Proposition: Conformal information does not fix calibration

The full cone $K_D$ determines the Lorentzian quadratic form (eq:quadratic) up to a positive scalar. It does not select a preferred Lorentz coordinate frame, a normalization of that scalar, or physical length and time units. Consequently it does not determine a numerical speed in physical units.

Proof. In Lorentz coordinates, any other quadratic form vanishing on the same null boundary and positive in the future interior has the form

Vanishing on $(1,u)$ and $(1,-u)$ for every unit $u$ gives $b=0$ and

Thus $C=-aI$, and positivity on $(1,0)$ gives $a>0$. This is the uniqueness calculation recorded in XXII-A CheungXXIIA (rem:conformal).

For every $s>0$, the map $s{T}_D$ still maps $K_D$ onto ${\mathcal{L}}_m$, but replaces $q_D$ by $s^2{q}_D$. Lorentz boosts also preserve ${\mathcal{L}}_m$. For example, on one temporal and one spatial coordinate,

does not select the same observer when $\eta \ne 0$.

Finally, if coordinate units are converted by

then the null equation becomes

The constants ${\ell }_0,{\tau }_0$ are not fixed by cone membership.