General Relativity and the barycenter

I'm self training in physics, trying to understand as much physics as possible despite having very basic math skills and understanding. Until recently I thought I had understood the basics of gravity: A given configuration of matter distorts spacetime geometry. This distorted geometry makes matter move in certain ways. The movement changes the matter configuration as the sources of gravity change their locations. If Einstein is right and curvature of spacetime gives rise to what we call gravity, I'm struggling to understand how the barycentre fits into the picture. When 2 bodies orbit each other they each follow the shortest path through in spacetime but just how does that lead to a barycentre? Each space time distortion is centred at the body center of mass but how does the interaction between the 2 space time distortions lead to the barycentre?

I would love to hear your answers on this since google results seem to give me anything but the answers to these questions (I get lost in results).

• Off the top of my head, I'm not sure that one can even define the notion of a barycentre in a general way in a curved space. The barycentre of a system is sort of a "weighted average" of the locations of its particles. But in a curved space, you can't really talk about adding the position vectors of multiple particles (which you'd need to do to add them): the "position vector" is not well-defined in curved space. I'll have to think about this more, though. – Michael Seifert Dec 8 '16 at 22:32
• Probably the pithiest answer I can give is that General Relativity implies Newton's Laws of Gravitation when gravity is relatively weak, which in turn implies the existence of a barycentre. But I admit that this answer is not entirely satisfying. – Michael Seifert Dec 8 '16 at 22:33

The concept of a barycenter, derived from Newtonian mechanics, does not generalize well to general relativity. In particular, the closed orbits you get by solving Lagrange's equation for two bodies are not solutions for full general relativity.

The most dramatic effect you get is that the orbiting bodies emit gravitational radiation, which will cause an inspiral of the bodies toward each other as the orbit loses energy. If the bodies have different masses, this radiation will have net momentum, which will create a net acceleration of the system. Typically, this will be quite small, but during the late stages of a black hole merger, you can "kick" up enough momentum on a black hole to accelerate it past escape velocity for a galaxy.

I am not sure if by barycentre you really mean centre of mass or centre of gravity, which are not, in general, necessarily the same. I will suppose you were asking about the centre of mass.

We would like to capture the classical picture of the point given by the mass weighted sum of the position vectors, the "middle" of the mass distribution. The main problem in the case of general relativity is that there are a priori no way to define equivalent calculation in the general case. To understand this bit, a little side discussion is needed.

To properly and consistently define translations in space and time, rotations in space and Lorentz boosts (or translations in momentum), what is really needed is what is called a representation of all those operations on the system under study. Those operations form what is called a group and a representation is really just describing how those physical transformations act on our mathematical models of reality. The important point here is that in the Newtonian/Special Relativity pictures, this is never a problem, whereas in General Relativity we only ask that this is implemented locally (precisely: that the tangent space at each points supports a representation).

We can now state the first obstacle to a good definition of a barycentre. A position vector is really just an encoding of a translation that leads to a point starting from a reference point. Likewise, the mass of an object can be associated to the effect of a Lorentz boost and a time translation. So, to carry out the barycentre computation, we would need the whole spacetime manifold to be a representation of the Lorentz group. This is not always possible as only infinitesimal transformations are always well defined.

When the spacetime is asymptotically flat (it essentially extends infinitely to a massless vacuum), there is a possibility to properly define those quantities. What is done is that very far from where there is mass, spacetime is locally assumed to be almost like the spacetime of Special Relativity, which is a representation of the Lorentz group. There, one can define a coordinate system and define the concepts mentioned above.

However, another difficulty can now appear. How do you treat massless particles such as photons? What is the position of a photon when the geodesic distance between two "separate" points on its path is zero? You could argue it has zero mass and is not relevant, but it does carry energy, and would, a the very least, certainly contribute to the centre of mass if in a bound state, for instance in a cavity. There are also ways to deal with this issue in some cases, but I hope you see the pattern here: dealing with General Relativity involves various subtleties that can get tricky. I myself made a terrible error in the first version of this post which lead me to rewriting it completely.

There are ways to systematically deal with the subtleties with the notion of intrinsic quantities, as defined in differential geometries. The general idea is to look for the quantities having a physical meaning that is invariant under the choice of coordinate system. This is to separate an artefact simply caused by a bad choice of coordinates from a truly physical phenomenon.

And now to give a final answer to your question, I need to introduce one last idea. There is a theorem that relates transformations, such as the ones above, to conserved quantities, when those transformations are also symmetries of the dynamic of a system. This a simple depiction of what is called Noether's theorem. For an isolated system, the Lorentz transformations are seen to be symmetries of reality.

There is one interesting conserved quantity associated to particular symmetry transformations that are of central relevance to your question. These transformations are the Lorentz boosts and the associated conserved quantity is what I think would be the best definition for a centre of mass in general relativity. It might not always concur with how we would picture the centre of mass, but it has the advantage of being tied deeply to the mathematical definition of the objects in the theory. Maybe a better name would be "centre of energy".

• This answer is completely incorrect. Local curvature of spacetime does not cancel out at the barycenter. Consider our own solar system, where the barycenter is very close to (and oftentimes inside) the Sun. Local curvature doesn't "cancel out" there. – David Hammen Dec 8 '16 at 17:40
• @DavidHammen You are right, what a mistake! I think I got confused with the "distortion" wording and myself calling the connection the curvature in my head (even if it totally isn't, using differential geometry terminology). Thank you for spotting the absurdity. I am correcting this now. – G. Bergeron Dec 8 '16 at 20:33
• @DavidHammen What do you think of this now? I think I learnt my lesson: do not post answers in a state of acute sleep deficit... – G. Bergeron Dec 8 '16 at 22:24