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The Newtonian Tidal Limit

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The Newtonian tidal tensor is the weak-field, slow-motion limit of the observer-projected Riemann tensor. With g00=−(1+2Φ)g_{00}=-(1+2\Phi), spacetime curvature reduces to the Hessian of the gravitational potential, and relativistic geodesic deviation becomes the familiar difference of gravitational accelerations at neighboring points. This limit fixes signs and dimensions while making clear that a uniform gravitational acceleration has no tidal curvature.

Required background. Affine and Variational Geodesics supplies affine geodesics and Jacobi fields; The Riemann Curvature Tensor supplies the curvature commutator and contractions; Observers, Rest Spaces, and Measured Components supplies local frame components.

Helpful background. Events, Intervals, and Minkowski Geometry reviews the low-velocity limit; Tensor Fields and Natural Operations reviews tensorial versus coordinate statements; The Newtonian Gravitational Field and Potential supplies g=−∇Φ\mathbf g=-\boldsymbol\nabla\Phi and Poisson’s equation.

Work in a nearly inertial chart over a region small compared with the background variation scale. Assume a stationary weak field, a dimensionless potential ∣Φ∣≪1|\Phi|\ll1, and test bodies with coordinate speed ∣v∣≪1|\mathbf v|\ll1. Retain first order in Φ\Phi and leading order in velocity. The time-time metric component is

g00=−(1+2Φ),g_{00}=-(1+2\Phi),

while the spatial metric corrections do not enter the leading slow-particle acceleration. The geodesic equation gives

d2xidt2≃−Γi00,Γi00≃∂iΦ,\frac{d^2x^i}{dt^2} \simeq-\Gamma^i{}_{00}, \qquad \Gamma^i{}_{00}\simeq\partial_i\Phi,

and therefore x¨i=−∂iΦ\ddot x^i=-\partial_i\Phi. This is a limit of metric gravity, not a claim that Φ\Phi is a generally covariant scalar potential in an arbitrary spacetime. The relation between curvature, nearby free fall, and tidal force is developed in Reall 2013, §§1.4 and 6.5, pp. 4–5 and 60–62, PDF.

Using the site’s Riemann convention and stationarity,

Ri0j0=∂jΓi00−∂0Γij0+O(Φ2)=∂i∂jΦ+O(Φ2).\begin{aligned} R^i{}_{0j0} &=\partial_j\Gamma^i{}_{00} -\partial_0\Gamma^i{}_{j0} +O(\Phi^2) \\ &=\partial_i\partial_j\Phi+O(\Phi^2). \end{aligned}

For a slowly moving reference observer, uμ≃(1,0)u^\mu\simeq(1,\mathbf0). The relativistic equation

D2ξμDτ2=−Rμανβuαξνuβ\frac{D^2\xi^\mu}{D\tau^2} =-R^\mu{}_{\alpha\nu\beta}u^\alpha\xi^\nu u^\beta

then reduces to

ξ¨i=−Hijξj,Hij≡∂i∂jΦ.\ddot\xi^i =-\mathcal H^i{}_j\xi^j, \qquad \mathcal H_{ij}\equiv\partial_i\partial_j\Phi.

The same result follows without relativity: expand gi(x+ξ)−gi(x)g^i(\mathbf x+\boldsymbol\xi)-g^i(\mathbf x) to first order and obtain −(∂i∂jΦ)ξj-(\partial_i\partial_j\Phi)\xi^j. Agreement of the two derivations is the independent sign check. The covariant derivation and the warning that relative acceleration is defined through covariant derivatives are given in MIT 2017, §13.1, pp. 54–57, PDF.

The Hessian Hij\mathcal H_{ij} is symmetric, so an orthonormal spatial basis diagonalizes it at an event. Its eigenvectors are principal tidal axes. A positive eigenvalue produces relative compression because the measured acceleration matrix is −Hij-\mathcal H^i{}_j; a negative eigenvalue produces stretching.

Poisson’s equation supplies the trace:

Hii=∇2Φ=4πGρ,tr⁡(−H)=−4πGρ.\mathcal H^i{}_i=\nabla^2\Phi=4\pi G\rho, \qquad \operatorname{tr}(-\mathcal H)=-4\pi G\rho.

More plainly, the sum of the three eigenvalues of the relative-acceleration matrix is −4πGρ-4\pi G\rho. In vacuum it is trace-free: stretching along one direction must be balanced by compression along the others. With c=1c=1, Φ\Phi is dimensionless, H\mathcal H and curvature have dimensions of inverse length squared, and Hξ\mathcal H\xi has the dimension of acceleration in geometric units.

For a point mass outside the source,

Φ=−GMr,Hij=GMr3(δij−3ninj).\Phi=-\frac{GM}{r}, \qquad \mathcal H_{ij} =\frac{GM}{r^3} \left(\delta_{ij}-3n_i n_j\right).

The radial Hessian eigenvalue is −2GM/r3-2GM/r^3 and each tangential eigenvalue is +GM/r3+GM/r^3. Consequently, an initially comoving freely falling ball of probes stretches radially and compresses in both tangential directions. The leading quadratic-in-time volume change vanishes because the tidal trace is zero. This is an instantaneous statement, not conservation of the cloud’s volume for all time: the induced shear can subsequently change its expansion. The probe-cloud diagram shows the same radial and transverse response.

Uniform fields and the limit of the approximation

Section titled “Uniform fields and the limit of the approximation”

Adding a constant or a linear term to Φ\Phi changes clock normalization or common acceleration but leaves ∂i∂jΦ\partial_i\partial_j\Phi unchanged. A uniform field can be removed locally by choosing a freely falling frame; its spatial variation cannot. This is precisely the boundary between the pointlike equivalence argument and tidal gravity Reall 2013, §1.4, pp. 4–5, PDF.

The reduction fails when time dependence, relativistic velocity, strong fields, large laboratory size, or appreciable gravitomagnetic terms enter. It also does not determine all spacetime curvature: the Hessian describes only the electric tidal block seen by the chosen slow observer.

1. Linear potential. Let Φ=gz\Phi=gz. Compute the acceleration and tidal tensor.

Solution

The common acceleration is g=−g ez\mathbf g=-g\,\mathbf e_z. Every second derivative of Φ\Phi vanishes, so Hij=0\mathcal H_{ij}=0 and nearby particles have no relative acceleration at this order. Common acceleration and tidal acceleration are therefore distinct.

2. Point-mass axes. For Φ=−GM/r\Phi=-GM/r, find the three eigenvalues of −Hij-\mathcal H^i{}_j.

Solution

Along n\mathbf n, H\mathcal H has eigenvalue −2GM/r3-2GM/r^3; on either orthogonal direction it has +GM/r3+GM/r^3. The relative-acceleration eigenvalues are therefore +2GM/r3,−GM/r3,−GM/r3+2GM/r^3,-GM/r^3,-GM/r^3. Their sum vanishes, as required in vacuum.

3. Infer a density. A measured relative-acceleration matrix has eigenvalues a1,a2,a3a_1,a_2,a_3 per unit separation. What Newtonian density does it imply?

Solution

Because the acceleration matrix is −H-\mathcal H, its trace is a1+a2+a3=−4πGρa_1+a_2+a_3=-4\pi G\rho. Thus ρ=−(a1+a2+a3)/(4πG)\rho=-(a_1+a_2+a_3)/(4\pi G). A positive sum would be incompatible with positive ordinary mass density within the stated Newtonian assumptions.

  • Guth, Alan H., lectures; Andrew P. Turner, notes. 8.962 General Relativity, Spring 2017. Massachusetts Institute of Technology, 2017, §13.1, pp. 54–57. PDF.
  • Reall, Harvey S. General Relativity. Cambridge Part III lecture notes, 2013, §§1.4 and 6.5, pp. 4–5 and 60–62. PDF.