# Curvature of reciprocal Euclidean space

1. Jul 1, 2004

### Loren Booda

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 $$1/ \gamma$$?

2. Jul 2, 2004

### robphy

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?

3. Jul 2, 2004

### Loren Booda

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?