Lets say i want to simulate the differential equations of GR with some numerical method. I can express the Einstein tensor in terms of the christoffel symbols which in turn can be expressed in terms of the metric and its first and second derivative.

Now i can impose a set of coordinates $[t, x, y, z]$ and set up a big cartesian grid. Each point contains information about the metric of that point, as seen by an observer at infinity. Initially the space is empty so the metric will reduce to the Minkowski-metric.

Now i place some mass at the origin, but with finite density. Assume that the mass encapsulates many grind points and the grid extends to a far distance so the metric at the end of the grid is approximatly flat.

Now i want to simulate what happens. To do this i rearrange the equations and solve for $\frac{\partial^2}{\partial t^2}g_{\mu\nu}$ which should govern the evolution of the system. Since i know the initial conditions of the metric and $T_{\mu\nu}$ i should be able to simulate the dynamics of what happens.

(I assume the metric outside would converge to the outer schwarzschild metric while the parts inside the mass would converge to the inner schwarzschild metric. Additionally a gravitational wave should radiate away because of the sudden appearance of a mass).

However, by doing so i have placed the spacetime itself on a background grid, which seems fishy to me.

Question 1: How does the coordinate system influences the equation? For example i could have choosen $[t, r, \phi, \theta]$ and would have gotten the same equations since it involves only ordinary derivatives. Do i assume correctly that the coordinate system properties only appear during numerical integration?

Question 2: What physical significance does this "cartesian grid" system have? If i look at a point near the surface of the mass after a long time, where is this point in the spacetime? A stationary observer would follow the curvature and may already have fallen into the mass. Does this mean my coordinate system itself "moves" along? How can i get a "stationary" (constant proper radial distance) coordinate system?

Question 3: Since i have the metric at every grid point i could calculate (numerically) the geodesic through this spacetime and find the path of an infalling observer, right?

Question 4: Would this work for simulating non-singular spacetimes? Or is this approach too easy?

edit1: To clarify question 1, different coordinate systems have different artefacts on their own. Instead of a grid i could use schwarzschild-coordinates. If i now expand the whole equation i would get the same form, because it is coordinate independent. However the resulting metric would look different (same metric, but expressed in another coordinate system). I'm a bit confused because the metric and the coordinate system are tied together. If i'm solving for the metric itself i still need to provide the coordinate system. And i dont understand how i introduce the coordinate system.

  • 2
    $\begingroup$ Have you looked at any papers on the subject of numerical relativity? $\endgroup$ – Kyle Kanos Dec 15 '16 at 12:12
  • $\begingroup$ It sounds as though you're doing this from first principles: have you read any numerical GR texts or background yet? It sounds like you are thinking about this intelligently, but GR is not trivial as far as numerical simulation goes - it can be as nasty as the most advanced fluid dynamics simulations and there are, as I understand it, a great many tricks of the trade which have emerged experimentally over quite a long time. $\endgroup$ – WetSavannaAnimal Dec 15 '16 at 12:12
  • $\begingroup$ @WetSavannaAnimalakaRodVance is absolutely right; numerical relativity is nasty and setting up the Cauchy problem is difficult let alone doing the numerics. There are also numerous constraints to consider, such as those related to the extrinsic curvature not in the Einstein equations. I recommend the book 'Numerical Relativity.' $\endgroup$ – JamalS Dec 15 '16 at 12:15
  • $\begingroup$ Could you say a bit more about (1): I don't think I follow. The equations are most assuredly not the same for different co-ordinate systems (when you expand them out as partial derivatives rather than in tensor form) - but this sounds too obvious a mistake from your level - hence I don't think I follow. (2) is a good conceptual question, to answer theoretically whether or not the answer is useful numerically, and probably worth asking as a conceptual question, framing the numerical simulation as simply a thought experiment. (3) seems trivial: once you have the metric, well, of course yes! $\endgroup$ – WetSavannaAnimal Dec 15 '16 at 12:55
  • 1
    $\begingroup$ Now I don't want to seem rude, but you probably are in over your head at this point. It might be better to take a few steps backwards. Perhaps solve a 1+1D problem first (plenty of examples for that, from my understanding) while you learn more diff geom $\endgroup$ – Kyle Kanos Dec 15 '16 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.