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In special relativity Einstein used Pythagorean theorem for proving Lorenz transformations. But in general relativity we discovered that space-time has curvature near massive objects, so the geometry near them isn't euclidean. The question is, how can we use Pythagorean theorem in special relativity, if general relativity proves that geometry in our world isn't fully Euclidean ? Or am I missing something?

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The reason that Special Relativity is called "special" is that it only deals with the special case where there are no large masses nearby--and therefore no gravity. With no gravity, spacetime is flat and the Pythagorean theorem holds.

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General relativity reduces to special relativity in the limit of short distances. Likewise, the error in the Pythagorean theorem is negligible on sufficiently short distance scales.

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    $\begingroup$ +1 To be more explicit, the whole point of special relativity is to make things simpler by not taking into account the effects that massive objects have on spacetime. It's an approximation of real life, and a good one when the ignored effects are small. $\endgroup$ – JiK Apr 4 '18 at 21:26
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Sorry made a mistake, they were right, specialty could handle some acceleration. I put some thought and the main debates here was about the Pythagorean theorem. Actually, the space in both GR and SR was assumed to be manifold, of which obay local euclidean geometry, and Pythagorean theorem holds on euclidean space. Even in curved space, the differential form of Pythagorean theorem holds i.e. $ds^2=dx^2+dy^2+dz^2$(only 3 space), which was essentially how they calculated the expression through Pythagorean in SR. Quote Ben's "negligible on sufficiently short distance scales", in infinitesimal, they were "equal".

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