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First teaching 2023

First exams 2025

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Lorentz Transformations (HL) (HL IB Physics)

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Ashika

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Ashika

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Lorentz Transformation Equations

  • To relate measurements from one reference frame to another in Galilean relativity, we used Galilean transformations 
  • However, Galilean transformations can no longer be used to describe distances, times and speeds for objects travelling close to the speed of light
  • Einstein's postulates of special relativity lead to the Lorentz transformation equations for the coordinates of an event in two inertial reference frames

The Lorentz Factor

  • Lorentz transformations are a correction of the Galilean transformations for speeds close to the speed of light, by multiplying by a scaling factor called the Lorentz factorgamma

gamma space equals space fraction numerator 1 space over denominator square root of 1 space minus space begin display style v squared over c squared end style end root end fraction

  • As v will always be less than c ( since nothing can travel faster than the speed of light), this means that gamma will always be greater than 1
  • This is especially important for time dilation and length contraction

Lorentz Transformation Equations

  • Again we have the reference frame S measuring with co-ordinates (xyt), and S' with co-ordinates (x', y', t')
  • Person F is moving away from Person E in their rocket ship at speed which is close to the speed of light c
    • Person E is a stationary observer on Earth
  • Both Person E and Person F witness a loud bang some distance away
    • Person F measures the loud bang to be a distance x' away
    • Person E measures the loud bang to be a distance x = γ(x' + vt') away

 1-5-5-lorentz-transformation

People E measures the loud bang to be at a different distance to person F

  • However, motion is relative
  • According to Person F, they are stationary and Person E is moving away from them at a velocity v in the opposite direction
  • So from Person F's point of view:
    • Person E's velocity is -v
    • Therefore, Person F measures the bang to happen at a distance x' = γ(x - vt) away
  • But, the time the bang happens is not the same in both frames of reference! t ≠ t
  • These are the exact same equations as the Galilean transformation equations, just with the added Lorentz factor
  • In summary, the Lorentz equations from frame S → S' are:

x apostrophe space equals space gamma open parentheses x space minus space v t close parentheses

t apostrophe space equals space gamma open parentheses t space minus space fraction numerator v x space over denominator c squared end fraction close parentheses

  • Where:
    • (xyzt) = the co-ordinates measured from one reference frame
    • (x'y'z't') = the co-ordinates measured from another reference frame moving at speed v relative to it

Galilean vs. Lorentz Transformations

Lorentz (v ≈ c)

Galilean (v << c)

x apostrophe space equals space gamma open parentheses x space minus space v t close parentheses

x apostrophe space equals space x space minus space v t

x space equals space gamma open parentheses x apostrophe space plus space v t apostrophe close parentheses

x space equals space x apostrophe space plus space v t

t apostrophe space equals space gamma open parentheses t space minus space fraction numerator v x space over denominator c squared end fraction close parentheses

t apostrophe space equals space t

t space equals space gamma open parentheses t apostrophe space plus space fraction numerator v x apostrophe space over denominator c squared end fraction close parentheses

t space equals space t apostrophe

y space equals space y apostrophe

y space equals space y apostrophe

z space equals space z apostrophe

z space equals space z apostrophe

  • Notice that time is different between reference frames t and t' for objects travelling close to the speed of light, whilst in Galilean transformations, time was absolute (it doesn't change) between reference frames

Worked example

A rocket of proper length 150 m moves to the right with speed 0.91c relative to the ground.

1-5-5-lorentz-transformation-worked-example-ib-2025-physics

A probe is released from the back of the rocket  at speed 0.45c relative to the rocket.

Determine the time it takes the probe to reach the front of the rocket according to an observer

(a)
At rest in the rocket.

(b)
At rest on the ground.
 

Answer:

(a)

Step 1: List the known quantities

  • Length of the rocket, l = 150 m 
  • Speed of the probe, v' = 0.40c

Step 2: Analyse the situation

  • In the reference frame of an observer at rest in the rocket, they are stationary
  • Therefore, the probe travels at a constant speed 0.45c across the full length of the rocket of 150 m

t apostrophe space equals fraction numerator space l over denominator v apostrophe end fraction space equals fraction numerator space 150 over denominator 0.45 c end fraction space equals space fraction numerator space 150 over denominator 0.45 space cross times space open parentheses 3 space cross times space 10 to the power of 8 close parentheses end fraction

t apostrophe space equals space 1.11 space cross times space 10 to the power of negative 6 end exponent space straight s

  • t' and v' are used because they are the times and velocity of the moving object in the reference frame of the observer at rest

(b)

Step 1: List the known quantities

  • Length of the rocket, l = 150 m
  • Speed of the rocket, v 0.91c
  • Time taken for the probe to reach the front of the rocket, t' = 1.11 × 10–6 s

Step 2: Analyse the situation

  • In reference to an observer at rest on the ground, they will see the probe taking longer to reach the front of the ship
  • Since object in question, the probe, is moving in both reference frames, we need to use a Lorentz transformation

Step 3: Calculate the gamma factor

gamma space equals space fraction numerator 1 over denominator square root of 1 space minus fraction numerator space v squared over denominator c squared end fraction end root end fraction space equals space fraction numerator 1 over denominator square root of 1 space minus fraction numerator space open parentheses 0.91 c close parentheses squared over denominator c squared end fraction end root end fraction

gamma space equals space fraction numerator 1 over denominator square root of 1 space minus space 0.91 squared end root end fraction space equals space 2.412

Step 3: Substitute values into the Lorentz transformation

  • Since the observer is at rest, the Lorentz equation for time t must be used

t space equals space gamma open parentheses t apostrophe space plus space fraction numerator v x apostrophe space over denominator c squared end fraction close parentheses

t space equals space 2.412 open parentheses open parentheses 1.11 cross times 10 to the power of negative 6 end exponent close parentheses space plus space fraction numerator open parentheses 0.91 c close parentheses open parentheses 150 close parentheses space over denominator c squared end fraction close parentheses space space equals space 2.412 open parentheses open parentheses 1.11 cross times 10 to the power of negative 6 end exponent close parentheses space plus space fraction numerator open parentheses 0.91 close parentheses open parentheses 150 close parentheses space over denominator 3 space cross times space 10 to the power of 8 end fraction close parentheses

t space equals space 3.8 space cross times space 10 to the power of negative 6 end exponent space straight s

Examiner Tip

Always check that your value of gamma is greater than 1.

You will often be given a speed v in terms of c e.g. v = 0.90c etc. When you put this value into the gamma factor, this is squared. Therefore, you do not need to put in 3.0 × 108 at all into your calculator, as the c2 will cancel.

E.g. gamma space equals space fraction numerator 1 over denominator square root of 1 space minus fraction numerator space v squared over denominator c squared end fraction end root end fraction space equals space fraction numerator 1 over denominator square root of 1 space minus fraction numerator space open parentheses 0.90 c close parentheses squared over denominator c squared end fraction end root end fraction space equals space fraction numerator 1 over denominator square root of 1 space minus fraction numerator space open parentheses 0.90 close parentheses squared open parentheses c close parentheses squared over denominator c squared end fraction end root end fraction

gamma space equals space fraction numerator 1 over denominator square root of 1 space minus fraction numerator space open parentheses 0.90 close parentheses squared up diagonal strike open parentheses c close parentheses squared end strike over denominator up diagonal strike c squared end strike end fraction end root end fraction space equals space fraction numerator 1 over denominator square root of 1 space minus open parentheses 0.90 close parentheses squared end root end fraction

The equations for x', t' and gamma are given in your data booklet, but you must remember the sign change if you want to calculate x or t (from the rest frame) instead! 

Some textbooks may go further into this for your understanding, you will not be expected to derive these equations in your exam. You will only be assessed on how to use them.

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Ashika

Author: Ashika

Expertise: Physics Project Lead

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.