In this blogpost here, I derived the Lorentz transformations from first principles. The derivation uses a simple thought experiment; two reference frames are moving relative to each other with a speed . The origins of coincide spatially at time . At this moment, a flash of light is created at their origins, and expands as a sphere.

The radius of this sphere of light will be in respectively. So, we can write

and

But, in reference frame , the distance that light travels in time is related to the speed of light , as , so for we can write that

We can do the same thing for reference frame . The distance that light travels in time in is related to the speed of light in , , so for we can write

.

However, crucially, Einstein said that the speed of light is *the same for all inertial observers.* This is the most important principle which underpins special relativity. It means that , and so we can write

which allows us to write Equations (1) and (2) as

and

Using the Galilean Transformations, which can be written as

we can substitute for , , and in Equ. (4) to give us

which is the same as Equ. (3), except for the two extra terms . This shows that Equations (3) and (4) are not equal under a Galilean transformation, even though the two equations should be equal (as both are equal to zero).

Let us now show that Equations (3) and (4) *are equal* using the Lorentz transformations. These can be written as

where is the Lorentz factor, and is defined as

We will substitute these expressions for and into Equ. (4). When we do this, we have

Multiplying this out, and dropping the and terms, we get

The two terms disappear and, gathering terms in and together, we can write

Remembering that , and changing the sign of the term, we can write

which is

which is

, exactly as in Equ. (3).

We have therefore shown that Equations (3) and (4) *are equal* if we use the Lorentz transformations.

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on 21/05/2019 at 15:17 |thesciencegeekGreat post,

But a few formatting issues

e.g. in the post it states:

latex x^{\prime} (=x – vt)$