Tidal and centripetal stress in a model human falling deep into a Kerr black hole

I calculate the tidal and centripetal stress in the Misner–Thorne–Wheeler model human—a rigid prism—falling into a high-spin supermassive Kerr black hole at a constant, near-equatorial latitude with high energy, enough to surmount the radial potential barrier near the throat of the black hole. The stress in the model human would be nonlethal.

vacuum. Any fall into a Schwarzschild black hole would be fatal; a fall into a Kerr black hole may not be.
Specifically, the example simulates the constant-latitude, near-equatorial fall of the MTW prism into a 4 million solar mass black hole at a polar angle θ=tan −1 (1155/68) from the spin axis, with enough energy (seven rest masses) to surmount the radial potential barrier near the throat of the black hole and continue falling to arbitrarily negative values of the radial coordinate r. The tidal torque on the prism is calculated, and the orientation of the prism is updated by integrating Eulerʼs equations of rigid-body motion. The normal tidal plus centripetal stress on the prism is also calculated; it would be well below the yield stress assumed by MTW.

Metric
In ingoing Kerr coordinates u r , , , [ ] q f and geometrized units in which G N =c=1 and length is measured in centimeters, Kerr spacetime is described by the line element [4]

Properties
The properties of Kerr spacetime have been reviewed by Wiltshire et al [5], Adamo and Newman [6], and Teukolsky [7], among others. Here I review only a few properties of Kerr spacetime and ingoing Kerr coordinates that may clarify the example presented here.

Class. Quantum Grav. 32 (2015) 245008 M B Callaham
Each event [u, r, θ, f] is assumed to be identified with the event [u, r, θ, f−2π]. Hence the coordinate f need only range over the open interval [0, 2 π). The coordinate θ ranges over [0, π]. The coordinates u and r range over the real numbers.
Let g 0 , g 1 , g 2 , and g 3 denote the coordinate basis vectors with covariant components The metric tensor components do not depend on u or f, so the coordinate basis vectors g 0 and g 3 are Killing vectors. The Killing vector g 0 is timelike at infinity, i.e., at large positive or negative r, so Kerr spacetime is stationary. The symmetry implied by the Killing vector g 3 is axial symmetry.
The coordinate basis vector g 2 is everywhere spacelike. The coordinate basis vector g 1 is everywhere null; nevertheless, a displacement in the direction g 1 is called a radial displacement and is measured in centimeters. The coordinate r is called the radius.
Some elements of the covariant metric tensor blow up as 0, 2 r  e.g., as r 0  in the equatorial plane θ=π / 2. Components of the Riemann curvature tensor R lmn k also blow up at ρ 2 =0. So does the Kretschmann curvature scalar K R R = klmn klmn [8], which does not depend on the choice of coordinate system. Hence the singularity at ρ 2 =0 is a physical singularity of Kerr spacetime curvature, not an artifact of the choice of ingoing Kerr coordinates.
Kerr spacetime depends on two parameters: m and a. The parameter m is called the mass parameter, or mass, and the parameter a is called the specific angular momentum parameter, or the spin parameter. These names describe the effects of the spacetime on the motion of test particles at large radius [3]. However, Kerr spacetime contains no matter, spinning or otherwise, and no nongravitational fields. If a=0, the metric describes Schwarzschild spacetime in ingoing Kerr coordinates, which in this case are ingoing Eddington-Finkelstein coordinates.
Analysis of the equations of motion for outgoing uncharged massless particles (e.g., photons) in Kerr spacetime reveals that, if a 2 m 2 , outgoing photons may remain, i.e. orbit, at radii r where Δ=0 [1]. The larger of these radii, r r m m a , 2 2 = º + -+ is called the outer horizon, and the smaller radius, r r m m a , 2 2 = º ---is called the inner horizon. Photons at r<r − may not propagate to any r>r − . Photons at r<r + may not propagate to any r>r + , hence the outer horizon should appear black to any outside observer. Moreover, analysis of the equations of motion (see below) for massive particles shows that massive particles at r<r + may not propagate to any r>r + . Hence the outer horizon has been called the event horizon [9], beyond which observers on the outside may acquire no information about events on the inside. However, Hawking [10] has conjectured that, when quantum effects are considered, information about the interior may leak out, so the outer horizon is an apparent horizon but not an event horizon.
If a 2 >m 2 , there are no horizons. The Kerr solution describes a spacetime with a 'naked' singularity, unclothed by a horizon, not a black hole [11,12].
Kerrʼs solution may not be a good model of the interior of a real spinning black hole; see discussion in section 8.

Curvature
The Riemann tensor describes the acceleration of separation between nearby free particles, a consequence of spacetime curvature. It is defined in terms of the Christoffel symbols and their derivatives [1]: The Christoffel symbols in turn may be expressed in terms of derivatives of metric tensor components: where the comma denotes ordinary differentiation, for example Explicit expressions for the Christoffel symbol components and the Riemann tensor components of Kerr spacetime are lengthy and, for brevity, are not displayed here. However, it is easy to display them using a computer algebra system, such as Maxima [13,14], that supports the symbolic calculation of the tensors of general relativity.

Parameters
For the numerical example we assume the Kerr mass parameter is m M 4 10 , 6 =´ where M e is the Solar mass. This is roughly the mass of the supermassive black hole at the center of the Milky Way, near the radio source Sgr A * [15][16][17].
We assume the Kerr spin parameter is a=0.998m, a plausible generic value for a supermassive black hole that grows by accretion from a thin disk [18]. The Sgr A * black hole may have a smaller spin [19][20][21]. In where μ, E, Φ, and K are constants of motion described in the following section, and The signs of R and Q may be positive or negative. The sign of R should be negative for a future-directed, radially inbound (r decreasing) portion of a geodesic. The sign of Q should be positive for a descending (θ increasing) portion of a geodesic and negative for an ascending (θ decreasing) portion of a geodesic.
These equations may be integrated numerically to find the geodesic followed by the MTW prism in this example.

Constants of motion
The constants of motion appearing in the four-velocity are μ, the rest mass; E, the energy; Φ, the angular momentum about the spacetime axis of symmetry; and K, Carterʼs fourth constant of motion [4], which is a generalized total angular momentum.
Test particles of different rest masses would fall in the same manner, so it is convenient to assume μ=1, instead of using the actual rest mass of the prism.
A constant-latitude fall requires dθ/dτ=0, which, from (17), requires Θ=0. Requiring the right-hand side of (19) to vanish implies Q a E cos sin , 23 After Hamilton [23], I choose Φ to stationarize Θ with respect to variations in θ: for the example geodesic. I choose E/μ=7, which is large enough for the infalling prism to fall monotonically to arbitrarily large negative values of r.

Initial conditions
I will impose initial conditions at r=0 and integrate the four-velocity forward and backward in proper time τ from there. At r=0, I assume:

Constructing a tetrad at r=0
At r=0 a tetrad is constructed using the Gram-Schmidt orthonormalization algorithm [24] adapted to signature -+++ spacetime. Let e 0 , e 1 , e 2 , and e 3 denote the basis vectors of the tetrad. Basis vector e i has contravariant components e i μ . The timelike basis vector is e 0 , the four-velocity of the test particle to which the tetrad frame is attached: The remaining basis vectors are obtained from the four-vectors g 3 , g 2 , and g 4 . One obtains e 97 56 3 ,  49 28 3   97 56 3 , 0, 0 , 31 If a basis vector e i at event x ν is parallel-propagated along a small displacement dx ν , its components change by small amounts x e e d d .
To parallel-propagate the tetrad basis vectors along the geodesic, these equations, for i=1, 2, 3, are integrated numerically simultaneously with Carterʼs first integrals, (15)-(18), of the geodesic equations. For i=0, (35) is just the geodesic equation, which may be integrated numerically, or one may use Carterʼs first integrals.

The tidal tensor
The Riemann tensor describes the acceleration of separation between nearby free particles. In Kerr spacetime, a vacuum spacetime, this deviation is solely a tidal effect: the Ricci curvature tensor, a trace of the Riemann tensor, vanishes, so the Weyl conformal tensor, the traceless part of the Riemann tensor, equals the Riemann tensor. The Weyl tensor describes tidal acceleration.
If parts of the MTW prism were not bound together by cohesive or, if compressed, repulsive electromagnetic forces, they would fall along separate geodesics, changing their distance from one another. But the parts of the prism are bound together by electromagnetic forces. The forces that are exerted to keep the distances between parts from changing may be calculated using Newtonʼs second law of motion in an inertial frame, such as the tetrad frame.
The partial derivatives of contravariant Kerr vector components with respect to tetrad vector components are just the tetrad basis vectors, so At each step of propagating the tetrad basis vectors along the geodesic, we have the tetrad basis vectors at the current event x ν . The derivatives x ,a a¢ (in general, x ,n m¢ ) may be calculated by matrix inversion: using the chain rule for derivatives, the identity tensor g λ ν may be expressed as the total derivative x x g , 3 9 , , and we may write Figure 2 shows the diagonal (in a matrix representation) components of the tidal tensor, which cause longitudinal tidal stress (tension if positive; compression if negative) in the MTW prism, to a degree that depends on the orientation of the prism in the tetrad frame. The diagonal components may also apply torques, accelerating rotation about the center of mass. Figure 3 shows the off-diagonal components of the tidal tensor, which cause shear stress in the MTW prism, causing acceleration of rotation about the center of mass.

The model human
MTW modeled the falling human as a large brick: a rigid right rectangular prism of depth D=20 cm, width W=20 cm, height H=180 cm, and mass M=7.5×10 4 g, uniform in composition, that yields-crushes or fractures-if internal pressure or tension reaches 100 atmospheres (1.01×10 8 dyn cm −2 ).

Rotation, tidal stress, and centripetal stress
An arbitrary orientation of the prism within the tetrad frame can be described in terms of Euler (or Eulerian) angles f, θ, and ψ [26]. We suppose that the prismʼs body-fixed axes x, y, and z initially coincide with the tetrad-frame axes x′, y′, and z′. The prism and its axes are then rotated right-handedly first through an angle f about the z axis, then through an angle θ about the x axis, then finally through an angle ψ about the z axis, which now points in a different direction within the tetrad frame.
A particle of the prism at position r x y z , , in body-frame coordinates is, before Euler rotations, at position r r x y z , , T [ ] ¢ º ¢ ¢ ¢ = in tetrad-frame coordinates. After right-   handed rotation through f about the z axis, then θ about the x axis, then ψ about the z axis, the tetrad-frame coordinates of the particle are ( ) f q y is the Euler transformation matrix, which depends on f, θ, and ψ. Figure 5 shows the orientation of the prism and its axes within the tetrad frame after rotations through f=0rad, then θ=π/2rad, and finally ψ=0rad. The x axis still coincides with the x′ axis, but the y axis now coincides with the z′ axis, and the z axis points in the −y′ direction. I assume this is the orientation of the prism at τ=0, and I assume rotation rates are zero at τ=0. Prior and subsequent orientations are obtained by integrating Eulerʼs equations [26] for the rotation of a rigid body: where N 1 , N 2 , and N 3 are the components of the torque N N N N , , about the x, y, and z axes (in this case applied by tidal stresses), an overdot denotes a derivative with respect to proper time τ, and sin sin cos , are the components of the instantaneous rotation rate ω in the x, y, and z directions. The torque N depends on the tidal tensor and the orientation of the prism in the tetrad frame-i.e., the Euler angles: Figure 5. The prism and its body-fixed axes [x, y, z], initially aligned with tetrad-frame axes [x′, y′, z′], have been rotated through Euler angles f=0rad, then θ=π/2rad, and finally ψ=0rad.  (3), is the mass density of the prism. Equation (57) follows from the equation N r F =´for the torque N exerted by a force F acting at a displacement r from the origin. In this case, we consider a vanishingly small volume of the prism, of measure dV=dxdydz, with mass dM=ρdV, at position r in the body frame. The position in the tetrad frame is E r , , .
A free-falling mass at this position would exhibit a tidal acceleration C E r , , , In the body frame, the force is The integral of this vanishingly small contribution to torque over the whole mass, i.e., over the whole volume, of the prism yields (57). The integrand is quadratic in x, y, and z, hence the integral can be expressed in closed form. The torque components are   Figure 7 shows the first (left) and last (right) frames of a supplementary animation that shows the rotation of the prism in the tetrad frame, starting about 3 s before reaching the radius (r=0 cm) of the singularity, and ending about 3 s after reaching the radius of the singularity. (Figure 5 showed the orientation of the prism at the radius of the singularity and is the middle frame of the animation.)  In a manner similar to the calculation of torques in (57), we calculate the aggregate tidal force exerted on the top half of the prism in the body frame: The integrand is linear in x, y, and z, so F  may be calculated using a lumped-mass model, i.e., as the tidal force on a point of mass M/2 located at the mean position of the mass, namely x=0, y=0, and z=H/4.
The aggregate tidal force exerted on the bottom half of the prism in the body frame is equal but opposite. If the top and bottom halves of the prism were not attached by cohesion, they would accelerate apart, if F z is positive, away from z=0. The tension that holds the top and bottom halves together must be F z . If F z is negative, then the force is repulsive; the prism is compressed.
The average normal tidal stress over the square where the z=0 plane bisects the MTW prism is The variation with proper time is plotted in figure 8. Positive stress is tensile; negative stress is compressive. The greatest tensile and compressive stresses are in the z direction, because of the orientation assumed at τ=0, and because the Euler angles do not change much near τ=0 (see figure 6).
The stress σ z is small until τ ≈−10 10 cm, a fraction of a second before the prism reaches r=0. Then σ z begins to increase, qualitatively as one would expect [1] in the Newtonian approximation or in the Schwarzschild limit a=0. Then σ z decreases suddenly and becomes negative, reaching 0.22 atm; z s »the prism is compressed along the z axis. The compression −σ z plotted is that required by the lower half of the prism to bear the weight of the upper half of the prism. The compression decreases rapidly and becomes tension after τ=0, reaching a peak of σ z ≈ 0.21atm before decreasing toward zero for large negative r. Hence the greatest normal tidal stress encountered in this scenario is the compression −σ z ≈ 0.22atm just before τ=0 and just outside r=0. This is much less than the yield stress of 100atm assumed by MTW. Thus we see that the prism would withstand the tidal stress in this scenario.
The prism, rotating in the inertial tetrad frame, would be subject to centripetal stress in addition to tidal stress. We now calculate the centripetal stress and the combined centripetal stress and tidal stress.
The components ω i of the angular velocity of the MTW prism, obtained by integrating Eulerʼs equations, are plotted in figure 9. The angular velocity component ω 3 about the long (z) axis of the MTW prism is much smaller than the other components, hence the rotation is about an axis perpendicular to the z axis, and the centripetal stress is in the z direction.
I estimate the maximum (over position) centripetal stress using the approximation for the maximum centripetal stress in a thin rod of uniform composition and cross-section, mass density ρ, and length H, rotating at angular velocity ω about an axis perpendicular to the rod and intersecting the rod at the center. The maximum stress occurs at the center of the rod, i.e., the waist of the MTW prism. The angular velocity history for this scenario is plotted in figure 10. The resulting centripetal stress plus tidal stress is shown in figure 11, along with tidal stress alone. The peak centripetal stress occurs at positive τ and overlaps the peak of tensile tidal stress, producing a peak centripetal stress plus tidal stress of σ z ≈ 0.26atm. This, too, is much less than the yield stress assumed by MTW. For a human, the stress would be momentarily uncomfortable, based on the discomfort criterion of Taylor and Wheeler [30]. axis of the prism is much smaller than the other components, hence the rotation is about an axis perpendicular to the z axis, which simplifies the estimation of centripetal stress.

Discussion
The calculations above show that an MTW prism falling deep into a Kerr black hole along a particular geodesic would withstand the theoretical tidal and centripetal stress. A prism falling into a real spinning black hole would be subjected to additional stresses and to radiation (e.g., [31]).  At the outer horizon, a falling object might [32], or might not [10], encounter a nonclassical firewall that destroys it. The debate seems unsettled; see, e.g., [33].
Inside the outer horizon, a prism that approaches the segment of the inner horizon at which the coordinate u blows up would encounter infinitely blueshifted radiation from the external Universe [9,34]. The prism in this example does not approach that segment of the inner horizon, hence it avoids such radiation as well as the weak, oscillatory, null curvature singularity that is predicted to form there [35][36][37] as a result of superposition of infalling metric peturbations and outflowing metric peturbations generated by matter that collapsed to form the black hole.
The prism is predicted to approach and cross the other segment of the inner horizon, the segment called the outgoing inner horizon by Marolf and Ori [38]. They predict that an infalling observer approaching the Kerr outgoing inner horizon would experience, to leading order in a linear expansion, a finite change in metric perturbation during a proper time interval that decreases exponentially as the coordinate time u (their v) at which the observer falls through the event horizon blows up. In this late-infall time limit, the proper time during which an observer would experience this finite metric change vanishes; the change becomes a step discontinuity in the perturbed metric, which Marolf and Ori call an effective gravitational shock wave. The prism avoids this shock by avoiding late infall. It would nevertheless be stressed to some extent that would depend on the details of collapse and accretion.
The prism would probably encounter infalling matter that collapsed to form the black hole or fell in later. Inward of this matter, the spacetime would not be Kerr, and the validity of this simulation would end, even if the prism does not collide with the matter (for example, if the matter consists of undisrupted stars). This end of validity would probably occur at some r>0 and τ<0 before some and perhaps before any of the interesting simulated dynamics and stresses occur.

Conclusion
I have simulated numerically the fall of the MTW prism into a high-spin (a/m=0.998) supermassive (m M 4 10 6 =´) Kerr black hole at a constant, near-equatorial latitude ( tan 1155 68 1 ( ) q = -) with enough energy (E=7 μ) to surmount the radial potential barrier near the throat of the black hole. I have also simulated the rotation of the prism within a comoving inertial frame of reference, driven by tidal torque, and calculated the tidal and centripetal stresses in the prism assuming a near-worst-case orientation of the prism at zero radius, and zero angular momentum there. The calculated maximum stress is about 0.26atmospheres, much less than the yield stress assumed by MTW.