# How do I choose my constants?

When solutions to Einstein's equations are found (in coordinates $t,r,\theta,\phi$), such as Schwarzschild, Kerr etc. the way they are found seems to be by starting with a generalised metric in terms of functions of the appropriate parameters such as $f(r)$ for Schwarzschild and $f(r,Q)$ for Reissner-Nordstrom and then of course the more complicated issue of cross terms for rotations, and then finding these functions such that Einstein's equations are satisfied.

However, when you go about this yourself you realise that the solutions to Einstein's equations are going to result in these functions having constants present simply because these are differential equations we're solving. Of course most people who have studied GR know this, but is it perhaps taken for granted how we set these constants appropriately?

How do I know what to set this constant to? Right now I'm assuming we use dimensional analysis and the boundary conditions (i.e. you usually want Minkowski space as $r \to \infty$), but what about choosing dimensionless parameters such as the $2$ in Schwarzschild's solution involving the integration constant $2M$? Then a slightly more complicated issue of $AdS$ space where the boundary conditions are different?

Right now, when Mathematica tells me I have a free constant that won't affect Einstein's equations being satisfied, I just set it to $M$, but surely there's more to it?

• what constants do you refer to exactly? personally i dont follow Jul 5 '14 at 21:05
• Mathematically, when you solve einstein's equations you get constants (as you do when you solve differential equations right). How do we determine these constants? Jul 5 '14 at 21:09
• by boundary/initial conditions Jul 5 '14 at 21:12
• Yes but it some spaces these are difficult for me to find. Also, why the 2 is schwarzschild's constant 2M? Jul 5 '14 at 21:30
• Indeed, e.g. a black brane solution (= planar black hole solution) in AdS is often written as $ds^2=-\left(-\frac{2M}{r}+\frac{r^2}{L^2}\right)dt^2+\left(-\frac{2M}{r}+\frac{r^2}{L^2}\right)dr^2+\frac{r^2}{L^2}(dx^2+dy^2)$, though the mass of the black brane is clearly not $M$ but infinite. Quantities like mass, charge, angular momentum are often defined by a surface integral which is the conserved charge corresponding to some symmetry (e.g. time translations for $M$). I think you might fix the constant in front of the surface integral by applying it to some configuration with known mass... Jul 6 '14 at 13:51

• This does indeed give the right mass term, but requiring Schwarzschild for the Kerr metric without angular momentum does not fix the constants. If we'd pick a constant $J'=2J$ instead of the usual angular momentum $J$ in the Kerr metric, we'd still get Schwarzschild for $J'=0$. Jul 6 '14 at 11:11