# GR two-body problem and summations of Christoffel symbols

• I
D.S.Beyer
TL;DR Summary
Can you approach the GR two body problem through summations of multiple Schwarzschild solutions?
Can you approach the GR two body problem through summations of multiple Schwarzschild solutions?
Specifically, by using the Schwarzschild metric for each body of mass, then adding the Christoffel symbols together, to arrive at a new geodesic equation.

Take point C between bodies A and B.
Solve the Schwarzschild metric for both A and B.
In relation to body A, point C has specific Christoffel symbols.
Similarly, in relation to body B, point C has different Christoffel symbols.
Before the geodesics are constructed, can you add these vectors together?
To get the average Christoffel symbols, and then use that to construct the geodesic?

Would such a geodesic help model the two body problem?
Or, and this is the most likely, is it worthless?

NOTE : I’m not a math guy (clearly) but I’m really enjoying thinking about and getting into the math. Thank you for your patience.

Mentor
Can you approach the GR two body problem through summations of multiple Schwarzschild solutions?

No, because the Einstein Field Equation is nonlinear, so adding together two solutions does not give another solution.

vanhees71, martinbn, Dale and 1 other person
Mentor
Summary:: Can you approach the GR two body problem through summations of multiple Schwarzschild solutions?

Can you approach the GR two body problem through summations of multiple Schwarzschild solutions?
As @PeterDonis said this approach will not work for general relativity. It would work for electromagnetism. The difference is that Maxwell’s equations are linear and Einstein’s field equation is not.

It is possible to write a linear approximation to GR. As with any approximation it will fail when its assumptions are violated.

vanhees71 and etotheipi

$$R \sim (\Gamma + \partial\Gamma)^2$$

And as you probably know, even if you're not a math guy,

$$R \sim (\Gamma + \partial\Gamma)^2 = (\Gamma)^2 + ( \partial\Gamma)^2 + 2 \Gamma \partial\Gamma$$

That last term ##2 \Gamma \partial\Gamma## spoils the linearity. It means that the curvature of black hole A plus the curvature of black hole B is not simply the sum of curvatures:

$$R_A + R_B \sim (\Gamma_A + \partial\Gamma_A)^2 + (\Gamma_B + \partial\Gamma_B )^2 \neq (\Gamma_A + \Gamma_B + \partial\Gamma_B + \partial\Gamma_B)^2 = R_{A+B}$$

Physically, the cross-terms can be interpreted as the self-interaction of the gravitational field. It's like two competing companies: if you put them next to each-other, they will compete, and as such their prices will be influenced by each-other, and hence different from the situation when they would be far apart, just doing their own business and asking the maximum prices their customers are happy to pay.

vanhees71 and Dale
D.S.Beyer

$$R \sim (\Gamma + \partial\Gamma)^2$$

And as you probably know, even if you're not a math guy,

$$R \sim (\Gamma + \partial\Gamma)^2 = (\Gamma)^2 + ( \partial\Gamma)^2 + 2 \Gamma \partial\Gamma$$

That last term ##2 \Gamma \partial\Gamma## spoils the linearity. It means that the curvature of black hole A plus the curvature of black hole B is not simply the sum of curvatures:

$$R_A + R_B \sim (\Gamma_A + \partial\Gamma_A)^2 + (\Gamma_B + \partial\Gamma_B )^2 \neq (\Gamma_A + \Gamma_B + \partial\Gamma_B + \partial\Gamma_B)^2 = R_{A+B}$$

Physically, the cross-terms can be interpreted as the self-interaction of the gravitational field. It's like two competing companies: if you put them next to each-other, they will compete, and as such their prices will be influenced by each-other, and hence different from the situation when they would be far apart, just doing their own business and asking the maximum prices their customers are happy to pay.

Thank you. This small bit of math is helping a lot.
Sorry I tagged this as 'intermediate' when it was clearly 'beginner'

vanhees71 and Dale
Mentor
Sorry I tagged this as 'intermediate' when it was clearly 'beginner'

Actually it's still "I" because of the subject matter, which is at least undergraduate level, not high school.

vanhees71
Homework Helper
A more interesting question I would have asked. Is the pure gravitational 2-body problem in GR fully solvable? As far as I know the answer is "no".

vanhees71
Mentor
Is the pure gravitational 2-body problem in GR fully solvable? As far as I know the answer is "no".

That is my understanding as well.

vanhees71
D.S.Beyer
That is my understanding as well.

That is, really, kinda mind blowing. I always assumed GR dealt with everything at the same time.
In some magical way. But really it can't even solve 2 bodies interacting.

It's still totally amazing, and I'm learning more everyday.
But this really makes me feel like we still don't really know how anything works for sure.

I'm getting a 'The more you know, the less you understand' vibe right now.

Thanks everyone for input. I feel like this thread can be closed.

Mentor
But really it can't even solve 2 bodies interacting.
It can, but numerically rather than analytically.

vanhees71
Mentor
I always assumed GR dealt with everything at the same time.

It does. It's just that, except for very simple, highly symmetric situations, its solutions can't be expressed as closed form equations. But as @Dale points out, you can still solve any situation numerically. That is how two-body solutions are done in practice--for example, in analyzing systems like binary pulsars or merging black holes, in order to compare model predictions with observations.

martinbn
Staff Emeritus
GR has been proven to have solutions, in that it's been shown, mathematically, to be a well-posed initial value problem. Wald discusses this in "General Relativity" if you want the details. Actually finding the solutions though isn't easy.

vanhees71 and martinbn
Mentor
Is the pure gravitational 2-body problem in GR fully solvable? As far as I know the answer is "no".

GR has been proven to have solutions, in that it's been shown, mathematically, to be a well-posed initial value problem.

It's probably worth clarifying that the words "solvable" and "solution" have two different possible meanings in this context: they can mean "there are known solutions that can be expressed in closed-form equations", or they can just mean "there is a well-posed initial value problem, so you can generate a solution numerically from valid initial data even if you can't express it in closed-form equations".

The key point for the original OP question is that, since the Einstein Field Equation is nonlinear, you can't add two valid "solutions" in either of the above senses to get another valid solution.

vanhees71