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 ( r , θ , ϕ , t ) {\displaystyle \left(r,\theta ,\phi ,t\right)} 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):
A spherically symmetric spacetime is one that is invariant under rotations and taking the mirror image. A static spacetime is one in which all metric components are independent of the time coordinate t {\displaystyle t} (so that ∂ ∂ t g μ ν = 0 {\displaystyle {\tfrac {\partial }{\partial t}}g_{\mu \nu }=0} ) and the geometry of the spacetime is unchanged under a time-reversal t → − t {\displaystyle t\rightarrow -t} . A vacuum solution is one that satisfies the equation T a b = 0 {\displaystyle T_{ab}=0} . From the Einstein field equations (with zero cosmological constant), this implies that R a b = 0 {\displaystyle R_{ab}=0} since contracting R a b − R 2 g a b = 0 {\displaystyle R_{ab}-{\tfrac {R}{2}}g_{ab}=0} yields R = 0 {\displaystyle R=0} . Metric signature used here is (+ + + −).
Diagonalising the metric The first simplification to be made is to diagonalise the metric. Under the coordinate transformation, ( r , θ , ϕ , t ) → ( r , θ , ϕ , − t ) {\displaystyle (r,\theta ,\phi ,t)\rightarrow (r,\theta ,\phi ,-t)} , all metric components should remain the same. The metric components g μ 4 {\displaystyle g_{\mu 4}} ( μ ≠ 4 {\displaystyle \mu \neq 4} ) change under this transformation as:
g μ 4 ′ = ∂ x α ∂ x ′ μ ∂ x β ∂ x ′ 4 g α β = − g μ 4 {\displaystyle g_{\mu 4}'={\frac {\partial x^{\alpha }}{\partial x^{'\mu }}}{\frac {\partial x^{\beta }}{\partial x^{'4}}}g_{\alpha \beta }=-g_{\mu 4}} ( μ ≠ 4 {\displaystyle \mu \neq 4} ) But, as we expect g μ 4 ′ = g μ 4 {\displaystyle g'_{\mu 4}=g_{\mu 4}} (metric components remain the same), this means that:
g μ 4 = 0 {\displaystyle g_{\mu 4}=0} ( μ ≠ 4 {\displaystyle \mu \neq 4} ) Similarly, the coordinate transformations ( r , θ , ϕ , t ) → ( r , θ , − ϕ , t ) {\displaystyle (r,\theta ,\phi ,t)\rightarrow (r,\theta ,-\phi ,t)} and ( r , θ , ϕ , t ) → ( r , − θ , ϕ , t ) {\displaystyle (r,\theta ,\phi ,t)\rightarrow (r,-\theta ,\phi ,t)} respectively give:
g μ 3 = 0 {\displaystyle g_{\mu 3}=0} ( μ ≠ 3 {\displaystyle \mu \neq 3} )
g μ 2 = 0 {\displaystyle g_{\mu 2}=0} ( μ ≠ 2 {\displaystyle \mu \neq 2} ) Putting all these together gives:
g μ ν = 0 {\displaystyle g_{\mu \nu }=0} ( μ ≠ ν {\displaystyle \mu \neq \nu } ) and hence the metric must be of the form:
d s 2 = g 11 d r 2 + g 22 d θ 2 + g 33 d ϕ 2 + g 44 d t 2 {\displaystyle ds^{2}=\,g_{11}\,dr^{2}+g_{22}\,d\theta ^{2}+g_{33}\,d\phi ^{2}+g_{44}\,dt^{2}}
where the four metric components are independent of the time coordinate t {\displaystyle t} (by the static assumption).
Simplifying the components On each hypersurface of constant t {\displaystyle t} , constant θ {\displaystyle \theta } and constant ϕ {\displaystyle \phi } (i.e., on each radial line), g 11 {\displaystyle g_{11}} should only depend on r {\displaystyle r} (by spherical symmetry). Hence g 11 {\displaystyle g_{11}} is a function of a single variable:
g 11 = A ( r ) {\displaystyle g_{11}=A\left(r\right)}
A similar argument applied to g 44 {\displaystyle g_{44}} shows that:
g 44 = B ( r ) {\displaystyle g_{44}=B\left(r\right)}
On the hypersurfaces of constant t {\displaystyle t} and constant r {\displaystyle r} , it is required that the metric be that of a 2-sphere:
d l 2 = r 0 2 ( d θ 2 + sin 2 θ d ϕ 2 ) {\displaystyle dl^{2}=r_{0}^{2}(d\theta ^{2}+\sin ^{2}\theta \,d\phi ^{2})}
Choosing one of these hypersurfaces (the one with radius r 0 {\displaystyle r_{0}} , say), the metric components restricted to this hypersurface (which we denote by g ~ 22 {\displaystyle {\tilde {g}}_{22}} and g ~ 33 {\displaystyle {\tilde {g}}_{33}} ) should be unchanged under rotations through θ {\displaystyle \theta } and ϕ {\displaystyle \phi } (again, by spherical symmetry). Comparing the forms of the metric on this hypersurface gives:
g ~ 22 ( d θ 2 + g ~ 33 g ~ 22 d ϕ 2 ) = r 0 2 ( d θ 2 + sin 2 θ d ϕ 2 ) {\displaystyle {\tilde {g}}_{22}\left(d\theta ^{2}+{\frac {{\tilde {g}}_{33}}{{\tilde {g}}_{22}}}\,d\phi ^{2}\right)=r_{0}^{2}(d\theta ^{2}+\sin ^{2}\theta \,d\phi ^{2})}
which immediately yields:
g ~ 22 = r 0 2 {\displaystyle {\tilde {g}}_{22}=r_{0}^{2}} and g ~ 33 = r 0 2 sin 2 θ {\displaystyle {\tilde {g}}_{33}=r_{0}^{2}\sin ^{2}\theta }
But this is required to hold on each hypersurface; hence,
g 22 = r 2 {\displaystyle g_{22}=\,r^{2}} and g 33 = r 2 sin 2 θ {\displaystyle g_{33}=\,r^{2}\sin ^{2}\theta }
An alternative intuitive way to see that g 22 {\displaystyle g_{22}} and g 33 {\displaystyle g_{33}} 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:
d s 2 = A ( r ) d r 2 + r 2 d θ 2 + r 2 sin 2 θ d ϕ 2 + B ( r ) d t 2 {\displaystyle ds^{2}=A\left(r\right)dr^{2}+r^{2}\,d\theta ^{2}+r^{2}\sin ^{2}\theta \,d\phi ^{2}+B\left(r\right)dt^{2}}
with A {\displaystyle A} and B {\displaystyle B} as yet undetermined functions of r {\displaystyle r} . Note that if A {\displaystyle A} or B {\displaystyle B} 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 ( 1 , 2 , 3 , 4 ) = ( r , θ , ϕ , t ) {\displaystyle (1,2,3,4)=(r,\theta ,\phi ,t)} . The sign ′ {\displaystyle '} denotes a total derivative of a function.
Γ i k 1 = [ A ′ / ( 2 A ) 0 0 0 0 − r / A 0 0 0 0 − r sin 2 θ / A 0 0 0 0 − B ′ / ( 2 A ) ] {\displaystyle \Gamma _{ik}^{1}={\begin{bmatrix}A'/\left(2A\right)&0&0&0\\0&-r/A&0&0\\0&0&-r\sin ^{2}\theta /A&0\\0&0&0&-B'/\left(2A\right)\end{bmatrix}}}
Γ i k 2 = [ 0 1 / r 0 0 1 / r 0 0 0 0 0 − sin θ cos θ 0 0 0 0 0 ] {\displaystyle \Gamma _{ik}^{2}={\begin{bmatrix}0&1/r&0&0\\1/r&0&0&0\\0&0&-\sin \theta \cos \theta &0\\0&0&0&0\end{bmatrix}}}
Γ i k 3 = [ 0 0 1 / r 0 0 0 cot θ 0 1 / r cot θ 0 0 0 0 0 0 ] {\displaystyle \Gamma _{ik}^{3}={\begin{bmatrix}0&0&1/r&0\\0&0&\cot \theta &0\\1/r&\cot \theta &0&0\\0&0&0&0\end{bmatrix}}}
Γ i k 4 = [ 0 0 0 B ′ / ( 2 B ) 0 0 0 0 0 0 0 0 B ′ / ( 2 B ) 0 0 0 ] {\displaystyle \Gamma _{ik}^{4}={\begin{bmatrix}0&0&0&B'/\left(2B\right)\\0&0&0&0\\0&0&0&0\\B'/\left(2B\right)&0&0&0\end{bmatrix}}}
Using the field equations to find A(r) and B(r) To determine A {\displaystyle A} and B {\displaystyle B} , the vacuum field equations are employed:
R α β = 0 {\displaystyle R_{\alpha \beta }=\,0}
Hence:
Γ β α , ρ ρ − Γ ρ α , β ρ + Γ ρ λ ρ Γ β α λ − Γ β λ ρ Γ ρ α β λ = 0 , {\displaystyle {\Gamma _{\beta \alpha ,\rho }^{\rho }}-\Gamma _{\rho \alpha ,\beta }^{\rho }+\Gamma _{\rho \lambda }^{\rho }\Gamma _{\beta \alpha }^{\lambda }-\Gamma _{\beta \lambda }^{\rho }\Gamma _{\rho \alpha {\vphantom {\beta }}}^{\lambda }=0\,,}
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:
R t t = − 1 4 B ′ A ( A ′ A − B ′ B + 4 r ) − 1 2 ( B ′ A ) ′ , {\displaystyle R_{tt}=-{\frac {1}{4}}{\frac {B'}{A}}\left({\frac {A'}{A}}-{\frac {B'}{B}}+{\frac {4}{r}}\right)-{\frac {1}{2}}\left({\frac {B'}{A}}\right)',}
R r r = − 1 2 ( B ′ B ) ′ − 1 4 ( B ′ B ) 2 + 1 4 A ′ A ( B ′ B + 4 r ) , {\displaystyle R_{rr}=-{\frac {1}{2}}\left({\frac {B'}{B}}\right)^{'}-{\frac {1}{4}}\left({\frac {B'}{B}}\right)^{2}+{\frac {1}{4}}{\frac {A'}{A}}\left({\frac {B'}{B}}+{\frac {4}{r}}\right),}
R θ θ = 1 − ( r A ) ′ − r 2 A ( A ′ A + B ′ B ) , {\displaystyle R_{\theta \theta }=1-\left({\frac {r}{A}}\right)^{'}-{\frac {r}{2A}}\left({\frac {A'}{A}}+{\frac {B'}{B}}\right),}
R ϕ ϕ = sin 2 ( θ ) R θ θ , {\displaystyle R_{\phi \phi }=\sin ^{2}(\theta )R_{\theta \theta },}
where the prime means the r derivative of the functions. Only three of the field equations are nontrivial (the fourth equation is just sin 2 θ {\displaystyle \sin ^{2}\theta } times the third equation) and upon simplification become, respectively:
4 A ′ B 2 − 2 r B ″ A B + r A ′ B ′ B + r B ′ 2 A = 0 , {\displaystyle 4A'B^{2}-2rB''AB+rA'B'B+rB'^{2}A=0,}
− 2 r B ″ A B + r A ′ B ′ B + r B ′ 2 A − 4 B ′ A B = 0 , {\displaystyle -2rB''AB+rA'B'B+rB'^{2}A-4B'AB=0,}
r A ′ B + 2 A 2 B − 2 A B − r B ′ A = 0 {\displaystyle rA'B+2A^{2}B-2AB-rB'A=0}
Subtracting the first and second equations produces:
A ′ B + A B ′ = 0 ⇒ A ( r ) B ( r ) = K {\displaystyle A'B+AB'=0\Rightarrow A(r)B(r)=K}
where K {\displaystyle K} is a non-zero real constant. Substituting A ( r ) B ( r ) = K {\displaystyle A(r)B(r)=K} into the third equation and tidying up gives:
r A ′ = A ( 1 − A ) {\displaystyle rA'=A(1-A)}
which has general solution:
A ( r ) = ( 1 + 1 S r ) − 1 {\displaystyle A(r)=\left(1+{\frac {1}{Sr}}\right)^{-1}}
for some non-zero real constant S {\displaystyle S} . Hence, the metric for a static, spherically symmetric vacuum solution is now of the form:
d s 2 = ( 1 + 1 S r ) − 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) + K ( 1 + 1 S r ) d t 2 {\displaystyle ds^{2}=\left(1+{\frac {1}{Sr}}\right)^{-1}dr^{2}+r^{2}(d\theta ^{2}+\sin ^{2}\theta \,d\phi ^{2})+K\left(1+{\frac {1}{Sr}}\right)dt^{2}}
Note that the spacetime represented by the above metric is asymptotically flat, i.e. as r → ∞ {\displaystyle r\rightarrow \infty } , 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
The geodesics of the metric (obtained where d s {\displaystyle ds} 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.)
0 = δ ∫ d s d t d t = δ ∫ ( K E + P E g ) d t {\displaystyle 0=\delta \int {\frac {ds}{dt}}dt=\delta \int (KE+PE_{g})dt}
(where K E {\displaystyle KE} is the kinetic energy and P E g {\displaystyle PE_{g}} is the Potential Energy due to gravity) The constants K {\displaystyle K} and S {\displaystyle S} are fully determined by some variant of this approach; from the weak-field approximation one arrives at the result:
g 44 = K ( 1 + 1 S r ) ≈ − c 2 + 2 G m r = − c 2 ( 1 − 2 G m c 2 r ) {\displaystyle g_{44}=K\left(1+{\frac {1}{Sr}}\right)\approx -c^{2}+{\frac {2Gm}{r}}=-c^{2}\left(1-{\frac {2Gm}{c^{2}r}}\right)}
where G {\displaystyle G} is the gravitational constant, m {\displaystyle m} is the mass of the gravitational source and c {\displaystyle c} is the speed of light. It is found that:
K = − c 2 {\displaystyle K=\,-c^{2}} and 1 S = − 2 G m c 2 {\displaystyle {\frac {1}{S}}=-{\frac {2Gm}{c^{2}}}}
Hence:
A ( r ) = ( 1 − 2 G m c 2 r ) − 1 {\displaystyle A(r)=\left(1-{\frac {2Gm}{c^{2}r}}\right)^{-1}} and B ( r ) = − c 2