The line element given above, with
f,
g regarded as undetermined functions of the Schwarzschild radial coordinate
r, is often used as a metric
ansatz in deriving static spherically symmetric solutions in general relativity (or other
metric theories of gravitation). As an illustration, we will indicate how to compute the connection and curvature using
Cartan's exterior calculus method. First, we read off the line element a
coframe field, \begin{align} \sigma^0 &= -a(r) \, dt \\ \sigma^1 &= b(r) \, dr \\ \sigma^2 &= r d\theta \\ \sigma^3 &= r \sin\theta \, d\phi \end{align} where we regard a \, b are as yet undetermined smooth functions of r. (The fact that our spacetime admits a frame having this particular trigonometric form is yet another equivalent expression of the notion of a Schwarzschild chart in a static, spherically symmetric Lorentzian manifold). Second, we compute the exterior derivatives of these cobasis one-forms: \begin{align} d\sigma^0 &= -a'(r) \, dr \wedge dt \\&= \frac{a'(r)}{b(r)} \, dt \wedge \sigma^1 \\[1ex] d\sigma^1 &= 0\, \\[1ex] d\sigma^2 &= dr \wedge d\theta \\[1ex] d\sigma^3 &= \sin\theta \, dr \wedge d\phi + r \, \cos\theta \, d\theta \wedge d\phi \\ &= -\left( \frac{\sin\theta\, d\phi}{b(r)} \wedge \sigma^1 + \cos\theta \, d\phi \wedge \sigma^2\right) \end{align} Comparing with Cartan's
first structural equation (or rather its integrability condition), d\sigma^\hat{m} = -{\omega^\hat{m}}_\hat{n} \, \wedge \sigma^\hat{n} we guess expressions for the
connection one-forms. (The hats are just a notational device for reminding us that the indices refer to our cobasis one-forms, not to the coordinate one-forms dt, \, dr, \, d\theta, d\phi.) If we recall which pairs of indices are symmetric (space-time) and which are antisymmetric (space-space) in {\omega^\hat{m}}_\hat{n}, we can confirm that the six connection one-forms are \begin{align} {\omega^0}_1 &= \frac{a'}{b}(r)\, dt, & {\omega^0}_2 &= 0, & {\omega^0}_3 &= 0, \\ {\omega^1}_2 &= -\frac{d\theta}{b(r)}, & {\omega^1}_3 &= -\frac{\sin\theta \, d\phi}{b(r)}, & {\omega^2}_3 &= -\cos\theta \, d\phi, \end{align} (In this example, only four of the six are nonvanishing.) We can collect these one-forms into a matrix of one-forms, or even better an SO(1,3)-valued one-form. Note that the resulting matrix of one-forms will not quite be
antisymmetric as for an SO(4)-valued one-form; we need to use instead a notion of transpose arising from the
Lorentzian adjoint. Third, we compute the exterior derivatives of the connection one-forms and use Cartan's
second structural equation {\Omega^\hat{m}}_\hat{n} = d{\omega^\hat{m}}_\hat{n} - {\omega^\hat{m}}_\hat{\ell} \wedge {\omega^\hat{\ell}}_\hat{n} to compute the curvature two forms. Fourth, using the formula {\Omega^\hat{m}}_\hat{n} = {R^\hat{m}}_{\hat{n}|\hat{\imath}\hat{\jmath}|} \, \sigma^\hat{\imath} \wedge \sigma^\hat{\jmath} where the
Bach bars indicate that we should sum only over the six
increasing pairs of indices (
i,
j), we can read off the linearly independent components of the
Riemann tensor with respect to our coframe and its dual
frame field. We obtain: \begin{align} {R^0}_{101} &= \frac{-a'' \, b + a' \, b'}{a \, b^3}(r) \\ {R^0}_{202} &= \frac{1}{r}\frac{-a'}{a \, b^2}(r) = {R^0}_{303} \\ {R^1}_{212} &= \frac{1}{r}\frac{b'}{b^3}(r) = {R^1}_{313} \\ {R^2}_{323} &= \frac{1}{r^2}\frac{b^2 - 1}{b^2}(r) \end{align} Fifth, we can lower indices and organize the components R_{\hat{m}\hat{n}\hat{i}\hat{j}} into a matrix \begin{bmatrix}R_{0101} & R_{0102} & R_{0103} & R_{0123} & R_{0131} & R_{0112} \\ R_{0201} & R_{0202} & R_{0203} & R_{0223} & R_{0231} & R_{0212} \\ R_{0301} & R_{0302} & R_{0303} & R_{0323} & R_{0331} & R_{0312} \\ R_{2301} & R_{2302} & R_{2303} & R_{2323} & R_{2331} & R_{2312} \\ R_{3101} & R_{3102} & R_{3103} & R_{3123} & R_{3131} & R_{3112} \\ R_{1201} & R_{1202} & R_{1203} & R_{1223} & R_{1231} & R_{1212} \end{bmatrix} = \begin{bmatrix} E & B \\ B^T & L \end{bmatrix} where E,L are symmetric (six linearly independent components, in general) and B is
traceless (eight linearly independent components, in general), which we think of as representing a linear operator on the six-dimensional
vector space of two forms (at each event). From this we can read off the
Bel decomposition with respect to the timelike
unit vector field \vec{X} = \vec{e}_0 = \frac{1}{a(r)} \, \partial_t. The
electrogravitic tensor is E[\vec{X}]_{11} = \frac{a'' \, b - a' \, b'}{a \, b^3}(r), \; E[\vec{X}]_{22} = E[\vec{X}]_{33} = \frac{1}{r}\frac{a'}{a \, b^2}(r) The
magnetogravitic tensor vanishes identically, and the
topogravitic tensor, from which (using the fact that \vec{X} is irrotational) we can determine the three-dimensional Riemann tensor of the spatial hyperslices, is L[\vec{X}]_{11} = \frac{1}{r^2}\frac{1-b^2}{b^2}(r), \; L[\vec{X}]_{22} = L[\vec{X}]_{33} = \frac{1}{r}\frac{-b'}{b^3}(r) This is all valid for any Lorentzian manifold, but we note that in general relativity, the electrogravitic tensor controls tidal stresses on small objects, as measured by the observers corresponding to our frame, and the magnetogravitic tensor controls any spin-spin forces on spinning objects, as measured by the observers corresponding to our frame. The dual
frame field of our coframe field is \begin{align} \vec{e}_0 &= \frac{1}{a(r)} \, \partial_t \\ \vec{e}_1 &= \frac{1}{b(r)} \, \partial_r \\ \vec{e}_2 &= \frac{1}{r} \, \partial_\theta \\ \vec{e}_3 &= \frac{1}{r \sin\theta} \, \partial_\phi \end{align} The fact that the factor \frac{1}{b(r)} only multiplies the first of the three orthonormal
spacelike vector fields here means that Schwarzschild charts are
not spatially isotropic (except in the trivial case of a locally flat spacetime); rather, the light cones appear (radially flattened) or (radially elongated). This is of course just another way of saying that Schwarzschild charts correctly represent distances within each nested round sphere, but the radial coordinate does not faithfully represent radial proper distance. ==Some exact solutions admitting Schwarzschild charts==