Length contraction of a rod and observed aging

Question

Consider an observer $O'$ at rest with respect to a rod and an observer $O$ moving away left from observer $O'$. Consider the below image I found. Observer $O'$ measures the length of the rod with events $0$ and $Q'$ simultaneous in his frame of reference whereas observer $O$ measures it with events $0$ and $Q$. From observer $O'$ view the event $Q$ happened before $Q'$.

Does this mean it would be possible to have the following scenario where from observer's $O'$ view if the rod were sitting in some atmosphere rusting away then from observer $O$ view he would see if going fast enough that the rod would rust (or age) completely at the end nearest him while simultaneously seeing the end furthest from him near rust free because $Q$ happens before $Q'$.

Answer 1

A picture of a long object (let‘s say your long rod) can be taken this way: you place a row of photo cameras along the rod and make clicks by these photo cameras simultaneously in your frame. Then you glue all these pictures into one.

This way, yes, different observers (photographers) moving with different velocities relatively to a measuring rod, would see different set of events. or different, not identical to each other rods. If the rod is covered with clocks, all these clocks will show different time, i.e. every infinitesimal part of it would appear to be of different age. So. if in the rest frame of the rod it starts rusting from the both ends simultaneously, in some other relatively moving frame it will appear almost untouched from one and almost completely rusted to the middle from the other end.

If there are many identical brothers at some distance apart from each others, let‘s say 30 years old in their rest frame, from the other relatively moving frame some of them may appear still sucking a pacifier while some others lying in a coffin already.

Obviously it is because of Einstein synchronization in each frame of reference.

To get rid of this rubbish all moving observers must choose another synchronization. They must take one arbitrary frame (that could be the frame of the rod) and take into account their velocity relatively to this frame and synchronize their clocks (clicks of photo cameras) accordingly. Obviously in these frames one way speed of light will not be c.

Imagine, that the rod rusts from the both ends in its rest frame simultaneously and disappears due to rusting. If you will describe remainder of the rod perfectly well, you will get its velocity.

That means, an abstract rod has ANY velocity. Even in Special Relativity a well – defined (or described) (remainder of the) rod has EXACT velocity, since there is one and only one frame, from which this exact remainder of the rod can be seen.

By the way: Feynman talking about definitions of objects - http://www.feynmanlectures.caltech.edu/I_12.html, about a chair, starting from: - Any simple idea is approximate; as an illustration, consider an object ...

Answer 2

I'll ignore the image because it's not clear to me what it illustrates. But yes, events that are simultaneous for $O'$ are not simultaneous for $O$. The Lorentz transformations translate from the $O$ coordinate system to $O'$:

$$ x' = \gamma (x-vt) \\ t' = \gamma (t-vx/c^2) $$

For the set of events for which $t'=0$, the time coordinate $t$ obviously must be $t = vx/c^2$. This means that if $O'$ has a rod that is 2 million light years long that is moving away at 1 m/s, the remote end will be ~22 days in the future (using the $O'$ sense of simultaneity) according to the $O$ coordinate system. So the near end will rust 22 days earlier than the remote end. But $O$ will of course have to wait 2 million years to see any evidence of this.

Note: length contraction and time dilation is negligible when the relative velocity is 1 m/s. I deliberately constructed my example to show how pure distance affects the relativity of simultaneity, even at very low speeds.