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I thought the spacetime diagrams were usual flat planes just with a new formula for distance. But the source I was reading suddenly started talking about hyperbolic angles in triangles ( I don't know what hyperbolic angles are). The triangles in those spacetime diagrams look like normal triangles (with straight edges), but the source says the inside angles are hyperbolic. Does that mean the triangles are actually residing on a hyperbolic surface and have curved edges?

Do I have to learn hyperbolic geometry to understand it?

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  • $\begingroup$ But the source I was reading... What source? $\endgroup$
    – user4552
    Jun 27, 2019 at 13:18
  • $\begingroup$ You have a misconception of what hyperbolic geometry is. $\endgroup$
    – safesphere
    Jun 28, 2019 at 3:14

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Do I have to learn hyperbolic geometry to understand it?

No

The triangles in those spacetime diagrams look like normal triangles (with straight edges), but the source says the inside angles are hyperbolic.

That seems to me a bad choice of wording. My remote viewing superpowers do not work unless I have the full name of the source. Nevertheless, those triangles are not 'normal' triangles. If they were normal triangles, the Pythagorean theorem would apply to them. Instead, the square of the hypotenuse is the difference of the squares of the sides.

Bottom line: do not sweat when some fancy professor enrolled in the genius protection program begins throwing obscure nomenclature in their notes. Usually, they do it when they are insecure about what they know and they want to impress.


'Hyperbolic Pythagorean Theorem' lulz. Fancy nomenclature to obscure simple concepts. No, the triangles are not curved. They are not normal in the sense that $c^2=a^2+b^2$ does not apply to them. The angles are hyperbolic in the sense that they will be fed to hyperbolic sine, cosine, tangent, etc. … you know the sinh, cosh, tanh in your calculator.

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  • $\begingroup$ web.csulb.edu/~wziemer/Papers/specialrelativity.pdf It's not on 13 and 14 page of this source. $\endgroup$
    – Ryder Rude
    Jun 27, 2019 at 11:26
  • $\begingroup$ I just thought the triangles were normal, but "distance" was defined differently, so the Pythagorean theorem wasn't applicable to this new definition of distance. Are the triangles really curved? $\endgroup$
    – Ryder Rude
    Jun 27, 2019 at 11:28
  • $\begingroup$ If this is your first time learning about Special Relativity and you want to become a Physicist, that 'Geometric Introduction To Spacetime And Special Relativity' is not your best bet. Read the original 1905 Einstein paper. Pay attention to the details there. You will learn how Einstein approaches Physics. $\endgroup$
    – user234994
    Jun 27, 2019 at 11:44
  • $\begingroup$ Distance is (to me) the layman term for norm (as in Banach spaces). In Special Relativity, the 'distance' is not positive definite, not even positive semi-definite. $\endgroup$
    – user234994
    Jun 27, 2019 at 11:49
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The spacetime of Special Relativity is flat---not curved like hyperbolic geometry. So, the geometry of Special Relativity satisfies the parallel postulate. (Triangle sides are straight.)

However, since the "angles" between timelike vectors is based on the unit hyperbola (playing the role of a circle in Minkowski spacetime geometry), Special Relativity uses hyperbolic trigonometry.

So, you don't need to learn hyperbolic geometry to understand the fundamentals of Special Relativity. [Hyperbolic geometry might be helpful to understand effects like the Thomas procession or velocity-composition in different spatial directions. The "space of rapidities (which is related to velocities)" is a hyperbolic space.]

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