What does 's' or 'ds' or $ds^2$ mean in the Schwarzschild equation(s)? [closed] And:   These are from Wikipedia on Schwarzschild metric, Derivation of the Schwarzschild metric and the last is from Science Direct, Schwarzschild metric. • What are the black bars meant to be. Nov 8 '21 at 22:37
• Forget the black bars.... I don't know why copying and pasting images from Wikipedia didn't work.... It works for text.... And I wasn't warned by any message or software that it would be blacked out.... Nov 9 '21 at 6:32
• You might actually link to the text. And quote the interesting parts as actual text. Nov 9 '21 at 11:12
• I think that sometimes if the png has an alpha channel the conversion to SE's pngs does this. If you can convert to jpeg first somehow (which does not have the alpha channel) that might fix it.
– uhoh
Nov 9 '21 at 11:40

ds refers to a distance in spacetime (this includes time). Just as, in classical physics, you see $$ds^2=dx^2+dy^2+dz^2$$ for any given line element, in spacetime you get similar things going on. For flat spacetime, this gives some good intuitive understanding. Depending on your signature, an element in flat (Minkowski) spacetime is $$ds^2=dt^2-dx^2-dy^2-dz^2$$ or $$ds^2=-dt^2+dx^2+dy^2+dz^2$$ Since the whole point of general relativity is the curvature of spacetime, a line element in curved space will be different. So for the Schwarzschild metric we see what you're seeing in the pictures you provided, except instead of cartesian spatial coordinates, spherical coordinates are used and since the Schwarzschild metric is non-rotating and spherically symmetric, we do not see any angular elements in the metric.