# How to conclude a vector is space-like?

I'm working with the Lorenztian inner product and would like to show that if a vector $v$ is lightlike, so $\langle v,v\rangle =0,$ and if $\langle v,w\rangle =0,$ then either $w$ is spacelike or $w$ is proportional to $v.$

So far I have that since $\langle v,v\rangle =0,$ then $$v_0^2 = v_1^2 +v_2^2 +v_3^3$$ and since $\langle v,w\rangle =0,$ then $$v_0w_0 = v_1w_1 + v_2w_2 + v_3w_3.$$ I think I can then conclude that since \begin{align*} v_0v_0 &= v_1v_1 +v_2v_2 +v_3v_3, \\ v_0w_0 &= v_1w_1 + v_2w_2 + v_3w_3 \end{align*} then $v_0=w_0,v_1=w_1, v_2=w_2, v_3=w_3.$ Does this mean that $v$ and $w$ are proportional? I'm not sure how to conclude that $w^a$ could be also be spacelike from the supposition.

• What is the difference (if any) between $v$ and $v^a$?
– user130529
Jan 29 '17 at 20:32
• There isn't a difference between $v$ and $v^a.$ I just didn't use the index notation inside the inner product. Jan 29 '17 at 20:35
• Hint: you can suppose that $v_2=v_3=0$ (by choosing an appropriate frame). Show then that $-w_0^2+w_1^2 = 0$, and conclude that $w$ is spacelike or lightlike.
– user130529
Jan 29 '17 at 21:09
• @claude chuber : I think that you must post your comment as an answer. Jan 29 '17 at 21:13
• If you understand any right answer to your question, don't be afraid to accept this as best upvoting it. Jan 29 '17 at 21:25

Hint: you can suppose that $v_2=v_3=0$ (by choosing an appropriate frame). Show then that $−w_0^2+w_1^2=0$, and conclude that $w$ is spacelike or lightlike.
• For $w$ to be spacelike, doesn't $-w_0^2 + w_1^2 >0?$ I thought that if it is equal to zero, then it is lightlike. Jan 29 '17 at 21:34
• @Setsss : $$-w_0^2+w_1^2=0 \tag{01}$$ $$w_2^2+w_3^2\ge0 \tag{02}$$ so $$-w_0^2+w_1^2+w_2^2+w_3^2\ge0 \tag{03}$$ Jan 29 '17 at 22:13