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Einstein's equations can also be solved on a computer using sophisticated numerical methods. [1] [2] [3] Given sufficient computer power, such solutions can be more accurate than post-Newtonian solutions. However, such calculations are demanding because the equations must generally be solved in a four-dimensional space.
Near the sonic speed the PG transformation features a singularity. The singularity is also called the Prandtl–Glauert singularity, and the flow resistance is calculated to approach infinity. In reality, aerodynamic and thermodynamic perturbations get amplified strongly near the sonic speed, but a singularity does not occur.
Given the difficulty of constructing explicit small families of solutions, much less presenting something like a "general" solution to the Einstein field equation, or even a "general" solution to the vacuum field equation, a very reasonable approach is to try to find qualitative properties which hold for all solutions, or at least for all ...
But if one requires an exact solution or a solution describing strong fields, the evolution of both the metric and the stress–energy tensor must be solved for at once. To obtain solutions, the relevant equations are the above quoted EFE (in either form) plus the continuity equation (to determine the evolution of the stress–energy tensor):
Solving the geodesic equations is a procedure used in mathematics, particularly Riemannian geometry, and in physics, particularly in general relativity, that results in obtaining geodesics. Physically, these represent the paths of (usually ideal) particles with no proper acceleration , their motion satisfying the geodesic equations.
Some solutions of a differential equation having a regular singular point with indicial roots = and .. In mathematics, the method of Frobenius, named after Ferdinand Georg Frobenius, is a way to find an infinite series solution for a linear second-order ordinary differential equation of the form ″ + ′ + = with ′ and ″.
Since z = 1 − x, the solution of the hypergeometric equation at x = 1 is the same as the solution for this equation at z = 0. But the solution at z = 0 is identical to the solution we obtained for the point x = 0, if we replace each γ by α + β − γ + 1. Hence, to get the solutions, we just make this substitution in the previous results.
(,) and (,) are given, 2. (,) is given and () is real on the real axis, 3. only (,) is given, 4. only (,) is given. He is really interested in problems 3 and 4, but the answers to the easier problems 1 and 2 are needed for proving the answers to problems 3 and 4.