Lorentz product on hyperboloid

[Woolyabyss]

Homework Statement

(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.)

Homework Equations

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

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.

 

[mighty2000]

Thank you for your question. I have reviewed your solution and I believe there are a few errors in your calculations. Here is a corrected solution for both parts (a) and (b):

(a) Let p(t) = (a(t), b(t), c(t)) be a smooth curve in S. Then p'(t) = (a'(t), b'(t), c'(t)) is also a vector in S, since the tangent vector to S at any point must lie in the tangent plane to S at that point. Therefore, p(t) .L p'(t) = a(t)a'(t) + b(t)b'(t) - c(t)c'(t). Since c(t) = √(a(t)^2 + b(t)^2 + 1), we have c'(t) = (a(t)a'(t) + b(t)b'(t)) / √(a(t)^2 + b(t)^2 + 1). Substituting this into the expression for p(t) .L p'(t), we get:

p(t) .L p'(t) = a(t)a'(t) + b(t)b'(t) - (a(t)a'(t) + b(t)b'(t)) = 0.

(b) Let p'(t) = (a'(t), b'(t), c'(t)) be a tangent vector to S at p(t). Then ||p'||^2 = p' .L p' = (a'(t))^2 + (b'(t))^2 - (c'(t))^2. By the Cauchy-Schwarz inequality, we have:

||p'||^2 = (a'(t))^2 + (b'(t))^2 - (c'(t))^2 ≥ 0.

Since c'(t) = (a(t)a'(t) + b(t)b'(t)) / √(a(t)^2 + b(t)^2 + 1), we have:

(c'(t))^2 = (a(t)a'(t) + b(t)b'(t))^2 / (a(t)^2 + b(t)^2 + 1).

Substituting this into the expression for ||p'||^2, we get:

||p'||^2 = (a'(t))^2 + (b'(t))^2 - (a(t)a'(t) + b(t)b