Null vector help!

[quantum123]

I read this in the notes:

Show that any vector that is orthogonal to a null vector must be either be:-
i) parallel to a null vector
ii) space-like

How??

[robphy]

Should this be in the howework section?
What's your starting point? What does it mean for two vectors to be orthogonal?

[quantum123]

Orthogonal means normal, I guess.
This is no homework. I am doing it as a hobby.
This comes out from GR notes describing the light cone, and a general description of the four linearly independent vectors that may or may not lie parallel to the light cone, which is a null surface, ie may or may not be null vectors, and the properties of the vectors that are not null vectors.

[George Jones]

I read this in the notes:

Show that any vector that is orthogonal to a null vector must be either be:-
i) parallel to a null vector
ii) space-like

How??

Let {e_0, e_1, e_2, e_3} be an arbitary orthonormal basis. Then, up to a constant multiple,

n = e_0 + e_1

is an arbitrary null vector. Let

v = v^0 e_0 + v^1 e_1 + v^2 e_2 + v^3 e_3

be an arbitrary 4-vector. If n and v are orthogonal, what does this give you?

[quantum123]

Thanks. I see.
<n,n>^2=-1^2+1^2=0
<n,v>=0 => =-v^0+v^1=0 => V^1=v^0
Therefore, <v,v>^2 = -v^0^2 + v^1^2 + V ^2^2 + v ^3^2 = -v^0^2 + v^0^2 + V ^2^2 + v ^3^2 = V ^2^2 + v ^3^2 >=0
If =0 , then null-like.
If >0 then space-like.
Correct?
But why can n = e_0 + e_1 for any arbitrary n such that <n,n>=0 where {e^v} is orthonormal basis?

[George Jones]

Thanks. I see.
<n,n>^2=-1^2+1^2=0
<n,v>=0 => =-v^0+v^1=0 => V^1=v^0
Therefore, <v,v>^2 = -v^0^2 + v^1^2 + V ^2^2 + v ^3^2 = -v^0^2 + v^0^2 + V ^2^2 + v ^3^2 = V ^2^2 + v ^3^2 >=0

Yes.

If =0 , then null-like.

For this case, there is one other, trivial, possibility. :smile:

But why can n = e_0 + e_1 for any arbitrary n such that <n,n>=0 where {e^v} is orthonormal basis?

Consider an arbitrary vector in a 2-dimensional spatial plane.

We are free to choose a basis that helps us. For example, we might choose: a basis such that the vector is in the direction of e_1; a basis such that the vector is in the direction of e_2; a basis such that the vector is halfway between e_1 and e_2, i.e., in the direction e_1 + e_2.

None of these choices restricts us.

[quantum123]

Thank you so much.
I guess you mean a vector subspace.
If we have 4 orthonormal vectors {e} that span a vector space, we can always find a 2 -D subspace which contains n, spanned by orthonormal basis f1, f2.
Hence if n is null, then <n,n>=0 => -n0^2 + n1^2 =0 => n1=n2.
And should be able to find f3 and f4 to complete the {f} for the total vector space T(M).