# Need an explanation for Null Vectors

1. Sep 8, 2014

### cy6nu5

So I have an issue understanding how to compute a matrix using the Minkowski metric as a null (read light-like) spacetime vector.
As best I can understand it, it is a vector which has all velocity in the spacial components and none in time.

Also, would a vector that transcends the speed of light mean that you are traveling through negative time? I don't really get how I should interpret a vector that is inside the light cone.

So I have
[-1 0 0 0]
[0 1 0 0 ]
[0 0 1 0 ]
[0 0 0 1 ]
for my matrix denoting the Minkowski metric (probably using wrong terminology)
what would a vector that determines a null vector be?

2. Sep 9, 2014

### sweet springs

Definition of Null Vevtor

$$\Lambda_{\mu\nu}$$ be your matrix, null vector $$x^\mu$$ satisfies the relation

$$x^\mu\Lambda_{\mu\nu} x^\nu=\Lambda_{\mu\nu} x^\nu x^\mu=0$$

Best

3. Sep 9, 2014

### pervect

Staff Emeritus
I'm not sure why you said velocity?, I hope I haven't misunderstood the question as a consquence of assuming you meant component.

A vector with only spacelike components and a zero time component would be a space-like vector, not a null vector.

A vector inside the lightcone would also be a spacelike vector, as in your previous example.

So I have
[-1 0 0 0]
[0 1 0 0 ]
[0 0 1 0 ]
[0 0 0 1 ]
for my matrix denoting the Minkowski metric (probably using wrong terminology)
what would a vector that determines a null vector be?[/QUOTE]

An example of a null vector for your metric (which is Minkowskii) would be [1,0,0,1].
If we let X be your vector, with the components of your vector be $x^0, x^1, x^2, x^3$ , and the nonzero components of your matrix (as above) being written in the following notation $g_{00} = -1, g_{11}=g_{22}=g_{33}=1$, then the length of your vector is

$$\sum_{\mu=0..3} \sum_{\nu=0..3} g_{\mu\nu} x^\mu x^\nu$$

which for Minkowskii space is equivalent to

$-(x^0)^2 + (x^1)^2 + (x^2)^2 + (x^3)^2$

and a null vector is just a vector with a length of 0, such as the vector with $x^0 = x^1 = 1$ mentioned previously.

4. Sep 9, 2014

### Meir Achuz

No. It has a time-like component that equals the magnitude of the space-like component.