 # Derivation of Lorentz Transformation

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 about Lorentz transformation which comes under linear transformation.

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.

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‘ = K[t - $\frac{x}{v}$(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 - $\frac{1}{kk‘}$)$^{2}$

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 - $\frac{vx}{c^{2}}$)

Lorentz Transformation Formula

The complete formula of Lorentz transformation can be obtained as;

t‘ = k(t - $\frac{vx}{c^{2}}$)x‘ = k(x - vt)y‘ = yz‘ = z

Where,

The coordinates of two frames are (x, y, z, t) and (t‘ , x‘ , y‘ , z‘) respectively

v = Velocity restricted to x – axis and

c = Speed of light

K = Constant

1. What is the Significance of Lorentz Transformation?

Ans- There are many relativistic effects that take place such as time dilation, which is noticeable. As we know Lorentz transformation is a relativistic transformation. So, it can be derived as the product of time dilation and Lorentz transformation written below:

T * K

• It can be transferred from one reference to the other reference.

• Lorentz transformation adopts the speed of light for the mentioning frame as that frame is constant

2. A Spaceship Sis on its Way to the Moon. Spacecraft S Passes at a Speed of C/4. A Signal is Sent that Lasts 2.4 Sec According to the Ship's Clock. So, What will be the Interval of Time of the Signal?

Ans-

Δt‘ = 2.4s Δx‘ = 0

So, the time, Δt = (2.4)/(√1 - (¼)²) = (2.4)/(0.97) = 2.47 sec

3. Proof of Lorentz Transformation

Lorentz transformation is tested and tried in our day today life experiments. It gives an accurate result. The equation is written below,

x’ = (1)/(√1-(v²/c²)) (x - vt) ; y’ = y ; z’ = z

And t’ = (1)/(√1-(v²/c²)) (t - (vx/c²))