Curvature of reciprocal Euclidean space

Loren Booda
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A triangle in Euclidean space can be described as having a hypotenuse of one, and legs of Lorentz parameters \beta and \gamma. What spatial curvature underlies a triangle with hypotenuse one, and legs 1/ \beta and 1/ \gamma?
 
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Loren Booda said:
A triangle in Euclidean space can be described as having a hypotenuse of one, and legs of Lorentz parameters \beta and \gamma.

If so, then 1^2=\beta^2+\gamma^2.
Since \gamma^2=\frac{1}{1-\beta^2}, you get 1-\beta^2=\frac{1}{1-\beta^2}. Then, \beta=0 and \gamma=1.
Am I misunderstanding something?
 
robphy,

Mea culpa.

Thank you for the reminder: \beta=v/c and \gamma=1/ \sqrt{1-v^2/c^2}

Revised: A triangle in Euclidean space can be described as having a hypotenuse of one, and legs of Lorentz parameters beta and 1/gamma. What spatial curvature underlies a triangle with hypotenuse one, and legs of 1/beta and gamma?
 
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