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Let us consider a Minkowski space of the form: $$ds^2 = -dt^2 + dx^2 + dy^2 +dz^2.$$

What would the linearly independent null vectors of this space be?

I am aware this is a trivial question but is something that has not been made clear to me and so causes me some confusion when reading some of the GR literature.

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  • $\begingroup$ The set of null vectors is not a vector space (it's the light cone), so it doesn't really make sense to ask for linearly independent vectors; there are infinitely many of them. $\endgroup$ – Javier Oct 20 at 22:48
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The light cone consists of all vectors $\mathbf V$ such that $\boldsymbol \eta(\mathbf V,\mathbf V) = 0$ (where $\boldsymbol \eta$ is the Minkowski metric). In Cartesian coordinates $(t,x,y,z)$, this means that

$$\boldsymbol \eta(\mathbf V,\mathbf V) = -(V^t)^2 + (V^x)^2+(V^y)^2+(V^z)^2 = 0$$

or $$V^t = \pm \sqrt{(V^x)^2+(V^y)^2+(V^z)^2}$$

As mentioned by Javier in the comments, the set of null vectors is not a vector space, so it's not clear what you mean when you ask for the linearly independent null vectors.

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    $\begingroup$ In particular if I add together (1, 1, 0, 0) and (1, -1, 0, 0) which are both null, I get (2, 0, 0, 0) which is timelike. One can create a "null tetrad", a basis of Minkowski space where all 4 basis four-vectors are null, if desired -- (1, 0, 0, -1), (1, 0, 0, 1), (1, 1, 0, 0), (1, 0, 1, 0) should be an example, Penrose-Newman formalisms instead choose (1, 0, 0, -1), (1, 0, 0, 1), (0, 1, i, 0) and (0, 1, -i, 0) but this is because they are happy for Minkowski space to be complexified. $\endgroup$ – CR Drost Oct 20 at 23:58

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