[Physics Class Notes] on Lorentz Transformation Derivation Pdf for Exam

Introduction

In physics, we study the magnetic electric force charge to find out its true nature while observing and researching. We analyze the force exerted into space with a constantly varying flow of current. 

 

The word electromagnetism can be elaborated as the summation of the electric and magnetic force acting simultaneously towards the space due to the motion of charged particles. The forces which are available around us can be studied through electromagnetism.

 

Lorentz transformation is put forward by the Dutch scientist Hendrik Antoon Lorentz. The frame of reference is any kind of that you are measuring something. For example, if you are standing on the floor and looking at some physical event such as a firecracker explosion or collision of two stones. that floor will become your frame of reference. If you are traveling inside the train and looking outside to a physical event, the train will be your frame of reference.

 

What is Lorentz Transformation?

Lorentz transformation can be elaborated as the linear transformation which consists of a rotation of space along with the constant distance between space and time. In physics, we can study Lorentz transformation which comes under linear transformation.

 

Lorentz transformations are a set of equations used in relativity physics to connect the space and time coordinates of two systems moving at the same speed. Lorentz transformations are used to describe extremely fast phenomena that approach the speed of light. They formally describe the relativity ideas of space and time not being absolute, length, time, and mass all being dependent on the observer’s relative motion, and the speed of light in a vacuum being constant and independent of the observer’s or source’s motion. In 1904, a Dutch physicist named Hendrik Antoon Lorentz devised the equations.

 

Lorentz Transformation Derivation

We are using mathematics to elaborate and predict the events that happen in the world. 

In general, an event indicates something that occurs at a given location in space and time. The Lorentz transformation transforms between two reference frames when one is moving with respect to the other. 

 

The Lorentz transformation can be derived as the relationship between the coordinates of a particle in the two inertial frames on the basis of the special theory of relativity.

 

The Lorentz transformations are exclusively related in terms of change in inertial frames. Furthermore, this link is frequently found in special relativity. This transformation is also a type of linear transformation in which mapping occurs between two modules involving vector spaces.

 

The operations of scalar multiplications and additions are preserved throughout linear transformation. Furthermore, this transition has some innate characteristics. An observer moving at a different velocity, for example, could measure event ordering, elapsed periods, and various distances, but the constraint here is that the speed of light should be the same in all inertial frames.

 

Here, 

 

S and S‘  = two inertial frames out of which S‘ is moving relative to S with v velocity along positive x-axis

 

At the beginning t = t‘ = 0

 

Origin O and O’ will coincide

 

Lorentz Transformation Equation Derivation

The wavefront of light emitted at t = 0 when reaches at P, the position and time observed by observers at O and O‘  are ( x, y, z , t ) and  (t‘,  x‘, y‘, z‘)respectively.

 

Time taken to reach O from P as observed in frame S is:

t = [frac{OP}{c}] = [frac{sqrt{x^{2} + y^{2} + z^{2}}}{c}]

or, x[^{2}] + y[^{2}] + z[^{2}] = c[^{2}]t[^{2}]

or, x[^{2}] + y[^{2}] + z[^{2}] – c[^{2}]t[^{2}] = 0

The same equation can be obtained for t‘ time taken by light to reach from O‘ to P,

 x[^{‘2}] + y[^{‘2}] + z[^{‘2}] – c[^{2}]t[^{‘2}] = 0

Since both equations represent the same spherical wavefront in S and S‘ frame, they can be equated as:

x[^{2}] + y[^{2}] + z[^{2}] – c[^{2}]t[^{2}] = x[^{2}] + y[^{2}] + z[^{2}] – c[^{2}]t[^{2}] ………….(1)

Since frame S and S‘ are moving relative to S along the x-axis, length in direction is perpendicular to direction of motion are unaffected.

i.e., y‘ = y and z‘ = z ……….(2)

From (1) and (2), we have  

x[^{2}] – c[^{2}]t[^{2}] = x[^{‘2}] – c[^{2}]t[^{‘2}] (a)

In the frame S

x = v*t

or, x – v*t = 0

Whereas for frame S‘

x‘ = k (x – v*t)   (3)

Since both are relative, we can assume s is moving relatively along with s‘ having a velocity v along the negative x – axis.

So, position of O at any instant of t‘ relative to observer is 

x‘ = – v ∗ t‘

or, x‘ + v ∗ t‘ = 0

Whereas position of O relative to observer O in frame S is x = 0 , so x and x‘ must be related as

x‘ = k‘ (x‘ – v ∗ t‘) (4)

Where, K‘  is another constant

Substituting the value of x‘ from equation 3, we get;

x = [K‘(x – vt) + vt‘]

t‘ = [(kt – [frac{x}{v}])(1 – [frac{1}{k‘}])] (5)

Putting x’ from eq.3 and t’ from eq.5, in eq.(1) we get 

[x^{2} – c^{2}t^{2} = K^{2}(x – vt)^{2} – c^{2}k^2 (t – [frac{x}{v}])(1 – [frac{1}{k‘}])^{2}]

By simplifying and equating coefficient of t² we get

K = [frac{1}{sqrt{1 – frac{v^{2}}{c^{2}}}}] (7)

 Similarly, we get,

K‘ = [frac{1}{sqrt{1 – frac{v^{2}}{c^{2}}}}]

Substituting the value of K and K’ in equation (3) and (5), we get,

x‘ = [frac{1}{sqrt{1 – frac{v^{2}}{c^{2}}}}] (x – vt) and 

t‘ = [frac{1}{sqrt{1 – frac{v^{2}}{c^{2}}}}] (t –  [frac{x}{v}] x  [frac{v^{2}}{c^{2}}])

= = [frac{1}{sqrt{1 – frac{v^{2}}{c^{2}}}}] (t – [frac{vx}{c^{2}}])

The Lorentz transformations are 

x‘ = [frac{1}{sqrt{1 – frac{v^{2}}{c^{2}}}}] (x – vt) ; y‘ = y ; z‘ = z

And t‘ = [frac{1}{sqrt{1 – frac{v^{2}}{c^{2}}}}] (t – [frac{vx}{c^{2}}])

What are the Differences Between Galilean and Lorentz Transformations?