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. ¨