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?
2 Answers
You actually described two inequivalent definitions of "period," both legitimate. The paper
- Geisler and McVittie (1965), "Orbital periods in the Schwarzschild space-time" (http://adsabs.harvard.edu/full/1965AJ.....70...14G)
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 $t(\tau)$.
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$\begingroup$ 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. $\endgroup$– GordonCommented Dec 9, 2018 at 0:42
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$\begingroup$ @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 $t(\tau)$? $\endgroup$ Commented Dec 9, 2018 at 0:44
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$\begingroup$ 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. $\endgroup$– GordonCommented Dec 9, 2018 at 0:48
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$\begingroup$ @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. $\endgroup$ Commented Dec 9, 2018 at 0:54
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$\begingroup$ 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). $\endgroup$– GordonCommented Dec 9, 2018 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 $$T_\phi = {2 \pi\,T_r \over 2\pi + \De\phi}.$$
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$\begingroup$ Thank you. How do you derive the final formula? $\endgroup$– GordonCommented Dec 10, 2018 at 12:53
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$\begingroup$ @Gordon Do you mean the one for $T_\varphi$? $\endgroup$ Commented Dec 10, 2018 at 13:29
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$\begingroup$ 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 $T_{\phi} = (2\pi/\Delta\phi) T_r$? $\endgroup$– GordonCommented Dec 10, 2018 at 17:27
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$\begingroup$ @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 $2\pi+\Delta\varphi$. $\endgroup$ Commented Dec 11, 2018 at 19:51