Prove dot product between two timelike vectors in Minkowski spacetimw [closed]

Question

In the Minkowski 4-dimensional space-time $(\mathbb{M}^4,\eta)$ the dot product is:

$a\cdot b = -a^0b^0 + a^1b^1 + a^2b^2 + a^3b^3 ~\qquad~ a = (a^0,a^1,a^2,a^3) ~~,~~ b = (b^0,b^1,b^2,b^3)$

Now consider 2 timelike vectors $a,b$ such that:

$a^2 = a\cdot a = -1$
$b^2 = b\cdot b = -1$

and

$-a\cdot b>0$

Then I need to prove that $-a\cdot b > 1$

I've tried everyhing and I can't find a proof for this :(

Answer 1

If $a$ is timelike, we can go to its rest frame, where $a^\mu = (1,0,0,0)$. We can then do a rotation to align $b$ with the $x$ axis, so that $b^\mu = (b^0, b^1, 0, 0)$ with $-(b^0)^2 + (b^1)^2 = -1$. In this frame, $a\cdot b = -b^0$, so $b^0 > 0$.

But then $b^0 = \sqrt{1+(b^1)^2}$, which is greater than one, which implies that $a\cdot b < -1$.