I am trying to figure out how to "see" things in relativity via a toy model. With a pinhole camera I'd like to capture a relativistic scene consisting of a vertical marked stick which is moving towards (or away from) me with at a speed $v$ comparable to the speed of light $c$:
The setting is the following. Since the problem is essentially 2-dimensional (in space), we can consider $\mathbb{R}^3$ as our ambient space, with points given by coordinates $(x,y,t)$ - the third coordinate is time. The pinhole camera is sitting at the spatial origin, i.e. is represented by the line $x=y=0$.
The marked vertical stick is represented by a segment $AB$, where $A=(d+vt,0,t)$ is the point at the base of the stick (touching the "floor") and $B=(d+vt,l,t)$ where $l$ is the length (height) of the stick. This setting implies that at time $t=0$ the base of the stick is at a distance $d$ from the camera. The world line -or should I say world $strip$?- of the stick is contained in the plane \begin{equation} \alpha:\qquad x-vt=d \end{equation}
Now, "taking a picture" at time $t=0$ means that we record all light rays which arrive at the origin at that instant: these rays are all contained in the "past" light cone originating at $(0,0,0)$, i.e.: \begin{equation} C:\qquad x^2+y^2-c^2t^2=0\qquad\qquad t<0 \end{equation} The light rays recorded by the camera which originated from the stick are then only coming from the intersection between this past light cone and the stick's world line, i.e. a subset of $\alpha\ \cap\ C$.
The result is a branch of a hyperbola; if the stick is moving towards us, the vertex of the hyperbola is the point $A$ (the base) which expresses the intuitive thought that since the base of the stick is nearer to the camera than the top edge ($B$), the recorded image of $A$ has a temporal delay smaller than that of $B$.
I could from this point calculate the image of the stick (by applying some kind of projection mimicking the pinhole camera principle).
My question is this: is my reasoning correct? Since there is no trace of the Lorentz transformations I am guessing that no, it has some pitfall somewhere; but on the other hand it might be right because the width of the stick is zero (I'm considering a $1$-dimensional stick).
Thought: if the stick had been leaning on the $x$-axis and moving along it maybe the Lorentz transformations would have been necessary. I am pretty confused.
EDIT: since it may not be clear enough, the observer sits at the (spatial) origin, and sees the vertical stick coming directly towards him or directly receding from him: what I expect to see in the "photograph" are the different spacings between the marks on the stick, which should differ from the spacings due only to perspective.
EDIT 2: I'm adding a picture here for more clarity. This time the stick is moving away from the observer.
EDIT 3 (last edit): just for fun, here's a still frame from a computer simulation I'm developing based on this idea, showing a square (slightly off-centered) coming towards the camera at about 50% light speed (right view), compared with a normal perspective view of the same square (left view).
The problem of visual perception of relativistic effects is in general quite difficult. It depends on whether the object is coming towards you in a straight line or you are seeing it go by from some distance, for example. It also depends on whether you want to consider things in two spatial dimensions or in three. There are papers about this you can find online, and even a cool animation of someone moving at high speed around a town somewhere (I saw this but I don't remember where, sorry).
The particular case you mention appears to be simple, although it is not quite clear what you have in mind. Is the stick coming right at you? Then you will notice nothing at all. Are you seeing the stick go by in front of you? Then what you see will depend on how high your eyes are. In any case, since the stick's velocity is orthogonal to its length the Lorentz transformation indeed plays no role.