Kepler's second law in General Relativity

Question

Kepler's second law, that planets in orbits sweep equal area in equal time, is a consequence of orbital angular momentum conservation. In the case of Schwarzschild spacetime, the angular orbital momentum is still conserved. Is Kepler's second also valid in Schwarzschild spacetime? Is there mathematical proof confirming the law or otherwise? Can any general remark be made regarding other astrophysical metrics.

Edit:

To make the question more precise and to clarify the points in the comments: if the following coordinate system is used, $$ds^2=-\left(1-\frac{2M}{r}\right)dt^2+\frac{1}{\left(1-\frac{2M}{r}\right)}dr^2+d\Omega^2$$

Then the time is the coordinate time and the area is the area on the $\theta=\frac{\pi}{2}$ plane and calculated with the induced metric on the plane.

Using Schwarzschild metric would give us correction to the Kepler's laws even for the Second law. I a wish to know how to go about calculating that correction. Hints will suffice.

Answer 1

You can find it in Chandrasekhar's textbook ("The mathematical theory of black holes" Chap 3, sec. 19).

Only you need is to show the conservation laws $r^2 d\phi/d\tau$ of particle in Schwarzschild metric.

Edit 1: How to derive approximate Kepler's second law from GR

The static metric on $\theta=\pi/2$ $$ ds^2=\frac{1}{1-\frac{1}{r}}dr^2+r^2 d\phi $$ the mesure $\sqrt{\gamma}\sim r$ as $r\to\infty$, thus the area element is $$ d A\sim \frac{1}{2}r^2 d\phi $$ which implies $2d A/d\tau\sim r^2 d\phi/d\tau ={\rm const}$.

Edit 2: There is no exact Kepler's second law in GR.

Now, you should consider the full integral $$ \int_{1}^{r_0} \frac{r dr}{\sqrt{1-\frac{1}{r}}} $$ If you want see the correction, I suggest to take a limit again, such that the sun can be treat as point (don't understand me wrong, I have used the metric of sun, not the black hole, even there is only a little tiny difference.) The full expansion of integrand is $$ \sum _{k=0}^{\infty } \frac{(2 k-1)!!}{\left(2^k k!\right) r^{k-1}} $$ after integration, you will see that the leading term is Kepler's second law, the remaining term is "correction".

Answer 2

A few simplifying assumptions are made: First, we assume that the contribution of the orbiting mass to the stress-energy tensor is negligible. Second, we will only consider motions over infinitesimal angles, so that the radial coordinate can be assumed constant. Third, we assume that any orbits take place at "large" distances (it will be made somewhat clearer what is meant by that). These are not necessary assumptions, but since this is more of a phenomenological study than a rigorous derivation, they should be justified.

We start by deriving the classical case. The Lagrangian of a particle of mass $m$ moving in a 2-dimensional central potential is, expressed in polar coordinates: \begin{equation} \mathcal{L} = \frac{m}{2}\left(\dot{r}^2+r^2\dot{\varphi}^2\right) - V(r) \end{equation} The conjugate momentum to $\varphi$ is then immediately found to be \begin{equation} J = \frac{\partial\mathcal{L}}{\partial\dot{\varphi}} = mr^2\dot{\varphi} \end{equation} which is seen to be conserved as $\mathcal{L}$ does not depend on $\varphi$. For a two-dimensional manifold with metric $g$, the area element is defined as a 2-form

\begin{equation}\label{eq:2D area}\tag{1} dA = \sqrt{\vert{\text{det }g}\vert}(de_1\land de_2) \end{equation}

Where $de_1$ and $de_2$ are coordinate 1-forms on the cotangent space of the manifold. If we now again consider $\mathbb{R}^2$ with the polar coordinate metric, we get

\begin{equation} \text{det }g = \begin{vmatrix} 1&0\\0&r^2 \end{vmatrix} = r^2 \end{equation}

and therefore

\begin{equation} dA = r(dr\wedge d\varphi) \end{equation}

Integrating over $r$ we get (using a somewhat nonstandard notation, but it will have to do) \begin{equation}\label{eq:cursd notation}\tag{2} d_\varphi A = \frac{r^2}{2}d\varphi \end{equation}

and, viewing everything as a function of the time $t$,

\begin{equation} \frac{dA}{dt} = \frac{r^2}{2}\frac{d\varphi}{dt} = \frac{J}{2m} \end{equation}

Which is Kepler's second law, assuming that $r$ can be taken as constant over the infinitesimal angle $d\varphi$. Note that we did not make any bold assumptions about the form of $V(r)$, meaning it holds true for a Newtonian potential with small general relativistic correction terms, as long as the underlying spacetime is assumed to be flat.

We are now interested in calculating the area in curved spacetime. We study what happens if we replace the metric in equation \eqref{eq:2D area} with the metric on the spacelike $r$, $\varphi$ slice of Schwarzschild spacetime. The metric on the $r$, $\varphi$ plane, denoted $\Sigma$, is of the form

\begin{equation} g_\Sigma = \left(1-\frac{2M}{r}\right)^{-1}dr^2 + r^2\sin^2\theta d\varphi^2 \end{equation}

Without loss of generality, we will set $\theta = \pi/2$ in the following. Restricting the metric like this is allowed since $\partial_t$ is a Killing field, i.e. a generator of an isometry, and no motion rakes place in the $d\theta$-direction. Another Killing field is $\partial_\varphi$, from which it follows that the quantity

\begin{equation} L = g_\Sigma(\partial_\varphi, u) = g_\Sigma(\partial_\varphi,\dot{\varphi}\partial_\varphi) = r^2\dot{\varphi} \end{equation}

is a constant of motion, which can be interpreted as angular momentum. The determinant of $g_\Sigma$ is

\begin{equation} \text{det }g_\Sigma = \frac{r^2}{\left(1-\frac{2M}{r}\right)} \end{equation}

which means that the area element is

\begin{equation} dA = \frac{r}{\sqrt{1-\frac{2M}{r}}}(dr\wedge d\varphi) \end{equation}

Again we want to integrate over $r$ to get an expression which only contains $d\varphi$. Using the notation from \eqref{eq:cursd notation} we get

\begin{equation}\label{eq:GR area}\tag{3} d_\varphi A = \int_{r_0}^r \frac{r^\prime}{\sqrt{1-\frac{2M}{r^\prime}}}dr^\prime\wedge d\varphi \end{equation} Where $r_0$ is taken to be larger than $2M$. Calculating this integral explicitly is a tedious task and does not yield much physical useful information in terms of how we need to "correct" Kepler's second law by adapting it to curved spacetime. Instead we choose to Taylor expand the integrand to first order around some large value $R$, where "large" means "closer to $r$". The first derivative of $\sqrt{\text{det }g_\Sigma}$ is: \begin{equation} \frac{d}{dr}\frac{r}{\sqrt{1-\frac{2M}{r}}} = \dfrac{r-3M}{\left(1-\frac{2M}{r}\right)^{3/2}r} = \frac{\left(\sqrt{\text{det }g_\Sigma}\right)^3}{r^2}\frac{r-3M}{r} \end{equation} So that we get \begin{align} \sqrt{\text{det }g_\Sigma} &\;= \sqrt{\text{det }g_\Sigma(R)} + \frac{\left(\sqrt{\text{det }g_\Sigma(R)}\right)^3}{R^2}\frac{R-3M}{R}(r-R) + \mathcal{O}((r-R)^2) = \\ &\; = \sqrt{\text{det }g_\Sigma(R)}\left[1 - \frac{\left(\sqrt{\text{det }g_\Sigma(R)}\right)^2}{R^2}\frac{R-3M}{R}(R-r)\right] + \mathcal{O}((R-r)^2) \end{align} Now making the further assumption that $R\gg2M$, meaning $\sqrt{\text{det }g_\Sigma(R)}\approx R$ we get \begin{equation} \sqrt{\text{det }g_\Sigma}\approx R\left[1-\frac{R-3M}{R}(R-r)\right] \end{equation} Plugging this back into \eqref{eq:GR area} yields \begin{align} d_\varphi A &\;\approx \int_{r_0}^rR\left[1-\frac{R-3M}{R}(R-r^\prime)\right]dr^\prime\wedge d\varphi \approx\\ &\;\approx \left[Rr - (R-3M)r + \frac{r^2}{2}\left(\frac{R-3M}{R}\right)\right]d\varphi \end{align} Where contributions from $r_0$ have been dropped as well. Now taking $R\rightarrow r$ we end up with \begin{equation} d_\varphi A\approx\left(\frac{1}{2}r^2 - \frac{3Mr}{2}\right)d\varphi \end{equation} Finally, differentiating with respect to the curve parameter $s$ we get \begin{equation}\label{eq:schwarzschild approx} \frac{dA}{ds} \approx \frac{r^2\dot{\varphi}}{2} - \frac{3Mr\dot{\varphi}}{2} = \frac{L}{2} - \frac{3Mr\dot{\varphi}}{2} \end{equation} Where we recognize the first term as Kepler's second law, and the second as the correction.

Answer 3

The answer is yes - with respect to clock time. With respect to coordinate time, however, there are qualifiers. Time is warped by gravity.

Not only will I show you the mathematical proof, I will actually show you what the general relativistic version of the equation $$\frac{d^2}{dt^2} = -\frac{μ}{r^3}\tag{Kepler}\label{1}$$ looks like - for both the Schwarzschild solution and Reissner-Nordström solution, so you can draw your own conclusions. For some reason, you don't see this anywhere in the literature (at least as far as I can find), even though it's obvious low-hanging fruit.

For Reissner-Nordström, we have the following line element: $$Φ dt^2 - α \left(\frac1Φ dr^2 + r^2 \left(dθ^2 + (\sin θ dφ)^2\right)\right) ∈ \left\{dτ^2, 0, -α dℓ^2\right\},$$ where $$Φ = 1 - \frac{2αμ}{r} + \frac{αϙ}{r^2},\quad μ = G M,\quad ϙ = \frac{μ_0}{4π} G q^2,$$ with $G$ being Newton's coefficient, $M$ the mass of the gravitating object, and $q$ its electrical charge. The three cases are those adapted to time-like geodesics, with clock-time $τ$, null geodesics with a zero line element and space-like geodesics, with proper length $ℓ$.

The parameter $α$ is used to distinguish the relativistic case $α = 1/c^2$ from the non-relativistic case $α = 0$. Strictly speaking, there is no sensible non-relativistic (i.e. $α → 0$) limit of the above line element ... however you can devise one, by pushing this up into the fifth dimension as a null line element, by just adopting a new coordinate $τ = t + α u$ for proper time. Substitute, bring everything over to one side, and divide out the $α$ and you get: $$ \left(\frac{2μ}{r} - \frac{ϙ}{r^2}\right) dt^2 + 2 dt du + α du^2 + \frac{r^2}{r^2 - 2αμr + αϙ} dr^2 + r^2 \left(dθ^2 + (\sin θ dφ)^2\right) = 0. $$ As $α → 0$, this approaches a null line element that can be converted to Cartesian form as: $$ \left(\frac{2μ}{r} - \frac{ϙ}{r^2}\right) dt^2 + 2 dt du + dx^2 + dy^2 + dz^2 = 0\quad \left(r = \sqrt{x^2 + y^2 + z^2}\right). $$ The fact that there isn't a clean-cut conversion to Cartesian form in the relativistic case $α ≠ 0$ is an indication that there is a warping of space, along the radial direction $r$, not just of time $t$. You could still write it in Cartesian form, by extracting out the Cartesian part of the line element $$ \left(\frac{2μ}{r} - \frac{ϙ}{r^2}\right) dt^2 + 2 dt du + α du^2 + α \frac{2μr - ϙ}{r^2 - 2αμr + αϙ} dr^2 + dx^2 + dy^2 + dz^2 = 0, $$ provided you set $$r = \sqrt{x^2 + y^2 + z^2},\quad dr = \frac{x dx + y dy + z dz}{r}.$$

If you work out the geodesic equation for the non-relativistic limit, you'll get the above-mentioned \ref{1} equation - in the case $ϙ = 0$ of zero charge. A footnote on that: had I set $αϙ = Gq^2/(4πε_0c^4) = α^2Gq^2/(4πε_0)$, then we would have gotten $ϙ → 0$ in the limit $α → 0$. So, there are actually two different non-relativistic limits, depending on how you write $ϙ$; one of them being non-trivial: a "cyberpunk" version of Reissner-Nordström. The non-trivial limit has the modified equation $$\frac{d^2}{dt^2} = -\frac{μ}{r^3} + \frac{ϙ}{r^4}\tag{Kepler EM}\label{2}.$$

If you work out the geodesic equation for the relativistic version ($α ≠ 0$) of the five-dimensional line element, you'll get the modified Kepler equations we're going to derive. In both cases, $du/dt$ reduces to minus the Lagrangian per unit mass for an orbiting body, and $u$ reduces to minus the action per unit mass, up to an additive constant.

For both \ref{1} and \ref{2}, $$_t = ×\frac{d}{dt},\tag{h0}\label{3}$$ is a constant of motion. The direction of $_t$ is perpendicular to the orbital plane (so the orbital plane is constant), and the magnitude $h_t = |_t|$ is also constant. That's twice the "areal velocity" of Kepler's Second Law.

For Reissner-Nordström, for time-like geodesics, this gets modified to: $$_τ = ×\frac{d}{dτ}.\tag{hα}\label{4}$$ It's a constant of motion. In contrast, the $_t$ vector of \ref{3} is only constant in direction, not in magnitude. The areal velocity changes with coordinate time $t$, because of time being warped by gravity, though it remains constant with respect to clock time $τ$.

In vector and polar form (from \ref{5}, below, with $s = τ$): $$\frac{{h_τ}^2}{r^2} = \left|\frac{d}{dτ}\right|^2 - \left(\frac{dr}{dτ}\right)^2 = r^2 \left(\left(\frac{dθ}{dτ}\right)^2 + \left(\sin θ \frac{dφ}{dτ}\right)^2\right)$$ for \ref{4}, and for \ref{3} with $τ = t$.

The Modified Kepler Equation

The Geodesic Equations

The Reissner-Nordström line-element $g_{μν} dx^μ dx^ν$ (the summation convention is being used here and below) has the following as the non-zero components for the metric: $$g_{00} = Φ,\quad g_{11} = -\frac{α}{Φ},\quad g_{22} = -α r^2,\quad g_{33} = -α (r \sin θ)^2,$$ with the coordinates indexed as $\left(x^0, x^1, x^2, x^3\right) = (t, r, θ, φ)$. The inverse metric has, as its sole non-zero components $$g^{00} = \frac1Φ,\quad g^{11} = -\frac{Φ}{α},\quad g^{22} = -\frac1{α r^2},\quad g^{33} = -\frac1{α (r \sin θ)^2}.$$ The Levi-Civita connection associated with the metric (using $∂_μ$ to denote $∂/∂x^μ$) $$ Γ^ρ_{μν} = \frac12 g^{ρσ} \left(∂_ν g_{μσ} - ∂_σ g_{νμ} + ∂_μ g_{σν}\right). $$ Here, the sole non-zero components are $$ Γ^0_{01} = Γ^0_{10} = \frac1Φ \left(\frac{αμ}{r^2} - \frac{αϙ}{r^3}\right),\quad Γ^1_{00} = Φ \left(\frac{αμ}{r^2} - \frac{αϙ}{r^3}\right),\quad Γ^1_{11} = -\frac1Φ \left(\frac{μ}{r^2} - \frac{ϙ}{r^3}\right),\\ Γ^2_{12} = Γ^2_{21} = \frac1r = Γ^3_{13} = Γ^3_{31},\quad Γ^1_{22} = -Φ r,\quad Γ^1_{33} = -Φ r \sin^2 θ,\\ Γ^3_{23} = Γ^3_{32} = \cot θ,\quad Γ^2_{33} = -\sin θ \cos θ. $$

The modified Kepler equation comes from the Geodesic Equation, which has the general form (which the link fails to note) $$\frac{d^2x^ρ}{ds^2} + Γ^ρ_{μν} \frac{dx^μ}{ds} \frac{dx^ν}{ds} = N_s \frac{dx^ρ}{ds},$$ where $s$ is an arbitrary parameter. When $s$ is either proper length $ℓ$ or proper time $τ$, $N_s = 0$. In contrast, for null geodesics, $N_s$ can't be so readily eliminated. Under a change to a different parameter $s' = s'(s)$, $N_s$ transforms to $$N_{s'} = \left(\frac{ds'}{ds}\right)^{-1} N_s - \left(\frac{ds'}{ds}\right)^{-2} \frac{d^2s'}{ds^2}.$$ A natural setting - at least for null and time-like geodesics - is to just set $s = t = x^0$ and extract $N$ from the equation for $d^2x^0/ds^2$.

The resulting geodesic equations are: $$ \frac{d^2t}{ds^2} + \frac2Φ \left(\frac{αμ}{r^2} - \frac{αϙ}{r^3}\right) \frac{dr}{ds} \frac{dt}{ds} = N_s \frac{dt}{ds},\\ \frac{d^2r}{ds^2} + Φ \left(\frac{μ}{r^2} - \frac{ϙ}{r^3}\right) \left(\frac{dt}{ds}\right)^2 -\frac1Φ \left(\frac{αμ}{r^2} - \frac{αϙ}{r^3}\right) \left(\frac{dr}{ds}\right)^2 - Φ r \left(\left(\frac{dθ}{ds}\right)^2 + \left(\sin θ \frac{dφ}{ds}\right)^2\right) = N_s \frac{dr}{ds},\\ \frac{d^2θ}{ds^2} + \frac2r \frac{dr}{ds} \frac{dθ}{ds} - \sin θ \cos θ \left(\frac{dφ}{ds}\right)^2 = N_s \frac{dθ}{ds},\\ \frac{d^2φ}{ds^2} + \frac2r \frac{dr}{ds} \frac{dφ}{ds} + 2 \cot θ \frac{dθ}{ds} \frac{dφ}{ds} = N_s \frac{dφ}{ds}. $$

For the three types of geodesics, from the line element, we also have: $$Φ \left(\frac{dt}{ds}\right)^2 - α \left(\frac1Φ \left(\frac{dr}{ds}\right)^2 + r^2 \left(\left(\frac{dθ}{ds}\right)^2 + \left(\sin θ \frac{dφ}{ds}\right)^2\right)\right) = L ∈ \left\{\left(\frac{dτ}{ds}\right)^2, 0, -α \left(\frac{dℓ}{ds}\right)^2\right\}.$$

First Derivative And Areal Velocity

The radius vector can be written in polar form as $ = r \hat{}$, in terms of the orthonormal triad: $$\hat{} = (\sin θ \cos φ, \sin θ \sin φ, \cos θ),\quad \hat{} = (\cos θ \cos φ, \cos θ \sin φ, -\sin θ),\quad \hat{} = (\sin φ, \cos φ, 0). $$ To find the derivatives $d/ds$ and $d^2/ds^2$, we may use the following: $$ \frac{d\hat{}}{ds} = \frac{dθ}{ds} \hat{} + \sin θ \frac{dφ}{ds} \hat{},\quad \frac{d\hat{}}{ds} = -\frac{dθ}{ds} \hat{} + \cos θ \frac{dφ}{ds} \hat{},\quad \frac{d\hat{}}{ds} = -\sin θ \frac{dφ}{ds} \hat{} - \cos θ \frac{dφ}{ds} \hat{}, $$ to write $$ \frac{d}{ds} = \frac{dr}{ds} \hat{} + r \frac{dθ}{ds} \hat{} + r \sin θ \frac{dφ}{ds} \hat{}. $$ Noting the relations $$ \hat{}×\hat{} = \hat{} = -\hat{}×\hat{},\quad \hat{}×\hat{} = \hat{} = -\hat{}×\hat{},\quad \hat{}×\hat{} = \hat{} = -\hat{}×\hat{}, $$ then, from this, we obtain $$ _s = ×\frac{d}{ds} = r^2 \left(\frac{dθ}{ds} \hat{} - \sin θ \frac{dφ}{ds} \hat{}\right),$$ $$\frac{{h_s}^2}{r^2} = \left|\frac{d}{ds}\right|^2 - \left(\frac{dr}{ds}\right)^2 = r^2 \left(\left(\frac{dθ}{ds}\right)^2 + \left(\sin θ \frac{dφ}{ds}\right)^2\right).\tag{hs}\label{5} $$

Substituting this into the geodesic equations for $θ$ and $φ$, we obtain: $$ \frac{d}{ds}\left((r \sin θ)^2 \frac{dφ}{ds}\right) = N_s (r \sin θ)^2 \frac{dφ}{ds},\quad \frac{d}{ds}\left({h_s}^2\right) = 2 N_s {h_s}^2. $$ For both time-like geodesics, with $s = τ$ and $N_τ = 0$, and space-like geodesics, with $s = ℓ$ and $N_ℓ = 0$, $h_s$ is a constant. For all geodesics, the direction of $_s$ will prove to be a constant, as well, as we'll see once we find the modified Kepler equation.

Before we get to that, though, we can also reduce the geodesic equation for $t$ to: $$\frac{d}{ds}\left(Φ \frac{dt}{ds}\right) = N_s Φ \frac{dt}{ds}.$$ So, if we set $s = t$, this leads to the following normalization: $$N_t = \frac1Φ \frac{dΦ}{dt}.$$ For $s = t$, this leads to the following equation for ${h_s}^2$: $$\frac{d}{dt}\left(\frac{{h_t}^2}{Φ^2}\right) = 0.$$

Thus, it is not the areal velocity that is constant, when referring to coordinate time $s = t$, but the time-warped version of it is: $$\frac{h_t}Φ = \frac{r^2}{r^2 - 2αμr + αϙ} \left|×\frac{d}{dt}\right|.$$

Second Derivative

Taking a second derivative, we have $$\begin{align} \frac{d^2}{ds^2} - N_s \frac{d}{ds} &= \left(\frac{d^2r}{ds^2} - r \left(\left(\frac{dθ}{ds}\right)^2 + \left(\sin θ \frac{dφ}{ds}\right)^2\right) - N_s \frac{dr}{ds}\right) \hat{}\\ &+ r \left(\frac{d^2θ}{ds^2} + \frac2{r} \frac{dr}{ds} \frac{dθ}{ds} - \sin θ \cos θ \left(\frac{dφ}{ds}\right)^2 - N_s \frac{dθ}{ds}\right) \hat{}\\ &+ r \sin θ \left(\frac{d^2φ}{ds^2} + \frac2{r} \frac{dr}{ds} \frac{dφ}{ds} + 2 \cot θ \frac{dr}{ds} \frac{dφ}{ds} - N_s \frac{dφ}{ds}\right) \hat{}. \end{align}$$ By the geodesic equations for $θ$ and $φ$, the $\hat{}$ and $\hat{}$ components cancel out, leaving behind only the $\hat{}$ component, which can be reduced - using the geodesic equation for $r$ and the line element, as well as \ref{5} - to the following: $$ \frac{d^2}{ds^2} - N_s \frac{d}{ds} = -\frac{μ}{r^3} \left(L + \frac{3α{h_s}^2}{r^2}\right) + \frac{ϙ}{r^4} \left(L + \frac{2α{h_s}^2}{r^2}\right),\tag{Kepler GR}\label{7} $$ where $$ L ∈ \left\{\left(\frac{dτ}{ds}\right)^2, 0, -α \left(\frac{dℓ}{ds}\right)^2\right\} $$ for the three geodesic types.

For the $_s$ vector, this leads to $$\frac{d_s}{ds} = N_s _s,$$ showing that its direction is constant ... and that it is a constant of motion in the cases $_τ$ of time-like geodesics and even $_ℓ$ of space-like geodesics, where $N_τ = 0$ and $N_ℓ = 0$. For $s = t$, after substituting in for $N_t$, this reduces to $$\frac{d}{ds}\left(\frac{_t}{Φ}\right) = ,$$ thus showing that $$\frac{_t}{Φ} = \frac{r^2}{r^2 - 2αμr + ϙ} ×\frac{d}{dt}$$ is a constant of motion.

For time-like geodesics, with $s = τ$ and $N_τ = 0$, \ref{7} reduces to: $$\frac{d^2}{dτ^2} = -\frac{μ}{r^3} \left(1 + \frac{3α{h_τ}^2}{r^2}\right) + \frac{ϙ}{r^4} \left(1 + \frac{2α{h_τ}^2}{r^2}\right).$$ To synchronize this to clock time, it has to be combined with the first integral for the geodesic equation for $t$: $$\frac{d}{dτ}\left(Φ \frac{dt}{dτ}\right) = 0 → \frac{dt}{dτ} = \frac{A}{Φ}$$ with the integration constant $A$ calibrated using the line element, for $s = τ$: $$\frac{A^2}{Φ} - \frac{α}{Φ} \left(\frac{dr}{dτ}\right)^2 - α \frac{{h_τ}^2}{r^2} = 1 → A = \sqrt{Φ \left(1 + α \frac{{h_τ}^2}{r^2}\right) + α \left(\frac{dr}{dτ}\right)^2}.$$

Finally, for the case $s = t$, after substituting in for $N_t$, \ref{7} reduces to $$ \frac{d}{dt}\left(\frac1Φ \frac{d}{dt}\right) = -\frac{μ}{Φr^3} \left(L + \frac{3α}{r^2}\left|×\frac{d}{dt}\right|^2\right) + \frac{ϙ}{Φr^4} \left(L + \frac{2α}{r^2}\left|×\frac{d}{dt}\right|^2\right), $$ where $$L ∈ \left\{\left(\frac{dτ}{dt}\right)^2, 0, -α \left(\frac{dℓ}{dt}\right)^2\right\}$$ for the three geodesic types.

This is the modified Kepler equation. This covers finite-speed faster-than-light space-like geodesics, in addition to null and time-like geodesics, but the infinite speed space-like geodesics (i.e. those in the hyperplane orthogonal to the $t$ axis) are outside this coverage and have to be handled separately (e.g. by setting $s = r$).

Finally, to close this out, we'll also need to the expression for $L$. Here, it will just be presented in the case of time-like geodesics: $$\begin{align} L = \left(\frac{dτ}{dt}\right)^2 &= Φ \left(\frac{dt}{dt}\right)^2 - \frac{α}{Φ} \left(\frac{dr}{dt}\right)^2 - α r^2 \left(\left(\frac{dθ}{dt}\right)^2 + \left(\sin θ \frac{dφ}{dt}\right)^2\right) \\ &= Φ - \frac{α}{Φ} \left(\frac{dr}{dt}\right)^2 - \frac{α}{r^2} \left|×\frac{d}{dt}\right|^2. \end{align}$$