Null vector help!

1. Jan 4, 2007

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??

2. Jan 4, 2007

robphy

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

3. Jan 4, 2007

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.

4. Jan 4, 2007

George Jones

Staff Emeritus
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?

5. Jan 4, 2007

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?

6. Jan 5, 2007

George Jones

Staff Emeritus
Yes.

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

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.

7. Jan 5, 2007

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).