# Keplerian orbits in special relativity

Based on PLANCKS 2019. Problem 4.

Also (Goldstein et al 2002, ex. 7.26). It is analyzed in depth in (Lemmon & Mondragon 2010). This post is based on this paper but is simplified and augmented to what the PLANCKS problem asks.

For a particle in potential field, the relativistic Lagrangian is1

$L = - m c^2 \gamma^{- 1} - V$

Utilizing the previous result, the relativistic Lagrangian for the sun-planet system, where the potential energy is2 the Newtonian gravitational energy, is

$L = - m c^2 \gamma^{- 1} + \mu m r^{- 1} \quad \mu = G M$

The problem is considered in polar coordinates $(r, \theta)$. That is

$v^2 = \dot{r}^2 + r^2 \dot{\theta}^2 \Rightarrow \gamma^{- 1} = (1 - v^2 / c^2)^{1 / 2} = (1 - (\dot{r}^2 + r^2 \dot{\theta}^2) / c^2)^{1 / 2}$

The equations of motions are derived from Lagrange’s equations

$\frac{d}{d t} \left( \frac{\partial L}{\partial \dot{q}_{i}} \right) - \frac{\partial L}{\partial q_{i}} = 0 \quad q_{i} \in \lbrace r, \theta \rbrace$

which take the form

$q_{i} = r \longrightarrow - m c^2 \frac{d}{d t} (\partial _{\dot{r}} \gamma^{- 1}) + m c^2 \partial _{r} \gamma^{- 1} + \mu m r^{- 2} = 0 \Rightarrow \frac{d}{d t} (\gamma \dot{r}) - \gamma r \dot{\theta}^2 + \mu r^{- 2} = 0$

$q_{i} = \theta \longrightarrow - m c^2 \frac{d}{d t} \partial _{\dot{\theta}} \gamma^{- 1} = 0 \Rightarrow m \frac{d}{d t} (\gamma r^2 \dot{\theta}) = 0 \Rightarrow \frac{d}{d t} (\gamma r^2 \dot{\theta}) = 0$

The first gives the equation of motion

$\gamma \ddot{r} + \dot{\gamma} \dot{r} + \mu r^{- 2} - \gamma r \dot{\theta}^2 = 0$

whereas the second a quantity that is conserved, analogous to Newtonian angular momentum

$l = \gamma r^2 \dot{\theta} =\operatorname{const}. \Rightarrow \gamma r \dot{\theta}^2 = \frac{l^2}{\gamma r^3}$

Eliminating time

$\dot{r} = \frac{d}{d \theta} \dot{\theta} r = \frac{d}{d \theta} \frac{l}{\gamma r^2} r = - \frac{l}{\gamma} \frac{d}{d \theta} r^{- 1} \Rightarrow \gamma \ddot{r} = - \dot{\gamma} \dot{r} - \frac{l^2}{\gamma r^2} \frac{d^2}{d \theta^2} r^{- 1}$

Substituting the last two in the equation of motion

$- \frac{l^2}{\gamma r^2} \frac{d^2}{d \theta^2} \frac{1}{r} + \mu \frac{1}{r^2} - \frac{l^2}{\gamma r^3} = 0 \Rightarrow \frac{d^2}{d \theta^2} \frac{l^2}{\mu} \frac{1}{r} + \frac{l^2}{\mu} \frac{1}{r} = \gamma$

During the nonrelativistic derivation the equation of motion takes the form

$\frac{d^2}{d \theta^2} \frac{r_{c}}{r} + \frac{r_{c}}{r} = 1$

with solution

$r^{- 1} = r_{c}^{- 1} (1 + e \cos (\theta - \theta _{0}))$

Comparing, we set $r_{c} = l \mu^{- 1}$ and $\lambda = \gamma - 1$. The previous is then written

$\frac{d^2}{d \theta^2} \frac{r_{c}}{r} + \frac{r_{c}}{r} = 1 + \lambda$

The $\lambda$ represents the correction due to SR. The planets are described by near-circular orbits ($e \ll 1 \Rightarrow \dot{r} \approx 0$) and small velocities ($v / c \ll 1$). Then the correction is approximated

$\lambda = \gamma - 1 \approx (1 / 2) (r \dot{\theta} / c)^2 \Rightarrow \lambda \approx (1 / 2) (l / r c)^2 (1 + \lambda)^{- 2} \Rightarrow \lambda \approx (1 / 2) (l / r c)^2$

Substituting to the previous

$\frac{d^2}{d \theta^2} \frac{r_{c}}{r} + \frac{r_{c}}{r} \approx 1 + \epsilon \left( \frac{r_{c}}{r} \right)^2 \quad \epsilon = \frac{1}{2} \left( \frac{\mu}{l c} \right)^2$

Let

$1 / s = r_{c} / r - 1 \ll 1 \Rightarrow (r_{c} / r)^2 \approx 1 + 2 / s$

The previous is rewritten

$\frac{d^2}{d \theta^2} \frac{1}{s} + \frac{1}{s} \approx 2 + \epsilon \left( 1 + \frac{1}{s} \right) \Rightarrow \frac{2}{\epsilon} \frac{d^2}{d \theta^2} \frac{1}{s} + \frac{2 (1 - \epsilon)}{\epsilon} \frac{1}{s} \approx 1$

Substituting $u = \theta \sqrt{1 - \epsilon}$

$\frac{d}{d u^2} \frac{s_{c}}{s} + \frac{s_{c}}{s} \approx 1 \quad s_{c} = \frac{2 (1 - \epsilon)}{\epsilon}$

with solution

$s^{- 1} = s_{c}^{- 1} (1 + \kappa \cos (u - u_{0}))$

Finally, restoring the original coordinates

$r^{- 1} = r_{0}^{- 1} (1 + \varepsilon \cos (\alpha (\theta - \theta _{0})))$

where (note that $\alpha \neq a$ where $a$ the semimajor axis)

$r_{0} \approx r_{c} (1 - \epsilon) = l \mu^{- 1} (1 - \epsilon) \quad \varepsilon \approx e (1 + \epsilon) \quad \alpha \approx 1 - \epsilon$

For a Keplerian orbit

$l^2 = \mu a (1 - e^2) \Rightarrow \epsilon = \frac{\mu}{c^2 a (1 - e^2)}$

The shift in perihelion is the angle

$\Delta \theta = 2 \pi (\alpha^{- 1} - 1) \approx 2 \pi \epsilon \Rightarrow \Delta \theta \approx \frac{2 \pi \mu}{c^2 a (1 - e^2)}$

Consider the case we’re given the mean radius $\bar{r}$ and orbital period $T$. The orbital period can give directly the semimajor through Kepler’s third law ($n = 2 \pi / P$ is the mean motion).

$\mu = n^2 a^3 \Rightarrow \frac{\mu}{4 \pi^2} P^2 \Rightarrow a = \left( \frac{\mu}{4 \pi^2} P^2 \right)^{1 / 3}$

The mean radius can be computed integrating over the orbit. The result is in terms of semimajor axis and eccentricity, which is then solved for eccentricity.

$\bar{r} = \frac{1}{2 \pi} \int_{0}^{2 \pi} r (\theta) d \theta = a \sqrt{1 - e^2} \Rightarrow e = \sqrt{1 - \bar{r}^2 / a^2}$

Having the semimajor axis and eccentricity the shift in perihelion can be computed.

## Bibliography

• Herbert Goldstein, Charles Poole, and John Safko. Classical mechanics. Addison Wesley, 2002. [URL]

• Tyler J Lemmon and Antonio R Mondragon. Kepler’s orbits and special relativity in introductory classical mechanics. Preprint arXiv:1012.5438, 2010.

1. See my previous post on relativistic Lagrangian

2. An alternative will be taking $V = \mu \gamma m r^{- 1}$. See (Lemmon & Mondragon 2010) where this case is also considered.