JD96 said:
when you write the length of the string stays constant in unprimed K, it's because each end of the string is always attached to one of the spaceships and since the problem is set up, so that the distance between the spaceship remains the same in K, the "actual" length of the string will also stay constant until it breaks.
Yes.
JD96 said:
when an observer in unprimed K wants to account for the breaking of the string using length contraction he should observe the unstressed length of the string be shortened by the lorentz factor, which I would argue means primed K should measure the unstressed rest length.
Nobody can directly measure the unstressed rest length of the string if the string is under stress. The string being under stress is an invariant; it doesn't depend on your choice of frame, and all observers agree on it.
When the observer in K says that the "unstressed length" of the string is shortened by the Lorentz factor, that's really a sloppy way of saying that if the string were not under stress, its measured length in K would be shortened by the Lorentz factor. But this is a counterfactual, not an actual measurement.
JD96 said:
how do we know the unstressed rest length stays constant
Because this is how the physics of strings works. More precisely, it's how the physics of solid objects in general works; their rest length when not under stress is an invariant property of the object.
It's true that this fact about objects is not something you can deduce purely from the postulates of SR. You also need knowledge about how solid objects are put together, physically. But for purposes of this scenario, you don't need to know any gory details of the microphysics of solid objects; you just need to know that that microphysics justifies the claim I made in the previous paragraph.
JD96 said:
with the ratio of the proper time I mean the following (now I am referring to the minkowski diagram in the insight post)...
Then what you are saying is not correct, at least not as an application of the equivalence principle. To apply the EP, the ships would have to be at rest relative to each other while they are accelerating, since the analogy is to objects at rest in a gravitational field at slightly different altitudes.
JD96 said:
an accelerating observer will justify him being at rest, despite feeling proper acceleration, via a gravitational field
Yes, this is true, as long as you interpret "gravitational field" correctly. See below.
JD96 said:
so that from an accelerating observer's perspective there's automatically spacetime curvature and other general relativistic effects involved
No. A "gravitational field" in this sense does not automatically imply spacetime curvature, because you are only sampling it locally. Locally, if you are feeling a 1 g acceleration, you can't tell whether it's because you're in a rocket in flat spacetime whose engine is producing 1 g of thrust ("gravitational field" present, but no spacetime curvature), or standing at rest on the surface of the Earth (gravitational field and spacetime curvature). But if you are allowed to make observations over a larger region of space and time, you will be able to tell the difference.