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I'm trying to get a relativistic solution to Kepler's equation starting with

$$\frac{d^2 u}{d\phi^2}+u = \frac{M}{L^2}+3Mu^2$$

by ignoring the higher order terms;

$$u(\phi, \epsilon)=u_0+\epsilon u_1$$

$$\frac{d u}{d\phi}= \frac{d u_0}{d\phi}+\epsilon \frac{d u_1}{d\phi} =u_0' + \epsilon u_1' \hspace{4mm},\hspace{4mm} \frac{d^2 u}{d\phi^2}= \frac{d^2 u_0}{d\phi^2}+\epsilon \frac{d^2 u_1}{d\phi^2}= u_0'' + \epsilon u_1''$$

$$u_0'' + \epsilon u_1'' + u_0 + \epsilon u_1 = \frac{M}{L^2}+3M(u_0 + \epsilon u_1)^2$$

where from the last term on the R.H.S of the equation I can ignore higher order terms but I'm not sure how to treat $\epsilon u_0 u_1$ term. I remember deriving $u$ for relativistic form of Kepler's equation a couple months back yet I don't fully recall how I proceeded. Later I carried out the calculation to see if there happens to be some trick I could get use of

$$\big(u_0'' + u_0 - 3M u_0^2\big) + \epsilon \big(u_1''+u_1-3Mu_0 u_1\big)-\frac{M}{L^2}=0$$

If there was no nonlinear term then solutions for $u_0$ would be in a superposition of some exponential forms and one could further deduce $u_1$ but here I got stuck.

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  • $\begingroup$ Is your question how to get from the first equation to the perihelion advance relation? Or to check where something may have gone wrong in your calculation? $\endgroup$
    – SG8
    Commented May 29, 2021 at 14:29
  • $\begingroup$ How to get from 1st equation to perihelion advance relation, more specifically how to treat above differential equation to get a solution. $\endgroup$
    – Monopole
    Commented May 29, 2021 at 15:22

1 Answer 1

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Derivation of the perihelion advance relation

Let's start with this equation

$$\frac{d^2 u}{d\phi^2}+u = \frac{M}{L^2}+3Mu^2, \tag 1$$

and let's go step-by-step. Remember that this equation is valid in units with $G=c=1$.

Step 1. First of all, since this is a perturbative analysis, we have to consider a circular orbit with a constant radius ($r_0$). This ($u_0=1/r_0$) obviously satisfies Eq. (1), i.e.,

$$u_c = \frac{M}{L^2}+3Mu_0^2, \tag 2$$

in which $\frac{d^2 u_0}{d\phi^2}=0$. You have made a mistake here, in fact, these derivatives are zero: $\frac{d^2 u_0}{d\phi^2}=\frac{d u_0}{d\phi}=0$.

Step 2. Next, we have to consider a slightly non-circular orbit as

$$u(\phi, \epsilon)=u_0+\epsilon u_0 u_1=u_0(1+\epsilon u_1), \tag 3$$

which is a perturbation of the previous circular orbit in Step 1. By inserting Eq. (3) into Eq. (1) together with the Eq. (2), you will obtain the following equation

$$\epsilon \frac{{{d^2}{u_1}}}{{d{\phi ^2}}} + \epsilon {u_1} = 6\epsilon M{u_0}{u_1} + 3{\epsilon ^2}M{u_0}u_1^2, \tag {4-A}$$

or equivalently

$$\frac{{{d^2}{u_1}}}{{d{\phi ^2}}} + {u_1} = 6 M{u_0}{u_1} + 3{\epsilon}M{u_0}u_1^2. \tag {4-B}$$

Now, since it's been assumed that $\epsilon u_1 \ll 1$, the second term on the R.H.S. of the Eq. (4) is much smaller than the first term, so ignore it. This results in

$$\frac{{{d^2}{u_1}}}{{d{\phi ^2}}} + \left( {1 - 6M{u_0}} \right){u_1} = 0. \tag 5$$

Step 3. Finally, we have ended up with the harmonic oscillator equation, Eq. (5). It's solution is simple, i.e.,

$${u_1} = A\cos \left( {\omega \phi + {\phi _0}} \right), \tag 6$$ where

$$\omega = \sqrt {1 - 6M{u_0}}. \tag 7$$

I think that you know the rest of typical small steps, in summary: a) when $u_1$ is a minimum, the orbit’s perihelion occurs b) then consider a change of the argument by $2\pi$ c) next, using the binomial approximation and assuming that $u_1 M \ll 1$, yielding

$$\omega \Delta \phi = 2\pi \Rightarrow \Delta \phi = 2\pi + 6\pi M{u_0}.$$

This is the famous relation of perihelion advance in general relativity.

Applying to the planets' orbits in the solar system

This derivation was performed in units with $G=c=1$. It is easy to convert this into SI units by performing this substitution: $M \to \frac{GM}{c^2}$. So, using this substitution together with $u_0=1/r_0$, one finds

$${\rm{perihelion}} \, {\rm{advance}}\,{\rm{relation:}}\,6\pi M{u_0} \to \frac{{6\pi GM}}{{{r_0}{c^2}}},$$

in which $r_0$ is the mean orbital radius (the mean distance from Sun). Considering Sun with the mass $1.989 \times 10^{30}$ kg and the Mercury with the mean distance from Sun about $5.79 \times 10^7$ km, one finds the famous result of $43$ $\frac{\rm{arc-seconds}}{\rm{century}}$. (data from Solar System Data)

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$ References

[1] T.A. Moore, A General Relativity Workbook, University Science Books (2012).
[2] S.M. Carroll, Spacetime and geometry: an introduction to general relativity, Addison-Wesley, San Francisco, USA (2004).
[3] M.P. Hobson, G.P. Efstathiou, and A.N. Lasenby, General relativity: an introduction for physicists, Cambridge University Press (2006).
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  • $\begingroup$ Yes, I unfortunately haven't considered derivatives to be zero that was the blunder. In $4-A$ last term should be $3 \epsilon^2 M u_1^2$ right? $\endgroup$
    – Monopole
    Commented May 29, 2021 at 15:46
  • $\begingroup$ @Monopole - I can perform a double check to be sure but it's not important since it is a second order approximation and will be ignored. $\endgroup$
    – SG8
    Commented May 29, 2021 at 15:52
  • $\begingroup$ Of course, I just wanted to point out. $\endgroup$
    – Monopole
    Commented May 29, 2021 at 15:55
  • $\begingroup$ @Monopole - I will check it soon (I think it's correct since I tried to be accurate but not sure completely :) ). The step 3 is clear enough? I intend to add a final remark in order to address the Mercury’s perihelion advance. $\endgroup$
    – SG8
    Commented May 29, 2021 at 15:59
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    $\begingroup$ @Monopole - Aha, I found it. I had started with this $u(\phi, \epsilon)=u_0+\epsilon u_0 u_1(\phi)$. Edited now. Yours is this $u(\phi, \epsilon)=u_0+\epsilon u_1(\phi)$. In fact, I tried to do this calculation with your notation. $\endgroup$
    – SG8
    Commented May 29, 2021 at 16:12

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