Physics in curved spacetimes

Recall the definition of the Ricci tensor and the Ricci scalar,

\[ \begin{align}\begin{aligned} \tensor{R}{_{\beta \nu}} = \tensor{g}{^{\alpha \mu}} \tensor{R}{_{\alpha \beta \mu \nu}} = \tensor{R}{^{\mu}_{\beta \mu \nu}}\\and\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} R = \tensor{g}{^{\beta \nu}} \tensor{R}{_{\beta \nu}}\\Note that the Ricci tensor is symmetric; by differentiating equation ,\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} 2 {R}_{\alpha \beta \mu \nu, \lambda} = \tensor{g}{_{\alpha \beta, \beta \mu \lambda}} - \tensor{g}{_{\alpha \mu, \beta \nu \lambda}} +\tensor{g}{_{\beta \mu, \alpha \nu \lambda}} - \tensor{g}{_{\beta \nu, \alpha \mu \lambda}}\\Recalling that partial derivatives do commute,\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:93} \tensor{R}{_{\alpha \beta \mu \nu, \lambda}} + \tensor{R}{_{\alpha \beta \lambda \mu, \nu}} + \tensor{R}{_{\alpha \beta \lambda, \mu}} = 0\\Since the Ricci tensor is derived in normal coordinates, where\end{aligned}\end{align} \]

\(\Gamma^{\mu}_{\alpha \beta} = 0\), so partial and covariant differentiation are equivalent, and so we get the Bianchi identities,

\[ \begin{align}\begin{aligned} \label{eq:93} \tensor{R}{_{\alpha \beta \mu \nu; \lambda}} + \tensor{R}{_{\alpha \beta \lambda \mu; \nu}} + \tensor{R}{_{\alpha \beta \lambda; \mu}} = 0\\Performing the Ricci contraction from equation on the Bianchi\end{aligned}\end{align} \]

identities,

\[ \begin{align}\begin{aligned} \label{eq:94} \tensor{R}{\beta \nu; \lambda} - \tensor{R}{_{\beta \lambda; \nu}} + \tensor{R}{^{\mu}_{\beta \nu \lambda; \mu}} = 0\\contracting these in turn,\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:95} \tensor{G}{^{\alpha \beta}_{; \beta}} = 0\\which is the contracted Bianchi identity, with :math:`\ten{G}`, the\end{aligned}\end{align} \]

Einstein tensor, defined

= - R

The equivalence principle

One statement of the equivalence principle is

All local, freely falling, non-rotating laboratories are fully equivalent for the performance of all physical experiments.

This doesn’t rule out the possibility that the complicated curvature of spacetime somehow reduces in a locally inertial frame. To preclude this we can reword the equivalence principle.

Any physical law which can be expressed in tensor notation in special relativity has exactly the same form in a locally inertial frame of curved spacetime.

This means that partial differentiation is basically just a special case of covariant differentiation for a locally inertial frame, that is, in tensor notation,

\[ \begin{align}\begin{aligned} \label{eq:96} \tensor{T}{^{\mu \nu}_{;\nu}} = 0\\This also tells us how matter is affected by spacetime; in SR a\end{aligned}\end{align} \]

particle moves along the timelike coordinate of a Minkowski diagram when at rest, and from the strong equivalence principle this must be the case with general relativity too; this picks out the curves generated by the timelike coordinate of a locally inertial frame, that is

Space tells matter how to move: free-falling particles move on timelike geodesics of the local spacetime.

Einstein’s Equations

Newton’s theory of gravity can be expressed in terms of a gravitational field, \(\phi\), and the force on a particle of mass \(m\) is \(f_i = -m \phi_{,i}\), and the source of the field has a mass density \(\rho\). The field equation connecting the two is

\[\label{eq:90} \tensor{\phi}{^{,i}_{,i}} = 4 \pi G \rho\]

which is Poisson’s equation; in a vacuum with a mass density \(\rho\),

\[\label{eq:91} \tensor{\phi}{^{,i}_{,i}} = 0\]

The acceleration of free-falling particles can be described, from equation as

\[ \begin{align}\begin{aligned}\dv[2]{\xi^i}{t} = - \tensor{\phi}{^{,i}_{,i}} \xi^j\\comparing this to equation we see both are equations of geodesic\end{aligned}\end{align} \]
deviation, implying that :math:`tensor{R}{^{mu}_{alpha nu

beta}} U^{alpha} U^{beta}` is analogous to

\(\tensor{\phi}{^{,i}_{,j}}\), where the indices of the Riemann tensor are swapped using its symmetries. The velocities of the particles, \(U^{\alpha}\) and \(U^{\beta}\) are arbitrary, so the \(\ten{\phi}\) of Poisson’s equation is analogous to \(\tensor{R}{_{\alpha \beta}} = \tensor{R}{^{\mu}_{\alpha \mu \beta}}\), meaning a good guess at the relativistic analogue of equation is

\[ \begin{align}\begin{aligned} \label{eq:97} \tensor{R}{_{\mu \nu}} = 0\\These are Einstein’s vacuum equations for general relativity; if\end{aligned}\end{align} \]

\(\tensor{R}{_{\mu \nu}} = 0\) then :math:`R = tensor{g}{^{mu nu}} R_{mu

nu} = 0`, thus

\[ \begin{align}\begin{aligned} \label{eq:98} G_{\mu \nu} = R_{\mu \nu} - \half R g_{\mu \nu} = 0\\But what if there’s matter? :math:`\rho` is frame-dependent, so the\end{aligned}\end{align} \]

energy-momentum tensor is more likely to be what the field is bound to. While \(R^{\mu \nu}= - \kappa T^{\mu \nu}\) seems plausible, but equation implies that :math:`tensor{R}{^{mu nu}_{;

nu}}=0`, which, via the Bianchi identity implies

:math:`tensor{R}{_{;

nu}} = 0`, and

\((g_{\alpha \beta} T^{\alpha \beta})_{; \nu} = 0\), so this would imply that all matter has constant density, which is not the case. What about

\[ \begin{align}\begin{aligned} \label{eq:99} \tensor{G}{^{\mu \nu}} = - \kappa \tensor{T}{^{\mu \nu}}\\numerous experiments have shown this to describe physical reality.\end{aligned}\end{align} \]

These are the Einstein field equations; ten second-order non-linear differential equations, which reduce to six independent equations when the Bianchi identities are used.

One variation of the field equations which is now being taken very seriously is

\[ \begin{align}\begin{aligned} \label{eq:100} \tensor{G}{^{\mu \nu}} + \Lambda \tensor{g}{^{\mu \nu}} = - \kappa \tensor{T}{^{\mu \nu}}\\which we can do because :math:`\tensor{g}{_{\alpha \beta; \mu}}=0`.\end{aligned}\end{align} \]

This contains an extra term, \(\Lambda\), the cosmological constant.

The Newtonian limit

In the weak-field approximation we can take the spacetime around a small object to be nearly Minkowskian, with

\[ \begin{align}\begin{aligned} \label{eq:101} \tensor{g}{_{\alpha \beta}} = \tensor{\eta}{_{\alpha \beta}} + \tensor{h}{_{\alpha \beta}}\\Thus :math:`g_{\alpha \beta}` is the result of a perturbation on flat\end{aligned}\end{align} \]

spacetime, and \(\ten{h}\), which encodes the perturbation is a tensor in Minkowskian space, and using this form in Einstein’s equations gives a mathematically tractable problem. In the Newtonian limit we have \(\abs{\phi} \ll 1\), and speeds \(\abs{\vec{v}} \ll 1\), so this implies \(\abs{T^{00}} \gg \abs{T^{0i}} \gg \abs{T^{ij}}\). We then identify \(T^{00} = \rho + \mathcal{O}(\rho v^2)\). Matching the resulting form of Einstein’s equation with Newton’s equation we fix the constant \(\kappa\), so, in geometrical units,

\[ \begin{align}\begin{aligned} \label{eq:102} G^{\mu \nu} = 8 \pi T^{\mu \nu}\\The solution in this approximation is then\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:103} h^{00} = h^{11} = h^{22} = h^{33} = -2 \phi\\So the Newtonian metric is then\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:104} \ten{g} \to \diag( -(1+ 2\phi), 1- 2\phi, 1-2\phi, 1 - 2 \phi)\\and its interval is then\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:105} \dd{s^2} = -(1+2 \phi) \dd{t}^2 + (1- 2 \phi) (\dd{x}^2 + \dd{y}^2 + \dd{z}^2 )\\The geodesic equation is :math:`\nabla_U U = 0`; this geodesic has an\end{aligned}\end{align} \]

affine parameter \(\tau\), but rescaling, \(\tau \to \tau/m\), we can express it in terms of momentum, \(p = m U\), so

\[ \begin{align}\begin{aligned} \label{eq:106} \nabla_p p = 0\\In component form this looks like\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:107} p^{\alpha} \tensor{p}{^{\mu}_{,\alpha}} + \Gamma^{\mu}_{\alpha \beta} p^{\alpha} p^{\beta} = 0\\Restricting the motion to non-relativistic particles,\end{aligned}\end{align} \]

\(\abs{p^0} \gg \abs{p^i}\), and so

\[ \begin{align}\begin{aligned} \label{eq:108} m \dv{\tau} p^{\mu} + \Gamma_{00}^{\mu} (p^0)^2 = 0\\The 0-0 symbols in this metric and approximation are\end{aligned}\end{align} \]
\[\begin{split}\begin{aligned} \label{eq:110} \Gamma^0_{00} &= \phi_{,0} + \mathcal{O}(\phi^2) \\ \label{eq:111} \Gamma_{00}^i &= -\half \tensor{(-2 \phi)}{_{,j}} \delta^{ij} \end{aligned}\end{split}\]

Thus,

\[\begin{split}\begin{aligned} \label{eq:113} \dv{p^0}{\tau} &= -m \pdv{\phi}{\tau} \\ \label{eq:114} \dv{p^i}{\tau} &= -m \phi^{,i} \end{aligned}\end{split}\]

Which is Newton’s law of gravitation.