Coordinates adapted to observer 4-velocity field

Difficulty: ★★★☆☆ undergraduate

Suppose you have a 4-velocity field $\mathbf u$ , which might be interpreted physically as observers or a fluid. It may be useful to derive a time coordinate $T$ which both coincides with proper time for the observers, and synchronises them in the usual way. Here we consider only the geodesic and vorticity-free case. Define:

$dT := -\mathbf u^\flat.$

The “flat” symbol is just a fancy way to denote lowering the index, so the RHS is just $-u_\mu$ . On the LHS, $dT$ is the gradient of a scalar, which may be expressed using the familiar chain rule:

$dT = \frac{\partial T}{\partial x^0}dx^0 + \frac{\partial T}{\partial x^1}dx^1 + \cdots,$

where $x^\mu$ is a coordinate basis. Technically $dT$ is a covector, with components $(dT)_\mu = \partial T/\partial x^\mu$ in the cobasis $dx^\mu$ . Similarly $-\mathbf u^\flat = -u_0dx^0 -u_1dx^1 -\cdots$ , so we must match the components: $\partial T/\partial x^\mu = -u_\mu$ . For our purposes we do not need to integrate explicitly, it is sufficient to know the original equation is well-defined. (No such time coordinate exists if there is acceleration or vorticity, which is a corollary of the Frobenius theorem, see Ellis+ 2012 book §4.6.2.)

The new coordinate is timelike, since $\langle dT,dT\rangle = \langle -\mathbf u^\flat,-\mathbf u^\flat\rangle = -1$ . One can show its change with proper time is $dT/d\tau = \langle dT,\mathbf u\rangle = 1$ . Further, the $T = \textrm{const}$ hypersurfaces are orthogonal to $\mathbf u$ , since the normal vector $(dT)^\sharp$ is parallel to $\mathbf u$ . This orthogonality means that at each point, the hypersurface agrees with the usual simultaneity defined locally by the observer at that point. (Orthogonality corresponds to the Poincaré-Einstein convention, so named by H. Brown 2005 book §4.6).

We want to replace the $x^0$ -coordinate by $T$ , and keep the others. What are the resulting metric components for this new coordinate? (Of course it’s the same metric, just a different expression of this tensor.) Notice the original components of the inverse metric satisfy $g^{\mu\nu} = \langle dx^\mu,dx^\nu\rangle$ . Similarly one new component is $g'^{TT} = \langle dT,dT\rangle = -1$ . Also $g'^{Ti} = \langle dT,dx^i\rangle = -u^i$ , where $i = 1,2,3$ . The $g'^{iT}$ are the same by symmetry, and the remaining components are unchanged. Hence the new components in terms of original components are:

$g'^{\mu\nu} = \begin{pmatrix} -1 & -u^1 & -u^2 & -u^3 \\ -u^1 & g^{11} & g^{12} & g^{13} \\ -u^2 & g^{21} & g^{22} & g^{23} \\ -u^3 & g^{31} & g^{32} & g^{33} \end{pmatrix}.$

The matrix inverse gives the new metric components $g'_{\mu\nu}$ . The 4-velocity components are: $u'_\mu = (-1,0,0,0)$ by the original equation. Also $u'^T = \langle dT,\mathbf u\rangle = 1$ , and the $u'^i = \langle dx^i,\mathbf u\rangle = u^i$ are unchanged. Hence $u'^\mu = (1,u^1,u^2,u^3)$ .

Anecdote: I used to write out $dT = -u_0dx^0 - u_1dx^1 - \cdots$ , rearrange for $dx^0$ , and substitute it into the original line element. This works but is clunky. My original inspiration was Taylor & Wheeler 2000 book §B4, and I was thrilled to discover their derivation of Gullstrand-Painlevé coordinates from Schwarzschild coordinates plus certain radial velocities. (I give more references in MacLaurin 2019 research paper (arXiv) §3.) I imagine that if a textbook presented the material above — given limited space and more formality — it may seem as if the more elegant approach were obvious. However I only (re?)-discovered it today by accident, using a specific 4-velocity from the previous post, and noticing the inverse metric components looked simple and familiar…

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