Using some elementary concepts, I've arrived at the apparent conclusion that all the space seen from a particle moving at the speed of light always collapses to a singularity at distance 0 from the particle itself. Looking for contrary proofs (or errors).
Knowing Lorentz factor is
$$\gamma(v) = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$
- $v\quad$ velocity of a moving object in reference to the observer
- $c\quad$ speed of light
Knowing that the velocity of a photon is always $c$ in every reference system
$$\Rightarrow \gamma(c) = \frac{1}{\sqrt{1 - \frac{c^2}{c^2}}} \rightarrow \infty$$
Using the Lorentz contraction formula
$$\Delta L' = \frac{\Delta L_0}{\gamma}$$
with
$\Delta L'$ the distance from the inertila reference frame seen by the moving reference frame
$\Delta L_0$ the distance from the moving reference system seen by the inertial reference system
$\gamma$ the Lorentz factor
$\Delta L' \rightarrow 0$ always
All the space$^*$ collapses to a singularity point overlying the reference frame moving at speed of light.
$^*$seen by the very reference frame moving at $c$
The computation is valid for an inertial reference system put in any point in space, for that I used the generalization "all space".
