How can the $v$ coordinate be null if $g_{vv}\neq 0$?

Question

I'm probably missing something very basic here. As far as I know, a coordinate is called null when its coordinate lines are null. This means that if $(M,g)$ is spacetime and $x^\mu$ a coordinate chart, the coordinate $x^\nu$ is null when $\partial_\nu$ is null. This means

$$g(\partial_\nu,\partial_\nu)=g_{\nu\nu}=0.$$

Now consider the Vaidya metric $$ds^2=-\left(1-\dfrac{2M(v)}{r}\right)dv^2+2dvdr+r^2(d\theta^2+\sin^2\theta d\phi^2)$$

People call the coordinate $v$ a null coordinate, but we clearly have $g_{vv}\neq 0$ in general.

The same happens in Minkowski spacetime in advanced coordinates $$ds^2=-dv^2+2dvdr+r^2(d\theta^2+\sin^2\theta d\phi^2).$$

Again $g_{vv}\neq 0$ and still the coordinate is called null.

So how can $v$ be a null coordinate when $g_{vv}\neq 0$? What am I missing here?

Answer 1

1. Definition: A function $f$ is called time-like, null/light-like, space-like on a pseudo-Riemannian manifold $(M,g)$ iff the covector $\mathrm{df}$ is time-like, null/light-like, space-like, respectively, i.e. it depends on the sign of$^1$ $$g^{ff}=g((\mathrm{df})^{\sharp},(\mathrm{df})^{\sharp}).$$ In plain English: It depends on the sign of the $ff$-component $g^{ff}$ of the inverse metric.

2. One reason why a partial derivative $\frac{\partial}{\partial f}$ [and therefore $g_{ff}=g(\frac{\partial}{\partial f},\frac{\partial}{\partial f})$] is not used is that it would depend on how the function $f$ is completed into a local coordinate system $(f, x^2,\ldots, x^n)$, i.e. it would depend on the choice of the remaining coordinates $( x^2,\ldots, x^n)$.

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$^1$ Here $\sharp$ denote the sharp map.

Answer 2

In the case of Minkowski space time, start with the line-element in the form

$$ ds^2 = -c^2dt^2+dr^2 + r^2(d\theta^2+\sin^2\theta\,d\phi^2)\,. $$

Consider a light ray which travels radially. In this case $\theta = const., \phi = const\,.$

This reduces the line element to

$$ ds^2 = -c^2dt^2+dr^2\,.$$

Now let $v = ct + r\,.$ The co-ordinate $v$ is null because only light travels along lines of constant $v$. To see this let $v=0$ and draw a Minkowski diagram in the co-ordinates $(ct,r).$

In terms of the coordinate $v$ the line-element reads $ds^2=-cdt^2+dr^2 = -dv^2+2dvdr\,.$

One can make the same substitution in the line-element of Minkowski space-time. In the case of the line-element describing the Vaidya space-time, $v$ is null for the same general reason.

I think Qmechanic's answer https://physics.stackexchange.com/q/439217 is technical reasoning behind my answer.