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Appendix G Solutions

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Medium

Medium

M-1. Derivation of \(\delta\sqrt{-g}\)

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Strategy: Take the variation of both sides of the fundamental metric relation \(g^{\mu\alpha}g_{\alpha\nu} = \delta^\mu_\nu\) to obtain the relationship between the variations of \(g^{\mu\nu}\) and \(g_{\mu\nu}\), then combine with Jacobi's formula \(\delta g = g \cdot g^{\mu\nu}\delta g_{\mu\nu}\) to find \(\delta\sqrt{-g}\).

Step 1: Derivation of \(g_{\mu\nu}\delta g^{\mu\nu} = -g^{\mu\nu}\delta g_{\mu\nu}\)

Take the variation of both sides of \(g^{\mu\alpha}g_{\alpha\nu} = \delta^\mu_\nu\). Since the right-hand side is a constant tensor, \(\delta(\delta^\mu_\nu) = 0\):

\(\delta(g^{\mu\alpha}g_{\alpha\nu}) = (\delta g^{\mu\alpha})g_{\alpha\nu} + g^{\mu\alpha}(\delta g_{\alpha\nu}) = 0\)

Multiply both sides by \(g^{\nu\beta}\) and simplify the contraction over \(\alpha, \nu\):

\(g^{\nu\beta}g_{\alpha\nu}\delta g^{\mu\alpha} + g^{\mu\alpha}g^{\nu\beta}\delta g_{\alpha\nu} = 0\)

\(\delta^\beta_\alpha \delta g^{\mu\alpha} + g^{\mu\alpha}g^{\nu\beta}\delta g_{\alpha\nu} = 0\)

\(\delta g^{\mu\beta} = -g^{\mu\alpha}g^{\nu\beta}\delta g_{\alpha\nu}\)

Multiply both sides by \(g_{\mu\beta}\) and contract:

\(g_{\mu\beta}\delta g^{\mu\beta} = -g_{\mu\beta}g^{\mu\alpha}g^{\nu\beta}\delta g_{\alpha\nu} = -\delta^\alpha_\beta g^{\nu\beta}\delta g_{\alpha\nu} = -g^{\nu\alpha}\delta g_{\alpha\nu}\)

Relabeling indices:

\(\boxed{g_{\mu\nu}\delta g^{\mu\nu} = -g^{\mu\nu}\delta g_{\mu\nu}}\)

Step 2: Derivation of \(\delta\sqrt{-g}\)

From Jacobi's formula:

\(\delta g = g \cdot g^{\mu\nu}\delta g_{\mu\nu}\)

Since \(\sqrt{-g} = (-g)^{1/2}\), by the chain rule:

\(\delta\sqrt{-g} = \frac{1}{2}(-g)^{-1/2} \cdot \delta(-g) = -\frac{1}{2\sqrt{-g}}\delta g = -\frac{1}{2\sqrt{-g}} \cdot g \cdot g^{\mu\nu}\delta g_{\mu\nu}\)

Since \(g < 0\) (Lorentzian metric), \(g = -(-g) = -(\sqrt{-g})^2\). Therefore \(g/\sqrt{-g} = -\sqrt{-g}\):

\(\delta\sqrt{-g} = -\frac{1}{2}(-\sqrt{-g})g^{\mu\nu}\delta g_{\mu\nu} = \frac{1}{2}\sqrt{-g}\,g^{\mu\nu}\delta g_{\mu\nu}\)

Substituting the result from Step 1, \(g^{\mu\nu}\delta g_{\mu\nu} = -g_{\mu\nu}\delta g^{\mu\nu}\):

\(\boxed{\delta\sqrt{-g} = -\frac{1}{2}\sqrt{-g}\,g_{\mu\nu}\delta g^{\mu\nu}}\)

Consistency check: In flat spacetime with \(g_{\mu\nu} = \eta_{\mu\nu} = \mathrm{diag}(-1,+1,+1,+1)\), we have \(g = -1\) and \(\sqrt{-g} = 1\). Considering a deviation from flat spacetime \(g_{\mu\nu} \to \eta_{\mu\nu} + h_{\mu\nu}\), we set \(\delta g_{\mu\nu} = h_{\mu\nu}\), and from the derived formula \(\delta\sqrt{-g} = \frac{1}{2}\sqrt{-g}\,g^{\mu\nu}\delta g_{\mu\nu}\), we obtain \(\delta\sqrt{-g} \approx \frac{1}{2}\eta^{\mu\nu}h_{\mu\nu} = \frac{1}{2}h\) (where \(h \equiv \eta^{\mu\nu}h_{\mu\nu}\) is the trace of the perturbation). This agrees with the known result in the weak-field expansion.

M-2. Newtonian Limit of Einstein-Hilbert

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Strategy: We impose three conditions—weak gravitational field, low velocity, and static—in sequence, and derive Newton's Poisson equation from the \(00\) component of the Einstein equation.

Step 1: The \(00\) component of the energy-momentum tensor

For non-relativistic matter, the energy density \(\rho c^2\) is far larger than the momentum density. Therefore the dominant component of \(T^{\mu\nu}\) is:

\(T^{00} \approx \rho c^2, \qquad T^{0i} \approx 0, \qquad T^{ij} \approx 0\)

Lowering the indices (with \(g_{00} \approx \eta_{00} = -1\)), \(T_{00} = g_{0\mu}g_{0\nu}T^{\mu\nu} \approx \rho c^2\).

The trace is:

\(T = g^{\mu\nu}T_{\mu\nu} \approx \eta^{\mu\nu}T_{\mu\nu} = -T_{00} + T_{ii} \approx -\rho c^2\)

Step 2: Rewriting the Einstein equation

Contracting both sides of the Einstein equation \(R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R = \frac{8\pi G}{c^4}T_{\mu\nu}\) with \(g^{\mu\nu}\):

\(R - 2R = \frac{8\pi G}{c^4}T \quad\Rightarrow\quad R = -\frac{8\pi G}{c^4}T\)

Substituting this back into the original equation and rearranging gives the trace-reversed form:

\(R_{\mu\nu} = \frac{8\pi G}{c^4}\left(T_{\mu\nu} - \frac{1}{2}g_{\mu\nu}T\right)\)

Step 3: The \(00\) component

\(R_{00} = \frac{8\pi G}{c^4}\left(T_{00} - \frac{1}{2}g_{00}T\right) \approx \frac{8\pi G}{c^4}\left(\rho c^2 - \frac{1}{2}(-1)(-\rho c^2)\right) = \frac{8\pi G}{c^4} \cdot \frac{\rho c^2}{2} = \frac{4\pi G \rho}{c^2}\)

Step 4: Substituting \(R_{00} \approx -\frac{1}{2}\nabla^2 h_{00}\) and \(h_{00} = -2\Phi/c^2\)

Using \(R_{00} \approx -\frac{1}{2}\nabla^2 h_{00}\), which was accepted as given in the problem:

\(-\frac{1}{2}\nabla^2 h_{00} = \frac{4\pi G\rho}{c^2}\)

Substituting \(h_{00} = -2\Phi/c^2\):

\(-\frac{1}{2}\nabla^2\!\left(-\frac{2\Phi}{c^2}\right) = \frac{4\pi G\rho}{c^2}\)

\(\frac{1}{c^2}\nabla^2\Phi = \frac{4\pi G\rho}{c^2}\)

\(\boxed{\nabla^2\Phi = 4\pi G\rho}\)

Supplement (Origin of \(R_{00} \approx -\frac{1}{2}\nabla^2 h_{00}\)): Computing the Christoffel symbols in the weak-field, static limit with metric signature \((-,+,+,+)\) gives \(\Gamma^i_{00} \approx -\frac{1}{2}\partial_i h_{00}\), from which \(R_{00} \approx \partial_i\Gamma^i_{00} = -\frac{1}{2}\nabla^2 h_{00}\) follows. In this problem, we accept this result as given and combine it with the \(00\) component of the Einstein equation to derive the Poisson equation—this is the essence of the problem.

Verification: In vacuum (\(\rho = 0\)), taking \(\Phi = -GM/r\) gives \(\nabla^2\Phi = 0\) (for \(r \neq 0\)), which is consistent with the right-hand side of the Poisson equation. In regions where a mass distribution is present, \(\nabla^2\Phi = 4\pi G\rho > 0\), and \(\Phi\) correctly reproduces gravity as an attractive potential (\(\Phi < 0\)).

M-3. Variation of the Cosmological Constant Term

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Strategy: Vary the action \(S = \frac{c^4}{16\pi G}\int (R - 2\Lambda)\sqrt{-g}\,d^4x + S_M\) with respect to \(g^{\mu\nu}\). The variation of the \(R\) part has already been computed in section G.2 of the main text, so we only need to compute the variation of the additional \(-2\Lambda\sqrt{-g}\) term.

Step 1: Variation of \(-2\Lambda\sqrt{-g}\)

Since \(\Lambda\) is a constant:

\(\delta(-2\Lambda\sqrt{-g}) = -2\Lambda \cdot \delta\sqrt{-g}\)

Substituting the result from Problem G.1, \(\delta\sqrt{-g} = -\frac{1}{2}\sqrt{-g}\,g_{\mu\nu}\delta g^{\mu\nu}\):

\(\delta(-2\Lambda\sqrt{-g}) = -2\Lambda \cdot \left(-\frac{1}{2}\sqrt{-g}\,g_{\mu\nu}\delta g^{\mu\nu}\right) = \Lambda\,g_{\mu\nu}\,\delta g^{\mu\nu}\,\sqrt{-g}\)

Step 2: Combining the variations

Combining with the result from section G.3, \(\delta(R\sqrt{-g}) = (R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R)\delta g^{\mu\nu}\sqrt{-g}\) (with boundary terms discarded):

\(\delta S_{EH+\Lambda} = \frac{c^4}{16\pi G}\int\left[\left(R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R\right) + \Lambda g_{\mu\nu}\right]\delta g^{\mu\nu}\sqrt{-g}\,d^4x\)

Defining the energy-momentum tensor from the variation of the matter action \(S_M\) as \(T_{\mu\nu} = -\frac{2}{\sqrt{-g}}\frac{\delta S_M}{\delta g^{\mu\nu}}\):

\(\delta S_M = -\frac{1}{2}\int T_{\mu\nu}\,\delta g^{\mu\nu}\sqrt{-g}\,d^4x\)

Step 3: Setting the total variation \(\delta S = 0\)

\(\delta S = \frac{c^4}{16\pi G}\int\left[R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R + \Lambda g_{\mu\nu} - \frac{8\pi G}{c^4}T_{\mu\nu}\right]\delta g^{\mu\nu}\sqrt{-g}\,d^4x = 0\)

Since \(\delta g^{\mu\nu}\) is arbitrary, the coefficient of the integrand must vanish:

\(\boxed{R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu}}\)

This is the Einstein field equation with cosmological constant.

Consistency check on physical meaning:

In vacuum (\(T_{\mu\nu} = 0\)), we have \(R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R + \Lambda g_{\mu\nu} = 0\). Taking the trace gives \(-R + 4\Lambda = 0\), i.e., \(R = 4\Lambda\). Substituting back yields \(R_{\mu\nu} = \Lambda g_{\mu\nu}\), which describes de Sitter spacetime (\(\Lambda > 0\)) or anti-de Sitter spacetime (\(\Lambda < 0\)). In our actual universe, \(\Lambda \approx 10^{-52}\ \mathrm{m}^{-2}\) (observed value) is the leading candidate for dark energy.

Alternative form (interpreting \(\Lambda\) as an energy-momentum tensor): Moving \(\Lambda g_{\mu\nu}\) to the right-hand side:

\(R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R = \frac{8\pi G}{c^4}\left(T_{\mu\nu} - \frac{\Lambda c^4}{8\pi G}g_{\mu\nu}\right) = \frac{8\pi G}{c^4}\left(T_{\mu\nu} + T^{(\Lambda)}_{\mu\nu}\right)\)

Here \(T^{(\Lambda)}_{\mu\nu} = -\frac{\Lambda c^4}{8\pi G}g_{\mu\nu}\) can be interpreted as the energy-momentum tensor of the vacuum. Reading off the \(T_{00}\) component in the rest frame (\(g_{00} = -1\)), the energy density is \(\rho_\Lambda c^2 = \frac{\Lambda c^4}{8\pi G}\) (i.e., mass density \(\rho_\Lambda = \frac{\Lambda c^2}{8\pi G}\)). From the spatial components \(T_{ij} = p\,g_{ij}\), the pressure is \(p_\Lambda = -\frac{\Lambda c^4}{8\pi G} = -\rho_\Lambda c^2\) (negative pressure). The equation of state parameter is \(w \equiv p/(\rho c^2) = -1\). This is the essence of "dark energy."