I Proper (and coordinate) times re the Twin paradox

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if it is not the same quantity then both equations are true
True in their respective geometries, yes. But they are different geometries. Euclidean 3-space is not the same geometry as 4-D Minkowski spacetime. Each one has its own formula for ##ds^2##. It makes no sense to say the formula for ##ds^2## in Euclidean 3-space is "true" in Minkowski spacetime.

It is still the same Spacetime
Euclidean 3-space is not spacetime. It's Euclidean 3-space.

Surely a2+b2+c2 is still the aggregate length in Minkowski Spacetime; while -ct2+a2+b2+c2 is the Spacetime interval and both are invariant intervals.
No, this is not correct. ##dx^2 + dy^2 + dz^2## is the invariant length in Euclidean 3-space. ##- c^2 dt^2 + dx^2 + dy^2 + dz^2## is the invariant length in 4-D Minkowski spacetime. This length is called a "spacetime interval" in the Minkowski spacetime, but that's just nomenclature.
 
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The Mathematics doesn't depend on how they are drawn.
Yes, but it does depend on which geometry you are in.

It just seems difficult to be sure what the terms mean when the same term ds2 means two different things...
It can be very confusing
Welcome to math and physics. :wink: Terminology will sometimes be confusing; you just have to learn how to figure out what is intended from context, and ask questions when something is not clear. What you should not do is assume that the same symbol must mean the same thing in a different context.
 
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The main difference in the two geometries is the number of dimensions, is it not?
In Euclidean 3-space, a2+b2+c2 is invariant because there is no time component.
So in Minkowski Spacetime the equivalent would be that a2+b2+c2 would be invariant at any single specific time.
Similarly, in classical mechanics using euclidean geometry, ds2=(ct')2+a2+b2+c2 where ct' is the time axis for a moving body
 
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The main difference in the two geometries is the number of dimensions, is it not?
No. That's one difference, but not the main one. The main difference is that Euclidean 3-space is Riemannian (the metric has +++ signature) while 4-D Minkowski spacetime is pseudo-Riemannian (the metric has -+++ signature). That means, as @Dale said, that while in Euclidean 3-space there is only one type of interval/length, in Minkowski spacetime there are three: spacelike, null, and timelike.

In Euclidean 3-space, a2+b2+c2 is invariant because there is no time component.
No, it's because there is no such thing as a "time component" in Euclidean 3-space. There are only three dimensions, and they're all spacelike because the metric has +++ signature.

in Minkowski Spacetime the equivalent would be that a2+b2+c2 would be invariant at any single specific time.
There is no such thing as "any single specific time" because there is no preferred inertial frame in Minkowski spacetime. An interval that has zero ##dt## in one frame will have nonzero ##dt'## in any other frame. And "invariant" means the equation has to hold in every frame, not just one, so your claim is wrong.

in classical mechanics using euclidean geometry, ds2=(ct')2+a2+b2+c2 where ct' is the time axis for a moving body
Wrong. There is no such spacetime interval in Newtonian mechanics.
 
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