Coordinates adapted to observer 4-velocity field

Difficulty:   ★★★☆☆   undergraduate general relativity

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 §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 §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 §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  §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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