Tetrad for Schwarzschild metric, in terms of e

Difficulty level:   ★ ★ ★

The following is a natural choice of orthonormal tetrad for an observer moving radially in Schwarzschild spacetime with “energy per unit rest mass” e:

    \begin{align*} e_{\hat 0}^\mu &= \left(e\Schw^{-1},\pm\eroot,0,0\right) \\ e_{\hat 1}^\mu &= \left(\pm\Schw^{-1}\eroot,e,0,0\right) \\ e_{\hat 2}^\mu &= \left(0,0,\frac{1}{r},0\right) \\ e_{\hat 3}^\mu &= \left(0,0,0,\frac{1}{r\sin\theta}\right) \end{align*}

The components are given in Schwarzschild coordinates. (The ± signs are not independent — they must be either both +1 or both -1. Note that e does not distinguish between inward and outward motion. There is additional freedom to define any of these vectors as their negative.)

We normally think of e as invariant, where there is a presumption of freely falling / geodesic motion, but even if not we can regard it as an instantaneous value.

\evec{0} is the 4-velocity computed previously. The other vectors can be obtained from substituting \gamma=e\Schw^{-1/2} and V=-\frac{1}{e}\sqrt{e^2-1+\frac{2M}{r}} into the tetrad here. \gamma is determined from -\fvec u\cdot\fvec u_{\rm obs}=\gamma and the equation for e above, then V follows from inverting \gamma\equiv(1-V^2)^{-1/2}. This orthonormal frame is useful for determining the object’s perspective, e.g. tidal forces, visual appearances, etc.

Tetrad for Schwarzschild metric

Difficulty level:   ★ ★ ★

Suppose an observer u moves radially with speed (3-velocity) V relative to “stationary” Schwarzschild observers, where we define V<0 as inward motion. Then one natural choice of orthonormal tetrad is:

    \begin{align*} (\evec{0})^\alpha &= \left(\gamma\Schw^{-1/2},V\gamma\Schw^{1/2},0,0\right) \\ (\evec{1})^\alpha &= \left(V\gamma\Schw^{-1/2},\gamma\Schw^{1/2},0,0\right) \\ (\evec{2})^\alpha &= \left(0,0,\frac{1}{r},0\right) \\ (\evec{3})^\alpha &= \left(0,0,0,\frac{1}{r\sin\theta}\right) \end{align*}

where the components are given in Schwarzschild coordinates. This may be derived as follows.

The Schwarzschild observer has 4-velocity

    \[\fvec u_{\rm obs}=\left(\Schw^{-1/2},0,0,0\right)\]

because the spatial coordinates are fixed, and the t-component follows from normalisation \fvec u_{\rm obs}\cdot\fvec u_{\rm obs}=-1 (Hartle §9.2).

Now the Lorentz factor for the relative speed satisfies -\fvec u\cdot\fvec u_{\rm obs}=\gamma, and together with normalisation \fvec u\cdot\fvec u=-1 and the assumption that the θ and φ components are zero, this yields \evec{0}\equiv\fvec u given above.

We obtain \evec{1} by orthonormality: \evec{1}\cdot\evec{0}=0 and \evec{1}\cdot\evec{1}=1, and again making the assumption the θ and φ components are zero. Note the negative of the r-component is probably an equally natural choice. Then \evec{2} and \evec{3} follow from simply normalising the coordinate vectors.

Strictly speaking this setup only applies for r>2M, because stationary timelike observers cannot exist inside a black hole event horizon! Yet remarkably the formulae can work out anyway (MTW …) .  An alternate approach is local Lorentz boost described shortly.

Hartle … Also check no “twisting” etc…

Radial motion in the Schwarzschild metric, relative to stationary observers

Difficulty level:   ★ ★ ★

Last time we derived the 4-velocity u of a small test body moving radially in the Schwarzschild geometry, in terms of e, the “energy per unit rest mass”. Another parametrisation is in terms of the 3-speed V relative to stationary observers. This turns out to be, in Schwarzschild coordinate expression,


To derive this, first consider the 4-velocity of stationary observers:


We know the “moving” body has 4-velocity u of form u^\mu=(u^t,u^r,0,0) since the motion is radial. The Lorentz factor \gamma\equiv(1-V^2)^{-1/2} for the relative speed is

    \[\gamma=-\fvec u\cdot\fvec u_{\rm{Schw}}\]

Evaluating and rearranging yields u^t=\gamma\Schw^{-1/2}. Normalisation \fvec u\cdot\fvec u=-1 leads to u^r=\pm V\gamma\Schw^{1/2}, after some algebra including use of the identity \gamma^2-1=V^2\gamma^2. We allow V<0 also, and define this as inward motion. Carefully considering the sign, this results in the top equation. (An alternate derivation is to perform a local Lorentz boost. Later articles will discuss this… The Special Relativity formulae cannot be applied directly to Schwarzschild coordinates.)

Some special cases are noteworthy. For V=0, γ=1, and u reduces to uSchw. This corresponds to e=-\fvec\xi\cdot\fveclabel{u}{Schw}=\Schw^{1/2}. Also we can relate the parametrisation by V (and γ) to the parametrisation by e via

    \[\gamma=e\Schw^{-1/2}\qquad V=-\frac{1}{e}\sqrt{e^2-1+\frac{2M}{r}}\]

where the leftmost equation follows from the definition e\equiv-\fvec\xi\cdot\fvec u, and subsequently the rightmost equation from γ=γ(V). For raindrops with e=1, the relative speed reduces to V=-\Schwroot.

We would expect the construction to fail for r\le 2M, as stationary timelike observers cannot exist there, and so the relative speed to them would become meaningless. But curiously, it can actually work for a faster-than-light V>1 “Lorentz” boost, as even the authorities MTW (§31.2, explicit acknowledgement) and Taylor & Wheeler (§B.4, implicitly vrel>1 for r<2M) attest. Sometime, I will investigate this further…

Radial motion in the Schwarzschild metric, in terms of e

Difficulty level:   ★ ★ ★

A nice way to parametrise the 4-velocity u of a small test body moving radially within Schwarzschild spacetime is by the “energy per unit rest mass” e:


For the “±” term, choose the sign based on whether the motion is inwards or outwards. All components are given in Schwarzschild coordinates (t,r,\theta,\phi). The result was derived as follows. In geometric units, the metric is:


By definition e\equiv-\fvec\xi\cdot\fvec u, where \fvec\xi\equiv\partial_t is the Killing vector corresponding to the independence of the metric from t, and has components \xi^\mu=(1,0,0,0) (Hartle §9.3). For geodesic (freefalling) motion e is invariant, however even for accelerated motion e is well-defined instantaneously and makes a useful parametrisation.

We want to find u^\mu=(u^t,u^r,u^\theta,u^\phi) say. Rearranging the defining equation for e gives u^t=e\Schw^{-1}. Radial motion means u^\theta=u^\phi=0, so the normalised condition \fvec u\cdot\fvec u=-1 yields the remaining component \abs{u^r}. The resulting formula is valid for all 0<r\ne 2M, and for e=1 the 4-velocity describes “raindrops” as expected.

Relative speed

Difficulty level:   ★ ★ ★

Suppose two observers at the same place and time (that is, “event”) move with 4-velocities u and v respectively, then they measure their relative speed as follows. The Lorentz factor is simply

    \[\gamma=-\fvec u\cdot\fvec v\]

(The dot is not the Euclidean dot product, but uses the metric: g_{\alpha\beta}u^\alpha v^\beta where the indices \alpha and \beta are summed over by the Einstein summation convention.) The proof is based on the axiom that some local inertial frame exists, although interestingly one does not need to explicitly construct it.

The relative 3-speed V, may then be recovered via:


See for instance Carroll (end of §2.5) who terms it “ordinary three-velocity”. Other sources express the first formula more indirectly, in terms of the energy and momentum measured by an observer \fvec u: E=-\fvec u\cdot\fvec p where \fvec p=m\fvec v is the 4-momentum of another observer/object, and combine this with E=m\gamma (MTW Exercise 2.5 in §2.8 term it “ordinary velocity”, or Hartle §5.6, and Example 9.1 in §9.3).


Difficulty level:   ★ ★ ★

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