Parallel Transport and Geodesic Deviation

Derivation

Parallel Transport and Geodesic Deviation

D-388 Home PU-403 Threads force · fields Depends on Covariant Derivative and the Metric Connection, Riemann Tensor from the Covariant Commutator

Statement

For a smooth one-parameter family (congruence) of timelike geodesics \(x^\mu(\tau,s)\) with unit tangent \(u^\mu=\partial x^\mu/\partial\tau\) and connecting (separation) vector \(\xi^\mu=\partial x^\mu/\partial s\), the second covariant derivative of the separation along the flow is fixed entirely by the Riemann tensor of spacetime: \(\dfrac{D^2\xi^\mu}{d\tau^2}=-R^{\mu}{}_{\nu\alpha\beta}\,u^\nu\,\xi^\alpha\,u^\beta\). The relative acceleration of neighbouring free-falling test particles is therefore a direct, coordinate-independent measurement of curvature — the geometric statement of tidal gravity.

Why it matters

A single freely-falling particle feels nothing: gravity can always be transformed away locally by going to a freely-falling frame (the equivalence principle). Curvature is what cannot be transformed away, and it only shows up when you compare two nearby trajectories. Geodesic deviation is the precise sense in which "tidal force" survives the equivalence principle and becomes the true, invariant signature of a gravitational field.

Operationally this is how curvature is measured. LIGO detects a passing gravitational wave as an oscillating relative acceleration between free test masses; the tidal stretching of an infalling body near a black hole ("spaghettification") is this equation evaluated in a strong field. It is also the bridge to the field equations: taking the trace of the deviation and using Einstein's equation ties the focusing of geodesics to the local energy density (Raychaudhuri).

Assumptions

Torsion-free (Levi-Civita) connection.With torsion, connecting vectors no longer satisfy \(\nabla_u\xi=\nabla_\xi u\); an explicit torsion term \(T^\mu{}_{\alpha\beta}u^\alpha\xi^\beta\) enters and the clean "curvature-only" reading is lost.

Each curve of the family is a geodesic, \(u^\alpha\nabla_\alpha u^\mu=0\).If the worldlines carry a non-gravitational four-force, the geodesic term does not vanish and the right side gains \(\xi^\alpha\nabla_\alpha a^\mu\), mixing genuine curvature with the force gradient.

A smooth congruence: exactly one worldline of the family through each nearby event, so \(u^\mu\) and \(\xi^\mu\) are coordinate vector fields of the map \((\tau,s)\mapsto x^\mu\).At a caustic (where neighbouring geodesics cross) \(\xi^\mu\to0\) and the field of tangents is singular; the derivation's commuting-field step breaks down.

Infinitesimal separation.The equation is the linearisation in \(\xi\). For finite separation there are corrections of order \(\xi^2\nabla R\) (tidal-tensor gradients); the linear form only measures curvature at the fiducial worldline.

Affine parametrisation shared across the family (proper time \(\tau\) on every curve, \(s\) labelling curves).A reparametrisation that varies from curve to curve reintroduces \(\tau\)-derivatives of the normalisation and spoils the identification of the two mixed derivatives.

Derivation

\[ u^\mu=\frac{\partial x^\mu}{\partial\tau},\qquad \xi^\mu=\frac{\partial x^\mu}{\partial s} \]

Coordinatise the congruence by \((\tau,s)\): \(\tau\) is proper time along each geodesic, \(s\) labels the geodesic. The tangent and the connecting vector are the two coordinate basis vectors of this map. A

\[ [u,\xi]^\mu=u^\alpha\partial_\alpha\xi^\mu-\xi^\alpha\partial_\alpha u^\mu=\frac{\partial^2 x^\mu}{\partial\tau\,\partial s}-\frac{\partial^2 x^\mu}{\partial s\,\partial\tau}=0 \;\Longrightarrow\; u^\alpha\nabla_\alpha\xi^\mu=\xi^\alpha\nabla_\alpha u^\mu \]

Mixed partials of the smooth map commute, so the Lie bracket vanishes. For a torsion-free connection \(\nabla_\alpha\nabla_\beta\) is symmetric on scalars, hence \([u,\xi]^\mu=u^\alpha\nabla_\alpha\xi^\mu-\xi^\alpha\nabla_\alpha u^\mu\); vanishing bracket gives the exchange identity. B

\[ \frac{Du^\mu}{d\tau}\equiv u^\alpha\nabla_\alpha u^\mu=0 \]

Each worldline of the family is a geodesic: its tangent is parallel-transported along itself. This is the free-fall condition that will kill a term at step 9. A

\[ \frac{D^2\xi^\mu}{d\tau^2}=u^\beta\nabla_\beta\!\left(u^\alpha\nabla_\alpha\xi^\mu\right) \]

Definition of the relative ("tidal") acceleration: apply the directional covariant derivative \(D/d\tau=u^\beta\nabla_\beta\) twice to the separation vector. This is the quantity an observer riding the fiducial geodesic actually measures. A

\[ \frac{D^2\xi^\mu}{d\tau^2}=u^\beta\nabla_\beta\!\left(\xi^\alpha\nabla_\alpha u^\mu\right) \]

Replace the inner factor using the exchange identity from step 2, \(u^\alpha\nabla_\alpha\xi^\mu=\xi^\alpha\nabla_\alpha u^\mu\). This is the pivotal move: it converts derivatives of \(\xi\) into derivatives of \(u\), whose second derivative we can commute. B

\[ \frac{D^2\xi^\mu}{d\tau^2}=\left(u^\beta\nabla_\beta\xi^\alpha\right)\nabla_\alpha u^\mu+u^\beta\xi^\alpha\,\nabla_\beta\nabla_\alpha u^\mu \]

Leibniz (product) rule for the covariant derivative acting on the product \(\xi^\alpha\,\nabla_\alpha u^\mu\). Metric compatibility is not even needed here — only linearity and the derivation property. B

\[ \left(u^\beta\nabla_\beta\xi^\alpha\right)\nabla_\alpha u^\mu=\left(\xi^\beta\nabla_\beta u^\alpha\right)\nabla_\alpha u^\mu \]

Apply the exchange identity (step 2) once more to the first term. We record this piece separately; it will cancel at step 10. B

\[ \nabla_\beta\nabla_\alpha u^\mu=\nabla_\alpha\nabla_\beta u^\mu+\big[\nabla_\beta,\nabla_\alpha\big]u^\mu=\nabla_\alpha\nabla_\beta u^\mu+R^{\mu}{}_{\nu\beta\alpha}\,u^\nu \]

Insert the commutator of covariant derivatives on the vector \(u^\mu\); by the prior result riemann-tensor-from-commutator-of-covariant-derivatives, \([\nabla_\beta,\nabla_\alpha]u^\mu=R^{\mu}{}_{\nu\beta\alpha}u^\nu\). This is where curvature enters. C

\[ u^\beta\xi^\alpha\,\nabla_\alpha\nabla_\beta u^\mu=\xi^\alpha\nabla_\alpha\!\big(u^\beta\nabla_\beta u^\mu\big)-\xi^\alpha\big(\nabla_\alpha u^\beta\big)\big(\nabla_\beta u^\mu\big)=-\,\xi^\alpha\big(\nabla_\alpha u^\beta\big)\big(\nabla_\beta u^\mu\big) \]

Rewrite \(\nabla_\alpha\nabla_\beta u^\mu\) using Leibniz in reverse, then apply the geodesic equation (step 3) \(u^\beta\nabla_\beta u^\mu=0\) to annihilate the first bracket. Only the "connection-squared" remainder survives. C

\[ \frac{D^2\xi^\mu}{d\tau^2}=\underbrace{\big(\xi^\beta\nabla_\beta u^\alpha\big)\nabla_\alpha u^\mu-\xi^\alpha\big(\nabla_\alpha u^\beta\big)\big(\nabla_\beta u^\mu\big)}_{=\,0}+R^{\mu}{}_{\nu\beta\alpha}\,u^\nu u^\beta\xi^\alpha \]

Collect step 7 and step 9: relabelling the dummy pair \(\alpha\leftrightarrow\beta\) in the first term makes it identical to the second, so the two non-curvature terms cancel exactly. What remains is pure Riemann. B

Result

\[ \frac{D^2\xi^\mu}{d\tau^2}=R^{\mu}{}_{\nu\beta\alpha}\,u^\nu u^\beta\xi^\alpha=-\,R^{\mu}{}_{\nu\alpha\beta}\,u^\nu\,\xi^\alpha\,u^\beta \]

Reading. The relative acceleration of two infinitesimally separated free-falling particles equals minus the Riemann tensor contracted twice with their common four-velocity and once with their separation. Flat spacetime (\(R^\mu{}_{\nu\alpha\beta}=0\)) gives zero relative acceleration — parallel free-fallers stay parallel, exactly the Newtonian statement that a uniform field produces no tides. A non-zero right-hand side is the invariant, frame-independent presence of a genuine gravitational field. The map \(\xi^\alpha\mapsto -R^{\mu}{}_{\nu\alpha\beta}u^\nu u^\beta\) is the relativistic tidal tensor; its symmetry (from the pair-symmetry of Riemann) guarantees principal tidal axes.

Units check. In SI, \(R^\mu{}_{\nu\alpha\beta}\) has dimension \(\mathrm{length}^{-2}=\mathrm{m^{-2}}\); the four-velocity \(u^\mu=dx^\mu/d\tau\) has dimension \(\mathrm{m\,s^{-1}}\); the separation \(\xi^\mu\) has dimension \(\mathrm{m}\). Hence the right side has \([\mathrm{m^{-2}}][\mathrm{m\,s^{-1}}]^2[\mathrm{m}]=\mathrm{m\,s^{-2}}\), an acceleration, matching \(D^2\xi^\mu/d\tau^2\) with \([\mathrm{m}]/[\mathrm{s}]^2=\mathrm{m\,s^{-2}}\). Both sides are accelerations.

Limiting cases

Flat spacetime: \(R^\mu{}_{\nu\alpha\beta}=0\Rightarrow D^2\xi^\mu/d\tau^2=0\). Neighbouring inertial worldlines have constant separation velocity — no tides.
Newtonian weak field: with \(u^\mu\approx(c,\mathbf 0)\), \(\tau\approx t\), and \(R^{i}{}_{0j0}\approx c^{-2}\partial_i\partial_j\Phi\), the equation reduces to \(\ddot\xi^{\,i}=-\partial_i\partial_j\Phi\,\xi^{\,j}\), the classical tidal-tensor equation.
Vacuum (Ricci-flat): \(R_{\mu\nu}=0\) makes the tidal tensor trace-free, so a small dust ball shears into an ellipsoid at fixed volume — pure stretch-and-squeeze, no focusing.
Static spherical field (Schwarzschild): orthonormal-frame tidal eigenvalues \(-2GM/r^3\) (radial stretch) and \(+GM/r^3\) (two transverse squeezes), summing to zero as vacuum requires.
Gravitational wave: transverse-traceless \(h_{ij}\) gives \(\ddot\xi^{\,i}=\tfrac12\ddot h_{ij}\xi^{\,j}\); the "plus" and "cross" quadrupolar strain patterns are read straight off this.

Breaks when

Caustics / geodesic crossings. When neighbouring geodesics converge and cross, \(\xi^\mu\to0\), the congruence stops being a smooth map, and the coordinate-vector-field construction (steps 1–2) fails. The linear equation cannot be continued through the focal point.
Finite separation in strongly varying curvature. The derivation linearises in \(\xi\). Where the tidal field changes appreciably over the separation (\(|\xi\,\nabla R|\sim|R|\)), the neglected \(O(\xi^2)\) terms dominate and the simple Riemann reading is wrong — relevant for extended bodies, not just test points.
Non-geodesic worldlines. If the particles are charged, thrusting, or otherwise forced, \(u^\alpha\nabla_\alpha u^\mu=a^\mu\neq0\); step 9 leaves \(\xi^\alpha\nabla_\alpha a^\mu\), and relative acceleration no longer measures curvature alone.
Curvature singularities. As \(r\to0\) in Schwarzschild the tidal eigenvalues \(\sim GM/r^3\) diverge; the equation stays formally valid but predicts unbounded relative acceleration — the test-particle idealisation and the classical geometry both cease to be physical.

Failure modes

Ordinary derivative for covariant. Writing \(d^2\xi^\mu/d\tau^2\) instead of \(D^2\xi^\mu/d\tau^2\): the coordinate second derivative is not a tensor and misses the Christoffel terms that build the curvature. Only the covariant \(D/d\tau=u^\beta\nabla_\beta\) gives a frame-independent result.
Dropping the exchange identity. Forgetting \(u^\alpha\nabla_\alpha\xi^\mu=\xi^\alpha\nabla_\alpha u^\mu\) (step 2) and treating \(u,\xi\) as independent — this is exactly the torsion-free, commuting-congruence input; without it the two spurious terms in step 10 do not cancel.
Sign / index-order slips on Riemann. Using \([\nabla_\alpha,\nabla_\beta]\) instead of \([\nabla_\beta,\nabla_\alpha]\) flips the sign; the physically fixed content is that radial tides stretch — check against \(-2GM/r^3\) in Schwarzschild.
Confusing tidal tensor with Ricci. The tidal tensor is the full \(R^i{}_{0j0}\); only its trace is Ricci-related. Students set it to zero in vacuum and wrongly conclude "no tides" — vacuum kills the trace, not the shear.
Assuming \(\xi\) stays orthogonal to \(u\) automatically. One must impose or verify \(u_\mu\xi^\mu=\) const (a gauge choice using \(s\)-freedom); otherwise part of the "separation" is just clock-offset along the same worldline.

Discussion

The deep content is that the equivalence principle and curvature are not in tension — they are two halves of one statement. At a point you can always pick Riemann normal coordinates where \(\Gamma^\mu{}_{\alpha\beta}=0\) and gravity vanishes; but you cannot make the second derivatives of the metric vanish, and those second derivatives are the Riemann tensor. Geodesic deviation is precisely the lowest-order gravitational effect that no local frame choice can remove, because it is a comparison across a finite (albeit small) separation. This is why tidal force, not "the field," is the physical essence of gravity in general relativity.

Structurally, the equation is a Jacobi equation: \(D^2\xi^\mu/d\tau^2+R^\mu{}_{\nu\alpha\beta}u^\nu u^\beta \xi^\alpha=0\) is a linear second-order ODE for \(\xi\) along the geodesic, and its solutions are Jacobi fields. Their zeros are conjugate points, the general-relativistic analogue of focal points in optics. This connects directly to singularity theorems: convergence of a geodesic congruence (focusing) is driven by the trace of the tidal tensor, and via Einstein's equation the trace is \(R_{\mu\nu}u^\mu u^\nu=8\pi G\,c^{-4}(T_{\mu\nu}-\tfrac12 T g_{\mu\nu})u^\mu u^\nu\), so ordinary (positive-energy) matter always focuses — the Raychaudhuri route to Penrose–Hawking.

The Newtonian correspondence is exact and instructive. The tidal tensor \(E_{ij}=\partial_i\partial_j\Phi\) is symmetric and, in vacuum, trace-free because \(\nabla^2\Phi=0\); its relativistic parent \(R^i{}_{0j0}\) inherits both properties from the symmetries of Riemann and from Ricci-flatness. The famous statement "you cannot feel gravity, only tides" is the assertion that observable gravity begins at the tidal tensor. A falling elevator is weightless; a tall falling elevator is not quite, because its ends sample slightly different tidal accelerations — and that residual is \(R^i{}_{0j0}\).

At the geometric level the result is a curvature identity in disguise. Parallel transport of a vector around an infinitesimal loop rotates it by an amount set by \(R^\mu{}_{\nu\alpha\beta}\); geodesic deviation is the "differential" version of that holonomy, in which the loop is spanned by \(u\,d\tau\) and \(\xi\,ds\). Equivalently, \(\xi^\mu\) is a section of the pullback tangent bundle over the fiducial geodesic and \(D/d\tau\) is the induced connection; the deviation equation states that its curvature-of-transport along \(\tau\), obstructed by the \(\xi\)-direction, is exactly \(R(u,\xi)u\). This is why the same tensor governs both the failure of parallel transport to be path-independent and the failure of neighbouring free-fallers to stay parallel — holonomy and tidal force are one object viewed two ways.

Common misconceptions. (i) "Tidal force is a Newtonian approximation, GR replaces it" — no, geodesic deviation is the exact relativistic tidal force, with Newton as its weak-field limit. (ii) "Curvature needs a source nearby" — the tidal tensor is non-zero in vacuum (outside any mass); only its trace requires local matter. (iii) "The separation vector must be spatial" — it need only be Lie-dragged; orthogonality to \(u\) is a convenient, imposable gauge, not an automatic fact.

Worked examples

Radial tidal stretch just above Earth's surface. Two coins are released, one 1 m directly above the other; find their relative acceleration.

Setup: use the Schwarzschild orthonormal-frame radial component \(R^{\hat1}{}_{\hat0\hat1\hat0}=-\,\dfrac{2GM}{c^2r^3}\). A

\[ \frac{D^2\xi^{\hat1}}{d\tau^2}=-R^{\hat1}{}_{\hat0\hat1\hat0}\,(u^{\hat0})^2\,\xi^{\hat1},\qquad u^{\hat0}=c \;\Rightarrow\; \frac{D^2\xi^{\hat1}}{d\tau^2}=+\frac{2GM}{r^3}\,\xi^{\hat1} \]

Contract the result with the static-observer four-velocity \(u^{\hat\mu}=(c,\mathbf0)\); the \(c^2\) cancels the \(1/c^2\) in the curvature. Positive sign ⇒ separation grows (stretch). B

\[ \frac{2GM}{r^3}=\frac{2(6.674\times10^{-11})(5.97\times10^{24})}{(6.37\times10^{6})^3}\ \mathrm{s^{-2}} \]

Insert \(GM_\oplus=3.986\times10^{14}\ \mathrm{m^3s^{-2}}\), \(r=R_\oplus=6.37\times10^6\ \mathrm{m}\); \(r^3=2.58\times10^{20}\ \mathrm{m^3}\). Numbers only after the symbolic form is fixed. A

\[ \frac{D^2\xi^{\hat1}}{d\tau^2}=(3.09\times10^{-6}\ \mathrm{s^{-2}})(1\ \mathrm{m})\approx 3.1\times10^{-6}\ \mathrm{m\,s^{-2}} \]

Reading. About \(3\times10^{-7}g\) of relative pull over 1 m — tiny, but this is the true measurable curvature at Earth's surface, and it stretches (does not compress) along the vertical.

Spaghettification: head-to-foot tidal acceleration on a 1.8 m astronaut falling feet-first at \(r=3r_s\) from a \(10\,M_\odot\) black hole.

Setup: same radial tidal law \(a_{\text{rel}}=\dfrac{2GM}{r^3}\,\xi\), with \(\xi=1.8\ \mathrm m\) and \(r=3r_s\), \(r_s=2GM/c^2\). A

\[ GM=(6.674\times10^{-11})(1.989\times10^{31})=1.327\times10^{21}\ \mathrm{m^3s^{-2}},\quad r_s=\frac{2GM}{c^2}=2.95\times10^{4}\ \mathrm m \]

Mass \(M=10\times1.989\times10^{30}\ \mathrm{kg}\); \(c^2=9\times10^{16}\ \mathrm{m^2s^{-2}}\). Then \(r=3r_s=8.85\times10^4\ \mathrm m\). A

\[ a_{\text{rel}}=\frac{2GM}{r^3}\,\xi=\frac{2(1.327\times10^{21})}{(8.85\times10^{4})^3}\,(1.8)\ \mathrm{m\,s^{-2}},\quad r^3=6.93\times10^{14}\ \mathrm{m^3} \]

Evaluate \(2GM/r^3=3.83\times10^{6}\ \mathrm{s^{-2}}\), then multiply by the body length. B

\[ a_{\text{rel}}\approx 6.9\times10^{6}\ \mathrm{m\,s^{-2}}\;\approx\;7\times10^{5}\,g \]

Reading. Even well outside the horizon of a stellar-mass black hole the head-to-foot tidal acceleration is hundreds of thousands of \(g\) — lethal. The effect scales as \(M/r^3\sim M^{-2}\) at fixed \(r/r_s\), so for a supermassive black hole (\(\sim10^6 M_\odot\)) the same \(3r_s\) tide is \(\sim10^{12}\) times gentler and an astronaut crosses the horizon intact.

Problems

(A) The exchange identity. For a torsion-free connection and commuting congruence, prove \(u^\alpha\nabla_\alpha\xi^\mu=\xi^\alpha\nabla_\alpha u^\mu\).
Solution
Torsion-free means \(\Gamma^\mu{}_{\alpha\beta}=\Gamma^\mu{}_{\beta\alpha}\). Write both covariant derivatives out: \(u^\alpha\nabla_\alpha\xi^\mu-\xi^\alpha\nabla_\alpha u^\mu=u^\alpha\partial_\alpha\xi^\mu-\xi^\alpha\partial_\alpha u^\mu+\Gamma^\mu{}_{\alpha\beta}(u^\alpha\xi^\beta-\xi^\alpha u^\beta)\). The Christoffel term vanishes by symmetry (it is contracted with the antisymmetric \(u^{[\alpha}\xi^{\beta]}\)). The remaining ordinary-derivative part is the Lie bracket \([u,\xi]^\mu=\partial^2 x^\mu/\partial\tau\partial s-\partial^2 x^\mu/\partial s\partial\tau=0\) since \(u,\xi\) are coordinate vector fields of a smooth map. Hence the difference is zero.
(B) Newtonian tidal tensor. From \(\Phi=-GM/r\), compute the eigenvalues of \(E_{ij}=\partial_i\partial_j\Phi\), show \(\mathrm{tr}\,E=0\), and evaluate the radial eigenvalue at 1 AU from the Sun.
Solution
\(\partial_i\partial_j\Phi=\dfrac{GM}{r^3}\left(\delta_{ij}-3\hat n_i\hat n_j\right)\), \(\hat n=\hat r\). Diagonalising: radial eigenvalue \((i=j\parallel\hat r)\) is \(\dfrac{GM}{r^3}(1-3)=-\dfrac{2GM}{r^3}\); the two transverse eigenvalues are \(+\dfrac{GM}{r^3}\) each. Trace \(=(-2+1+1)GM/r^3=0\), consistent with \(\nabla^2\Phi=0\) in vacuum. At 1 AU: \(GM_\odot=1.327\times10^{20}\ \mathrm{m^3s^{-2}}\), \(r=1.496\times10^{11}\ \mathrm m\), \(r^3=3.35\times10^{33}\ \mathrm{m^3}\), so radial eigenvalue \(=-2(1.327\times10^{20})/3.35\times10^{33}=-7.9\times10^{-14}\ \mathrm{s^{-2}}\) (stretch toward/away from Sun).
(B) ISS tides. Two instruments 10 m apart along the radial direction orbit at 400 km altitude. Find their relative tidal acceleration.
Solution
\(a_{\text{rel}}=\dfrac{2GM_\oplus}{r^3}\,\xi\), \(r=R_\oplus+h=6.371\times10^6+4.0\times10^5=6.771\times10^6\ \mathrm m\), \(r^3=3.104\times10^{20}\ \mathrm{m^3}\). Then \(2GM/r^3=2(3.986\times10^{14})/3.104\times10^{20}=2.57\times10^{-6}\ \mathrm{s^{-2}}\). Times \(\xi=10\ \mathrm m\): \(a_{\text{rel}}=2.57\times10^{-5}\ \mathrm{m\,s^{-2}}\approx2.6\times10^{-6}g\). This is the microgravity residual that governs gradient-sensitive experiments.
(C) Trace and focusing. Show that the trace of the tidal tensor over spatial directions equals \(R_{\mu\nu}u^\mu u^\nu\), and hence that a small dust ball's volume obeys \(\ddot V/V\big|_0=-R_{\mu\nu}u^\mu u^\nu\). What does this give in vacuum?
Solution
The tidal tensor is \(K^i{}_j=R^i{}_{0j0}\) in the rest frame \(u^\mu=(1,\mathbf0)\) (units \(c=1\)). Its trace \(K^i{}_i=R^i{}_{0i0}=R^\mu{}_{0\mu0}=R_{00}=R_{\mu\nu}u^\mu u^\nu\) (the \(\mu=0\) term vanishes by antisymmetry). Each principal separation obeys \(\ddot\xi^{i}=-K^i{}_j\xi^j\); for a small sphere \(V\propto\prod\xi^i\), so to lowest order \(\ddot V/V=-\sum_i K^i{}_i=-R_{\mu\nu}u^\mu u^\nu\). In vacuum \(R_{\mu\nu}=0\Rightarrow\ddot V/V=0\): the ball momentarily preserves volume, shearing sphere→ellipsoid. This is the seed of the Raychaudhuri equation and the focusing theorem.
(C) Neutron-star surface tide. Estimate the head-to-foot (1 m) tidal acceleration at the surface of a \(1.4\,M_\odot\), 12 km neutron star, and compare to Earth's surface gravity.
Solution
\(M=1.4\times1.989\times10^{30}=2.785\times10^{30}\ \mathrm{kg}\), \(GM=1.859\times10^{20}\ \mathrm{m^3s^{-2}}\), \(R=1.2\times10^4\ \mathrm m\), \(R^3=1.728\times10^{12}\ \mathrm{m^3}\). Radial tide \(2GM/R^3=2(1.859\times10^{20})/1.728\times10^{12}=2.15\times10^{8}\ \mathrm{s^{-2}}\). Over \(\xi=1\ \mathrm m\): \(a_{\text{rel}}\approx2.2\times10^{8}\ \mathrm{m\,s^{-2}}\approx2\times10^{7}g\). The tidal acceleration across a 1 m body already exceeds twenty million times Earth gravity — any macroscopic object is torn apart long before reaching the surface. (The Newtonian estimate is within tens of percent of the exact strong-field Schwarzschild value here, since \(2GM/c^2R\approx0.34\).)