Einstein's famous mass-energy equivalence is also a Pythagorean relationpic.twitter.com/JtXG4VipBD
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What would the mass-energy relation look like on a curved surface?
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Well, I guess it's basically the same since at the point you're measuring these quantities everything is locally flat. Although splitting into E and p is less clear because it's observer (coordinate) dependent, right... So
Of course the lengths are measured with the metric 1/2 -
and all bets are off as soon as you want to want to include gravitational field energy. Incidentally, I recently heard someone explain special relativity as "a minus sign in Pythagoras" and this triangle is a fun way to make your eye twitch at first, but then view it this way 2/2
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It should be exactly the same, as the mass-energy relation is a relation on the norm of (co-)vectors in the (co-)tangent space of the manifold, which is always isomorphic to Minkowski space in appropriate coordinates.
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Momentum 4 vector (E,p). In an object's rest frame, linear momentum,p, is 0 and E is the same as mc^2. So magnitude of the 4 vector is -mc^2. Since a 4 vector is a tensor, it's magnitude is invariant in any coordinates and hence the relation -E^2 +p^2 = -mc^2.
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