- #1

cianfa72

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*[Moderator's note: Spin off from another thread due to topic change.]*

I was thinking about the following: can we take as a basis vector a null (i.e. lightlike) vector to write down the metric ?

Call ##v## such a vector and add to it 3 linear independent vectors. We get a basis for the tangent space (the first vector in the basis is the ##v## vector itself).

Then in such a basis the metric coefficient ##g_{00}## should be zero since the vector ##v## has coordinate ##(1,0,0,0)## and by definition it has zero length. Hence we are not allowed to use it, I believe.