Electric and Magnetic Parts of Curvature

An observer splits the ten components of the four-dimensional Weyl tensor into two spatial, symmetric, trace-free tensors. The electric part is the trace-free tidal field; the magnetic part governs differential frame-dragging and the complementary half of free curvature. The split is observer dependent, just as electric and magnetic fields mix under electromagnetic boosts, but the combined pair is equivalent to the Weyl tensor at the event.

Required background. We use the Riemann and Weyl curvature tensors and the operational domain of local gravitational measurement.

Helpful background. The construction assumes special-relativistic observer splits, tensor projections, oriented tetrads, and the relation between curvature and Jacobi fields.

Observer projection and dual convention

Let $u^a$ be a future-directed unit timelike vector, $u^au_a=-1$, and let

$$h_{ab}=g_{ab}+u_au_b$$

project into the observer’s rest space. Choose the spacetime volume form with $\epsilon_{0123}=+\sqrt{-g}$ in an oriented coordinate basis and define the left dual by

$$({}^\star C)_{abcd} =\frac12\epsilon_{ab}{}^{ef}C_{efcd}.$$

The electric and magnetic Weyl tensors are

$$E_{ab}=C_{acbd}u^cu^d, \qquad B_{ab}=({}^\star C)_{acbd}u^cu^d.$$

This explicit placement of the dual fixes the sign of $B_{ab}$. In four dimensions the left and right duals of the Weyl tensor agree when index positions and the volume-form convention are kept consistent. Reversing the orientation changes the sign of $B_{ab}$; moving timelike indices before dualizing must also include the metric signs.

Maartens and Bassett give the covariant $1+3$ split, its inverse, and its measurable interpretation Maartens and Bassett 1998, §3, arXiv p. 7, Eqs. (29)–(30), PDF.

Spatial, symmetric, and trace-free

Antisymmetry of $C_{acbd}$ in $a,c$ gives

$$E_{ab}u^a=0,$$

because it contracts an antisymmetric pair with $u^au^c$. The same argument applied to the dual shows $B_{ab}u^a=0$. Pair exchange and the algebraic Bianchi identity imply $E_{ab}=E_{ba}$ and $B_{ab}=B_{ba}$. Finally, every trace of the Weyl tensor vanishes, so

$$E^a{}_a=0, \qquad B^a{}_a=0.$$

Each tensor is therefore a symmetric trace-free $3\times3$ spatial matrix: six symmetric entries minus one trace, or five independent components. Their five-plus-five components match the ten independent components of Weyl curvature in four dimensions.

The electric part of the full Riemann tensor, $R_{acbd}u^cu^d$, is the complete tidal matrix. $E_{ab}$ is only its Weyl, trace-free contribution; matter-associated Ricci terms must be added when they are present.

Reconstructing the Weyl tensor

To display the inverse split, define the observer’s spatial alternating tensor by

$$\varepsilon_{abc}=\epsilon_{abcd}u^d.$$

In an orthonormal tetrad with $u=e_{(0)}$ and $\epsilon_{0123}=+1$, this definition gives $\varepsilon_{123}=-1$. Stating that sign prevents an unnoticed reversal when replacing spatial tensor contractions by three-dimensional component notation.

With the dual convention above, the Weyl tensor is reconstructed by

$$\begin{aligned} C_{abcd}={}&4\left(u_{[a}u_{[c}+h_{[a[c}\right)E_{b]d]}\\ &+2\varepsilon_{abe}u_{[c}B_{d]}{}^e +2u_{[a}B_{b]e}\varepsilon_{cd}{}^e. \end{aligned}$$

Projecting this expression twice along $u^a$ returns $E_{ab}$; left-dualizing and projecting returns $B_{ab}$. These two contractions are the cleanest independent checks of the signs and factors. They also show that no Weyl information remains outside the pair once $u^a$ and the orientation are specified.

Relative to an orthonormal tetrad with time leg $u$, $E_{ij}$ supplies $C_{i0j0}$. Mixed time–space components are spatial-dual combinations of $B_{ij}$, and purely spatial components are spatial-double-dual combinations of $E_{ij}$. This is why different kinds of local transport experiments can recover the two matrices.

Tides, differential precession, and boosts

$E_{ab}$ maps a small spatial separation into the trace-free Weyl contribution to relative free-fall acceleration. Its eigenvectors give principal tidal directions for that observer. $B_{ab}$ appears in differential gyroscope precession and in the coupling between the evolution of electric curvature and spatial curls through the Bianchi identities.

The electromagnetic names are analogical, not a claim that gravity contains two vector fields. Both $E_{ab}$ and $B_{ab}$ are rank-two tensors. A boost changes $u^a$, and hence mixes the two parts while leaving $C_{abcd}$ unchanged. Statements such as “the magnetic Weyl tensor vanishes” must therefore name the observer family.

For static observers in a static spherically symmetric vacuum field, $B_{ab}=0$ while $E_{ab}\ne0$. The geometry still has nonzero Weyl curvature and tidal forces. This example prevents the magnetic part from being treated as a necessary marker of all gravitational fields. Conversely, both parts can be nonzero in radiative situations, but nonzero $B_{ab}$ alone is not a sufficient radiation criterion.

Checks and limitations

The component count gives a strong algebraic check: any proposed split with a trace or antisymmetric part contains redundant or inconsistent information. Reversing spacetime orientation flips $B_{ab}$ but not $E_{ab}$, as expected for a magnetic-type quantity. In a conformally flat spacetime $C_{abcd}=0$, so both tensors vanish for every observer even though Ricci tidal curvature may remain.

The split is pointwise. Propagation, constraints, and wave interpretation require the differential Bianchi identities and the observer congruence’s expansion, shear, acceleration, and vorticity. Those additional equations should not be inferred from the algebraic definitions alone.

Exercise

For static observers outside a spherical source, suppose an orthonormal frame gives

$$E_{\hat i\hat j}=q\,\operatorname{diag}(-2,1,1), \qquad B_{\hat i\hat j}=0, \qquad q>0.$$

Check the Weyl component count represented by this special case, identify the relative acceleration of radial and transverse separations, and state what reverses when the spacetime orientation is reversed.

Solution

The trace $-2q+q+q=0$, so this is a valid symmetric trace-free electric tensor. Spherical symmetry has reduced its five possible components to one scalar $q$, and static spherical symmetry sets all five magnetic components to zero; this specialization does not change the general five-plus-five count. Release a freely falling reference probe momentarily comoving with the static observer and compare nearby geodesics in its parallel frame. For a separation eigenvector, $\ddot\xi^{\hat i}=-E^{\hat i}{}_{\hat j}\xi^{\hat j}$ then gives radial stretching $\ddot\xi^{\hat r}=2q\xi^{\hat r}$ and transverse compression $\ddot\xi^{\hat\theta}=-q\xi^{\hat\theta}$, with the same result in the $\hat\phi$ direction. An array held static also has support-force and accelerated-frame terms and cannot use this free-fall equation directly. Orientation reversal sends $B\mapsto-B$ and leaves $E$ fixed; here $B=0$ remains zero.

References

• Maartens, Roy, and Bruce A. Bassett. “Gravito-electromagnetism.” Classical and Quantum Gravity 15 (1998): 705–717. DOI; arXiv.