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The spacetime interval $ds$ is the limit of this expression as $\Delta x^a \rightarrow 0$. I still treat that like a scalar because the finite $\Delta s$ was a scalar. … It is the limit of taking smaller and smaller segments of a path between two spacetime points, which connects to your third question below. How can we interpret the proper time as tensor? …

Whether making a 3+1 split is the "right thing to do" depends on the problem you want to solve. There are a wide range of interesting astrophysical problems where this makes sense as a starting point …

The exponential function only returns positive numbers. Writing it this way ensures that the $dt$ term has a negative coefficient (because of the explicit minus sign) and the other terms have positiv …

That's done all of the time, using the fancy-sounding name "spacetime diagram". … It is true that lengths under the spacetime metric behave differently, but that doesn't prevent the drawing of the diagram and, with some practice, many people do generate intuition for these types of …

A proper scalar arising from contraction requires the vectors to be defined at the same location in spacetime. … In a general, curved spacetime this will break down. …

The first time that you do this experiment, you have six unknowns, the speed along each of the three legs of your triangle ($c_{SO}$, $c_{SM}$, and $c_{MO}$) plus the time take to transit each leg ($t …

Leaving a clock "motionless" in space is already science fiction. To do what you suggest, clock A would need substantial rockets running continuously to prevent it from falling directly into the sun o …

The experimental way to determine dimensionality of space is to look at things like the power-law for the gravitational potential or electrical potential. The "usual" power for fall-off in radius $r$ …

The book cited in the question (contrary to much of the discussion in the comments) is about simulating physics on computers. When you go to a discrete grid and simulate with finite differences inste …

The video goes on to say that proper time is the time as measured by a clock carried by the apple (the object that's moving in the video) along its world line, which is correct. As you've correctly ob …

With the discovery of special relativity and the associated notion of spacetime, it becomes clear that two distinct sets of units aren't quite right. …

You have misunderstood the geodesic. There's no physical process of "straightening." The geodesics are extrema of distance - usually shortest distance - between two points on the manifold. That's t …

Second, there's the question of making the spacetime diagrams that you raise, which could be extended to other conveniences in how to write quantities in special relativity. … So for some circumstances, using $ct$ instead of $t$ is handy because it emphasizes that space and time are different pats of a coherent spacetime. …

Maybe there’s some context to this statement that somehow saves it, but taken alone it looks wrong: the "law of geodesic motion" is actually a mathematical law, not a physical one There’s some freed …

This is a consequence of the fact that there is no globally inertial frame of reference in this spacetime. … In that sense that spacetime is not dynamical such that some sort of "twisting" has a cumulative effect on the curvature. …