Static models with spherical symmetry

Orthogonal Metrics

An orthogonal metric takes the form

\[ \begin{align}\begin{aligned} \label{eq:78} \tensor{g}{_{\alpha \beta}} = 0 \qquad\forall\alpha \neq \beta\\implying that there are no cross-terms in the invariant interval. Thus\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:79} \dd{s}^2 = \tensor{g}{_{\alpha \alpha}} (\dd{x}^{\alpha})^2\\The components of a metric will not be orthogonal in every coordinate\end{aligned}\end{align} \]

system; suppose that \(\tensor{g}{_{\alpha \beta}}\) are the metric components in a particular coordinate system where the metric is orthogonal. Let \(\tensor{g'}{_{\mu \nu}}\) be the components of the metric in another coordinate system, then

\[\begin{split}\begin{aligned} \label{eq:80} \tensor{g'}{_{\mu \nu}} &= \pdv{x^{\alpha}}{x'^{\mu}} \pdv{x^{\beta}}{x'^{\nu}} \tensor{g}{_{\alpha \beta}} \\ &= \pdv{x^{\alpha}}{x'^{\mu}} \pdv{x^{\alpha}}{x'^{\nu}} \tensor{g}{_{\alpha \alpha}} \\ { \mathrel{{\ooalign{\hidewidth$\not\phantom{=}$\hidewidth\cr$\implies$}}}}\tensor{g'}{_{\mu \nu}} &= 0 \qquad \forall \mu' \neq \nu' \nonumber\end{aligned}\end{split}\]

The orthogonal metric components are closely related to the question of whether the coordinate system has orthogonal basis vectors. If we have a coordinate system with basis vectors \(\set{\vec{e}_i}\), and two vectors \(\vec{A} = A^i \vec{e}_i\), \(\vec{B} = B^i \vec{e}_i\), then

\[ \begin{align}\begin{aligned}\vec{A} \vdot \vec{B} = (A^i \vec{e}_i) \vdot (B^j \vec{e}_j) = A^i B^j (\vec{e}_i \vdot \vec{e}_j)\\so it follows\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\tensor{g}{_{ij}} = \vec{e}_i \vdot \vec{e}_j\\and :math:`\tensor{g}{_{ij}} = 0` iff :math:`\vec{e}_i` and\end{aligned}\end{align} \]

\(\vec{e}_j\) are orthogonal.

We normally want to choose a coordinate system in which the metric coefficients are orthogonal, to simplify the expressions for geometrical objects; if the contravariant metric components are orthogonal then the diagonal terms are simply the reciprocal of the covariant diagonal terms. To see this, we know

\[ \begin{align}\begin{aligned}g^{\gamma \beta} g_{\gamma\gamma} = \delta^{\beta}_{\gamma} \quad\therefore \quad g^{\gamma\gamma} = \frac{1}{g_{\gamma \gamma}}\\The Christoffel symbols also take a simple form in an orthogonal\end{aligned}\end{align} \]

metric,

\[\begin{split}\begin{aligned} \label{eq:109} \Gamma^{\lambda}_{\mu \nu} &= 0 \qquad \text{for } \lambda, \mu, \nu \text{ all different.} \\ \Gamma^{\lambda}_{\lambda \mu} = \Gamma^{\lambda}_{\mu \lambda} &= \quad \frac{g_{\lambda\lambda , \lambda}}{2 g_{\lambda \lambda}}\\ \Gamma^{\lambda}_{\mu \mu} &= - \frac{g_{\mu \mu, \lambda}}{2 g_{\lambda \lambda}} \\ \Gamma^{\lambda}_{\lambda \lambda} &= \quad \frac{g_{\lambda \lambda, \lambda}}{2 g_{\lambda \lambda}} \end{aligned}\end{split}\]

For an affine parameter \(s\) the geodesic equation takes the form

\[ \begin{align}\begin{aligned}\dv{s} \qty( g_{\lambda \nu} \dv{x^{\nu}}{s} ) - \half \pdv{g_{\mu \nu}}{x^{\lambda}} \dv{x^{\mu}}{s} \dv{x^{\nu}}{s} = 0\\for an orthogonal metric this reduces to\end{aligned}\end{align} \]
\[\label{eq:112} \dv{s} \qty( g_{\lambda \lambda} \dv{x^{\lambda}}{s} ) - \half \pdv{g_{\mu \mu}}{x^{\lambda}} \qty( \dv{x^{\mu}}{s} )^2 = 0\]

Spherically symmetric metrics

Spherically symmetric solutions to the field equations are suitable for describing the spacetime inside and around stars. In flat Minkowski spacetime a polar coordinate system can be used to give an invariant interval

\[ \begin{align}\begin{aligned} \label{eq:115} \dd{s}^2 = - \dd{t}^2 + \dd{r}^2 + r^2 \qty( \dd{\theta}^2 + \sin^2{\theta} \dd{\phi}^2 )\\Surfaces of constant :math:`r` and :math:`t` have the geometry of a\end{aligned}\end{align} \]

2-sphere with an interval

\[ \begin{align}\begin{aligned} \label{eq:116} \dd{\ell}^2 = r^2 \qty( \dd{\theta}^2 + \sin^2{\theta} \dd{\phi}^2 )\\Thus a spacetime is spherically symmetric if every point in the\end{aligned}\end{align} \]

spacetime lies on a 2D surface which is a 2-sphere. Labelling the coordinates \((r', t, \theta, \phi)\) then every point in a spherically symmetric spacetime lies on a 2D surface, given by

\[ \begin{align}\begin{aligned} \label{eq:117} \dd{\ell}^2 = f(r', t) \qty[ \dd{\theta}^2 + \sin^2{\theta} \dd{\phi}^2]\\with :math:`\sqrt{f}` the radius of curvature of the 2-sphere.\end{aligned}\end{align} \]

In curved spacetime there is no trivial relation between the angular coordinates of the two-sphere and the remaining coordinates at each point in spacetime, but if we define

\[ \begin{align}\begin{aligned} \label{eq:118} r^2 = f(r', t)\\and we can line-up the origins of the 2-sphere coordinate systems\end{aligned}\end{align} \]

\((\theta, \phi)\) for points in spacetime with different values of \(r\). Spherical symmetry requires that any radial path in the space is orthogonal to the 2D spheres on which the points along the radial path lie, implying

\[ \begin{align}\begin{aligned} \label{eq:119} g_{r \theta} = g_{r \phi} = 0\\So the spacetime metric is reduced to the form\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\begin{split} \label{eq:120} \begin{split} \dd{s^2} = g_{tt} \dd{t^2} + 2 g_{tr} \dd{r}\dd{t} + 2 g_{t \theta} \dd{\theta}\dd{t} \\+ g_{rr} \dd{r^2} + r^2 \qty(\dd{\theta^2} + \sin[2](\theta) \dd{\phi^2}) \end{split}\end{split}\\Considering the curve with :math:`r`, :math:`\theta`, and :math:`\phi`\end{aligned}\end{align} \]

are constant, which is a worldline of a particle in spacetime with constant spatial coordinates; this curve must also be orthogonal to 2-spheres on which each point lies, otherwise there would be a preferred direction in spacetime. Thus

\[ \begin{align}\begin{aligned} \label{eq:121} g_{t\theta} = g_{t \phi} = 0\\so the general form for a metric in a spherically symmetric spacetime\end{aligned}\end{align} \]

is

\[ \begin{align}\begin{aligned}\begin{split} \label{eq:122} \begin{split} \dd{s^2} = g_{tt} \dd{t^2} + 2 g_{tr} \dd{r}\dd{t} + g_{rr} \dd{r^2} \\+ r^2 \qty( \dd{\theta^2} + \sin[2](\theta) \dd{\phi^2}) \end{split}\end{split}\\For :math:`g_{tt}`, :math:`g_{tr}`, and :math:`g_{rr}` arbitrary\end{aligned}\end{align} \]

functions of \(r\) and \(t\).

Static Spacetime

In s static spherically symmetric spacetime we can find a time coordinate \(t\) where

  1. all metric components are independent of \(t\)

  2. the metric is unchanged under a time-reversal operation, :math:`t to

    -t`.

The second property implies \(g_{tr}=0\), meaning that the interval is

\[ \begin{align}\begin{aligned} \label{eq:123} \dd{s^2} = - e^{\nu} \dd{t^2} + e^{\lambda} \dd{r^2} + r^2 \qty(\dd{\theta^2} + \sin[2](\theta) \dd{\phi^2})\\which is orthogonal. The functions :math:`\nu(r)` and\end{aligned}\end{align} \]

\(\lambda(r)\) replace \(g_{tt}\) and \(g_{rr}\),since the exponential function is strictly positive for all \(r\), this is legitimate, provided \(g_{tt}<0\) and \(g_{rr}>0\).

The Christoffel symbols for this spacetime are

\[\begin{split}\begin{aligned} \label{eq:124} \Gamma^t_{rt}=\Gamma^t_{tr} &= \half \nu' & \Gamma^{\theta}_{r \theta} = \Gamma^{\theta}_{\theta r} &= \frac{1}{r} \\ \Gamma^r_{tt} &= \half \nu' e^{\nu-\lambda} & \Gamma^{\theta}_{\phi \phi} &= - \sin(\theta) \cos(\theta) \\ \Gamma^r_{rr} &= \half \lambda' & \Gamma^{\phi}_{r \phi} = \Gamma^{\phi}_{\phi r} &= \frac{1}{r} \\ \Gamma^r_{\theta \theta} &= - r e^{- \lambda} & \Gamma^{\phi}_{\theta \phi} = \Gamma^{\phi}_{\phi \theta} &= \cot(\theta) \end{aligned}\end{split}\]
\[\Gamma^r_{\phi\phi} = -r e^{-\lambda} \sin[2](\theta)\]

The Ricci tensor is given by

\[ \begin{align}\begin{aligned} \label{eq:125} R_{\lambda \nu} = \Gamma^{\tau}_{\lambda \nu} \Gamma^{\sigma}_{\tau\sigma} - \Gamma^{\tau}_{\lambda \sigma} \Gamma^{\sigma}_{\tau\nu} + \Gamma^{\sigma}_{\lambda\nu,\sigma} - \Gamma^{\sigma}_{\lambda\sigma,\nu}\\so\end{aligned}\end{align} \]
\[\begin{split}\begin{aligned} \label{eq:126} R_{tt} &= \half e^{\nu-\lambda} \qty( \nu'' + \half \nu'^2 - \half \nu' \lambda' + \frac{2}{r} \nu') \\ \label{eq:129} R_{rr} &= - \half \qty( \nu'' + \half \nu'^2 - \half \nu' \lambda' - \frac{2}{r} \lambda') \\ \label{eq:130} R_{\theta\theta} &= 1 - e^{-\lambda} \qty[1+\frac{r}{2}(\nu'-\lambda')] \\ \label{eq:131} R_{\phi\phi}&= R_{\theta\theta} \sin[2](\theta) \end{aligned}\end{split}\]

The Schwarzschild metric

We can derive the metric for the spacetime exterior to a star from the static spherically symmetric metric; the Schwarzchild metric; if the star is in an isolated region of space we can assume all components of the Ricci tensor to be zero, so

\[ \begin{align}\begin{aligned} \label{eq:127} e^{\lambda-\nu}R_{tt}+R_{rr} = \frac{\nu'+\lambda'}{r}=0\\which implies that :math:`\nu+\lambda` is constant. At a large distance\end{aligned}\end{align} \]

from the star the metric should reduce to special relativity, so as

\[ \begin{align}\begin{aligned}r \to \infty, \quad e^{\nu}\to 1, e^{\lambda} \to 1 \implies \nu \to 0, \lambda \to 0\\and so :math:`\nu+\lambda=0`, giving\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:128} e^{\nu} = e^{-\lambda}\\This lets us eliminate :math:`\nu` from the :math:`R_{\theta\theta}`\end{aligned}\end{align} \]

expression, equation , so

\[ \begin{align}\begin{aligned} \label{eq:132} e^{-\lambda} (1-\lambda'r) = 1 \implies \dv{r} \qty(r e^{-\lambda})=1\\Integrating this we get\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:133} e^{\nu} = e^{-\lambda} = 1 + \frac{\alpha}{r}\\where :math:`\alpha` is a constant.\end{aligned}\end{align} \]

Consider a material test particle, with so little rest mass that it does not disturb the metric, which is released from rest, then

\[ \begin{align}\begin{aligned} \label{eq:134} \dd{x^j}{\tau} = 0 \quad j = 1,2,3\\for :math:`\tau` the proper time, and\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:135} \dv{x^0}{\tau} \equiv \dv{t}{\tau} \neq 0\\Realling that\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}g_{\alpha\beta} \dv{x^{\alpha}}{\tau} \dv{x^{\beta}}{\tau} = -1\\then\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:136} \dv{t}{\tau} = e^{- \frac{\nu}{2}}\\Applying the geodesic equations, equation , at the instance that the\end{aligned}\end{align} \]

particle is released this reduces to

\[ \begin{align}\begin{aligned} \label{eq:137} \dv[2]{r}{\tau} + \Gamma_{tt}^r \qty( \dv{t}{\tau})^2 =0\\Substituting the Christoffel symbol and equation we obtain\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:138} \dv[2]{r}{t} = \frac{\alpha}{2 r^2}\\In the limit of a weak field this reduces to Newtonian gravity,\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:139} \dv[2]{r}{t} = - \frac{GM}{r^2}\\for :math:`M` the mass of the star, meaning :math:`\alpha = -2GM =-2M`\end{aligned}\end{align} \]

for \(G=1\). Thus the invariant interval is

\[\begin{split}\label{eq:140} \begin{split} \dd{s^2} = - \qty(1-\frac{2M}{r}) \dd{t^2} + \frac{\dd{r^2}}{\qty(1-\frac{2M}{r})} \\ + r^2 \dd{\theta^2} + r^2 \sin[2](\theta) \dd{\phi^2} \end{split}\end{split}\]