# Lorentz product on hyperboloid

1. Oct 30, 2016

### Woolyabyss

1. The problem statement, all variables and given/known data
(a) Let p(t) be a smooth curve in the hyperboloid S = {(x, y, z) ∈ R3 |z^2 − x^2 − y^2 = 1, z > 0}. Prove that p(t) .L p'(t) = 0, where .L is the Lorentz product.
(b) Prove that any non-zero tangent vector to S has positive (Lorentz)length. (Hint: Use the Cauchy-Schwarz inequality for the dot product.)

2. Relevant equations
x .L y = x1y1+x2y2-x3y3 (the lorentz product)

3. The attempt at a solution
(a)
Let p(t) = (a(t),b(t),c(t))
c(t) =(a^2+b^2+1)^.5
p'(t) = (a',b',.5*(2aa'+2bb')*(a^2+b^2+1)^-.5)

p(t) .L p'(t) = aa'+bb'-((a^2+b^2+1)^.5)*(aa'+bb')*((a^2+b^2+1)^-.5)
=aa'+bb'-aa'-bb'=0

(b)
|<p,p'>| <= ||p||*||p'||
from (a)
0 <= ||p||*||p'||
and we know p' is non zero so p must also
so 0 < ||p||*||p'||
now ||p||^2 = p .L p = aa+bb -(aa+bb+1) = -1

By doing the lorentz product of p with itself I was hoping to show that ||p|| > 0
and hence that ||p'|| must also be greater than 0 to satisfy the inequality, but it seems I've instead shown that ||p|| = -i.

Any help would be appreciated.

2. Nov 5, 2016