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## Expanding spheres of light

In this blogpost here, I derived the Lorentz transformations from first principles. The derivation uses a simple thought experiment; two reference frames $S \text{ and } S^{\prime}$ are moving relative to each other with a speed $v$. The origins of $S \text{ and } S^{\prime}$ coincide spatially at time $t=t^{\prime}=0$. 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 $r \text{ and } r^{\prime}$ in $S \text{ and } S^{\prime}$ respectively. So, we can write

$x^{2} + y^{2} + z^{2} = r^{2} \text{ in } S \text{ (1) }$
and
$x^{\prime}{^2} + y^{\prime}{^2} +z^{\prime}{^2} = r^{\prime}{^2} \text{ in } S^{\prime} \text{ (2) }$

But, in reference frame $S$, the distance $r$ that light travels in time $t$ is related to the speed of light $c$, as $c = r/t$, so for $S$ we can write that

$r = ct$

We can do the same thing for reference frame $S^{\prime}$. The distance $r^{\prime}$ that light travels in time $t^{\prime}$ in $S^{\prime}$ is related to the speed of light $c^{\prime}$ in $S^{\prime}$, $c^{\prime} = r^{\prime} / t^{\prime}$, so for $S^{\prime}$ we can write

$r^{\prime} = c^{\prime} t^{\prime}$.

However, crucially, Einstein said that the speed of light $c$ is the same for all inertial observers. This is the most important principle which underpins special relativity. It means that $c^{\prime} = c$, and so we can write

$r^{\prime} = c t^{\prime}$

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

$x^{2} + y^{2} + z^{2} -c^{2}t^{2} = 0 \text{ in } S \text{ (3) }$

and

$x^{\prime}{^2} + y^{\prime}{^2} + z^{\prime}{^2} - c^{2}t^{\prime}{^2} = 0 \text{ in } S^{\prime} \text{ (4) }$

Using the Galilean Transformations, which can be written as

we can substitute $x$for $x^{\prime} (=x - vt)$, $y^{\prime}=y$, $z^{\prime}=z$ and $t^{\prime} =t$ in Equ. (4) to give us

$(x-vt)^{2} + y^{2} + z^{2} - c^{2}t^{2} = x^{2} - 2vxt +v^{2}t^{2} + y^{2} + z^{2} - c^{2}t^{2}$

which is the same  as Equ. (3), except for the two extra terms $-2vxt \text{ and } v^{2}t^{2}$. 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 $\gamma$ is the Lorentz factor, and is defined as

$\gamma = \frac{ 1 }{ \sqrt { \left( 1 - v^{2}/c^{2} \right) } }$

We will substitute these expressions for $x^{\prime},y^{\prime},z^{\prime}$ and $t^{\prime}$ into Equ. (4). When we do this, we have

$\left( \gamma (x-vt) \right)^{2} + y^{2} + z^{2} - c^{2} \left( \gamma (t - vx/c^{2}) \right)^{2}$

Multiplying this out, and dropping the $y^{2}$ and $z^{2}$ terms, we get

$\gamma^{2}(x^{2} -2vxt +v^{2}t^{2}) -c^{2}\gamma^{2}\left(t^{2} - \frac{2vxt}{c^{2}} + \frac{v^{2}x^{2}}{c^{4}}\right)$

$\gamma^{2}x^{2} -2\gamma^{2}vxt + \gamma^{2}v^{2}t^{2} - c^{2}\gamma^{2}t^{2} +2\gamma^{2}vxt - (\gamma^{2}v^{2}x^{2})/c^{2}$

The two terms $2\gamma^{2} vxt$ disappear and, gathering terms in $x^{2}$ and $t^{2}$ together, we can write

$\gamma^{2}x^{2} - \left( \frac{ (\gamma^{2}v^{2}) }{ c^{2} } \right) x^{2} + \gamma^{2}v^{2}t^{2} - \gamma^{2}c^{2}t^{2}$

$\gamma^{2}x^{2} (1 - v^{2}/c^{2}) + \gamma^{2}t^{2}(v^{2}-c^{2})$

Remembering that $\gamma^{2} = \frac{ 1 }{ ( 1 - v^{2}/c^{2} ) }$, and changing the sign of the $t^{2}$ term, we can write

$\frac{ 1 }{ (1 - v^{2}/c^{2} ) } \cdot (1 - v^{2}/c^{2}) \cdot x^{2} - \frac{ 1 }{ (1 - v^{2}/c^{2} ) } \cdot (c^{2} - v^{2}) \cdot t^{2}$

which is

$x^{2} - \frac{ 1 }{ (1 - v^{2}/c^{2} ) } \cdot c^{2}(1 - v^{2}/c^{2}) \cdot t^{2}$

which is

$x^{2} - c^{2}t^{2}$, exactly as in Equ. (3).

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