Proof that two timelike vectors cannot be orthogonal

[Arman777]

For fun, I decided to prove that two timelike never can be orthogonal. And for this, I used the Cauchy Inequality for that. Such that

The timelike vectors defined as,

$$g(\vec{v_1}, \vec{v_1}) = \vec{v_1} \cdot \vec{v_1} <0$$
$$g(\vec{v_2}, \vec{v_2}) = \vec{v_2} \cdot \vec{v_2} <0$$

And the ortogonality is,

$$g(\vec{v_1}, \vec{v_2}) = \vec{v_1} \cdot \vec{v_2} = 0$$

where $g(\vec{e}_{\mu}, \vec{e}_{\nu})=g_{\mu \nu} = \eta_{\mu \nu} = diag(-1,1,1,1)$

So I tried something like,

$\vec{v_1} \cdot \vec{v_1} = -(v_1^0)^2 + (v_1^1)^2 + (v_1^2)^2 + (v_1^3)^2 < 0$
So
$$(v_1^0)^2 > (v_1^1)^2 + (v_1^2)^2 + (v_1^3)^2 = V_1^2 ~~(1)$$
Similarly
$$(v_2^0)^2 > (v_2^1)^2 + (v_2^2)^2 + (v_2^3)^2 = V_2^2 ~~(2)$$
And

$\vec{v_1} \cdot \vec{v_2} = -(v_1^0v_2^0) + (v_1^1v_2^1) + (v_1^2v_2^2)+ (v_1^3v_2^3)$Then I multiplied $(1)$ and $(2)$ to get

$$(v_1^0)^2(v_2^0)^2 > V_1^2 V_2^2$$

However if we stated that two timelike vectors are orthogonal we would get,

$$(v_1^0)^2(v_2^0)^2 = (V_1 \cdot V_2)^2$$

So since this contradicts the Cauchy-Schwarz Inequality we can say that two-timelike vectors cannot be orthogonal. Is there any more elegant proof that you know?

 

[PeroK]

Consider a frame of reference in which one of the vectors has zero spatial components.

 

[George Jones]

For fun, I decided to prove that two timelike never can be orthogonal.

How do you prove an axiom?

Is there any more elegant proof that you know?

This property is part of an elegant, basis-independent definition of Minkowski (vector) space.

Minkowski spacetime $\left( V,g \right)$ is a 4-dimensional real vector space $V$ together with a symmetric, non-degenerate, bilinear mapping $g:V\times V\rightarrow\mathbb{R}$. A vector in $V$ is called a 4-vector, and a 4-vector $v$ is called timelike if $g\left(v,v\right) < 0$, lightlike if $g\left(v,v\right) = 0$, and spacelike if $g\left(v,v\right) > 0$.

$\left( V,g \right)$ is such that:

1) timelike vectors exist;
2) if $u$ and $v$ are such $u$ is timelike and $g\left(u,v\right) = 0$, then $v$ is spacelike.

 

[Arman777]

How do you prove an axiom?

I did not know that this was an axiom. Logically indeed it makes sense.

 

[PeterDonis]

$\left( V,g \right)$ is such that:

1) timelike vectors exist;
2) if $u$ and $v$ are such $u$ is timelike and $g\left(u,v\right) = 0$, then $v$ is spacelike.

Can 2), or perhaps something weaker such as the OP's proposition, that if $u$ and $v$ are timelike, $g\left(u,v\right) \neq 0$, not be proven from 1) and the other properties that are given?

 

[George Jones]

Can 2), or perhaps something weaker such as the OP's proposition, that if $u$ and $v$ are timelike, $g\left(u,v\right) \neq 0$, not be proven from 1) and the other properties that are given?

I am not sure what "weaker" means here. The original post uses an implicit definition of Minkowski space (given below) that is equivalent to the definition of Minkowksi space that I gave in post #3, i.e., each implies the other.

Minkowski spacetime $\left( V,g \right)$ is a 4-dimensional real vector space $V$ together with a symmetric, non-degenerate, bilinear mapping $g:V\times V\rightarrow\mathbb{R}$ such there exists a basis $\left\{e_0 , e_1, e_2 , e_3 \right\}$ for $V$ with:

a) $g\left(e_\mu , e_\nu \right) = 0$ when $\mu \ne \nu$;
b) $1 = -g\left(e_0 , e_0 \right) = g\left(e_1 , e_1 \right) = g\left(e_2 , e_2 \right) = g\left(e_3 , e_3 \right)$.

Given the definition of Minkowski space that I gave in post #3, it is possible to prove (using something like Gram-Schmidt) the existence of basis with poperties a) and b). Given the definition of Minkowski space in this post, it is possible to prove (using Cauchy-Schwarz) property 2) in post #3.

I did not know that this was an axiom. Logically indeed it makes sense.

Well, I used a smile in postc #3, because, in that post, I used something like 'bait and switch', i.e., I changed the definition of Minkowski space from the (perfectly fine) definition that you used in the original post.

 

[robphy]

1) timelike vectors exist;
2) if $u$ and $v$ are such $u$ is timelike and $g\left(u,v\right) = 0$, then $v$ is spacelike.

I think $v$ has to be nonzero.

 

[George Jones]

I think $v$ has to be nonzero.

Yes, this needed to be stated explicitly.

 

[vanhees71]

How do you prove an axiom?

It's not an axiom but follows from the signature (3,1) of the fundamental form of Minkowski space.

 

[George Jones]

It's not an axiom but follows from the signature (3,1) of the fundamental form of Minkowski space.

It is an axiom in in the basis-free formulation that for the signature condition that I gave in post #3. See my post #6.