Vijay-Chawan / Schwarzchild-Black-Hole-Solution

Einstein field equation solution for a static and sperically symmetric object kept in vaccume using Mathematica.

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Schwarzchild-Black-Hole-Solution

Deriving the Einstein field equation solution for static and sperically symmetric object kept in vacuum using Mathematica.

Background: We will start with a static and spherically symmetric metric. Only gtt, grr, gθθ gphi-phi terms survives. Out of which gthetha-thetha and gphi-phi are -r^2 and -r^2Sin^2(thetha), this is because for a t=constant and r=constant surface the object must have a line element of a two-sphere. So what left is to calculate gtt and grr. will call gtt and grr terms to be A(r) and B(r) respectively. As the object is kept in vaccume, we will have T_mu-nu to be zero, this gives R_mu-nu to be zero. In general R_mu-nu is has 10 components as it is a symmetric tensor, but because of symmetries we invoked in the line element we will see that only the digonal terms are non-zero. All the non-diagonal terms will vanishs.

Procedure: We start calculating Affine connection, Riemann tensor, and Ricci tensor. We will then set the components of Ricci tensor to be zero, this will result four equations(R11=0=R22=0=R33=0=R44). We will see that the fourth equation is a dependent equation. We will solve the three equations and invoke conditions such as the space is asymptotically flat and the gtt component must give Newton-gravity in weak field limit to calulate A(r) and B(r). Solving R11 = 0 and R22 = 0 simultaneously, gives B'/A' = -B/A(prime is derivative with respect to r), which gives, A(r)*B(r) = constant. We will use the condition that the space is asymptotically flat, thus lim r tends to infinity both A(r) and B(r) must be 1(line element of flat space in). This gives A(r)B(r) = 1. Next, we will use B'/A' = -B/A in R33, from here we will obtain A + rA' = 1, which results Ar = r + constant, i.e. A = 1 + constant/r. The value of this constant comes by invoking the weak-field limit, under wich gtt = (1 - 2GM/rc^2) comparing the two we seee, constant = -2GM/c^2.

Final Solution is A(r) = 1 - (2GM/rc^2) and B(r) = 1/A(r) = (1 - (2GM/rc^2))^-1. Thus by solving three equations and invoking conditions such as space is asymptotically flat and gtt component must give Newton-gravity in weak field limit, we calcuate the line element.

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Einstein field equation solution for a static and sperically symmetric object kept in vaccume using Mathematica.