Lorentz Transformations

Lorentz factor, often known as the Lorentz term, is a measurement that describes an object's measurements of time, length, and other physical properties, which vary when it moves. The expression occurs in derivations of the Lorentz transformations and is found in a number of special relativity equations.

It is named after the Dutch physicist Hendrik Lorentz, the term originates from its earlier use in Lorentzian electrodynamics. 

Lorentz Factor Definition

Lorentz factor is the factor that describes the dilated time of a moving clock evaluated in a stationary frame in the time dilation formula.

There are two frames of reference, which are:

• Inertial Frames – Motion with a constant velocity
• Non-Inertial Frames – Rotational motion with constant angular velocity, acceleration in curved paths

The Lorentz factor, which is typically represented by the Greek letter gamma (γ), is equal to:

γ = 1 / √(1-(v/c)²) 

where,
γ is Lorentz Factor
v is Relative velocity of two observers
c is the speed of light in a vacuum

Since the quantity (v/c) is often denoted by the β symbol, the equation above can be simplified as follows:

γ = 1 / √(1-(β)²) 

Inertial Frame of Reference

• An inertial frame of reference is defined as a frame that is either stationary or moves at a constant speed relative to an imaginary inertial reference frame.
• Within an inertial frame, a hypothetical inertial coordinate system is established, surrounded by an environment that adheres to Newton's laws of motion.
• Newton's laws hold true in inertial frames, meaning objects within these frames will remain at rest or continue moving uniformly unless acted upon by an external force.
• A reference frame serves as an environment to measure the motion of objects, and if Newton's laws are valid, any uniformly rotating reference system must also follow the law of inertia.
• Examples of inertial reference systems include platforms at rest, such as a train station platform, where objects satisfy the law of inertia due to the absence of acceleration within the frame.

Non-Inertial Frame of Reference 

• A frame that undergoes acceleration relative to an assumed inertial frame is termed non-inertial, where Newton's laws do not hold true.
• To apply Newton's laws in a non-inertial frame, we introduce a fictitious force known as a pseudo force.
• Non-inertial frames are characterized by varying velocities or accelerations, unlike inertial frames which move at constant speeds.
• Examples of non-inertial frames include vehicles traveling along a circular road or accelerating in a straight line, where objects experience acceleration.
• Compared to the default inertial coordinate system, non-inertial coordinate systems are accelerated, often indicated by non-zero accelerometer readings.
• For instance, when a car accelerates from a standstill at a traffic light, it enters a non-inertial frame until reaching a constant velocity, during which accelerometers detect acceleration.

Difference between Inertial Frame of Reference and Non-Inertial Fames of Reference

Check: Inertial and Non-Inertial Frame of Reference

Lorentz Transformation

The Lorentz transformations are a one-parameter family of linear transformations from a frame in spacetime that is in a fixed position to a frame that is moving with constant speed. These transformations are named after a Dutch physicist, Hendrik Lorentz.

The formula for Lorentz transformation can be given as,

t' = γ(t - (vx)/c²)

x' = γ(x - vt)

y' = y 

z' = z

where, 
(t, x, y, z) and (t', x', y', z') are the coordinates of event in two frames.
v is restricted velocity to x-direction.
c is speed of light

Since the Galilean transformation cannot explain why observers traveling at various speeds measure different distances and experience events in a different sequence even if light travels at the same speed in all inertial reference frames, the Lorentz transformations were developed from it.

From the Galilean transformation, we can derive Lorentz transformation as,

x' = a₁x + a₂t

y' = y 

z' = z

t' = b₁x + b₂t

With speed v in non-inertial frame S, the origin of the inertial frame is x' = 0. Let x = vt represent the position in non-inertial frame S at time t for the light beam.

Therefore, x' = 0 = a₁x + a₂t ⇒ x = -(a₂/a₁) t = vt

where, 

a₂/a₁ = -v

Now, the above equation can be written as,

x' = a₁x + a₂t = a₁(x + (a₂/a₁)t) = a₁ (x - vt)

a₁²(x - vt₂) + y'² + z'² - c²(b₁x + b₂t)² = x² + y² + z² - c²t²

a₁²x² - 2a₁²xvt + a₁²v²t² - c²b₁²x² - 2c²b₁b₂xt - c²b₂²t² = x² - c²t²

(a₁² - c²b₂)x² = x²  OR  a₁² - c²b₁² = 1

(a₁²v² - c²b₂²)t² = -c²t²  OR  c²b₂² - a₁²v² = c²

(2a₁²v + 2b₁b₂c²)xt = 0  OR  b₁b₂c² = -a₁²v

b₁²c² = a₁² - 1

b₂²c² = c² + a₁²v²

b₁²b₂²c⁴ = (a₁² - 1) (c² + a₁²v²) = a₁⁴v²

a₁²c² - c² + a⁴v² - a₁²v² = a₁⁴v²

a₁²c² - a₁²v² = c²

a₁²(c² - v²) = c²

a₁² = c²/(c² - v²) = 1/(1 - v²/c²)

a₂ = -v(1 / √(1 - v²/c²))

b₁²c² = (1/(1 - v²/c²) - 1)

b₁²c² = (1-(1 - v²/c²))/(1 - v²/c²) = (v²/c²)/(1-(v²/c²)) = v²/c²(1/1-(v²/c²))

b₁² = v²/c⁴ (1/1-(v²/c²))

b₁ = -v/c² (1/√(1-(v²/c²)))

b₂²c² = (c² + v²(1/1-(v²/c²)) = c²(1 - v²/c²) + v² / 1-(v²/c²) = c²-v²+v²/1-(v²/c²) = c² / 1 - (v²/c²)

b₂² = 1/1-(v²/c²)

b₂ = 1/√1-(v²/c²)  (b₂ is close to a₁)

γ = 1 / √1 - (v²/c²)

the equation can also be written as,

a₁ = γ

a₂ = -γv

b₁ = -(v/c²)γ

b₂ = γ

The final Equation of Lorentz transformation is: 

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

Time Dilation

Either a difference in gravitational potential between their locations or the relative velocities between the two frames of reference produce time dilation (gravitational time dilation taken from general relativity). "Time dilation" describes the velocity-related effect, when it cannot be determined.

Assume that in the reference frame, the time interval between the events is denoted by the symbol Δt0 and is known as proper time or one-position time. In another reference frame (i.e. the observers' reference frame) the time interval between two events is denoted by the symbol Δt. the observer time will always be higher than the proper time. This is what we refer to as time dilation.

The time dilation formula can be written as,

Δt = Δt₀ / √(1-(v/c)²)

where,
Δt is Observer time or two-position time
Δt₀ is Proper time or one position time
v is Relative velocity of two observers
c is the speed of light in a vacuum

Properties of Lorentz Factor

Following are the properties of Lorentz Factor

• The value of the Lorentz Factor is always greater than 1. (γ > 1)
• The Lorentz factor is very close to 1, If the clock speed (v) is slow in comparison to the speed of light (c).
• If the clock speed, v, approaches the speed of light, c, the Lorentz factor increases significantly.

Also, Check

• Relative Motion
• Special Theory of Relativity
• Lorentz Force

Solved Examples on Lorentz Factor

Problem 1: If the relative velocity between the two observers is 120 m/s, Determine the Lorentz factor. (Speed of light is 3 x 108 m/s).

Solution:

Given:

Relative Velocity (v) = 120 m/s

Speed of light (c) = 3 x 108 m/s

Therefore, Lorentz factor is given as,

γ = 1 / √(1-(v/c)²)

γ = 1 / √(1-(120/3 x 108)²)  

= 1 / √(1 - (14400 / 9 x 1016))

= 1 

Problem 2: If the relative velocity between the two observers is 300 m/s, Determine the Lorentz factor. (Speed of light is 2.99 x 108 m/s).

Solution:

Given:

Relative Velocity (v) = 300 m/s

Speed of light (c) = 2.99 x 108 m/s

Therefore, Lorentz factor is given as,

γ = 1 / √(1-(v/c)²)

γ = 1 / √(1 - (300/3 x 108)²)  

= 1 / √(1 - (90000 / 8.9401 x 1016))

= 1

Problem 3: The ratio of v to c is given as 26.7 x 10-8, Determine the Lorentz factor. (Speed of light is 2.99 x 108 m/s).

Solution:

Given:

The ratio of v to c (v/c) = β = 26.7 x 10-8

Therefore, Lorentz factor is given as,

γ = 1 / √(1-(v/c)²)

γ = 1 / √(1-(26.7 x 10^-8)²)  

= 1

Problem 4: If the time interval is 25 seconds and the observer velocity is 30,000 m/s, Find the relative time.

Solution:

Given,

Time interval (Δt₀) = 25 seconds

Observer velocity (v) = 30,000 m/s

Relative time Δt = Δt₀ / √(1 - v²/c²)

= 25 / √(1 - 30,000²/299,792,4582)

= 25 sec

Therefore, the relative time is 25 seconds.

Problem 5:  Find the relative time, If the time interval is 32 seconds and the observer velocity is 50,000 m/s.

Solution:

Given,

Time interval (Δt₀) = 32 seconds

Observer velocity (v) = 50,000 m/s

Relative time Δt = Δt₀ / √(1 - v²/c²)

= 32 / √(1 - 50,000²/299,792,4582)

= 32 sec

Therefore, the relative time is 32 seconds.