A Proving $g(u,v)≠0$ with Linear Independence

[isaacdl]

TL;DR

Contraction of a lightlike vector with another vector in a free coordinate scheme.

I'm trying to prove that there exist always a vector w whose contraction with a lightlike vector u (g(u,u)=0) it's always different from zero:
$g(u,v)≠0$I know how to do this with coordinates, but in a free cordinate scheme I'm totally lost.

Any help?

PD: Both vectors are linearly independent.

 

[Orodruin]

What have you tried so far?
How did you do it in a coordinate system?
Can you generalize that to work without coordinates?

 

[isaacdl]

In a coordinates it's straight forward. If u=(1,1,0,0) and w=(1,0,0,0) timelike or same to ligthlike, it's not difficult to prove they are colinear. But I don't know how to traduce it in a free coordinate way.

 

[ergospherical]

pseudo-Riemannian metrics are non-degenerate; can you argue by contradiction?

 

[isaacdl]

Mmm, I think I have to use the fact w and u are linearly independent. My first idea was to use the properties of bilinearity of the metric. So by contradiction you mean to consider g(u,v)=0 and get to u is not lightlike?

 

[ergospherical]

As in, if there were no $\mathbf{w}$ giving $g(\mathbf{u},\mathbf{w}) \neq 0$ that’s the same as saying $g(\mathbf{u},\mathbf{w}) = 0$ for all $\mathbf{w} \in \mathbf{R}^4$. How does that mesh with the non-degeneracy condition on $g$?

 

[robphy]

Consider the common plane for the two vectors…. Given the lightlike vector, can you write the “other vector” using it?

 

[isaacdl]

As in, if there were no $\mathbf{w}$ giving $g(\mathbf{u},\mathbf{w}) \neq 0$ that’s the same as saying $g(\mathbf{u},\mathbf{w}) = 0$ for all $\mathbf{w} \in \mathbf{R}^4$. How does that mesh with the non-degeneracy condition on $g$?

Ok, I was using proof by contradiction wrong. Ok, if $g(\mathbf{u},\mathbf{w}) = 0$ for all $\mathbf{w} \in \mathbf{R}^4$, by non-degenacy condition if S is non-degenerated u is a null vector u=0. But I'm not sure if the space S or equiv, g | S is non-degenerate. The only info I have is that $S=(v_1,...,u,..._w,...,v_m)$ is a linearly independent set.

What I have to find now is that $u$ is not lightlike, right?PD: I don't know why LATEX is not showing properly.

 

[isaacdl]

Consider the common plane for the two vectors…. Given the lightlike vector, can you write the “other vector” using it?

Yes, I can see it graphically, but my problem is to show it in a free cordinate way.

 

[ergospherical]

@isaacdl if $g(\mathbf{u}, \mathbf{w}) = 0$ for all possible $\mathbf{w}$ then the non-degeneracy of $g$ would imply that $\mathbf{u} = 0$, but this is not true by assumption ($\mathbf{u}$ is a non-zero lightlike vector).

 

[isaacdl]

@isaacdl if $g(\mathbf{u}, \mathbf{w}) = 0$ for all possible $\mathbf{w}$ then the non-degeneracy of $g$ would imply that $\mathbf{u} = 0$, but this is not true by assumption ($\mathbf{u}$ is a non-zero lightlike vector)

Yeah totally understood that part, but what I said in the previous post is that I don't know if g in S is non-degenerated. It's not given in the exercise.

PD: Wait, u are totally right, we can use it, it's given in a previous part of the exercise! Thanks a lot.

 

[PeterDonis]

Moderator's note: A subthread on a different topic has been moved to a separate thread:

https://www.physicsforums.com/threads/can-a-basis-vector-be-lightlike.1014841/