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

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?

• $v$ is a null coordinate because the vector $\partial_v$ is null. This means that worldlines along which only $v$ is changing are null. – Prahar Nov 6 '18 at 2:42

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.

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.