Cauchy-Schwarz for two spacelike vectors

[bcrowell]

In Euclidean geometry, the Cauchy-Schwarz inequality is $|e\cdot f| \le |e||f|$. In a Minkowskian signature, the inequality is reversed for timelike vectors. Apparently for spacelike vectors it depends on whether the two vectors span the light cone:

http://math.stackexchange.com/questions/283073/cauchy-schwarz-for-metrics-with-arbitrary-signatures

Does anyone know a proof of this?

 

[martinbn]

You can use the idea of the "usual" proof for the Euclidean case. If $v$ and $w$ are space-like then for any real number $t$ consider the vector $x=v+tw$ and its inner product with itself. You have $(x,x)=(v+tw,v+tw)=|v|^2+2(v,w)t+|w|^2t^2$. The inequality holds if and only if the discriminant of the quadratic polynomial is negative, if and only if the polynomial has only positive values. So if the inequality hold if and only if the span of the two vectors consists of space-like vectors.

 

[bcrowell]

Thanks, martinbn! I'll have to work that out and make sure I understand it.

 

[bcrowell]

I have a discussion of Cauchy-Schwarz and triangle inequalities now in section 1.5 of my SR book, http://www.lightandmatter.com/sr/ . The case we discussed here is relegated to a homework problem, where I suggest the idea of martinbn's argument (with credit to martinbn) and ask the reader to carry it out.