The orbital Lense–Thirring precession in a strong field

Abstract

We study the exact evolution of the orbital angular momentum of a massive particle in the gravitational field of a Kerr black hole. We show analytically that, for a wide class of orbits, the angular momentum’s hodograph is always close to a circle. This applies to both bounded and unbounded orbits that do not end up in the black hole. Deviations from the circular shape do not exceed $\approx 10\%$ and $\approx 7\%$ for bounded and unbounded orbits, respectively. We also find that nutation provides an accurate approximation for those deviations, which fits the exact curve within $\sim 0.01\%$ for the orbits of maximal deviation. Remarkably, the more the deviation, the better the nutation approximates it. Thus, we demonstrate that the orbital Lense–Thirring precession, originally obtained in the weak-field limit, is also a valid description in the general case of (almost) arbitrary exact orbits. As a by-product, we also derive the parameters of unstable spherical timelike orbits as a function of their radii and arbitrary rotation parameter a and Carter’s constant Q. We verify our results numerically for all the kinds of orbits studied.

Appendices

A Numeric solution of geodesic equations

The geodesic equations can be recast into a form that is more convenient for numeric simulations [43,44,45]. Although the number of equations increases, now there is no need to track signs of the square roots present in the initial form. The equations to be solved are

$$\begin{aligned} \frac{dr}{d\gamma }= & {} r_1 \end{aligned}$$

(68)

$$\begin{aligned} \frac{d r_1}{d\gamma }= & {} \frac{1}{2E}\frac{\,\mathrm{d}R(r)}{\,\mathrm{d}r} \nonumber \\= & {} 2r^3 + r\left( a^2 - \xi ^2 - \eta \right) + \left( a-\xi \right) ^2 + \eta - \frac{r}{E^2}\left( 2r^2 - 3r + a^2\right) \end{aligned}$$

(69)

$$\begin{aligned} \frac{d\theta }{d\gamma }= & {} \theta _1 \end{aligned}$$

(70)

$$\begin{aligned} \frac{d\theta _1}{d\gamma }= & {} \frac{1}{2E}\frac{\,\mathrm{d}\varTheta (\theta )}{\,\mathrm{d}\theta }=\cos \theta \left[ a^2(E^{-2}-1)\sin \theta +\frac{\xi ^2}{\sin ^3\theta }\right] \end{aligned}$$

(71)

$$\begin{aligned} \frac{d\phi }{d\gamma }= & {} \frac{\xi }{\sin ^2\theta } - a +\frac{a(r^2+a^2-\xi a)}{r^2-2r+a^2}, \end{aligned}$$

(72)

where $d\tau /d\gamma =\rho ^2/E$, $\xi = L_z/E$, and $\eta =Q/E^2$. Also, the following relations hold between the functions:

$$\begin{aligned} r_1^2= & {} r^4 + \left( a^2 - \xi ^2 - \eta \right) r^2 \nonumber \\&+ 2\left[ (a-\xi )^2 + \eta \right] r - a^2\eta - \frac{r^2}{E^2}\left( r^2-2r+a^2\right) , \end{aligned}$$

(73)

$$\begin{aligned} \theta _1^2= & {} \eta - \cos ^{2}{\theta }\left[ a^2(E^{-2}-1) + \frac{\xi ^2}{\sin ^{2}{\theta }}\right] . \end{aligned}$$

(74)

This set of differential equations was numerically solved with either the classical or variable step [21] fourth-order Runge–Kutta method. That numeric solution was then used to draw the hodographs in Figs. 2, 3, 5, and 6. For convenience and reproducibility, we summarize the parameters for those hodographs in Table 2. Together with $r_1$ and $\theta _1$ evaluated through relations (73) and (74), they yield initial conditions for equations of motion (68)–(72). Also, we always choose $r_1<0$ and $\theta _1>0$ at the start.

Table 2 Initial conditions for orbits with noncircular hodographs

B Behavior of functions E(y, a) and $L_z(y,a)$ (the parameters of stable circular equatorial orbits)

Consider functions E(y, a) and $L_z(y,a)$, which are given by (16) and (17), in the range $y\in (0,1/r_{\mathrm{ph}}^{\pm })$. Their derivatives are

$$\begin{aligned} \frac{\partial E}{\partial y}= & {} -\frac{1 - 6y \pm 8ay^{3/2} - 3a^2y^2}{2\left( 1 - 3y \pm 2ay^{3/2}\right) ^{3/2}}, \end{aligned}$$

(75)

$$\begin{aligned} \frac{\partial L_z}{\partial y}= & {} \mp \frac{(1 - 6y \pm 8ay^{3/2} - 3a^2y^2)(1\pm ay^{3/2})}{2\left[ y\left( 1 - 3y \pm 2ay^{3/2}\right) \right] ^{3/2}}. \end{aligned}$$

(76)

They vanish simultaneously at $y = 1/r_{\mathrm{isco}}^{\pm }$, which satisfies [18]

$$\begin{aligned} 1 - 6y \pm 8ay^{3/2} - 3a^2y^2 = 0. \end{aligned}$$

(77)

Besides, $E\rightarrow 1$ and $L_z\rightarrow \pm \infty$ as $y\rightarrow 0$, and $E\rightarrow +\infty$ and $L_z\rightarrow \pm \infty$ as $y\rightarrow y_{\pm } = 1/r_{\mathrm{ph}}^{\pm }$. Therefore, at $y = 1/r_{\mathrm{isco}}^{\pm }$, E(y, a) must have a minimum, and $L_z(y,a)$ must have a minimum (maximum) at $y = 1/r_{\mathrm{isco}}^{+}$ ($y = 1/r_{\mathrm{isco}}^{-}$). In other words, both E(y, a) and $L_z^{2}(y,a)$ have a minimum at $y = 1/r_{\mathrm{isco}}^{\pm }$.

C Parameters of unstable spherical orbits with $E>1$ and arbitrary Q

Consider conditions (26) and (27) without omitting terms $\propto q\equiv 1/Q$:

$$\begin{aligned}&\epsilon ^{2} - 2 \epsilon x\cdot a y^{2} - x^{2}y^{2}(1 - 2 y) - y^{2}\varDelta _y- q\varDelta _y = 0, \end{aligned}$$

(78)

$$\begin{aligned}&- 2 \epsilon x \cdot a y - x^{2}y(1- 3 y) - y (1 - 3 y + 2 a^{2} y^{2}) + q(1-a^2 y) = 0 , \end{aligned}$$

(79)

where $\varDelta _y\equiv a^2y^2 - 2y + 1 > 0$.

Solving the second of the equations for $\epsilon$ and subtituting the solution to the first equation leads to a quadratic equation with respect to $x^2$:

$$\begin{aligned} Ax^4 + 2Kx^2 + C =0, \end{aligned}$$

(80)

where

$$\begin{aligned} A\equiv & {} -(4 a^{2} y^{3} - 9 y^{2} + 6 y - 1)<0, \qquad 1/r_{\mathrm{ph}}^{-}< y < 1/r_{\mathrm{ph}}^{+}, \end{aligned}$$

(81)

$$\begin{aligned} K\equiv & {} 2a^2y^2\varDelta _y + A - q\left( a^2 + a^2y - \frac{3y-1}{y}\right) , \end{aligned}$$

(82)

$$\begin{aligned} C\equiv & {} \left( 2a^2y^2 - 3y + 1 + \frac{q(a^2y - 1)}{y}\right) ^2\ge 0. \end{aligned}$$

(83)

Let us now study how solutions to the quadratic equation behave at $y_{\pm }=1/r_{\mathrm{ph}}^{\pm }$. First, note that

$$\begin{aligned} \lim \limits _{y\rightarrow y_{\pm }}{A(y)} =\lim \limits _{y\rightarrow y_{\pm }}{[(3y-1)^2 - 4a^2y^3]}=0. \end{aligned}$$

(84)

Since $C > 0$ and $A\rightarrow 0^{-}$ as $y\rightarrow y_{\pm }$, the roots of (80) have opposite signs. One of them tends to infinity and the other tends to a finite number. Their specific signs depend on the sign of K. If $K>0$, the infinite root is positive and the finite root is negative. Otherwise, the infinite root is negative and the finite root is positive. If $K=0$, both roots tend to infinities of opposite signs.

If $q=0$ or, by continuity, $q<<1$, then $\lim \limits _{y\rightarrow y_{\pm }}{K}>0$, and at both $y_{-}$ and $y_{+}$ there is a positive infinite root. This root is the one that leads to solution (30)–(32) obtained under assumption $q<<1$.

For arbitrary $q>0$, it proves helpful to actually solve equation (80). A positive root is found from relation

$$\begin{aligned} -x^2 A= & {} (2a^2y^2 - 3y + 1)^2 - q\left( a^2y+a^2 - \frac{3y-1}{y}\right) \nonumber \\&+ 2a\varDelta _y\left[ \frac{\sqrt{\left( 2q + y(3y-1)\right) ^2 - y^2 A}}{2\sqrt{y}}-ay^2\right] . \end{aligned}$$

(85)

At $y_{+}$, which satisfies $3y-1 = 2ay^{3/2}>0$, the coefficient that multiplies q is reduced as follows:

$$\begin{aligned}&-a^2y-a^2 + \frac{3y-1}{y} + \frac{2a\varDelta _y}{y^{1/2}} = -a^2y-a^2 - 2ay^{1/2} + 2a^3y^{3/2} + \frac{2a}{y^{1/2}} \nonumber \\&\quad = -a^2y-a^2 - 2ay^{1/2} + a^2(3y-1) + \frac{2a}{y^{1/2}} = \frac{2a(1-y)(1-ay^{1/2})}{y^{1/2}} \nonumber \\&\quad = \frac{ay(1 + y - 2ay^{1/2})}{y^{3/2}} =\frac{a(y(1+y)-3y+1)}{y^{3/2}} = \frac{a(1-y)^2}{y^{3/2}}. \end{aligned}$$

(86)

Hence, as soon as $0<a<1$,

$$\begin{aligned} x^2 = -\frac{(2a^2y^2 - 3y + 1)^2 + \frac{a(1-y)^2}{y^{3/2}}q}{A} \rightarrow +\infty \quad \text{ as } \quad A\rightarrow 0^{-}. \end{aligned}$$

(87)

At $y_{-}$, which satisfies $3y-1 = -2ay^{3/2}<0$, the result depends on the magnitude of q. If $q\ge -y(3y-1)/2 = ay^{5/2}$, a similar calculation leads to a right-hand side of (85) that vanishes as $A\rightarrow 0^{-}$. Thus, in this case the root tends to a finite positive value. Otherwise, the root tends to an infinite positive value.

The function $a[y_{-}(a)]^{5/2}$ grows in the range $0\le a\le 1$ reaching its maximal value 1 / 32 at $a=1$. Consequently, for a given $a\in (0,1)$, there is a critical value of q that is below 1 / 32, which separates the cases of bounded and unbounded growth of $x^2$ in a neighborhood of $y_{-}$. Also, if q is sufficiently small, the growth is always unbounded, which corresponds to the limiting case represented by (30)–(32).

Another feature of $x^2$ is that, for sufficiently small q, it vanishes at a point $y_0\in (y_{-},y_{+})$ where it also reaches a minimum. Indeed, for $q=0$ it is trivial to verify this. If $q<<1, q\ne 0$, the minimum value must simultaneously satisfy equations

$$\begin{aligned}&Ax^4 + 2Kx^2 + C =0, \end{aligned}$$

(88)

$$\begin{aligned}&x^4\frac{\partial A}{\partial y} + 2x^2\frac{\partial K}{\partial y} + \frac{\partial C}{\partial y} =0. \end{aligned}$$

(89)

Since C and $\partial C/\partial y$ have a common root, $x^2=0$ satisfies this last set of equations at that common root and, by continuity, is a minimum of the non-negative function $x^2$. The case of q that is large enough for the common root to become complex is beyond the scope of this article.

To ensure that $\epsilon >0$, the following signs must be chosen for x:

$$\begin{aligned} x = \left\{ \begin{array}{ll} \sqrt{x^2}, &{} \quad y_0\le y< y_{+}, \\ -\sqrt{x^2}, &{} \quad y_{-}<y<y_0. \end{array} \right. \end{aligned}$$

(90)

Finally, the x found is used to evaluate $\epsilon$ and $\lambda _z$:

$$\begin{aligned} \epsilon= & {} x\frac{3y-1}{2a} -\frac{1}{2ax}\left( \frac{q(a^2 y-1)}{y} + 2a^2y^2 - 3y+1\right) , \end{aligned}$$

(91)

$$\begin{aligned} \lambda _z= & {} x + a\epsilon . \end{aligned}$$

(92)

Note that $\epsilon (y_0)$ is a well-defined number, because term 1 / x multiplies an expression that vanishes at that point. From (78), $\epsilon (y_0) = \sqrt{(y_0^2 + q)\varDelta _y(y_0)}$.

As it was pointed out, for q that is small enough and $0<a<1$, $\lim \limits _{y\rightarrow y_{\pm }}{x^2}=+\infty$. Therefore,

$$\begin{aligned}&\lim \limits _{y\rightarrow y_{\pm }}{\epsilon }=+\infty , \end{aligned}$$

(93)

$$\begin{aligned}&\lim \limits _{y\rightarrow y_{+}}{\lambda _z} = +\infty , \quad \lim \limits _{y\rightarrow y_{-}}{\lambda _z} = -\infty , \end{aligned}$$

(94)

$$\begin{aligned}&\lim \limits _{y\rightarrow y_{\pm }}{\frac{\lambda _z}{\epsilon }} = \lim \limits _{y\rightarrow y_{\pm }}{\left. \frac{\lambda _z}{\epsilon }\right| _{q=0}} = -\left. \frac{1 - 3 y + a^{2} y^{2} + a^{2} y^{3} }{a y^{2} \left( 1 - y \right) }\right| _{y=y_{\pm }}. \end{aligned}$$

(95)

In the last equation we have used the fact that the ratio does not depend on q. Indeed, if we divide (78) by $x^2$, terms that contain q vanish as $x^2\rightarrow +\infty$ and have no impact on the value of the limit.