PeterDonis
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I think this is a bit misstated.Sagittarius A-Star said:Under these conditions, SR postulate 2 (invariance of the speed of light in vacuum) is not needed to derive the LT.
You can find such a derivation of the LT here
The units of the constant ##\alpha## in the derivation you refer to are inverse velocity squared. So assuming that ##\alpha > 0## is equivalent to assuming that there is a finite invariant speed. It is true that this, by itself, does not tell you that that finite invariant speed is the speed of light; but since there can be only one finite invariant speed, anything that has an invariant speed must have that invariant speed, including light. With that taken into account, assuming ##\alpha > 0## is really equivalent to assuming invariance of the speed of light.
The interesting part of the derivation is showing that assuming ##\alpha > 0## implies ##t' \neq t##. But you only know that by already having the transformation equations in terms of ##\alpha##. So I don't think this is quite the same as deriving the LT from relativity of simultaneity without using the invariance of the speed of light; that would imply that relativity of simultaneity by itself, without using the invariance of the speed of light, or something logically equivalent to it (like ##\alpha > 0##), could tell you the form of the transformation equations. But the derivation of the transformation equations in terms of ##\alpha## doesn't use relativity of simultaneity at all.