.

Now, let's find out what time interval Δt Alice measures for the same event. We can use the Lorentz transformation for time. If the event occurs at x'₁ and x'₂ in Bob's frame, then:

• t'₁ = γ(t₁ - vx₁/c²)
• t'₂ = γ(t₂ - vx₂/c²)

So, Δt' = t'₂ - t'₁ = γ( (t₂ - t₁) - v(x₂ - x₁)/c² ).

If the event occurs at the same location in Bob's frame, then x'₁ = x'₂, which means Δx' = 0. Using the inverse Lorentz transformation for x':

• x₁ = γ(x'₁ + vt'₁)
• x₂ = γ(x'₂ + vt'₂)

Since x'₁ = x'₂, we have x₁ - x₂ = γv(t'₂ - t'₁) = γvΔt'.

Now, let's consider Alice's measurement of the time interval Δt = t₂ - t₁ for the same event. The event occurs at different locations in Alice's frame, x₁ and x₂.

Using the Lorentz transformation for t:

• t'₁ = γ(t₁ - vx₁/c²)
• t'₂ = γ(t₂ - vx₂/c²)

Subtracting these gives: Δt' = t'₂ - t'₁ = γ( (t₂ - t₁) - v(x₂ - x₁)/c² ) = γ( Δt - v(x₂ - x₁)/c² ).

Substitute x₂ - x₁ = γvΔt' into this equation:

Δt' = γ( Δt - v(γvΔt')/c² )

Δt' = γΔt - γ²v²Δt'/c²

Δt'(1 + γ²v²/c²) = γΔt

This path is getting complicated. Let's use a simpler approach considering a light clock.

The Light Clock Thought Experiment

Imagine a clock consisting of two mirrors placed a distance L apart, with a photon bouncing between them. One bounce (up and down) represents one tick of the clock.

For an observer at rest with the clock (Bob), the photon travels a distance 2L. The time for one tick (Δt' or Δτ) is 2L/c.

Now, consider Alice observing this clock moving with velocity 'v' horizontally. For Alice, the photon does not travel straight up and down. As the clock moves, the photon travels diagonally. Let the time Alice measures for one tick be Δt.

In the time Δt/2, the clock moves a horizontal distance v(Δt/2). The vertical distance is still L. The distance the photon travels diagonally is c(Δt/2).

Using the Pythagorean theorem for the right triangle formed:

(cΔt/2)² = L² + (vΔt/2)²

c²(Δt)²/4 = L² + v²(Δt)²/4

(Δt)²(c² - v²)/4 = L²

(Δt)² = 4L² / (c² - v²) = (4L²/c²) / (1 - v²/c²)

We know that Δt' = 2L/c, so (Δt')² = 4L²/c².

Therefore, (Δt)² = (Δt')² / (1 - v²/c²).

Taking the square root:

Δt = Δt' / √(1 - v²/c²) = γΔt'

This equation shows that the time interval Δt measured by Alice is longer than the time interval Δt' measured by Bob. In other words, Alice observes Bob's clock ticking slower than her own. This phenomenon is called time dilation.

Example: Muon Decay Muons are subatomic particles created in the Earth's upper atmosphere by cosmic rays. They have a very short average lifetime (about 2.2 microseconds) when at rest. According to classical physics, even traveling at nearly the speed of light, most muons should decay before reaching the Earth's surface. However, a significant number of muons are detected at ground level. This is explained by time dilation. From our perspective on Earth, the muons' internal clocks are running slower due to their high speed, extending their effective lifetime and allowing them to travel much farther than expected.

Length Contraction

Another fascinating consequence of the Lorentz transformation is length contraction. It states that the length of an object moving relative to an observer is measured to be shorter along the direction of motion than its proper length (the length measured in the object's rest frame).

Consider a rod of proper length L₀ at rest in Bob's frame S'. Alice, in frame S, observes this rod moving with velocity 'v' along the x-axis. The proper length L₀ is the length measured when the rod is stationary relative to the observer, so L₀ = L' in Bob's frame.

To measure the length of the moving rod, Alice must measure the positions of its two ends simultaneously in her frame S. Let the positions of the ends be x₁ and x₂ at time t. The length measured by Alice is L = x₂ - x₁.

Using the Lorentz transformation for x:

• x₁ = γ(x'₁ + vt'₁)
• x₂ = γ(x'₂ + vt'₂)

Subtracting these gives: L = x₂ - x₁ = γ( (x'₂ - x'₁) + v(t'₂ - t'₁) ).

Alice measures the positions of the ends *simultaneously* in her frame, so t₁ = t₂ = t. This means Δt = 0 for Alice's measurement of the rod's length.

Now let's look at the corresponding times in Bob's frame. Using the inverse Lorentz transformation for t':

• t'₁ = γ(t₁ - vx₁/c²)
• t'₂ = γ(t₂ - vx₂/c²)

Since t₁ = t₂, we have t'₂ - t'₁ = -γv(x₂ - x₁)/c² = -γvL/c².

Substitute this back into the equation for L:

L = γ( (x'₂ - x'₁) + v(-γvL/c²) )

L = γL' - γ²v²L/c²

Here, L' = x'₂ - x'₁ is the length of the rod as measured in Bob's frame. Since the rod is at rest in Bob's frame, L' is its proper length, L₀.

L = γL₀ - γ²v²L/c²

Rearrange to solve for L:

L (1 + γ²v²/c²) = γL₀

This is also becoming complicated. Let's use the Lorentz transformation directly on the positions.

Alice measures the ends of the rod at x₁ and x₂ at the same time t.

x₁ = γ(x'₁ + vt'₁)

x₂ = γ(x'₂ + vt'₂)

L = x₂ - x₁ = γ( (x'₂ - x'₁) + v(t'₂ - t'₁) )

Now, let's consider the measurement of length in Bob's frame. Bob measures the ends at x'₁ and x'₂ at the same time t'. L₀ = x'₂ - x'₁.

The key is that Alice measures the positions *simultaneously* in her frame (t₁ = t₂). This does *not* mean that Bob measures them simultaneously in his frame (t'₁ ≠ t'₂ unless v=0 or the rod is perpendicular to motion).

Let's use the definition of length measurement carefully. Alice measures the positions of the ends of the rod at a single instant of time 't' in her frame. Let these positions be x₁ and x₂. So, L = x₂ - x₁.

Using the Lorentz transformation x = γ(x' + vt'):

x₁ = γ(x'₁ + vt'₁)

x₂ = γ(x'₂ + vt'₂)

L = x₂ - x₁ = γ [ (x'₂ - x'₁) + v(t'₂ - t'₁) ]

Here, x'₁ and x'₂ are the positions of the ends of the rod in Bob's frame. For the rod to be of length L₀ in Bob's frame, L₀ = x'₂ - x'₁ (assuming the ends are measured simultaneously in Bob's frame, t'₁ = t'₂).

If Bob measures the ends simultaneously (t'₁ = t'₂), then L₀ = x'₂ - x'₁. In this case, Alice's measurement involves times t'₁ and t'₂ which are not simultaneous.

Let's restart with the definition of length measurement. Alice wants to measure the length of the rod moving along the x-axis. She marks the position of the front end (x₂) and the rear end (x₁) *at the same time* t in her frame. So L = x₂ - x₁.

Using the Lorentz transformation:

• x₁ = γ(x'₁ + vt'₁)
• x₂ = γ(x'₂ + vt'₂)

L = x₂ - x₁ = γ [ (x'₂ - x'₁) + v(t'₂ - t'₁) ]

Now, consider the rod in Bob's frame. Its proper length is L₀. Let's say Bob measures its ends at x'₁ and x'₂ *simultaneously* in his frame (t'₁ = t'₂). Then L₀ = x'₂ - x'₁.

If Bob measures the ends simultaneously, then t'₂ - t'₁ = 0. In this case, L = γ(x'₂ - x'₁) = γL₀. This would mean Alice measures a *longer* length, which is incorrect.

The critical point is that Alice must measure the positions of the two ends *at the same time t* in her frame. These positions x'₁ and x'₂ in Bob's frame will correspond to *different times* t'₁ and t'₂ in Bob's frame.

Let's use the inverse Lorentz transformation for time:

• t'₁ = γ(t - vx₁/c²)
• t'₂ = γ(t - vx₂/c²)

Subtracting these: t'₂ - t'₁ = -γv(x₂ - x₁)/c² = -γvL/c².

Now substitute this expression for (t'₂ - t'₁) back into the equation for L:

L = γ [ (x'₂ - x'₁) + v(-γvL/c²) ]

L = γ(x'₂ - x'₁) - γ²v²L/c²

Let L₀ be the proper length of the rod, measured in Bob's frame where it is at rest. So, L₀ = x'₂ - x'₁.

L = γL₀ - γ²v²L/c²

L (1 + γ²v²/c²) = γL₀

Let's use the definition of γ: γ² = 1 / (1 - v²/c²). So, γ²v²/c² = v²/c²(1 / (1 - v²/c²)) = (v²/c²) / (1 - v²/c²).

1 + γ²v²/c² = 1 + (v²/c²) / (1 - v²/c²) = (1 - v²/c² + v²/c²) / (1 - v²/c²) = 1 / (1 - v²/c²) = γ².

So, L (γ²) = γL₀.

L = (γL₀) / γ² = L₀ / γ

Since γ ≥ 1, L ≤ L₀. This means Alice measures a length L that is shorter than the proper length L₀. This phenomenon is called length contraction.

Length contraction only occurs in the direction of motion. Dimensions perpendicular to the direction of motion are unaffected.

Example: Spacecraft Travel If a spacecraft travels at 0.99c relative to Earth, an observer on Earth would measure the spacecraft's length to be contracted by a factor of γ = 1 / √(1 - 0.99²) ≈ 7.09. So, a spacecraft that is 70.9 meters long in its own rest frame would appear only 10 meters long to an observer on Earth. Conversely, the observer on Earth would appear length-contracted to the astronauts on the spacecraft.

It's important to note that length contraction is a real physical effect, not an optical illusion. It affects the measurement of distances and dimensions for moving objects.

Velocity Addition Law

In classical mechanics, velocities simply add up. If Alice is stationary and Bob throws a ball forward with velocity u relative to him, and Bob is moving towards Alice with velocity v, Alice would measure the ball's velocity as u + v. This is the Galilean velocity addition.

However, this simple addition breaks down at speeds approaching the speed of light. The Lorentz transformation leads to a relativistic velocity addition law that ensures the speed of light remains constant for all inertial observers.

Let's consider three frames: S (stationary), S' (moving with velocity v relative to S along the x-axis), and S'' (moving with velocity u' relative to S' along the x-axis). We want to find the velocity u of S'' as measured by S.

The velocity of S' relative to S is v. The velocity of S'' relative to S' is u'. We want to find the velocity of S'' relative to S, which we'll call u.

From the Lorentz transformation equations:

• x = γ(x' + vt')
• t = γ(t' + vx'/c²)

We know that velocity is dx/dt. Let's differentiate these equations with respect to time t in frame S.

dx/dt = u (the velocity we want to find)

We need to express dx and dt in terms of dx' and dt'.

From x = γ(x' + vt'), dx = γ(dx' + v dt').

From t = γ(t' + vx'/c²), dt = γ(dt' + v dx'/c²).

Now, u = dx/dt = [γ(dx' + v dt')] / [γ(dt' + v dx'/c²)].

u = (dx' + v dt') / (dt' + v dx'/c²).

Divide the numerator and denominator by dt':

u = ( (dx'/dt') + v ) / ( 1 + v(dx'/dt')/c² ).

We know that dx'/dt' is the velocity of S'' relative to S', which is u'.

So, the relativistic velocity addition law is:

u = (u' + v) / (1 + u'v/c²)

This formula applies to velocities along the direction of relative motion. Similar formulas exist for transverse velocities, but they are simpler. If u_y is the velocity component perpendicular to the motion, then u_y = u'_y / [γ(1 + u'v/c²)].

Example 1: Two spaceships Spaceship A travels towards Earth at 0.8c relative to Earth. Spaceship B travels away from Earth in the same direction at 0.6c relative to Earth. What is the velocity of spaceship A relative to spaceship B?

Let Earth be frame S. Velocity of A relative to S (v_A/S) = 0.8c. Velocity of B relative to S (v_B/S) = 0.6c. We want v_A/B. This is equivalent to: Frame S' is spaceship B moving at v = -0.6c relative to Earth (S). We want to find the velocity of A (u') in this frame. v_A/S = u = 0.8c. v_B/S = v = 0.6c. We want v_A/B. Let's set up the frames differently. Let S be Earth's frame. Let S' be spaceship B's frame, moving at v = 0.6c relative to S. We want the velocity of spaceship A relative to B. Let this be u'. The velocity of A relative to S is u = 0.8c. Using the velocity addition formula: u = (u' + v) / (1 + u'v/c²) 0.8c = (u' + 0.6c) / (1 + u'(0.6c)/c²) 0.8c (1 + 0.6u'/c) = u' + 0.6c 0.8c + 0.48u' = u' + 0.6c 0.2c = u' - 0.48u' = 0.52u' u' = 0.2c / 0.52 = (20/52)c = (5/13)c ≈ 0.385c. So, spaceship A is moving at approximately 0.385c relative to spaceship B.

Example 2: Speed of light If Bob throws a ball at speed c relative to him (u' = c), and Bob is moving at speed v relative to Alice, what speed does Alice measure for the ball?

u = (c + v) / (1 + cv/c²) = (c + v) / (1 + v/c) = c(1 + v/c) / (1 + v/c) = c.

This confirms that if something is moving at the speed of light relative to one observer, it is moving at the speed of light relative to all inertial observers, regardless of their relative motion.

Momentum and Energy in Relativistic Mechanics

In classical mechanics, momentum is given by p = mv and kinetic energy by KE = ½mv². These definitions must be modified in the relativistic domain to remain consistent with the postulates of special relativity.

Relativistic Momentum

The relativistic momentum is defined as:

p = γmv

where 'm' is the rest mass of the particle (the mass when it is at rest), 'v' is its velocity, and γ is the Lorentz factor, γ = 1 / √(1 - v²/c²).

Notice that as v approaches c, γ approaches infinity. This means that the momentum of a particle also approaches infinity as its speed approaches the speed of light. This implies that an infinite amount of energy would be required to accelerate a particle with mass to the speed of light, which is impossible. This is why massive objects cannot reach the speed of light.

At low speeds (v << c), γ ≈ 1, so the relativistic momentum p ≈ mv, which reduces to the classical momentum.

Relativistic Energy

Einstein's famous equation E = mc² relates mass and energy. In special relativity, this equation represents the total energy of a particle when 'm' is the rest mass. However, the full picture involves the concept of rest energy and kinetic energy.

The total energy (E) of a particle with rest mass 'm' and velocity 'v' is given by:

E = γmc²

This total energy can be split into two parts: the rest energy (E₀) and the kinetic energy (KE).

The rest energy is the energy a particle possesses due to its mass alone, when it is at rest (v=0, γ=1):

E₀ = mc²

The kinetic energy is the energy of motion. It is the difference between the total energy and the rest energy:

KE = E - E₀ = γmc² - mc² = mc²(γ - 1)

Let's expand γ for low speeds: γ = (1 - v²/c²)^-1/2. Using the binomial expansion (1+x)ⁿ ≈ 1 + nx for small x:

γ ≈ 1 + (-1/2)(-v²/c²) = 1 + ½v²/c²

So, KE ≈ mc²( (1 + ½v²/c²) - 1 ) = mc²(½v²/c²) = ½mv².

This shows that the relativistic kinetic energy reduces to the classical kinetic energy at low speeds.

Mass-Energy Equivalence

The equation E = mc² is perhaps the most famous equation in physics. It signifies that mass and energy are interchangeable. Mass can be converted into energy, and energy can be converted into mass. This principle is fundamental to understanding nuclear reactions (like those in stars and nuclear power plants) and particle physics.

There is also a very useful relationship between total energy, momentum, and rest mass:

E² = (pc)² + (mc²)²

This equation is particularly useful when dealing with particles that have zero rest mass, like photons. For a photon, m=0, so E = pc.

Example: Electron Acceleration An electron (rest mass m_e ≈ 9.11 × 10^-31 kg) is accelerated to a speed of 0.99c. Calculate its relativistic momentum and total energy.

First, calculate γ: γ = 1 / √(1 - (0.99c)²/c²) = 1 / √(1 - 0.9801) = 1 / √(0.0199) ≈ 1 / 0.141 ≈ 7.09.

Relativistic Momentum (p): p = γm_ev = (7.09) × (9.11 × 10^-31 kg) × (0.99 × 3 × 10⁸ m/s) p ≈ (7.09) × (9.11 × 10^-31) × (2.97 × 10⁸) kg m/s p ≈ 1.92 × 10^-22 kg m/s.

Total Energy (E): E = γm_ec² = (7.09) × (9.11 × 10^-31 kg) × (3 × 10⁸ m/s)² E = (7.09) × (9.11 × 10^-31) × (9 × 10¹⁶) Joules E ≈ 5.81 × 10^-13 Joules.

Rest Energy (E₀): E₀ = m_ec² = (9.11 × 10^-31 kg) × (9 × 10¹⁶ m²/s²) E₀ ≈ 8.20 × 10^-14 Joules.

Kinetic Energy (KE): KE = E - E₀ ≈ 5.81 × 10^-13 J - 8.20 × 10^-14 J ≈ 4.99 × 10^-13 Joules.

Note how the kinetic energy is significantly larger than the rest energy for this highly relativistic electron.

Centre-of-Mass System for Two Relativistic Particles

The concept of the centre-of-mass (CM) system is crucial for analyzing collisions and interactions between particles. In classical mechanics, the CM system is defined as the inertial frame in which the total momentum of the system is zero. This simplifies the analysis of collisions, as the particles move towards each other and then away from each other with their momenta reversed (in elastic collisions).

In relativistic mechanics, we extend this concept. For a system of two particles with rest masses m₁ and m₂, and four-momenta p₁ = (E₁/c, **p**₁) and p₂ = (E₂/c, **p**₂), the total four-momentum of the system is P = p₁ + p₂.

The relativistic centre-of-mass system (CM frame) is defined as the inertial frame in which the total *spatial* momentum is zero.

Let the total four-momentum in an arbitrary frame S be P = (E/c, **P**), where E = E₁ + E₂ is the total energy and **P** = **p**₁ + **p**₂ is the total spatial momentum.

In the CM frame, let the total four-momentum be P* = (E*/c, **P***). By definition of the CM frame, **P*** = 0.

The energy E* in the CM frame is related to the total energy E in the arbitrary frame S by the Lorentz transformation. However, a more direct way to find E* is through the invariant mass of the system.

Invariant Mass

A fundamental concept in relativity is the invariant mass (or rest mass) of a system. The invariant mass M of a system is defined such that M²c² = P ⋅ P, where P is the total four-momentum and ⋅ denotes the Minkowski inner product.

P ⋅ P = (E/c)² - **P** ⋅ **P**

In the CM frame, **P*** = 0, so P* ⋅ P* = (E*/c)².

Since the inner product P ⋅ P is invariant under Lorentz transformations, it is the same in all inertial frames, including the CM frame. Therefore:

M²c² = P ⋅ P = P* ⋅ P* = (E*/c)²

This gives us the total energy in the CM frame:

E* = Mc

The invariant mass M is related to the individual particle energies and momenta in the arbitrary frame S:

M²c² = (E₁ + E₂)² - (**p**₁ + **p**₂)²c²

Since E_i = γ_im_ic² and **p**_i = γ_im_i**v**_i, where γ_i = 1/√(1 - v_i²/c²).

M² = (E₁/c² + E₂/c²)² - (**p**₁/c + **p**₂/c)²

M² = (γ₁m₁ + γ₂m₂)² - (γ₁m₁**v**₁/c + γ₂m₂**v**₂/c)²

This expression for M can be calculated in any frame. The resulting M is the invariant mass of the system.

Properties of the CM Frame

In the CM frame (where **P*** = 0), the two particles move in opposite directions with equal and opposite momenta. Let the momenta be **p***₁ and **p***₂.

**P*** = **p***₁ + **p***₂ = 0 => **p***₁ = -**p***₂

Let p* = |**p***₁| = |**p***₂|.

The energies of the particles in the CM frame are E*₁ = √( (p*c)² + (m₁c²)² ) and E*₂ = √( (p*c)² + (m₂c²)² ).

The total energy in the CM frame is E* = E*₁ + E*₂ = Mc.

So, Mc = √( (p*c)² + (m₁c²)² ) + √( (p*c)² + (m₂c²)² )

This equation allows us to determine the magnitude of the momentum p* in the CM frame, given the invariant mass M and the rest masses m₁ and m₂.

Example: Particle Annihilation Consider the annihilation of a particle and its antiparticle, each with rest mass M, into two photons. Let the initial four-momentum of the particle be p₁ = (E₁/c, **p**₁) and the antiparticle be p₂ = (E₂/c, **p**₂). The total four-momentum is P = p₁ + p₂. The invariant mass squared is M_inv² = P ⋅ P / c² = (E₁+E₂)²/c⁴ - |**p**₁+**p**₂|²/c². If the particle and antiparticle are at rest initially, then **p**₁=0, **p**₂=0, E₁=Mc², E₂=Mc². Then P = (2Mc, 0). The invariant mass is M_inv = 2M. The final state consists of two photons, each with zero rest mass. Let their four-momenta be p₃ = (E₃/c, **p**₃) and p₄ = (E₄/c, **p**₄). The total four-momentum of the photons is P' = p₃ + p₄. The invariant mass of the photon system is M'_inv² = P' ⋅ P' / c² = (E₃+E₄)²/c⁴ - |**p**₃+**p**₄|²/c². By conservation of four-momentum, P = P', so M_inv = M'_inv. Thus, 2M = M'_inv. In the CM frame of the initial particles (which is also the CM frame of the final photons), the total spatial momentum is zero: **p**₃ + **p**₄ = 0 => **p**₃ = -**p**₄. The energies are E₃ = |**p**₃|c and E₄ = |**p**₄|c. So, E₃ = E₄ = p*c, where p* = |**p**₃|. The total energy in the CM frame is E* = E₃ + E₄ = 2p*c. The invariant mass M'_inv corresponds to E*/c. So, M'_inv = 2p*. Equating the invariant masses: 2M = 2p* => p* = M. Therefore, each photon has momentum p* = Mc, and energy E* = Mc². This shows how the CM frame simplifies the analysis by setting the total spatial momentum to zero, and the total energy in this frame is directly related to the invariant mass of the system.