Category: General Relativity
Hits: 3509


               After a quick introduction to the Schwarzschild metric solution, it is now time to derive it.

According to his letter from 22 december 1915, Schwarzschild started out from the approximate solution in Einstein’s “perihelion paper”, published November 25th.

We will go through a more formal derivation, which could be broken down into the following steps:

- simplify the metric for  a static and spherically symmetric solution with some coefficients as functions f(r), i.e depending on r only.

- deduce the corresponding Christoffel symbols

- calculate the Ricci tensor components

- deduce the exact form of the above coefficient functions f(r)  by setting the above Ricci tensor components as null as they should be in vacuum.


Step 1 - Expression of the metric tensor for a static and spherically symmetric solution

We recall that in space-time the distance interval has the following form

In spherical coordinates t, r, θ, φ (which makes sense in case of a spherical solution..), the spacetime interval can be expanded as below:

First of all, because the solution is static, it should not depend on time. In particular, a change from t -> -t  should not change anything (time reversible), So that we should not have any croseed term in the form of dtdr, dtdθ or dtdφ, but only of the form dt2.

Then, taking in account the spherical symmetry, we can start out by the Minkowski metric written in spherical coordinates:

We are free to multiply the terms by any arbitrary r-dependent coefficients, leading to:

where α(r), β(r) and γ(r) represent some unkwnown functions of r.


In this metric, the coordinate r is the usual radial distance from the origin (the center of the source mass). I we assume that γ(r)=1, we have changed r and can no longer assume that it is the simple radial distance from the origin. This step makes sense, however, as we can rewrite the metric as simply:

We have just absorbed γ(r) by a redefnition of r (replace r by example by r'=√γ(r)r)

We note that this proposed metric has the following properties:

- none of the metric components depend on time - meaning the metric is static.

- if we let r and t be constant, then dt2 = 0 and dr2 = 0 and the metric becomes

      ds2 = r2(dθ2 + sin2θdφ2)

which is the line element for the surface of a sphere - meaning the metric is spherically symmetric.

- the functions α(r) and  β(r) must be consistent with Tμν = 0 (empty spacetime in the vicinity of a source mass)

- both α(r) and  β(r) must approach 1 as r approaches infinity, to become the Minkowski metric in spherical coordinates (the metric should be asymptotic flat)


Step 2 - Finding the connection coefficients

We recall the expression of the connection coefficient relative to the metric components gμν:


as well as the meaning of the superscript and subscript variables of the connection coefficient (refer to the article Christoffel Symbol or Connection coefficient for more details)

In the below image, the subscript index α specifies the basis vector for which the derivative is being taken, the index β denotes the coordinates being varied to induce this change in the αth basis vector, and the index γ identifies the direction in which this component of the derivatives points:


Thus the below equation should be read as follows in two-dimensionna space: the change in er caused by a change in θ has zero magnitude in the er direction, and the change in er caused by a change in θ varies inversely with distance in the eθ direction.

 We have to find the four following different coefficients:

with μ and ν taking themselves the four values t, r, Θ and φ hence a total of 4 x 16 = 64 values.

But actually, as we recall from the Christoffel symbol definition, by symmetry of the lower indices, we just have 10 independant values for each coefficient, so that the 64 values reduce to only 40 independant values (10 for each Christoffel symbol).

Let's start with the expression of Γtμν:



Writing the proposed metric in its matrix form for more lisibility

we see easily that g equates gtt as all the other terms of the first line are null; so that the Christoffel symbol can be written as

Replacing gtt by its value A(r) and knowing that gμν,t is null (nothing depends on time in the metric), we get

 which by replacing each μ and ν gives, in its matrix form


 We know already that every term gxy,t  is null as no term depends on time. We  note dA(r)/dr = A' and dB(r)/dr = B'


Let's focus now on Γrμν

We can write






 or by plugging the values of gμν and gμν

 In the same manner, we can derive Γθμν

Finally, we are left with the coefficient Γφμν


Step 3 - Calculating the Ricci tensor components

We first recall the expression of the Riemann curvature tensor

 As we know from the article Bianchi identity and Ricci tensor we contract the first and last indices of the Riemann tensor to give

where Rμν is the Ricci tensor.

Let us first calculate the component Rrr


If we now expand the first product term, we get:


 Doing the same for the last product term, we obtain:

 Coming back to the expression of Rrr:

 By deriving the first component and rearranging the terms

 In the very same way, it can easily be show that the Rtt component of the Ricci tensor is:


Step 4 - Finding the functions A(r) and B(r)

These equations look horribly complicated but thanksfully things start to get easier by combining in a clever way the two expressions: it can be shown as per below that AB = cste


To find this constant, we should bear in mind that the proposed metric must approximate to the Minkowski flat space metric as r approaches infinity.


so that

Going back to Rrr expression of the Ricci tensor, we can deduce the A function


 And as per above, keeping in mind that the proposed metric must approximate to the Minkowski flat space metric as r approaches infinity, we have k2 = c2 so that we can write A(r) as

 so that B(r) is now easily deduced

so that our proposed metric takes the following form

From our previous article Geodesic equation in the Newtonian Limit, we know that at the Newtonian limit

so that

and will retieve the final form of the Schwarzschild metric as introduced in our previous article


If we magic away our star so that M=0, then Rs=0 and the Schwarzschild metric reverts again to the flat Minkowski metric of special relativity expressed in spherical coordinates.

Only when M=0 do the coordinates t and r represent real clock-time and radial distance from the center of the mass. If we increase M we start to curve spacetime and we can no longer assume that t and r correspond to these measurable quantities.


Step 6 - Calculating the Earth and the Sun Schwarzschild radius

Assuming that they are both spherically symmetric bodies, we can  calculate the Schwarzschild radius of the Earth and the Sun.

Starting with the Earth, using the mass of the Earth = 5.97 x 1024 kg, and the gravitational constant G = 6.67 x 10-11 Nm2kg-2  we have:


or about 9 millimeters. The actual radius of the Earth is about 6370km.

Given the mass of the Sun  = 1.99 x 1030 kg, the same calculus gives

or about 3km. The actual radius of the Sun is about 696,000 km.


The most noticeable property of the Schwarzschild radius is that if all of the mass M could be squeezed inside of sphere of radius Rs, light would not be able to escape from the object and we would have created a black hole.

From the previous calculus we can see that if we wanted to transform the Earth into a black hole, we would need to squash it down into a  9mm radius sphere.