Derivation of The Schwarzschild Solution - Wikipedia

Derivation of the
Schwarzschild
solution
The Schwarzschild solution describes

spacetime under the influence of a
massive, non-rotating, spherically
symmetric object. It is considered by
some to be one of the simplest and most
useful solutions to the Einstein field
equations .
Assumptions and notation
Working in a coordinate chart with
coordinates labelled 1 to 4
respectively, we begin with the metric in
its most general form (10 independent
components, each of which is a smooth
function of 4 variables). The solution is
assumed to be spherically symmetric,
static and vacuum. For the purposes of
this article, these assumptions may be
stated as follows (see the relevant links
for precise definitions):
1. A spherically symmetric spacetime

is one that is invariant under
rotations and taking the mirror
image.
2. A static spacetime is one in which
all metric components are
independent of the time coordinate
(so that ) and the
geometry of the spacetime is
unchanged under a time-reversal
.
3. A vacuum solution is one that
satisfies the equation .
From the Einstein field equations
(with zero cosmological constant),
this implies that since
contracting
yields .
4. Metric signature used here is (+,+,+,
−).
Diagonalising the metric

The first simplification to be made is to
diagonalise the metric. Under the
coordinate transformation,
, all metric
components should remain the same.
The metric components ( )
change under this transformation as:
But, as we expect (metric

components remain the same), this
means that:
( )
Similarly, the coordinate transformations

and
respectively
give:
( )
( )
Putting all these together gives:
( )
and hence the metric must be of the

form:
where the four metric components are
independent of the time coordinate (by
the static assumption).
Simplifying the components

On each hypersurface of constant ,
constant and constant (i.e., on each
radial line), should only depend on
(by spherical symmetry). Hence is a
function of a single variable:
A similar argument applied to shows

that:
On the hypersurfaces of constant and
constant , it is required that the metric
be that of a 2-sphere:
Choosing one of these hypersurfaces

(the one with radius , say), the metric
components restricted to this
hypersurface (which we denote by
and ) should be unchanged under
rotations through and (again, by
spherical symmetry). Comparing the
forms of the metric on this hypersurface
gives:
which immediately yields:
and
But this is required to hold on each

hypersurface; hence,
and
An alternative intuitive way to see that

and must be the same as for a
flat spacetime is that stretching or
compressing an elastic material in a
spherically symmetric manner (radially)
will not change the angular distance
between two points.
Thus, the metric can be put in the form:

with and as yet undetermined
functions of . Note that if or is
equal to zero at some point, the metric
would be singular at that point.
Calculating the Christoffel

symbols
Using the metric above, we find the
Christoffel symbols, where the indices
are . The sign
denotes a total derivative of a function.
Using the field equations to
find A(r) and B(r)
To determine and , the vacuum field
equations are employed:
Hence:
where a comma is used to set off the

index that is being used for the
derivative. The Ricci curvature is diagonal
in the given coordinates:
where the prime means the r derivative of
the functions.
Only three of the field equations are

nontrivial and upon simplification
become:
(the fourth equation is just times
the second equation). Subtracting the
first and third equations produces:
where is a non-zero real constant.

Substituting into the
second equation and tidying up gives:
which has general solution:
for some non-zero real constant .

Hence, the metric for a static, spherically
symmetric vacuum solution is now of the
form:
Note that the spacetime represented by

the above metric is asymptotically flat,
i.e. as , the metric approaches
that of the Minkowski metric and the
spacetime manifold resembles that of
Minkowski space.
Using the weak-field
approximation to find K and
S
This diagram gives the route to find the Schwarzschild solution by using the weak field approximation. The equality on
the second row gives g44 = −c2 + 2GM/r, assuming the desired solution degenerates to Minkowski metric when the
motion happens far away from the blackhole (r approaches to positive infinity).
The geodesics of the metric (obtained

where is extremised) must, in some
limit (e.g., toward infinite speed of light),
agree with the solutions of Newtonian
motion (e.g., obtained by Lagrange
equations). (The metric must also limit to
Minkowski space when the mass it
represents vanishes.)
(where is the kinetic energy and

is the Potential Energy due to
gravity) The constants and are fully
determined by some variant of this
approach; from the weak-field
approximation one arrives at the result:
where is the gravitational constant,

is the mass of the gravitational source
and is the speed of light. It is found
that:
and
Hence:
and
So, the Schwarzschild metric may finally

be written in the form:
Note that:
is the definition of the Schwarzschild
radius for an object of mass , so the
Schwarzschild metric may be rewritten in
the alternative form:
which shows that the metric becomes

singular approaching the event horizon
(that is, ). The metric singularity
is not a physical one (although there is a
real physical singularity at ), as can
be shown by using a suitable coordinate
transformation (e.g. the Kruskal–
Szekeres coordinate system).
Alternate derivation using
known physics in special
cases
The Schwarzschild metric can also be
derived using the known physics for a
circular orbit and a temporarily stationary
point mass.[1] Start with the metric with
coefficients that are unknown
coefficients of :
Now apply the Euler–Lagrange equation

to the arc length integral
Since
is constant, the integrand can be
replaced with because the
E–L equation is exactly the same if the
integrand is multiplied by any constant.
Applying the E–L equation to with the
modified integrand yields:
where dot denotes differentiation with

respect to
In a circular orbit so the first

E–L equation above is equivalent to
Kepler's third law of motion is
In a circular orbit, the period equals

implying
since the point mass is negligible

compared to the mass of the central
body So and
integrating this yields
where is an
unknown constant of integration. can
be determined by setting in
which case the spacetime is flat and
So and
When the point mass is temporarily

stationary, and The
original metric equation becomes
and the first E–L
equation above becomes
When the point
mass is temporarily stationary, is the
acceleration of gravity, So
Alternative form in isotropic
coordinates
The original formulation of the metric
uses anisotropic coordinates in which
the velocity of light is not the same in the
radial and transverse directions. Arthur
Eddington gave alternative forms in
isotropic coordinates.[2] For isotropic
spherical coordinates , , ,
coordinates and are unchanged, and
then (provided )[3]

,
, and
Then for isotropic rectangular

coordinates , , ,
The metric then becomes, in isotropic

rectangular coordinates:
Dispensing with the static
assumption – Birkhoff's
theorem
In deriving the Schwarzschild metric, it
was assumed that the metric was
vacuum, spherically symmetric and
static. The static assumption is
unneeded, as Birkhoff's theorem states
that any spherically symmetric vacuum
solution of Einstein's field equations is
stationary; the Schwarzschild solution
thus follows. Birkhoff's theorem has the
consequence that any pulsating star that
remains spherically symmetric does not
generate gravitational waves, as the
region exterior to the star remains static.
See also
Karl Schwarzschild
Kerr metric
Reissner–Nordström metric
References
1. Brown, Kevin. "Reflections on
Relativity" (http://www.mathpages.co
m/rr/s5-05/5-05.htm) .
2. A S Eddington, "Mathematical Theory

of Relativity" (https://books.google.c
om/books?id=Hhg0AAAAIAAJ&pg=P
A93) , Cambridge UP 1922 (2nd
ed.1924, repr.1960), at page 85 (http
s://books.google.com/books?id=Hhg
0AAAAIAAJ&pg=PA85) and page 93
(https://books.google.com/books?id
=Hhg0AAAAIAAJ&pg=PA93) .
Symbol usage in the Eddington
source for interval s and time-like
coordinate t has been converted for
compatibility with the usage in the
derivation above.
3. Buchdahl, H. A. (1985). "Isotropic

coordinates and Schwarzschild
metric". International Journal of
Theoretical Physics. 24 (7): 731–
739. Bibcode:1985IJTP...24..731B (ht
tps://ui.adsabs.harvard.edu/abs/198
5IJTP...24..731B) .
doi:10.1007/BF00670880 (https://do
i.org/10.1007%2FBF00670880) .
S2CID 121246377 (https://api.seman
ticscholar.org/CorpusID:12124637
7) .
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Derivation of the

The Schwarzschild solution describes

1. A spherically symmetric spacetime

Diagonalising the metric

But, as we expect (metric

Similarly, the coordinate transformations

Putting all these together gives:

and hence the metric must be of the

Simplifying the components

A similar argument applied to shows

Choosing one of these hypersurfaces

But this is required to hold on each

An alternative intuitive way to see that

Thus, the metric can be put in the form:

Calculating the Christoffel

where a comma is used to set off the

Only three of the field equations are

where is a non-zero real constant.

which has general solution:

for some non-zero real constant .

Note that the spacetime represented by

The geodesics of the metric (obtained

(where is the kinetic energy and

where is the gravitational constant,

So, the Schwarzschild metric may finally

which shows that the metric becomes

Now apply the Euler–Lagrange equation

where dot denotes differentiation with

In a circular orbit so the first

In a circular orbit, the period equals

since the point mass is negligible

When the point mass is temporarily

then (provided )[3]

Then for isotropic rectangular

The metric then becomes, in isotropic

2. A S Eddington, "Mathematical Theory

3. Buchdahl, H. A. (1985). "Isotropic

This page was last edited on 12 December 2022,

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