# Step by step algorithm to solve Einstein's equations

I cannot completely understand what is a regular method to solve Einstein's equations in GR when there are no handy hints like spherical symmetry or time-independence.

E.g. how can one derive Schwarzschild metric starting from arbitrary coordinates $x^0, x^1, x^2, x^3$? I don't even understand the stress-energy tensor form in such a case - obviously it should be proportional to $\delta(x - x_0(s))$, where $x_0(s)$ is a parametrized particle's world-line, but if the metric is unknown in advance how do I get $x_0(s)$ without any a priori assumptions?

• There is the perturbation method leading to the weak field metric. – Torsten Hĕrculĕ Cärlemän Dec 29 '13 at 19:33
• @TorstenHĕrculĕCärlemän, what to do if the field is not weak? – xaxa Dec 29 '13 at 19:56
• Well, you can always see the field equations as a system of partial differential equations and solve them numerically. Of course, there would be simplifications like the Bianchi identites etc,. – Torsten Hĕrculĕ Cärlemän Dec 29 '13 at 20:55
• Not my field, but I believe "singularities" like point masses are replaced with analytical solutions (Schwartzschild, and so on) near the mass and the numerical simulation is done in the discretised spacetime outside this reagion: the analytical solution sets the boundary conditions at the chosen bounding surface that "excises" the mass. – Selene Routley Dec 30 '13 at 0:29
• Look at the ADM formalism for a 3+1 decomposition with explicit Cauchy problem formulation/evolution equations for the metric. Version more suitable for numeric calculations is called BSSN formalism. – user23660 Dec 30 '13 at 6:18

Second, Einstein's equations don't determine the metric uniquely. Even with well-defined initial/boundary conditions, they only determine the solution (metric tensor field) up to a general coordinate transformation (which may be determined by 4 functions $X^\mu(x^\nu)$ of the old coordinates). It means that out of the 10 components of the symmetric metric tensor, only 6 functions are really independently physical. When we impose 4 "gauge-fixing" conditions on the metric tensor field, we effectively define the "right" coordinates and we are left with 6 independent equations for the remaining 6 functions that determine the metric tensor as a function of the coordinates. Einstein's equations are superficially 10 equations but 4 of them (more precisely 4 equations constructed out of the derivatives of these equations and the equations themselves), the covariant divergence $\nabla_\mu (G^{\mu\nu} - K\cdot T_{\mu\nu})=0$, are obeyed identically so they don't constrain the metric.
Third, general relativity may also contain point masses, the point-like sources of the gravitational field that indeed add a delta-function of a sort to the metric tensor. If that's so, general relativity is a coupled system of mutually interacting Einstein's partial differential equations and ordinary differential equations for the world lines which may be parameterized e.g. by $t(x^i)$ or otherwise (e.g. using an auxiliary time parameter along the world line – which requires us to deal with a one-dimensional coordinate transformation redundancy analogous to the four-dimensional above). Alternatively, matter may be described by electromagnetic, Klein-Gordon, Dirac, and other fields. In that case, we deal with a coupled system of many partial differential equations – Einstein's equations plus Maxwell's equations plus the Dirac equation(s) and Klein-Gordon equation(s) with various source terms.
• I'm not expecting to find an analytic solution - I don't quite understand how to state the problem so that it would be a complete system of equations + boundary conditions. Having chosen 4 constrains on $g_{\mu\nu}$ how do I proceed to connect coordinates with $T_{\mu\nu}$? – xaxa Dec 29 '13 at 21:35
• Dear xaxa, $g_{\mu\nu}$ and $R_{\mu\nu}$ and $T_{\mu\nu}$ are just tensors i.e. packages of 10 functions of the four coordinates $x^\lambda$; the curvature tensors are expressed in terms of the metric tensor and their derivatives using the standard formulae. So Einstein's equations are sets of partial differential equations like any other set. The tensors' being collections of functions of coordinates is how they are "connected" with the coordinates - any other "connection" you are thinking about probably means that you don't understand the concept of a differential equation. – Luboš Motl Dec 30 '13 at 17:20
• That is all very fine, but if the meaning of coordinates is unknown how do I find $T_{\mu\nu}(x)$? This is a question of connection between physics and math. For example, in Schwarzschild coordinates $r$ is "distance", however under the horizon it turns into "time". So initial physical statement that point particle is at rest in the origin is not actually correct. But in this particular case there is a "guiding" principle of spherical symmetry. What is a procedure in general case? – xaxa Dec 30 '13 at 19:15
• Dear xaxa, $T_{\mu\nu}$ is just another tensor field, a set of functions of x. It is determined in terms of other, more fundamental degrees of freedom. For the electromagnetic field, it is expressed using the $FF$ products, and similarly for other fields. For idealized point masses, $T_{\mu\nu}$ has the form of the mass times the delta-function localized at the right place $x^\mu(\tau)$, and so on. The more fundamental degrees of freedom such as $F_{\mu\nu}(x^\lambda)$ or $x^\mu(\tau)$ are functions that are constrained by the (Newton or Maxwell or Dirac analogous) differential equations, too! – Luboš Motl Dec 31 '13 at 11:05
• Dear @Luboš, HNY to you! It seems you don't quite get my question. I understand what a tensor is and how $G_{\mu\nu}$ is built from $g_{\mu\nu}$. My question is more about dependence of $T_{\mu\nu}$ on $x$. Even if we have a plain EM-field, so that $T$ is constructed from $F$, still somewhere there are currents $j$ and/or boundary conditions. Example problem: Two wires are separated by distance $a$ and constant current $j_0$ floats through them. Question: find grav. field. – xaxa Jan 3 '14 at 16:43