Why Is Minkowski Spacetime Non-Euclidean?

[Sagittarius A-Star]

What is valid is that, for a fixed $c$, as $v \to 0$ then the LT does tend to the Galilean transformation.

If you mean $\lim_{v \rightarrow 0}{}$, then you get only a special case of the GT

$\lim_{v \rightarrow 0}{(\gamma (x-vt))} = x-0 \cdot t = x, \ \ \ \ \$ $\lim_{v \rightarrow 0}{(\gamma (t-vx/c^2))} = t$

 

[PeroK]

If you mean $\lim_{v \rightarrow 0}{}$, then you get only a special case of the GT

$\lim_{v \rightarrow 0}{(\gamma (x-vt))} = x-0 \cdot t = x, \ \ \ \ \$ $\lim_{v \rightarrow 0}{(\gamma (t-vx/c^2))} = t$

You're missing the point. If $v$ is small, then $\gamma \approx 1$. And you can study a subset of SR where $v$ is small and you have approximately the Galilean transformation.

If $c$ is large, then not all gamma factors are approximately $1$. You don't have an approximation to Galilean relativity.

 

[Sagittarius A-Star]

If $v$ is small, then $\gamma \approx 1$. And you can study a subset of SR where $v$ is small and you have approximately the Galilean transformation.

That's correct, except if the $x$ in $\gamma (t-vx/c^2)$ is very large, as @PAllen mentioned in #50.

 

[Dale]

as c is taken to approach infinity, the error of any observation compared to Galilean approaches zero

This is really the scientifically important fact. For SR to be a valid theory the differences between it and Galilean relativity must be smaller than the experimental uncertainty in any domain where Galilean relativity has been experimentally validated. You can express that requirement with several different limits, but the important fact is that in any of those limits the difference must approach zero.