# Talk:Two-body problem in general relativity

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## Wrong Lagrangian and other issues

To WillowW (talk · contribs): I have several issues with this article:

• First, a minor issue of nomenclature. $ds\!$ as defined in the article is the differential of proper time multiplied by the speed of light to convert it to meters. It would seem more reasonable to either give it as time in seconds or to change the sign of the square and give the differential of proper distance in meters.
• Second, what you call the "kinetic energy", T, is just identically one-half. If you divide both sides of the equation for the metric by $ds^2$, then you will see that you get that 1 is equal to the right hand side of your equation for 2T. In any case, I do not see how you can justify calling that expression "kinetic energy".
• Third, there is gravitational potential energy in general relativity. However, it is not a tensor. See chapter 32, "Explicit expression for the pseudo tensor", in "General Theory of Relativity" by P.A.M.Dirac, Princeton University Press. Notice that the gravitational force field (the Christoffel symbol) is also not a tensor.
$- m c^2 \frac{d \tau[t]}{d t} = - m c \sqrt{\left( 1 - \frac{r_{s}}{r[t]} \right) c^{2} - \frac{1}{1 - \frac{r_{s}}{r[t]}}(\frac{d r[t]}{d t})^2 - r^{2}[t] (\frac{d \phi[t]}{d t})^{2}}$

Of course, if the Lagrangian is wrong, one cannot expect the resulting Euler-Lagrange equations to be correct, except by accident. I hope you can clarify these problems. JRSpriggs 05:23, 17 June 2007 (UTC)

As a geometer, I spot here the difference between the length functional and the energy functional for geodesics (except you are parameterizing worldlines by coordinate time instead of proper time). For curves parameterized by arclength, the critical points of these two functionals are the same (geodesics), and the derivation of the first variation formula which shows this is elementary. So I don't see a problem for worldlines of massive particles parameterized by proper time, but the physical derivation of lightlike geodesics may be more subtle. Geometry guy 10:30, 17 June 2007 (UTC)

Let me clarify/correct a few points about what I said. The action integral is:

$S = \int {- m c^2 \frac{d \tau[t]}{d t} d t} = - m c^2 \int {d \tau[t]} = - 2 m c \int {\frac{1}{2} d s[t]} = - 2 m c \int {T[t] d s[t]}$

So perhaps it does not matter which way one does this. However, aside from a constant multiple, the proper time is the only simple (e.g. leaving out squaring the acceleration) invariant which could be used as the action of a free particle. (Is this what Geometry guy was saying?) Also that to which I referred above as gravitational potential energy is actually the energy of the gravitational field which is needed to make total energy conserved (using a partial derivative rather than a covariant derivative). The equivalent of the negative of the Newtonian gravitational potential energy of the particle is approximated by expanding the square root in a series and looking at the term:

$m c^2 \frac{r_{s}}{2 r[t]} = \frac{G M m}{r[t]}$

In any case, more motivation is needed for the steps of the derivation in the article, including indicating which variable one is varying to get each of the equations of motion. JRSpriggs 05:05, 18 June 2007 (UTC)

Both of you seem to have hit the nail on the head! :) The article is still under active development — it's rather sketchy and muddled at the moment — but I really appreciate your pointing out things we need to work on.
It's basically a dilemma of teaching, guessing which explanation will be most readily understood by our readers. I wanted to derive the geodesic equations without resorting to the Christoffel symbols, and was betting that people with a little physics training (but less math, like me) would be more comfortable understanding a geodesic as a Lagrangian equation of motion. So, in other words, I'm exploiting the geodesic variational definition as a kind of Hamilton's principle (I think).
The choice of ds versus the proper time is, as you say, a matter of taste, et de gustibus nihil disputandum est. ;) I chose ds because I wanted to keep it closer to its geodesic roots and it reduces the number of c's flying around. Of course, the proper time is attractive because it's closer to the idea of an equation of motion. You say "potayto", I say "potahto" — let's call the whole thing "off". ;)
The "potential energy" issue will go away if I re-word that derivation better. I do understand that Christoffel symbols are not tensors. I'll try to improve the article today, although I'm likely to be very busy doing inventory. Please be patient with me! :) Gotta run, Willow 12:05, 18 June 2007 (UTC)
PS. I'm going to tinker with other derivations; we can decide later which one to keep foremost in the article, being swiftest to explain and easiest to understand, and which others to relegate to an appendix section near the end of the article. Willow 20:21, 18 June 2007 (UTC)

I agree that bypassing the Christoffel symbols, in a concrete case like this where it is possible, is highly desirable. And I expected that you would know that they are not tensors. What I was trying to do by mentioning that fact was to deflect any criticism of my claim that there is an energy of the gravitational field. Some people have been known to say that if it is not a tensor, then it does not exist or is unphysical. Obviously, I think that that gravitational energy, like the its force field, does exist. JRSpriggs 05:28, 19 June 2007 (UTC)

## Perihelion shift plot

In Gravitation, a simple expression in polar coordinates is derived for a planet's perihelion shift:

\begin{align} r &{}= \frac {(1-e^2) a} {1 + e \cos \left( (1 - \frac{\delta}{2\pi}) \phi \right)} , \\ \delta &{}= \frac {6 \pi M} {(1-e^2) a} , \end{align}

where e is the eccentricity and a is the semimajor axis for a Kepler orbit, and where M is the mass of the sun.

We should replace the jaggy PNG with a pretty SVG plot, and this gives an easy way to do it; probably gnuplot will suffice. Note that Mercury has e = 0.206, a = 0.39 AU, and only a tiny shift (plus, in reality, many larger perturbations); as a reality-based example, it's useless for an informative plot. I may get around to making an SVG — or not; meanwhile, assuming there are no objections to this formulation, perhaps someone else would like to try. --KSmrqT 08:29, 21 June 2007 (UTC)

Thank you, that would be great if you made the SVG! :) I'm not very handy with graphing programs, and I've never used gnuplot, although I have heard of it.
The formula for the precession angle is derived already in the article (albeit with G and c included), but the (approximate) r formula is not. Should we try to work it in? Willow 09:44, 21 June 2007 (UTC)
I am going to give it a shot, but this will be my first graph made with gnuplot.--Cronholm144 11:26, 21 June 2007 (UTC)
Well, I graphed the function, but it turned out something that is very cool but that is also not remotely correct. (sheepish grin) Oh well, I will keep working with it, but advice would be appreciated.--Cronholm144 12:36, 21 June 2007 (UTC)
Great! First get it working with δ = 0. The output should be simple: an ellipse with one focus at the origin. Then experiment with a very small δ. You'll need enough eccentricity to stretch the ellipse visibly beyond perfect circularity, but keep 0 < e < 1. Choose a to fit the scale of the plot (or vice versa). I believe you'll need to use t for the angle, never explicitly mention r, and do "set polar" as well as "set size ratio -1" (and "set zeroaxis"). Try "plot cos(3*t)" for practice; you should see a pretty rose with 3-fold rotation symmetry.
We don't necessarily need to explain the form used for the graph; that can be our little secret, if we like. The key question is, what will help the reader understand the concepts of interest. If they get to the point where they can say, "I think I understand how curved space can cause perihelion shift; but how did you make that cool plot?", then we've won. :-) --KSmrqT 17:45, 21 June 2007 (UTC)
Thanks, Cronholm! I think that e=0.9 is a good choice, but the precise value probably doesn't matter.
May I ask for two other Figures that would really help the article?
• It's be wonderful to have a graph with a family of plots of V(r) as a varies from zero to large. It'd be nice if the r-axis were measured in units of rs, and if you could mark the extrema rinner and router with two different symbols. I'd like our readers to get a qualitative feeling for the behavior of the orbits, and how rinner is unstable and router is stable, and how below a certain value of a, the particle falls straight into r=0.
• It'd also be nice to have a plot for a few archetypical functions G(ζ) that show the three roots, and illustrate how orbits can oscillate between e1 to e2 or diverge starting from over e3. At least I think this graph would be useful; what do you all think?
Thanks muchly! :) Willow 20:28, 21 June 2007 (UTC)

Hmm... I can't seem to get the plotter to go beyond 0 to 2π. I have set the rrange many different ways but it is not working...--Cronholm144 00:59, 22 June 2007 (UTC)

You're stuck on the default range for t (think "theta"), not r (think "radius"). Either use "set trange [0:6*pi]" (for example), or include the range in the plot command, as in "plot [0:6*pi] f(t)". --KSmrqT 04:11, 22 June 2007 (UTC)
Try the following to get started:
set polar
set size ratio -1
set zeroaxis
set sample 600
unset key
unset tics
unset border
ek=0.667
a=1.0
s=(1-ek**2)*a
dist(t)=s/(1+ek*cos(t*(1-dd/(2*pi))))
dd=0.15
plot [0.5*pi:8.5*pi] dist(t)

Then you can change eccentricity (ek), δ (dd), and so on as you like. --KSmrqT 07:12, 22 June 2007 (UTC)

Thanks again! :)--Cronholm144 07:35, 22 June 2007 (UTC)

Okay, I have set it to what Willow's picture looks like now, except with blue coloration and no tics(I can't get them to look good). Is a copy what we want? Or do we want something else? BTW I think I am finally getting a hang of this plotter :)--Cronholm144 09:21, 22 June 2007 (UTC)

I have copied your version here. (Changing what I wrote confuses the thread.) Per Cronholm144:
1. Yep, gnuplot achieves a nice blend of capability and usability; looks like you're catching on.
2. I think I preferred my settings, except that I'd now use a trange of [0.75*pi:8.75*pi].
3. When I try your version, the plot spills out of the requested range.
4. To my eye, an eccentricity of 0.9 is too much.
5. Instead of drawing zero axes, I'm thinking it might be nice to plot a "Sun" as a small circle/disk around the origin.
6. I don't think the idea is necessarily to reproduce Willow's picture, especially since I don't know what limitations she was working under and what she really wanted. I'd just go for the best illustration of the concept. To that end, it might be nice if the plot showed (in different colors) both a Kepler orbit and a precessing orbit (if it's not too cluttered).
Anyway, now that you're more familiar with gnuplot, you can experiment easily. (Maybe we'll even convince Willow to try.) Make an SVG that you think best represents the idea, and put it here so everyone can see a picture. Then try one of Willow's other requests! --KSmrqT 04:51, 23 June 2007 (UTC)
1. Turn off the zero axes with "unset zeroaxes"; they're meaningless in space and only add clutter in the figure.
2. Don't give explicit xrange and yrange data; gnuplot is good at choosing these automatically.
3. Don't put a title in the image; that's best left to a separate caption, partly because the image is then language-independent, for use by any Wikipedia.
4. Generate an SVG, not a PNG.
Apparently I didn't mention the incantation for making an SVG, "set term svg" (and, of course, set the output). It will choose some defaults and tell you what they are. Probably the main option of interest is the dimensions. --KSmrqT 06:57, 23 June 2007 (UTC)
• I agree and I turned it off in my code already(see above).
• If I don't give an explicit range the autoset fuction scales each image separately
• I will rm it
• When I do set term svg (set terminal svg)? horribly confusing things happen... it produces hundreds of lines of code when I plot a graph. :(
• By the way I just used a screen-snip tool to make the png, I didn't even save the graph. :)

--Cronholm144 07:14, 23 June 2007 (UTC)

Hmm. I think there is a way to get an autoset range for one plot then use it for all. I had assumed you produced a PNG using the output option; that's the meaning of my "of course, set the output" comment! You need to redirect the output to a file. First, choose a directory using "cd 'your/path/here'". Then, after setting the terminal type but before plotting, direct the output to a file using "set output 'Orbit_test.svg'". Otherwise, the generated SVG contents will be dumped to the console. --KSmrqT 07:39, 23 June 2007 (UTC)
Yep, gnuplot is so clever. Set the range(s) to be automatic, but use the writeback option before plotting the precession. Then when you plot the Kepler orbit and the Sun, they'll use the automatically set range. --KSmrqT 08:03, 23 June 2007 (UTC)

I used the "set yrange [] writeback" and xrange bu the pic came out strangely, take a look. I also rewrote the code to reflect the picture. --Cronholm144 09:21, 23 June 2007 (UTC)

Relativistic precession by 0.15
You're so close. For various reasons I decided a single parametric plot without writeback would be better, one reason being a nicer Sun. It goes like this (see image, right):
set parametric
set size ratio -1
set sample 600
unset key
unset tics
unset border
ek=0.667
a=1.0
s=(1-ek**2)*a
dist(t)=s/(1+ek*cos(t*(1-dd/(2*pi))))
dd=0.15
aff(t)=4*t+0.75*pi
set trange [0:2*pi]
plot \
0.05*cos(t),0.05*sin(t) with filledcurves lt rgb "#fec200", \
dd=0, -dist(t)*cos(t),-dist(t)*sin(t) with lines lt rgb "#bc1e47" lw 3, \
dd=0.15, -dist(aff(t))*cos(aff(t)),-dist(aff(t))*sin(aff(t)) with lines lt rgb "#0081cd" lw 3
set term svg size 420 390
cd 'your/path/here'
set output 'Orbit_test.svg'

I edited the output to replace the Sun path with a circle and the Kepler orbit with an ellipse, since SVG can easily do those perfectly and cleanly. So, is everybody happy?

## Re-write of Lagrangian section

I like parts of the "massive rewrite" of the Langrangian approach section, but others not. For instance, I like the way it goes quickly from the general gμν to the explicit Schwarzschild metric, and also the way it discusses the individual variations. What I don't like is the way it ignores the difference between varying a quantity and its square root, and the way it dispenses with the Lagrangian equations. The original motivation for that section was to help people who might know a little physics (Lagrange equations) but not much math. As a technical point, it seems to me that the assumption that a particle of negligible mass follows a geodesic in space-time is more fundamental than the mathematical form of the action; the action was derived to produce the geodesics, not the reverse. As a compromise, shall we keep both subsections, to offer both viewpoints to our readers? Willow 09:44, 21 June 2007 (UTC)

I didn't like the rewrite much. The former version explained what was going on conceptually; in the latter, the concepts are submerged in calculation. The individual variations are not particularly helpful and are difficult to read. So I found it a step backwards in comprehensibility, and don't see much point in keeping it. There's no reason to have two sections explaining the same thing. Geometry guy 10:20, 21 June 2007 (UTC)
The reason I rewrote it is because I think that WillowW's version is very confusing. Minor problems: (1) You injected an unbalanced factor of c in your first equation since you say that the square root of 2T is c rather than 1, and yet you do not compensate for that when you put the square root in. (2) Why is there a factor of 2 in 2T? No reason is given for this arbitrary insertion.
Major problem: You are trying to vary the proper time while holding the proper time fixed. At first, you use τ as a variable which is presumably going to change (since it is the only thing there which can change). Then you hold it fixed while you vary the coordinates upon which it depends. One of the reasons I introduced q as the independent variable was to avoid this source of confusion. The other reason is that just as the Lagrangian should be invariant with respect to changes in the coordinate system for space-time, it should also be invariant with respect to the "coordinate" q (or whatever) of the particle's world-line. It is best to make this explicit by using a distinct variable. Of course, later, if there were any need (which there is not), you could choose q to be the same as t, r, φ, or τ.
Perhaps it would be a good thing to fill in more steps in my derivation. This is a difficult decision because more details would leave some people unable to see the forest for the trees. Personally, I like to derive things from first principles in most cases. This is why I took a short cut by varying the Lagrangian rather than applying the Euler-Lagrange equations. But it amounts to the same thing. JRSpriggs 06:26, 22 June 2007 (UTC)

It would probably be good for both of us to improve our derivations. :) I'm not sure, though, if I understand the "major problem" that you outline above. The proper time is not being held constant; rather, we are saying that variations about a geodesic path through space-time should produce no first-order variations in the proper time

$0 = \delta s = c \delta \tau = \delta \int c d\tau = \delta \int \sqrt{2T} d\tau$

However, that was converted into a different requirement (no first-order variations in a pseudo-action ʃTdτ) by the following line of reasoning (found in the Pauli reference and, I think, the Adler-Bazin-Schiffer reference)

$0 = \delta \int \sqrt{2T} d\tau = \sqrt{2} \int \frac{\delta T}{\sqrt{2T}} d\tau = \frac{\sqrt{2}}{c} \delta \int T d\tau$

Application of the Euler-Lagrange equations to this new requirement yields the desired results, don't you agree? It's written up pretty much this way in the Whittaker reference. But maybe there's something I'm overlooking? Hoping that we can come to a happy consensus, Willow 15:48, 22 June 2007 (UTC)

I also hope that the merits of the two approaches can be combined into such happy consensus! I think forest-for-trees was my main concern with the second derivation, so I wouldn't support adding more details. Concerning the major problem with the first derivation, yes, this is a subtle point. A related subtle point is the fact that one must vary among curves with fixed endpoints to get the geodesic equation. Thus if γ:[a,b] -> M is a curve with endpoints γ(a) and γ(b), the length is
$\int_a^b |\dot\gamma| dt.$
If the curve is parameterized by arclength (which is essentially proper time in this context), then $|\dot\gamma|=1$ and the integral is just b-a. Hence to allow any variation at all in the length, you have to vary a and b (but with γ(a) and γ(b) fixed), which is rather awkward! The solution I sometimes adopt is a compromise between arbitrary parameterizations and parameterizations by arclength: I require that curves are parameterized by a constant multiple of arclength. This makes the derivation of the first variation formula, and hence the geodesic equation much easier, and along the way one can derive the basic fact that for curves parameterized in this way, the geodesics are not only critical for length, but also the "energy"
$\int_a^b |\dot\gamma|^2 dt.$
In this setting, the analogue would be parameterizing a constant multiple of proper time (or perhaps, in more physical terms, proper time without specifying the units!). Geometry guy 17:24, 22 June 2007 (UTC)

Salut, V. E.! :)

Forgive me, but now I'm really confused! Hopefully you can help me understand all this; this is what I get for trying to take a few days off — there's no rest for the wicked! ;)

Can we divide the question into two parts, for me? The first part is whether you can vary the arc-length of a path through space-time while holding its endpoints fixed. I would think that you can, by varying the path. As a trivial example, consider the straight line along the x-axis from x=0 to x=π; if we vary the coordinates about that path (say, to y = ε sin x) then the arc-length gets slightly longer (growing from π to π (1 + ε2/4 + ...), right? That suggests to me that the first equation above is correct

$0 = \delta s = c \delta \tau = \delta \int c d\tau = \delta \int \sqrt{2T} d\tau$

in that a geodesic is defined as a path in space-time that is unaffected to first order by variations about that path while holding the end-points fixed. If I've understood that incorrectly, please let me know!

The second, and I'm guessing more subtle, part is the transition to the second equation

$0 = \delta \int \sqrt{2T} d\tau = \sqrt{2} \int \frac{\delta T}{\sqrt{2T}} d\tau = \frac{\sqrt{2}}{c} \delta \int T d\tau$

Are we allowed to still keep the endpoints fixed for the second requirement, that is, when finding the extremal paths of ʃTdτ? I thought that we could, because 2T = c2 everywhere and so we can just pull it out of the integral. But perhaps it does so only along the geodesics? Perhaps 2T doesn't exactly equal c2 along our varied paths? This is where I'm confused, I think. Thanks for your help! :) Befuddled Willow 18:30, 22 June 2007 (UTC)

My students say I'm confusing all the time, so you are not alone! It seems to me that you have got the point, though. The first part of your reply looks fine: you can indeed vary the length of a curve while fixing its endpoints. However, you can't do it if you parameterize the curve by arclength and fix the domain of the parameter. In your example, the parameter is x and the domain is [0,π], but the curve (x,ε sin x) is not parameterized by arclength unless ε=0.
In the second part, yes the endpoints should still be fixed when varying ʃTdτ. It seems to be precisely the subtle point that is confusing you. Indeed, if 2T were equal to c2 along our varied paths, then δT would be zero, and we would learn nothing. So we must allow T to vary depending on the path.
In Riemannian geometry it is common to make the variation explicit, by introducing a two parameter family α(s,t), where γ(t)=α(0,t) is the curve being varied (in your example s is ε). Then 2T would be equal to c2 at s=0, but not necessarily for nonzero s. However, the first order variation is obtained by taking the derivative with respect to s at s=0, so the argument you use to pull 2T out of the integral is correct, because everything has been evaluated at s=0. Correct, but subtle ;) Geometry guy 19:01, 22 June 2007 (UTC)

My garden is full of blooming flowers, but none are so beautiful as the one that just opened up in my brain. :D I think I get it! The problem lay in choosing τ (the arclength) as the integration variable instead of a space-time coordinate such as t, right? We can and should hold the endpoint events in space-time fixed, but the endpoints in tau must vary (to second order) as we vary the path in space time. Exquisitely beautiful. Thank you as always, Willow 19:26, 22 June 2007 (UTC)

## Required citation

Astoundingly, the "Bible" is not cited:

• Misner, Charles W.; Thorne, Kip S.; Wheeler, John Archibald (1973), Gravitation, San Francisco: W. H. Freeman, ISBN 978-0-7167-0344-0 Unknown parameter |month= ignored (|date= suggested) (help)

The size may be intimidating, but the style is like a friendly conversation. If I have time to add actual content, I'll put this in. Meanwhile, it's recommended reading. --KSmrqT 08:56, 22 June 2007 (UTC)

It might be easier just to link to Gravitation (book). JRSpriggs 09:30, 22 June 2007 (UTC)
Best to do both. Anyway, this article is only a couple of weeks old, so it is bound to have gaps. Since I see no reason to be shy about adding references, so I have copied it over. Geometry guy 11:23, 22 June 2007 (UTC)
It's by no means astounding that I should be ignorant of the classic references in the field. ;) Sorry! It was probably checked out of the library, or maybe they don't have it there. Anyway, please include the appropriate page numbers in the reference; I've been trying to be scrupulous about that, since one of Wikipedia's chief advantages seems to be directing people to good outside sources. Thank you all very much for filling in the gaps; it's much needed! :) Willow 15:55, 22 June 2007 (UTC)
The "Gravitation (book)" article is a sad little stub, but at least it's there. I came across it when I was writing the above note, and found it lacked even a proper citation, just an ISBN-10! (I did insert the above citation there.)
As for page numbers, that looks to be a major undertaking. The last page of the book is numbered 1279. The bibliography alone consumes pages 1221 to 1254! Discussions are scattered throughout. Many have illustrations. For example: all of Chapter 25 (pp. 636–687) is devoted to Particle motion in Schwarzschild geometry; §33.5, Equations of motion for test particles, discusses a charged particle and a black hole, deriving four constants of the motion; §40.5, Perihelion shift and periodic perturbations in geodesic orbits, gives the formula I mentioned recently. It is important to observe the notice on page 36, that the book adopts “geometrized units” in which the speed of light c, Newton's gravitational constant G, and Boltzman's constant k are all equal to unity.
A quotation from the overview of Chapter 25 (pp. 636–7) may be of interest:

The principal new result can be stated in a single sentence: The particle is governed by an “effective potential” (Figure 25.2 and §§25.5, 25.6) that possesses not only (1) the long distance −M/r attractive behavior and (2) the shorter distance (angular momentum)2/r2 repulsive behavior of Newtonian gravitational theory, but also (3) at still shorter distances a pit in the potential, which (1) captures a particle that comes too close; (2) establishes a critical distance of closest approach for this black-hole capture process; (3) for a particle that approaches this critical point without crossing it, lengthens the turn-around time as compared to Newtonian expectations; and thereby (4) makes the period for a radial excursion longer than the period of a revolution; (5) causes an otherwise Keplerian orbit to precess; and (6) deflects a fast particle and a photon through larger angles than Newtonian theory would predict.

Whew! It's quite the tome. I believe much of its unique style derives from Wheeler, who gave us the term "black hole". He's the third author in alphabetical order, but certainly not in importance. Although it's been over three decades since its publication, all but one of the other works cited are older still.
Incidentally, I always make it a point to use the current ISBN standard, which switched to 13 digits as of January 2007. It seems counterproductive to insist on ISBN-10, as was done when the citation was inserted into the article. --KSmrqT 19:29, 22 June 2007 (UTC)
It is a great book, isn't it? I have no objections to using ISBN-13, as long as it is done consistently for all the references. Geometry guy 20:05, 22 June 2007 (UTC)

As my good deed for the day, here are the current references with ISBN-13, filling in missing ones, and with other improvements. (For those who don't already know, there's a handy checker/converter available.) Use 'em or lose 'em, as you wish.

See also page 163 of this item in the Internet Archive, in a translation of Einstein's seminal paper of 1916.

Enjoy. --KSmrqT 22:08, 22 June 2007 (UTC)

Many thanks for this good turn! I have added the ISBNs (and ISSN), and the invaluable biblical page references to the article. Geometry guy 22:48, 22 June 2007 (UTC)

## Problem with Flamm's paraboloid?

Is the problem with the two Flamm's paraboloid paragraphs that I forgot to consider the time curvature in considering the path of particles? The path of a particle should minimize the proper time, not merely be a spatial geodesic, which is implied by the present description. That discrepancy had been nagging at me, but I forgot about fixing it until just now. If that's indeed the problem, I'd still favor fixing it up rather than eliminating it altogether. Wondering Willow 19:36, 29 September 2007 (UTC)

See my edits to the version at Schwarzschild metric. My main concern is that the text makes the paraboloid sound too much like a gravity well, which is a totally unrelated concept from Newtonian physics. E.g. it says that the dimple is downwards in w, when in fact only the intrinsic geometry matters, and the dimple could just as well point upwards. The image itself, aside from pointing downwards like they all do, has an inFlammable central portion which looks suspiciously like the central portion of the Newtonian potential as shown in the gravity-well article, so I'm not too happy about it either.
Though Flamm's paraboloid certainly deserves a section in the Schwarzschild metric article, I'm not convinced that it has any bearing on orbital mechanics. The motivation for including it here seems to be that it's an intuitive way of seeing what causes test particles to orbit massive bodies. But it's not intuitive. The intuition that most people have -- of objects moving in a gravity well or rubber sheet -- is dead wrong, and the crawling-ant picture of geodesic motion is not that much better, given that actual (helical) orbits have no obvious counterpart in geodesics on the sheet. All you're left with is some tachyonic worldlines that look a bit like hyperbolic orbits, and which most people will misinterpret in rubber-sheet terms unless specifically warned not to. I can't help feeling that the whole thing does more harm than good. -- BenRG 00:52, 30 September 2007 (UTC)
I removed the section again, but I'm still happy to talk about it. -- BenRG 14:49, 16 October 2007 (UTC)

## Humongous animation

Hi Willow,

I noticed that the animation is 3 Megabytes. I have downloaded that animation, and I have used optimisation software to reduce the size to 1.6 Megabytes, that's lossless reduction. (The optimisation took 5 seconds or so on a PIII 800 MHz computer.)

I intend to replace the 3 MB version with the 1.6 MB version. Preferably I want you to approve first. I have uploaded the 1.6 MB version to this temporary location.

It says on the image description page that the animation was generated with Blender. It appears that the authors of Blender allow Blender to generate astronomically bloated GIF-animations. To give you an idea of what GIF-animation is capable of, please check out the animations in the rotational-vibrational coupling article. Those animations are vivid and they are between 30 and 40 KB in size.

My assumption is that in many parts of the world internet access is through dial-in connection; internet via plain old telephone line. In particular in developing countries I think dial-in connection is most common. I believe that wikipedia editors should strive to keep wikipedia material accessible to users in developing countries. --Cleonis | Talk 00:47, 23 March 2008 (UTC)

Evil Willow thwarts the education of developing counties by neutralizing the articles of her good twin Willow. Mwa ha ha! ;)
Dear Cleonis,
Of course I have no objection if you manage to shrink my animation to a smaller size; thank you! :) At least I won't, once you teach me how to do it myself. ;) I have a lot of other animations to shrink, such as those at Newton's theorem of revolving orbits.
I get the impression that you're a little annoyed with me for having posted such a large file; the words "astronomically bloated" were a hint. ;) But having known me so long, you really should give me the benefit of the doubt, no? At least I don't remember having any intention to cripple the education of people in developing countries, especially on articles that I've been trying to write. ;) Instead of looking for zebras in a herd of horses, you might assume that I was an honestly inept newbie to animations? Indeed, I taught myself Blender just last Christmas, reading late into the night in my little garret when everyone else downstairs was fast asleep and, I assure you, with only good intentions. I'm very sorry for posting such large animations and if you could teach me better, I would be very grateful. I've been conscious that they were unhappily large, but I haven't known how to fix that problem while maintaining the quality of the animation. My basic approach has been to generate JPG image files using Blender and then to combine them using the GIMP. As an aside, I want to — umm, have to — use freeware; other image-editing software is beyond my means. :( Thanks for your help and, oh by the way, you should consider contributing to this. :) Talk to you soon, Willow (talk) 10:08, 23 March 2008 (UTC)

Dear Willow,

I know you well enough to know that you only need a slight hint to sense disgruntlement. :-) When I started making animations I was likewise inexperienced. I deferred uploading any animations until I was satisfied that the animations gave good value for money. Can I possibly persuade you to replace the animation with a diagram? My criterium is that I will use an animation when the animation adds an element of vividness that cannot be conveyed by a diagram. I think that in this particular case a diagram can convey the same information just as vividly.

Of course there was never any intention on your part to make wikipedia less accessible.

I have showed you that it is possible to manufacture vivid GIF-animations (with 60 frames) that are around 50 KB in size. I admit that the 50 KB size for a 60 frames animation is the very optimum, I squeezed really hard. Still, the kind of animation that you uploaded for the Kepler problem in GTR article can be realized equally vivid with a size in the order of 100 or 200 KB.

For GIF-optimatisation I use the program GIF Movie Gear. If I remember correctly a license for GIF Movie Gear can be purchased for something like 25 dollars. I use graphics programs to generate the frames, storing them as PNG-files. Then I use GIF movie gear to generate the animation.

For comparison, the animation that is in this sandbox article
It is tempting to use a miniature Earth and a miniature Sun in the animation, but with GIF such fine details are expensive. In GIF-format mono-color surfaces can be coded efficiently, but the cloud pattern of even a miniature Earth is expensive. The animation in that sandbox article is 82 KB (for 120 frames, if memory serves me), showing that even when using strong colors a file size under 100 KB is possible, as long as fine details are kept to a minimum.

You write that you store the frames generated by Blender as JPG files. I'm sure that that is the main cause of the GIF-bloating. The artifacts that are introduced by the JPG-encoding make the frames very, very expensive to code as GIF-format.
If Blender allows storing frames as bitmap or PNG then please do so. That should make a lot of difference. --Cleonis | Talk 11:30, 23 March 2008 (UTC)

## A friendly challenge

L is defined implicitly, it should be defined in words "the total angular momentum of the system".

L is also defined incorrectly. Unfortunately, I've yet to see a relativist who can solve the central force problem. I need to believe that there is one out there somewhere, so I'm leaving this as a challenge problem. First one to fix it gets a barnstar. --Norbeck (talk) 06:54, 29 January 2010 (UTC)

## Advantages of reduced mass in exposition?

It's not clear to me why introducing the reduced mass is advantageous, but I'm willing to learn better. :) I'll present my concerns, and hope to hear another point of view shortly.

In the typical Schwarzschild solution, it is assumed (as we say in the article) that the gravitational field is generated entirely by the larger mass M, which does not move. The smaller "test particle" moves in this static gravitational field, and we can solve for its motion exactly, as shown in the article. In this case, there is no need for the reduced mass, because m/M is effectively zero. I feel that, for simplicity, we should present this idealized solution; i think that it will be easier for the reader to follow. It also seems historical.

If we allow the smaller mass m to contribute to the gravitational field, then the Schwarzschild solution is no longer valid. In that case, we would need to use the post-Newtonian expansion or solve Einstein's equations numerically. Luckily, the former has been done for us. In a 1938 paper ("The Two Body Problem in General Relativity", Annals of Mathematics, 39, 101–104, link), H. P. Robertson solved the post-Newtonian problem for two bodies to first order and showed that "the orbit of a double star in general relativity differs, in its secular behavior, from the classical orbit only in an advance of periastron equal to that which an infinitesimal planet, describing the same relative orbit, would undergo in the field of a star whose mass is the sum of the two components of the double star."

Therefore, it seems reasonable that, in the section on the post-Newtonian approximation, we might introduce M+m into the formulae for M, most prominently in the definition of the Schwarzschild radius rs and the precession angle δφ per revolution. As described in the section on elliptical functions, the shape of Robertson's "relative orbit" is determined by rs and the length-scales a and b. I suppose that you could introduce the reduced mass by adjusting the values of L and E to keep a and b constant, but I'm not sure why that would be advantageous, especially so late in the article? I see L and E as merely constants of motion, and a scaled constant is still a constant. But I'm open to hearing another perspective! Perhaps you want to have a consistent classical limit? Willow (talk) 04:43, 7 July 2010 (UTC)

## Magnitude of reduced-mass correction

This article has been revised to say that, when the mass of Mercury is not neglected, the rate of anomalous precession is "more than twice as large as the relativistic effect." Could you please help me to understand that?

My own calculations suggest that, when the mass m of Mercury is not neglected compared to the mass M of the Sun, the rate of precession is given by the formula

$\delta \varphi \approx \frac{6\pi G(M+m)}{c^2 A \left( 1 - e^{2} \right)}$

You will find the same formula derived in the Robertson article cited in the previous section of this Talk page, and in section 106 (pp. 342–343) of Volume 2 of Landau and Lifshitz (The Classical Theory of Fields). Since this formula is proportional to the total mass M + m. the relative error introduced by neglecting m is equal to m/M, which is roughly 1.6×10-7, 1 part in 6 million. Am I missing something? if I don't hear back, I'll amend the article to reflect this formula. Thanks! :) Willow (talk) 02:39, 10 July 2010 (UTC)

## Are we solving the Kepler problem, or the central force problem?

Perhaps this article should be split or renamed. Currently the article switches arbitrarily between the central force approximation and the full 2-body (Kepler) problem. I've seen tons of solutions of the central force problem in GR, but never a (correct) solution to the 2 body problem in GR.

If we decide that we are actually solving the "Kepler problem", then the only advantage I can think of for using the reduced mass is that you get the correct answers. When solving the two-body (Kepler) problem you have two choices, use the reduced mass to reduce the problem to a central force problem, or keep track of the two bodies separately. It's much simpler (in Newtonian mechanics) to use the reduced mass. The greatly increased accuracy of the 2-body solution comes with such an insignificant increase in complexity the only rational reason to use central force solution is ignorance of the 2-body solution. Equations without the reduced mass are non-physical and should seem glaringly incorrect. The tension on a string between two spinning masses is proportional to the reduced mass, any formula that neglects this fact is not "simpler", just wrong. If you intend to calculate the tension on the string and you know the correct formula, you would never ever use any other formula. The same is true in Newtonian gravitation.

It was me who provided the formula you posted: $\frac{6\pi G(M+m)}{c^2 A \left( 1 - e^{2} \right)}$. It is my intention to amend the article to support this very formula. Previously the derivations promptly lead to the utterly incorrect formula: $\frac{6 \pi GM}{c^2 A \left( 1 - e^{2} \right)}$. It is irrelevant how insignificant the difference is between the two formulas, the old formula and it's derivation were wrong. The current formula is correct, but it's derivation still needs correction and clarification.

You are correct, the error in the apsidal precession rate introduced by the central force approximation is insignificant, but what about much much simpler quantities, such as the relative gravitational acceleration? The acceleration of the Sun due to Mercury is about 720 ngals = $\frac{G m}{p^2}$ (where "p" is the semi-latus rectum). The additional acceleration due to General Relativity is about 350 ngals = $\frac{ 3 G M h^2}{c^2 p^4}$. Ergo, neglecting the Sun's acceleration introduces an error twice as large as the relativistic effect.

The central force approximation assumes that the Sun is located at barycenter of the Sun/Mercury system. In reality the Sun is offset by 9207 meters = $\frac{m}{M+m}p$. This offset is about 3.14 times larger than the gravitational radius. The central force solution is inappropriate for the planet Mercury because the errors introduced for most quantities are the same order of magnitude as their relativistic perturbations.

NOrbeck (talk) 08:57, 10 July 2010 (UTC)

Sorry, I didn't mean to give the impression that I questioned the validity or usefulness of the reduced mass in classical physics. I'm a big fan; if you consult the revision history of the (classical) two-body problem, you'll see that I agree how useful it is, allowing us to transform a two-body problem into an equivalent (and often soluble) one-body problem.
However, it's not obvious to me — and I'm afraid that it won't be obvious to the typical reader, but I'm willing to be corrected — why the classical reduced mass pertains to the theory of general relativity. There are no forces anymore, Newton's laws don't pertain; we're solving relativistic field and geodesic equations instead. To add a classical quantity into a general relativistic argument seems to be making a chimera that might confuse the reader, especially if we don't motivate it strongly. I do see that a few sections might benefit from a consistent classical limit, such as the section on the "effective potential"; it's conceivable that we could spin that classical limit and the Equivalence Principle into a hand-waving argument for the reduced mass. But that would be original research; I'm not aware of any reliable sources that adopt that approach. All the sources I've seen (those listed on the page) begin with the Schwarzschild solution (the "central force problem", as you put it) and then discuss the post-Newtonian correction, which is the approach I've adopted here.
Regretfully, I do not understand your argument that the reduced-mass solution is correct and the Schwarzschild solution is incorrect. First, I'm not sure what you mean by the reduced-mass solution; I'm guessing that you mean the first-order post-Newtonian solution? Second, you surely know that neither solution is perfectly correct. Both are approximations, albeit very accurate approximations. In the case of Mercury, the 0.000016% change from m to μ is, as far as I know, not detectable by any experiment and will likely remain so for the foreseeable future. There are other corrections that are far greater (but still below the threshold of detectability) such as the ζ term in the post-Newtonian expansion (produced by the Sun's rotation), the terms deriving from the oblateness of the Sun, and the post-Newtonian corrections of the other planets. (You can find a discussion of those corrections on pages 230–233 (section 9.5) of the cited Weinberg textbook.) On the other hand, I grant you that the Schwarzschild solution is not accurate when dealing with a binary star; that's where the post-Newtonian solution comes in, as discussed in the article already.
May I propose a path out of our labyrinth? Let's eliminate the discussion of mass altogether, in keeping with the Equivalence Principle, and discuss the solutions only in terms of the length-scales rs, a and b, which are constants of the motion. Then we can relate those length scales to the classical conceptions of force, energy, angular momentum and, yes, reduced mass. Hoping you and the others here favor that approach, Willow (talk) 14:40, 10 July 2010 (UTC)
Whoah, I never claimed the Schwarzschild solution is wrong and that classical physics is right. I didn't just fall off of the turnip truck, my personal ephemeris uses the PPN n-body Einstein–Infeld–Hoffmann equation, the Chebychev implementation of the TE405 time dilation integral, the EGM96 spherical harmonic geopotential to 70th order, the gtopo30 global digital elevation model, the Unified Lunar DEM, and 312,000 minor planets from the MPCORB database. It matches observed lunar laser ranges to within 10 centimeters.
Anyway, the definition of "the Kepler problem" is "a special case of the 2-body problem where the force varies as the inverse square of the distance". Since the gravitational force in GR is not inverse square, there is no analog to the Kepler problem in GR.
I think Wikipedia could use two articles. One on the Schwarzschild solution, and one on "the 2-body problem in GR". Currently the Schwarzschild solution redirects to the Schwarzschild metric, which seems backwards.
"There are no forces anymore". Oh no, one of the most dreaded Relativity clichés ever! At least you didn't say "Mass tells space how to curve and space tells mass how to move". Lol. If I don't quote Einstein right now, I'll have to drink twice. "Time is what is measured by a clock, and a force is what is measured by a force gauge". The article says, "General relativity introduces a third force that attracts the particle slightly more strongly than Newtonian gravity". As you well know, in addition to contact and electromagnetic forces, there are fictitious forces in GR, and gravity happens to be one of them. NOrbeck (talk) 13:14, 12 July 2010 (UTC)
Hi Norbeck, I didn't mean to suggest that you believed that the classical solution was absolutely correct; it's just that you seemed to be saying above that the first-order PPN solution, the Einstein–Infeld–Hoffmann equation, was absolutely correct. As I wrote, I knew that you knew better, but I was just trying to remind you that the EIH equation is not an exact solution of the 2-body problem in GR. If it were, then there would be no trouble in simulating the merger of two black holes.
I agree that "Two-body problem in general relativity" is a better name, and I'll be glad to move the page. I'd been thinking of doing that already, since most of the sources refer to it that way. I chose to call it the "Kepler problem" when I started the article, because I wanted to keep a close connection with the classical problem, for the benefit of readers who might be going back and forth. "Kepler problem" is nice, too, because it connotes that only gravity is at work, not other interactions; I don't believe that it's restricted to 1/r2 forces. Strictly speaking, the "two-body problem in GR" would have to include the possibility of, e.g., electromagnetic interactions, and I had wanted to avoid such complications for this article.
I understand why you'd like to split the article, and I definitely agree that the Schwarzschild solution is poorly redirected. But since the two-body problem reduces to the Schwarzschild solution in the first-order PPN, albeit with the combined mass, it seems sensible to keep them together, don't you agree? If we find that the article becomes too long, then of course we should split it into two, no question.
I've been re-designing the page for the past few days, while I was awaiting your reply, and I think I have a good solution that will satisfy your concerns. If it doesn't, well, then we'll keep tinkering until it does. Have a little patience with me? I think you'll like what you see. :) Willow (talk) 15:00, 12 July 2010 (UTC)
I moved the page as suggested, but now I'm beginning to think that it is a good idea to split the pages, since the Schwarzschild solution pertains to more than just the two-body problem. As you say, it really is a central-force problem, which could in principle describe many moving masses, such as the Solar System. It also has the "no interactions except gravity" built in, which is an advantage. But splitting won't leave much in this article, once the Schwarzschild material is taken out. I suppose that would give us more space to cover gravitational radiation and the merger of black holes... :) Please let me think about it for a few hours; I'll try to do something reasonable tonight. Willow (talk) 15:27, 12 July 2010 (UTC)
Please feel free to revert any of my edits. I broke the live version of the page, and visitors are currently being subjected to a draft without so much as a warning. The important thing is that the correct formula for the precession rate is given: 6piG(M+m)h^2/c^2p. Every book I own and 99.9% of sources still give the incorrect formula. For equal mass binaries the classic formula predicts half the actual precession. Norbeck (talk) 12:21, 13 July 2010 (UTC)

### Force exists in GR

To WillowW: You said "There are no forces anymore, ...". No, there is a notion of force in general relativity; it is just not a tensor. The total force on a test particle is
$\frac{d p_\mu}{d t} = \Gamma^{\lambda}_{\mu \nu} p_{\lambda} \frac{d x^\nu}{dt} + F_{\mu \nu} q \frac{d x^\nu}{dt} + \text{ other forces } \,$
where t is a time coordinate for the particle (not necessarily the same as either x0 or τ). The linear momentum, pμ, of a particle is a covariant tensor. For particles which have mass, linear momentum is
$p_{\alpha} = m \, g_{\alpha \beta} \, \frac{d x^{\beta}}{d \tau} \,.$
JRSpriggs (talk) 01:23, 12 July 2010 (UTC)
Hi, JR, and thank you for pointing that out! I'd never seen that calculation before. :) I definitely agree that dp/dt exists in GR. I was only trying to point out that Newton's laws of motion (which led to the introduction of the reduced mass for the classical two-body problem) and Newton's law for the gravitational force don't hold in GR; therefore, it's not obvious to me why μ is the right quantity to introduce when the masses of the two bodies become comparable, unless you use the hand-wavy Equivalence Principle idea I sketched above. Rather, we have to solve the Einstein field equations, which is quite difficult even for a pair of black holes. I'll be adding more on recent numerical simulations of coalescing black holes shortly — stay tuned! :) Willow (talk) 02:35, 12 July 2010 (UTC)
For the derivation of the formula above and more information, please see User:JRSpriggs/Force in general relativity. JRSpriggs (talk) 23:54, 6 June 2012 (UTC)

### Correspondence to Classical Physics

The introduction of the Gaussian gravitational constant by Gauss in Theoria Motus was the 19th Century's revolution in astronomy. The constant expression which he used to define k is arrived at by equating h from the conservation of angular momentum with that in p = h2/μ from the solution for the orbit in the two-body problem where μ = GM(1+m). Classically n2a3 = GM(1+m) = k2(1+m) and this is the result that GR has to agree with. The correct Newtonian potential also involves the sum of the masses and the term GM(1+m). Both seem to suggest the GM → G(m1 + m2). What seems to be a little strange is that the metrics for the two bodies are combined into one for the system due to their motions being coupled. The GR equation for the orbit only appears to have an altered true anomaly. The parameter p approximately equals the classical value. It is probably better to think in terms of a precessing frame of reference for the orbit in GR rather than an osculating orbit. All this suggests that there should be an analog of Gauss' constant expression in GR for a body in orbit about the Sun. --Jbergquist (talk) 13:01, 21 November 2011 (UTC)

## Boundary condition for absence of incoming gravitational radiation

Something needs to be said about how to formulate a boundary condition specifying that there is no incoming gravitational radiation. Without that condition, nothing can be said about the stability of the two body system or whether there will be any outgoing radiation. In particular, the time-reversal of any solution is also a solution. So if there are solutions with no incoming radiation, then there are solutions with no outgoing radiation. JRSpriggs (talk) 05:51, 16 July 2010 (UTC)

## Lineal gravity

Something is missing in this article. The metric for the case of 2 comparable masses (e.g. binary stars) cannot be obtained in closed form and so one has to resort to e.g. Post-Newtonian approximations valid in the weak field approximation and it's the non-linear aspects of GRT which are interesting. The latter is questionable because nobody really understand what it means to "perturb the metric". However, if instead of space-time i.e. 3 spatial dimensions + one time dimension (3+1), you restrict yourself to only one spatial dimension + one time dimension (1+1), you can actually get the metric in closed-form -well almost - in this case an exact solution in terms of the Lambert W function and its generalization. This was found out by Robert Mann and co-workers in the late 1990s. The physics in 1+1 dimensions, called "lineal gravity" (N.B. "lineal" not "linear" - "lineal" means on the line) is different though: gravitational energy exchanged between the 2 bodies is done via dilatons NOT gravitons which require 3+1 dimensions to propagate. Apart from the dimensional reduction, it fully contains a one-dimensional limit of the Einstein Field equations - namely R=T theory. If nobody has any objections, I would like to a comment or two on this aspect. It's instructive because it's one of the few many-body problems in gravity theory with an exact analytical solution! TonyMath (talk) 23:00, 18 November 2010 (UTC)

## Nice Job

Just one little thing... I thought this article quite well written. Clear, informative, and to the point in simple language. Very Einsteinian. Bravo. There is one little point... someone has written that Einstein "gradually groped" his way to the general theory. I have corrected the unreferenced hyperbole to "worked toward". Anyone thinking "gradually groped" is correct should discuss it before reverting. LeChiff200 (talk) 20:40, 1 January 2011 (UTC)

Just saying "worked toward" makes it sound as if he was solving a straight-forward problem. If that were the case, he (being a genius) would have arrived at the solution much sooner. He did not know at first what criteria the new theory of gravity would have to satisfy. Many possibilities were either too restrictive and thus gave a contradictory result, or they were not sufficient to give a usable theory. He lacked experimental results bearing on the regime where special relativity and Newtonian gravity conflicted (i.e. high speeds and large masses together). The theoretical constructs which might be included were all of uncertain validity. Calculations of trial solutions could not be carried out exactly because of the nonlinear nature of the theory. So I would say that "groping" (trying to find your way slowly by feel in the darkness) was a good description. JRSpriggs (talk) 12:21, 2 January 2011 (UTC)

## roughly 1.75 arcseconds, roughly one millionth part of a circle

The following in 'Bending of light by gravity' is not clear: "roughly 1.75 arcseconds, roughly one millionth part of a circle". The reason it is not clear: a circle with 2 * pi radians and a "one millionth part of a circle" are two different things. — Preceding unsigned comment added by Reddwarf2956 (talkcontribs) 13:14, 12 July 2012 (UTC)

The amount of deflection of light just grazing the Sun is 1.75 arcseconds which is
$1.75 \cdot \frac{1}{60} \cdot \frac{1}{60} \cdot \frac{1}{360} = 1.35 \cdot 10^{-6} \,$
of a circle. This is 0.00000848 radians. So what is your problem? JRSpriggs (talk) 15:47, 12 July 2012 (UTC)
Sorry for the delay. Yes, your comments here are correct. But, 0.00000848 radians times 10^6 makes 8.48 micro-radians. This is how I misunderstood -- That number is almost 10 times larger than 10^-6. It might be nice to also state the units in micro-radians or as micro-turns. — Preceding unsigned comment added by Reddwarf2956 (talkcontribs) 01:48, 10 August 2012 (UTC)
You can add the value in whichever units you prefer to the article. I am happy with arcseconds. JRSpriggs (talk) 06:48, 10 August 2012 (UTC)

## Where is h defined?

This equation has a variable h:

$\left( \frac{dr}{d\tau} \right)^{2} = \frac{E^{2}}{m^{2}c^{2}} - \left( 1 - \frac{r_{s}}{r} \right) \left( c^{2} + \frac{h^{2}}{r^{2}} \right).$

What is h? It is later defined that a=h/c. It also says that a is a length metric. The variable a appears in this equation:

$r_{\mathrm{outer}} \approx \frac{2a^{2}}{r_{s}}$

This would also imply that h has units of length. This means that it's certainly not Heisenberg's constant. I don't think it's defined in the article. I really want to know what this is, I can't understand anything without it. -Theanphibian (talkcontribs) 19:48, 10 December 2012 (UTC)

I think that h is the orbital angular momentum L divided by the reduced mass μ. See Schwarzschild geodesics which used to be part of the same article before it was split in two. It must have units of length squared divided by time. JRSpriggs (talk) 05:45, 11 December 2012 (UTC)

This article is a mess. First of all, what it describes is NOT what is called the two-body problem, it is the ONE-body problem. The two-body problem is for two non-negligible masses. It is well known that there is no known exact solution for the two-body problem in general relativity. (There is, of course, an exact solution in Newtonian physics.) The Schwarzschild solution in general relativity is the essentially unique solution of the ONE-body problem. It represents the field of a single body. The paths of any number of test particles with negligible mass-energy can be found in this field, but that does not make it a solution of the 2-body problem, let alone the N-body problem. Now, it IS possible to deal numerically with the two-body (and even the N-body) problem in general relativity, but this article doesn't correctly describe any of that. I think the article should be completely re-written by someone who knows something about the subject.Urgent01 (talk) 13:00, 27 May 2013 (UTC)

I agree with Urgent01. When I first read this article, I wonder where is two body problem in GR in it. Maybe the section "Corrections to the Schwarzschild solution" describes something about two body, the rest is similar to the Schwarzschild geodesics article. Many reliable sources like this and [1] can be used for rewriting this article and it needs attention from an experts in the subject.Earthandmoon (talk) 05:01, 28 May 2013 (UTC)
I agree that the article is rather unbalanced. However, I would disagree that the test mass limit of the extreme mass-ratio regime (which is what 90% of this article is effectively describing) is not part of the relativistic two body problem. In fact, for a better written version of this article it is probably still a good idea to start with this to introduce some concepts like relativistic precision, orbital frequencies, etc. and then move one to radiation reaction and end up with the comparable mass regime. If I have some time to waste in the near future, I might have a go at this.TR 12:28, 16 December 2013 (UTC)

## It needs to explain the Geodesic equation

This article needs to explain the geodesic equation, as it is important in solving the two body problem. This article needs to talk about the geodesic equation, as it is important in solving the two body problem. The geodesic equation is given by $\frac{d^2x^a}{d\tau ^2}=-\Gamma ^a _{bc} \frac{dx^b}{d\tau } \frac{dx^c}{d\tau }$. This is a very important equation, as it describes the acceleration of an object in a gravitational field. I found it, but it gives no explanation of how it relates to the two body problem. This article also needs to explain the metric when there is two bodies affecting the gravitational field. — Preceding unsigned comment added by Theendercreeper (talkcontribs) 16:16, 23 February 2014 (UTC)

The article already mentions this equation in Two-body problem in general relativity#Geodesic equation and provides a link to Schwarzschild geodesics which gives more details about the geodesic equation for the most relevant particular metric. JRSpriggs (talk) 00:46, 24 February 2014 (UTC)

## \pi missing in Rate of period decrease

I found a \pi missing in the period decrease formula, so I added it. I assume the equation was obtained by equating the rate of losing energy and differentiation of Newtonian gravitational potential. — Preceding unsigned comment added by 188.230.159.89 (talk) 14:21, 10 August 2015 (UTC)