Note 7. Proof of Lemma 3

I have already shown that , since events in have zero hyperbolic interval with both , and is the orthogonal bisector plane. It will be enough to show that , and then argue from symmetry to prove the second statement.

There is no loss of generality if we assume that that is the origin and . Since , we may construct the linear isomorphism (analogous to the Euclidean inversion across a plane)

Clearly, leaves points in -- events Z such that -- fixed. Also, , , and . Actually, is simply the "hyperbolic inversion" across the plane orthogonal to .

A straightforward calculation shows that

so preserves the hyperbolic interval. Thus, it is clear that maps the light cone at the origin -- events such that -- into itself.

Further, it maps the light cone at to the light cone at : If , then

since

so

This proves that for , since leaves invariant and maps .

That is, if , then , but

Fot the second part of the proof, we consider the cases depending on the sign of .

Suppose that = . Assuming again that is the origin and that , this implies that . Then we want to verify that the intersection of the plane

with the light cone at ,

is an ellipse.

Suppose that has coordinates . And suppose we represent the coordinates of as . Then we are assuming that . If is in the light cone at , then the argument is similar to the one we made in Planes Intersecting Cones for the tangent plane to the hyperboloid

and

Therefore

and on substitution this becomes

or

The coefficients of and are both positive, and the discriminant

is clearly negative, since .

The argument for the hyperbolic case is similar. ¨