I have been thinking about non-circular orbits in the Schwarzschild spacetime. How would you define a period of one orbit? I was thinking, in terms of proper time, for $r$, how long it takes to go from one apogee to another. For $\phi$, again in terms of $\tau$, how long it takes to cover $2\pi$. What about $t$, though? Is my reasoning wrong?

You actually described two inequivalent definitions of "period," both legitimate. The paper

considers essentially the same two definitions of "period" that you described. One is the time from one perihelion to the next; they call this the anomalistic period. The other is the lapse between two successive passages across $\phi=0$; they call this the sidereal period. The two periods are not the same. Both periods may be expressed either in terms of the object's own proper time $\tau$, or in terms of the coordinate time $t$. These again are not the same.

The key is to specify the worldline by expressing all of the coordinates as functions of a shared parameter, which we can take to be the object's proper time. Then we have functions $r(\tau)$, $\phi(\tau)$, and $t(\tau)$. We need all of these functions anyway to solve the free-fall equations that define what "orbit" means. Given those functions, we can use $r(\tau)$ to compute what those authors called the anomalistic period, or we can use $\phi(\tau)$ to compute what those authors called the sidereal period, both in terms of the object's proper time. To relate those proper-time periods to coordinate-time periods, we can use the function $t(\tau)$.

  • I realise that each of these defines a period for the given coordinate and that these are not equivalent. I don't really like the formalism of this paper, I might try to re-derive the periods in terms of other quantities. – Gordon Dec 9 at 0:42
  • @Gordon Okay, I guess I misunderstood the purpose of the question. Were you asking if there should be a third period, one corresponding to the function $t(\tau)$? – Dan Yand Dec 9 at 0:44
  • No worries, your answer helps a lot. I was asking 1. if this is a legitimate way of thinking about periods of Schwarzschild orbits, 2. how one would go about calculating them. – Gordon Dec 9 at 0:48
  • @Gordon Yes, you are thinking about the orbits correctly. Thanks to the symmetry of the Schwarzschild metric, these definitions that at first seem to be coordinate-dependent can actually be turned into (less concise) coordinate-independent definitions. So they are legitimate. As far as how one would calculate them, the hard part is determining which worldlines correspond to orbits (free-fall). I don't have any good hints to offer. The farthest I've ever gone with non-circular orbits is to write down the free-fall eqn's for $r,\phi,t$ that need to be solved, but I haven't actually solved them. – Dan Yand Dec 9 at 0:54
  • Thank you. Just so we are on the same page, what do you mean by free-fall equations? I presume the equations of motion for the metric. As to which world-lines correspond to orbits, I am considering non-circular geodesics: these lie in a plane, and precess between two radii (the perigee and the apogee). – Gordon Dec 9 at 1:00

$\let\a=\alpha \let\b=\beta \let\phi=\varphi \let\De=\Delta \def\D#1#2{{d#1\over d#2}} \def\dr{\dot r} \def\dt{\dot t} \def\dx{\dot x} \def\dphi{\dot\phi} \def\half{{\textstyle {1 \over 2}}}$ If I could assume you can read Italian I would have an easy life - had only to give a link. But since I find it unlikely, I'll write a synthesis of essential points.

Consider a general metric $$d\tau^2 = g_{\a\b}\,dx^\a dx^\b.\tag1$$ We are interested in timelike geodesics, which can be parametrized with proper time $\tau$. Then coordinates are functions of $\tau$: $$x^\a = x^\a(\tau) \qquad dx^\a = \D {x^\a}\tau\,d\tau$$ and from (1) we have $$g_{\a\b}\,\dx^\a \dx^\b = 1.\tag2$$

It can be shown that geodesics obey a variational principle with lagrangian $$W = \half\,g_{\a\b}\,\dx^\a \dx^\b.$$ Eq. (2) shows that $W$ is a constant of the motion, with $2W=1$ on a timelike geodesics.

Schwarzschild's metric, restricted to the plane $\theta=\pi/2$, is $$d\tau^2 = \left(\!1 - {1\over r}\!\right) dt^2 - {dr^2\over 1 - 1/r} - r^2 d\phi^2$$ where units were so chosen $G=1$, $c=1$, $2M=1$ ($M$ Sun's mass). Then $$2W = {r - 1 \over r}\,\dt^2 - {r \over r - 1}\,\dr^2 - r^2 \dphi^2 = 1$$ Since $W$ doesn't depend on $t$ and on $\phi$, we have the constants of the motion $${r - 1 \over r}\,\dt = E \qquad r^2 \dphi = J.\tag3$$

Substituting (3) into (2) we have $$\dr^2 = E^2 - \left(\!1 - {1 \over r}\!\right)\! \left(\!1 + {J^2 \over r^2}\right)\!.\tag4$$ Integrating eq. (4) by separation of variables we get $\tau(r)$ and the radial period. Unfortunately an elliptic integral is involved.

As to $\phi$, from the second of (3) and (4) we have $${J^2 \over r^4} \left(\!\D r\phi\!\right)^{\!\!2} = E^2 - \left(\!1 - {1 \over r}\!\right)\! \left(\!1 + {J^2 \over r^2}\right)$$ which gives $\phi(r)$, again as an elliptic integral. An approximation is possible to deduce perihelion precession (I'll not show how to do it).

Instead I find a problem if the azimuthal period is of interest. The reason is the following. Let's start form perihelion: when $\phi$ increases by $2\pi$ we are not yet arrived at another perihelion, because of precession. This shows qualitatively that azimuthal period is less than radial one. But a further increment by $2\pi$ brings us still farther from perihelion, and I expect that the latter $2\pi$ variation of $\phi$ takes a different time from the former (larger or smaller?) So it seems that a well definite azimuthal period doesn't exist.

Edit. But an average period can be defined. let $T_r$ be the radial period, $\De\phi$ the perihelion advance in time $T_r$. Then in the average $\phi$ advances by $2\pi$ in time $$T_\phi = {2 \pi\,T_r \over 2\pi + \De\phi}.$$

  • Thank you. How do you derive the final formula? – Gordon Dec 10 at 12:53
  • @Gordon Do you mean the one for $T_\varphi$? – Elio Fabri Dec 10 at 13:29
  • Yes. I can see how it works, apart for the extra $2\pi$ in the denominator. If over a $T_r$ period $\phi$ advances by $\Delta\phi$, then isn't $T_{\phi} = (2\pi/\Delta\phi) T_r$? – Gordon Dec 10 at 17:27
  • Any ideas @Elio Fabri? – Gordon 2 days ago
  • @Gordon Sorry I'd missed your reply. If over a $T_r$ period $\varphi$ advances by $\Delta\varphi$, then isn't $T_\varphi=(2\pi/\Delta\varphi)\,T_r$? No, because $\Delta\varphi$ is perihelion advance. $\varphi$ increases by $2\pi+\Delta\varphi$. – Elio Fabri 2 days ago

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