What does it mean that an observer can provide a quantitative temporal order only to the events on his worldline?

Question

I'm currently reading the introduction to Naber's The Geometry of Minkowski Spacetime, and in this post I'm writing down a few silly questions that keep popping into my head. I have near-zero formal training in Physics behind a freshman mechanics course (I'm a Mathematics student), so please be kind.

Naber starts by describing a class of "observers" (of spacetime) which he calls admissible. For us, an observer is to be pictured as a distinguished material particle sitting at the origin of an orthonormal right-handed 3-dimensional coordinate system. Each of such (admissible) observers is capable of giving a quantitative temporal order to the events. In the author's words:

Each admissible observer is provided with an ideal standard clock based on an agreed unit of time with which to provide a quantitative temporal order to the events on his worldline.

The emphasis is mine. I'm confused by this last requirement. What does it mean for an event to "be" on the worldline of an observer? Suppose the event $ \mathcal A $ lies on the worldline of the observer $ O $. Does this mean exactly that, at some point in time, the spatial coordinates of $ \mathscr A $ and $ O $ where the same? (I wrote "coordinates" but I did not want to imply that a coordinate system had been chosen).

Going further, there's another postulate:

Any two admissible observers agree on the temporal order of any two events on the worldline of a photon, i.e., if two such events have coordinates $ (x^1,x^2,x^3,x^4) $ and $ (x_0^1,x_0^2,x_0^3,x_0^4) $ in $ \mathcal S $ and $ (\hat x^1,\hat x^2,\hat x^3,\hat x^4) $ and $ (\hat x_0^1,\hat x_0^2,\hat x_0^3,\hat x_0^4) $ in $ \hat{\mathcal S} $ then $ x^4 − x_0^4 $ and $ \hat x^4 - \hat x_0^4 $ have the same sign.

Again, what does it mean for an observer to be on the worldline of a photon? Could you provide me an example situation where this happens?

Finally, let's get to the mathematics. At some point it is claimed that

Since photons propagate rectilinearly with speed $ 1 $, two events on the worldline of a photon have coordinates in $ \mathcal S $ which satisfy $$ x^i - x_0^i = v^i(x^4 - x_0^4) $$ for some constants $ v^1 $, $ v^2 $ and $ v^3 $ with $ (v^1)^2 + (v^2)^2 + (v^3)^2 = 1 $.

How? This seems trivial and I feel I understood nothing about worldlines.

Answer 1

To the first question: If event A lies on the world line of observer O, then the event A and observer O are at the same point in space time.

To the second question: because of the relativity postulate, viz. the constancy of the speed of light for all observers, the world lines of photons have a simple geometry; that is photons always travel along a null path. This null path is always a subset of the set of points in Minkowski space that have zero distance between them. In two plus one dimensional, flat space-times, this is easy to visualize, it is the familiar "light cone". If an observer is on the world line of a photon, then either that observer has received a photon, or has sent a photon, hence their world lines intersect at this point.

To the third: solving problems concerning the distance between two points in space-time requires that one write out the Minkowski metric or distance function for the two points concerned, however, if the two points lie on the world line of a photon, then set the resulting distance to zero, since the two points are on the null path, and then solve for whatever particulars you desire. It is the indefinite nature of the Lorentzian metric that gives rise to so many interesting phenomena in relativity.