Orthonormal Tetrads and Measured Components

An orthonormal tetrad replaces coordinate components by components in a local Lorentz frame. Its timelike leg can be chosen as an observer’s four-velocity, so contractions with the tetrad give the energy, spatial momentum, and tensor components resolved by that observer. The metric is reconstructed exactly from the coframe, while different tetrads related by Lorentz transformations describe the same geometry.

Required background. Pseudo-Riemannian Metrics and Signature supplies Lorentzian orthogonality and causal normalization; Affine Connections and Covariant Derivatives supplies the general connection and covariant-derivative framework, which this page applies to frame fields.

Helpful background. Calculus and Linear Algebra for Relativity supplies changes of basis and bilinear forms; Exterior Algebra and Differential Forms supplies the one-form and wedge algebra used for coframes. Killing Fields and Isometry Groups supplies the spacetime-isometry side of the distinction from a local frame change.

Orthonormal Lorentz frames

On a four-dimensional Lorentzian manifold, an orthonormal tetrad $e_{\hat a}$ and dual coframe $\theta^{\hat a}$ satisfy

$$g(e_{\hat a},e_{\hat b})=\eta_{\hat a\hat b}, \qquad \theta^{\hat a}(e_{\hat b})=\delta^{\hat a}{}_{\hat b},$$

where $\eta_{\hat a\hat b}=\mathrm{diag}(-1,1,1,1)$. Hatted indices identify frame components. In coordinates,

$$e_{\hat a}=e_{\hat a}{}^\mu\partial_\mu, \qquad \theta^{\hat a}=\theta^{\hat a}{}_{\mu}dx^\mu.$$

The inverse matrices reconstruct the metric and inverse metric:

$$g_{\mu\nu} =\eta_{\hat a\hat b}\theta^{\hat a}{}_{\mu}\theta^{\hat b}{}_{\nu}, \qquad g^{\mu\nu} =\eta^{\hat a\hat b}e_{\hat a}{}^\mu e_{\hat b}{}^\nu.$$

These formulas are not approximations at a point; they hold throughout the frame patch. Carroll gives the tetrad relations and conversion of components in Carroll 1997, §3, pp. 88–92, eqs. (3.114)–(3.129).

Components resolved by an observer

Choose a future-directed unit timelike vector $u=e_{\hat0}$. For any vector $V$,

$$V^{\hat a}=\theta^{\hat a}{}_{\mu}V^\mu, \qquad V^\mu=e_{\hat a}{}^\mu V^{\hat a}.$$

If $p$ is a future-directed four-momentum, the energy resolved by $u$ is

$$E_{(u)}=-p\cdot u=p^{\hat0}.$$

The spatial components $p^{\hat i}$ are its momentum components along the three spacelike tetrad legs. For a rank-two covariant tensor,

$$T_{\hat a\hat b} =e_{\hat a}{}^\mu e_{\hat b}{}^\nu T_{\mu\nu}.$$

This is a change of basis, not a claim that all coordinate effects disappear over a finite region. A tetrad is locally inertial at a point only when it is chosen appropriately and the Levi-Civita connection is considered there; curvature prevents one frame from removing connection effects throughout a generic neighborhood.

All tensor operations can be carried out with the constant internal metric once components have been converted. For example, $V^2=\eta_{\hat a\hat b}V^{\hat a}V^{\hat b}$ equals the coordinate contraction $g_{\mu\nu}V^\mu V^\nu$. A chosen oriented tetrad also fixes the sign of $\theta^{\hat0}\wedge\theta^{\hat1}\wedge\theta^{\hat2}\wedge\theta^{\hat3}$ relative to the spacetime orientation. Reversing one leg may preserve orthonormality while reversing orientation or time orientation, so those properties require separate checks.

A reliable construction method

Given a metric and a chosen unit timelike field, first set $e_{\hat0}=u$. Choose trial fields $w_1,w_2,w_3$ such that $u,w_1,w_2,w_3$ form a basis on the patch, and project them into the observer’s rest space:

$$w_i^\perp=w_i+g(u,w_i)u, \qquad g(u,w_i^\perp)=0.$$

The projected fields remain independent precisely because the original four vectors form a basis. Apply Gram–Schmidt to them using the positive-definite metric on $u^\perp$. Then invert the tetrad component matrix to obtain the coframe. Finally check both inverse relations and reconstruct $g_{\mu\nu}$ from $\eta_{\hat a\hat b}$. Merely choosing three individually nonparallel vectors to $u$ would not guarantee independence after projection.

Starting with the timelike leg is not cosmetic. Naive Gram–Schmidt in an indefinite vector space can encounter a null intermediate vector, which cannot be normalized. Once $u$ is fixed, its orthogonal complement is positive definite and ordinary spatial orthonormalization is safe. A different $u$ selects a different observer-adapted family, related pointwise by a boost and spatial rotation.

The method is local. It can fail at a coordinate singularity because the chosen trial vectors become dependent even when another tetrad is regular. It can also fail globally because a single smooth tetrad trivializes the tangent bundle. Multiple patches with Lorentz-valued transition functions are then required.

Orientation and time orientation reduce the allowed frame changes to the proper, orthochronous Lorentz group when those structures are chosen. Before those choices, all components of $O(1,3)$ may relate orthonormal tetrads.

Boost check and frame dependence

In a local Minkowski frame, boost with rapidity $\chi$ in the $1$ direction:

$$\begin{aligned} e'_{\hat0}&=\cosh\chi\,e_{\hat0}+\sinh\chi\,e_{\hat1},\\ e'_{\hat1}&=\sinh\chi\,e_{\hat0}+\cosh\chi\,e_{\hat1}. \end{aligned}$$

Direct contraction gives $g(e'_{\hat0},e'_{\hat0})=-1$, $g(e'_{\hat1},e'_{\hat1})=1$, and $g(e'_{\hat0},e'_{\hat1})=0$. Thus the tetrad changes while the reconstructed metric does not.

For a photon with $p=E(e_{\hat0}+e_{\hat1})$, the boosted observer measures

$$E'=-p\cdot e'_{\hat0}=E(\cosh\chi-\sinh\chi)=Ee^{-\chi}.$$

The energy component is frame-dependent, while the null condition $p^2=0$ is invariant. This calculation distinguishes a local Lorentz change of components from a coordinate transformation and from an isometry that moves points on the manifold.

Scope and limitations

A tetrad supplies algebraic components at each event. Connecting readings made at separated events additionally requires a transport rule and a path. Dynamics, detector response, and operational protocols are separate questions. Nor does choosing a tetrad fix its six local Lorentz degrees of freedom; the next pages describe the associated connection and gauge transformation.

Curvature components such as $R_{\hat0\hat i\hat0\hat j}$ are likewise frame components of an invariant tensor. Their numerical entries change under a boost, although scalar contractions and the tensor equation do not. Interpreting a particular component requires the associated tetrad, its normalization, and its orientation; quoting a bare number erases the information needed to reproduce the projection.

Exercises

Let $V=V^{\hat0}e_{\hat0}+V^{\hat1}e_{\hat1}$ in the unprimed frame. Derive its primed components under the boost above and verify that $-(V^{\hat0})^2+(V^{\hat1})^2$ is unchanged.

Solution

Because the basis is boosted by $\Lambda$, vector components transform by $\Lambda^{-1}$:

$$V'^{\hat0}=\cosh\chi\,V^{\hat0}-\sinh\chi\,V^{\hat1}, \qquad V'^{\hat1}=-\sinh\chi\,V^{\hat0}+\cosh\chi\,V^{\hat1}.$$

Expanding $-(V'^{\hat0})^2+(V'^{\hat1})^2$, the cross terms cancel and $\cosh^2\chi-\sinh^2\chi=1$ leaves $-(V^{\hat0})^2+(V^{\hat1})^2$. The component change is nontrivial, but the Lorentzian norm is unchanged.

References

• Carroll, Sean M. Lecture Notes on General Relativity. arXiv:gr-qc/9712019 (1997), arXiv.