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Appendix D Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Schwarzschild \(\Gamma^r_{\ tt}\)

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Solution Strategy

In a diagonal metric, among the summed index \(\alpha\) in the definition of the Christoffel symbol, only \(\alpha = \mu\) gives \(g^{\mu\alpha} \neq 0\). Since \(\mu = r\), only \(\alpha = r\) contributes.

Detailed Calculation

Substituting \(\mu = r\), \(\nu = \sigma = t\) into the definition:

\[ \Gamma^r{}_{tt} = \frac{1}{2}g^{r\alpha}\left(\partial_t g_{\alpha t} + \partial_t g_{\alpha t} - \partial_\alpha g_{tt}\right) \]

Since the metric is diagonal, only \(\alpha = r\) contributes:

\[ \Gamma^r{}_{tt} = \frac{1}{2}g^{rr}\left(\underbrace{\partial_t g_{rt}}_{=0} + \underbrace{\partial_t g_{rt}}_{=0} - \partial_r g_{tt}\right) = -\frac{1}{2}g^{rr}\,\partial_r g_{tt} \]

Calculating \(g^{rr}\):

\[ g^{rr} = \frac{1}{g_{rr}} = 1 - \frac{2M}{r} \]

Calculating \(\partial_r g_{tt}\):

\[ g_{tt} = -\left(1 - \frac{2M}{r}\right) = -1 + \frac{2M}{r} \]
\[ \partial_r g_{tt} = \partial_r\left(-1 + \frac{2M}{r}\right) = -\frac{2M}{r^2} \]

Substitution:

\[ \Gamma^r{}_{tt} = -\frac{1}{2}\left(1 - \frac{2M}{r}\right)\left(-\frac{2M}{r^2}\right) = \frac{M}{r^2}\left(1 - \frac{2M}{r}\right) \]

Final Answer

\[ \boxed{\Gamma^r{}_{tt} = \frac{M}{r^2}\left(1 - \frac{2M}{r}\right)} \]

Verification

In the far-field limit \(r \gg 2M\), we have \(\Gamma^r{}_{tt} \approx M/r^2\). This agrees with the Newtonian gravitational acceleration \(GM/r^2\) (with \(G=1\)). ✓

B-2. Schwarzschild \(\Gamma^r_{\ rr}\)

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Solution Strategy

Substitute \(\mu = \nu = \sigma = r\). Since the metric is diagonal, \(\Gamma^r{}_{rr} = \frac{1}{2}g^{rr}\,\partial_r g_{rr}\).

Detailed Calculation

Substituting into the definition:

\[ \Gamma^r{}_{rr} = \frac{1}{2}g^{rr}\left(\partial_r g_{rr} + \partial_r g_{rr} - \partial_r g_{rr}\right) = \frac{1}{2}g^{rr}\,\partial_r g_{rr} \]

Calculating \(\partial_r g_{rr}\):

For \(g_{rr} = \left(1 - \frac{2M}{r}\right)^{-1}\), let \(u = 1 - \frac{2M}{r}\) so that \(g_{rr} = u^{-1}\).

\[ \frac{du}{dr} = \frac{2M}{r^2} \]
\[ \partial_r g_{rr} = -u^{-2}\frac{du}{dr} = -\left(1 - \frac{2M}{r}\right)^{-2}\cdot\frac{2M}{r^2} \]

Substitution:

\[ \Gamma^r{}_{rr} = \frac{1}{2}\left(1 - \frac{2M}{r}\right)\cdot\left[-\left(1 - \frac{2M}{r}\right)^{-2}\cdot\frac{2M}{r^2}\right] \]
\[ = -\frac{M}{r^2}\left(1 - \frac{2M}{r}\right)^{-1} \]

Final Answer

\[ \boxed{\Gamma^r{}_{rr} = -\frac{M}{r^2}\left(1 - \frac{2M}{r}\right)^{-1}} \]

Verification

Checking the product of \(\Gamma^r{}_{rr}\) and \(\Gamma^r{}_{tt}\): \(\Gamma^r{}_{tt}\cdot\Gamma^r{}_{rr} = \frac{M}{r^2}(1-2M/r)\cdot\left(-\frac{M}{r^2}\right)(1-2M/r)^{-1} = -M^2/r^4\). This is consistent with the values in the formula reference. ✓

B-3. \(\Gamma^\theta_{\ \varphi\varphi}\) in Minkowski Spherical Coordinates

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Calculation Details

Substituting \(\mu = \theta\), \(\nu = \sigma = \varphi\). Since the metric is diagonal, only \(\alpha = \theta\) contributes:

\[ \Gamma^\theta{}_{\varphi\varphi} = \frac{1}{2}g^{\theta\theta}\left(\underbrace{\partial_\varphi g_{\theta\varphi}}_{=0} + \underbrace{\partial_\varphi g_{\theta\varphi}}_{=0} - \partial_\theta g_{\varphi\varphi}\right) = -\frac{1}{2}g^{\theta\theta}\,\partial_\theta g_{\varphi\varphi} \]

Calculation of each quantity:

\[ g^{\theta\theta} = \frac{1}{r^2} \]
\[ g_{\varphi\varphi} = r^2\sin^2\theta \quad \Longrightarrow \quad \partial_\theta g_{\varphi\varphi} = 2r^2\sin\theta\cos\theta \]

Substitution:

\[ \Gamma^\theta{}_{\varphi\varphi} = -\frac{1}{2}\cdot\frac{1}{r^2}\cdot 2r^2\sin\theta\cos\theta = -\sin\theta\cos\theta \]

Final Answer

\[ \boxed{\Gamma^\theta{}_{\varphi\varphi} = -\sin\theta\cos\theta} \]

Verification

In standard formula collections for the Schwarzschild metric, \(\Gamma^\theta{}_{\varphi\varphi} = -\cos\theta\sin\theta\) as well, and since this component depends only on the angular part, the results agree. ✓

B-4. FRW \(\Gamma^r_{\ tr}\)

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Calculation Details

Substituting \(\mu = r\), \(\nu = t\), \(\sigma = r\). For a diagonal metric, only \(\alpha = r\) contributes:

\[ \Gamma^r{}_{tr} = \frac{1}{2}g^{rr}\left(\partial_t g_{rr} + \partial_r g_{rt} - \partial_r g_{tr}\right) = \frac{1}{2}g^{rr}\,\partial_t g_{rr} \]

(Since \(g_{rt} = 0\), the 2nd and 3rd terms vanish.)

Calculation of each quantity:

For \(k=0\), \(g_{rr} = a^2(t)\), so

\[ g^{rr} = \frac{1}{a^2(t)}, \qquad \partial_t g_{rr} = 2a\dot{a} \]

Substitution:

\[ \Gamma^r{}_{tr} = \frac{1}{2}\cdot\frac{1}{a^2}\cdot 2a\dot{a} = \frac{\dot{a}}{a} \]

Final Answer

\[ \boxed{\Gamma^r{}_{tr} = \frac{\dot{a}}{a} = H(t)} \]

Verification

This is equal to the Hubble parameter \(H(t)\). It represents the "dragging effect" on comoving coordinates due to the expansion of the universe, which is physically reasonable. It also agrees with standard reference values. ✓

B-5. FRW \(\Gamma^t_{\ \theta\theta}\)

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Calculation Details

Substituting \(\mu = t\), \(\nu = \sigma = \theta\). For a diagonal metric, only \(\alpha = t\) contributes:

\[ \Gamma^t{}_{\theta\theta} = \frac{1}{2}g^{tt}\left(\underbrace{\partial_\theta g_{t\theta}}_{=0} + \underbrace{\partial_\theta g_{t\theta}}_{=0} - \partial_t g_{\theta\theta}\right) = -\frac{1}{2}g^{tt}\,\partial_t g_{\theta\theta} \]

Computation of each quantity:

\[ g^{tt} = -1, \qquad g_{\theta\theta} = a^2(t)\,r^2 \]
\[ \partial_t g_{\theta\theta} = 2a\dot{a}\,r^2 \]

Substitution:

\[ \Gamma^t{}_{\theta\theta} = -\frac{1}{2}(-1)\cdot 2a\dot{a}\,r^2 = a\dot{a}\,r^2 \]

Final Answer

\[ \boxed{\Gamma^t{}_{\theta\theta} = a\dot{a}\,r^2} \]

Verification

This agrees exactly with the reference value \(\Gamma^t{}_{\theta\theta} = a\dot{a}\,r^2\). ✓

B-6. Verification of Schwarzschild Orthonormal Basis

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Calculation Details

From the formula reference:

\[ (\mathbf{e}_{\hat{r}})^\alpha = \left[0,\; \left(1 - \frac{2M}{r}\right)^{1/2},\; 0,\; 0\right] \]

Computing the inner product:

\[ g(\mathbf{e}_{\hat{r}},\, \mathbf{e}_{\hat{r}}) = g_{\alpha\beta}\,(\mathbf{e}_{\hat{r}})^\alpha\,(\mathbf{e}_{\hat{r}})^\beta \]

The only non-zero component is \(\alpha = \beta = r\):

\[ = g_{rr}\cdot\left[\left(1 - \frac{2M}{r}\right)^{1/2}\right]^2 = \left(1 - \frac{2M}{r}\right)^{-1}\cdot\left(1 - \frac{2M}{r}\right) = 1 \]

Final Answer

\[ \boxed{g(\mathbf{e}_{\hat{r}},\, \mathbf{e}_{\hat{r}}) = +1} \]

Verification

Similarly, we can confirm that \(g(\mathbf{e}_{\hat{t}},\, \mathbf{e}_{\hat{t}}) = g_{tt}\cdot(1-2M/r)^{-1} = -(1-2M/r)\cdot(1-2M/r)^{-1} = -1\), satisfying the orthonormal basis condition \(\eta_{\hat{\alpha}\hat{\beta}}\). ✓

B-7. Christoffel Symbol Verification for General Spherical Symmetry

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Calculation Details

From the formula collection for the general spherically symmetric metric:

\[ \Gamma^r{}_{tt} = \frac{\nu'}{2}\,e^{\nu - \lambda} \]

For Schwarzschild, \(e^{\nu} = 1 - \frac{2M}{r}\), \(e^{-\lambda} = 1 - \frac{2M}{r}\), so \(e^{\lambda} = (1-2M/r)^{-1}\).

Calculation of \(\nu'\):

\[ \nu = \ln\left(1 - \frac{2M}{r}\right) \]
\[ \nu' = \frac{d\nu}{dr} = \frac{1}{1 - 2M/r}\cdot\frac{2M}{r^2} = \frac{2M}{r^2}\left(1 - \frac{2M}{r}\right)^{-1} \]

Calculation of \(e^{\nu - \lambda}\):

\[ e^{\nu - \lambda} = e^{\nu}\cdot e^{-\lambda} = \left(1 - \frac{2M}{r}\right)\cdot\left(1 - \frac{2M}{r}\right) = \left(1 - \frac{2M}{r}\right)^2 \]

Substitution:

\[ \Gamma^r{}_{tt} = \frac{1}{2}\cdot\frac{2M}{r^2}\left(1 - \frac{2M}{r}\right)^{-1}\cdot\left(1 - \frac{2M}{r}\right)^2 = \frac{M}{r^2}\left(1 - \frac{2M}{r}\right) \]

Final Answer

\[ \boxed{\Gamma^r{}_{tt} = \frac{M}{r^2}\left(1 - \frac{2M}{r}\right)} \]

This agrees with the result from D1. ✓

Verification

The reduction from the general formula to the special case was carried out correctly. The intermediate result \(\nu' = \frac{2M}{r^2(1-2M/r)}\) gives \(\nu' \to 2M/r^2 \to 0\) as \(r \to \infty\) (approaching flat spacetime), which is reasonable. ✓

B-8. Application of Riemann Tensor Symmetries

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Calculation Details

From the formula collection: \(R_{\hat{r}\hat{t}\hat{r}\hat{t}} = -2M/r^3\).

Method 1: Pair Exchange Symmetry of First and Second Pairs

\[ R_{\hat{t}\hat{r}\hat{t}\hat{r}} = R_{\hat{t}\hat{r}\hat{t}\hat{r}} \]

Applying pair exchange symmetry \(R_{\alpha\beta\gamma\delta} = R_{\gamma\delta\alpha\beta}\):

\[ R_{\hat{t}\hat{r}\hat{t}\hat{r}} = R_{\hat{t}\hat{r}\hat{t}\hat{r}} \]

This just returns the same quantity, so we use a different method.

Method 2: Applying Antisymmetry Twice

Antisymmetry in the 1st and 2nd indices:

\[ R_{\hat{t}\hat{r}\hat{t}\hat{r}} = -R_{\hat{r}\hat{t}\hat{t}\hat{r}} \]

Antisymmetry in the 3rd and 4th indices:

\[ -R_{\hat{r}\hat{t}\hat{t}\hat{r}} = -(-R_{\hat{r}\hat{t}\hat{r}\hat{t}}) = R_{\hat{r}\hat{t}\hat{r}\hat{t}} \]

Therefore:

\[ R_{\hat{t}\hat{r}\hat{t}\hat{r}} = R_{\hat{r}\hat{t}\hat{r}\hat{t}} = -\frac{2M}{r^3} \]

Final Answer

\[ \boxed{R_{\hat{t}\hat{r}\hat{t}\hat{r}} = -\frac{2M}{r^3}} \]

Verification

Direct confirmation via pair exchange symmetry: \(R_{\hat{t}\hat{r}\hat{t}\hat{r}} = R_{\hat{t}\hat{r}\hat{t}\hat{r}}\) (trivial). Applying antisymmetry once gives \(R_{\hat{t}\hat{r}\hat{t}\hat{r}} = -R_{\hat{r}\hat{t}\hat{t}\hat{r}} = +R_{\hat{r}\hat{t}\hat{r}\hat{t}}\) (antisymmetry in the 3rd and 4th indices). The result is consistent. ✓

B-9. Schwarzschild \(R_{tt} = 0\)

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Solution Strategy

Expand the definition of the Ricci tensor \(R_{\hat{t}\hat{t}} = R^{\hat{\rho}}{}_{\hat{t}\hat{\rho}\hat{t}}\), noting that in an orthonormal basis, indices are raised and lowered with \(\eta_{\hat{\alpha}\hat{\beta}}\).

Detailed Calculation

\[ R_{\hat{t}\hat{t}} = R^{\hat{t}}{}_{\hat{t}\hat{t}\hat{t}} + R^{\hat{r}}{}_{\hat{t}\hat{r}\hat{t}} + R^{\hat{\theta}}{}_{\hat{t}\hat{\theta}\hat{t}} + R^{\hat{\varphi}}{}_{\hat{t}\hat{\varphi}\hat{t}} \]

Term 1: \(R^{\hat{t}}{}_{\hat{t}\hat{t}\hat{t}} = \eta^{\hat{t}\hat{t}}R_{\hat{t}\hat{t}\hat{t}\hat{t}} = (-1)\cdot 0 = 0\)

(By the antisymmetry of the Riemann tensor, \(R_{\hat{t}\hat{t}\hat{t}\hat{t}} = 0\))

Term 2: \(R^{\hat{r}}{}_{\hat{t}\hat{r}\hat{t}} = \eta^{\hat{r}\hat{r}}R_{\hat{r}\hat{t}\hat{r}\hat{t}} = (+1)\cdot\left(-\frac{2M}{r^3}\right) = -\frac{2M}{r^3}\)

Term 3: \(R^{\hat{\theta}}{}_{\hat{t}\hat{\theta}\hat{t}} = \eta^{\hat{\theta}\hat{\theta}}R_{\hat{\theta}\hat{t}\hat{\theta}\hat{t}} = (+1)\cdot\frac{M}{r^3} = \frac{M}{r^3}\)

(From the reference formulas, \(R_{\hat{\theta}\hat{t}\hat{\theta}\hat{t}} = M/r^3\))

Term 4: \(R^{\hat{\varphi}}{}_{\hat{t}\hat{\varphi}\hat{t}} = \eta^{\hat{\varphi}\hat{\varphi}}R_{\hat{\varphi}\hat{t}\hat{\varphi}\hat{t}} = (+1)\cdot\frac{M}{r^3} = \frac{M}{r^3}\)

Sum:

\[ R_{\hat{t}\hat{t}} = 0 - \frac{2M}{r^3} + \frac{M}{r^3} + \frac{M}{r^3} = 0 \]

Final Answer

\[ \boxed{R_{\hat{t}\hat{t}} = 0} \]

Verification

Since the Schwarzschild spacetime is a vacuum solution (\(G_{\mu\nu} = 0\)), we have \(R_{\mu\nu} = 0\). The result \(R_{\hat{t}\hat{t}} = 0\) is consistent with this. ✓

B-10. Scalar Curvature of FRW

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Solution Strategy

Take the trace of the Einstein tensor definition \(G_{\hat{\mu}\hat{\nu}} = R_{\hat{\mu}\hat{\nu}} - \frac{1}{2}\eta_{\hat{\mu}\hat{\nu}}R\).

Detailed Calculation

Computing the trace:

\[ G^{\hat{\alpha}}{}_{\hat{\alpha}} = \eta^{\hat{\alpha}\hat{\beta}}G_{\hat{\alpha}\hat{\beta}} = -G_{\hat{t}\hat{t}} + G_{\hat{r}\hat{r}} + G_{\hat{\theta}\hat{\theta}} + G_{\hat{\varphi}\hat{\varphi}} \]

On the other hand, the trace of the definition:

\[ G^{\hat{\alpha}}{}_{\hat{\alpha}} = R^{\hat{\alpha}}{}_{\hat{\alpha}} - \frac{1}{2}\delta^{\hat{\alpha}}{}_{\hat{\alpha}}R = R - \frac{1}{2}\cdot 4\cdot R = R - 2R = -R \]

Therefore:

\[ R = -G^{\hat{\alpha}}{}_{\hat{\alpha}} = G_{\hat{t}\hat{t}} - G_{\hat{r}\hat{r}} - G_{\hat{\theta}\hat{\theta}} - G_{\hat{\varphi}\hat{\varphi}} \]

FRW Einstein tensor components (from the formula collection):

\[ G_{\hat{t}\hat{t}} = 3\frac{\dot{a}^2 + k}{a^2} \]
\[ G_{\hat{r}\hat{r}} = G_{\hat{\theta}\hat{\theta}} = G_{\hat{\varphi}\hat{\varphi}} = -\frac{2\ddot{a}}{a} - \frac{\dot{a}^2 + k}{a^2} \]

(By isotropy, all spatial components are equal.)

Substitution:

\[ R = 3\frac{\dot{a}^2 + k}{a^2} - 3\left(-\frac{2\ddot{a}}{a} - \frac{\dot{a}^2 + k}{a^2}\right) \]
\[ = 3\frac{\dot{a}^2 + k}{a^2} + \frac{6\ddot{a}}{a} + 3\frac{\dot{a}^2 + k}{a^2} \]
\[ = 6\frac{\dot{a}^2 + k}{a^2} + \frac{6\ddot{a}}{a} \]

Final Answer

\[ \boxed{R = 6\left(\frac{\ddot{a}}{a} + \frac{\dot{a}^2 + k}{a^2}\right)} \]

Verification

For \(k = 0\), \(a = \text{const}\) (Minkowski), we get \(R = 0\). ✓

For de Sitter spacetime (\(a \propto e^{Ht}\), \(k=0\)), we have \(\dot{a}/a = H\), \(\ddot{a}/a = H^2\), so \(R = 6(H^2 + H^2) = 12H^2 = 4\Lambda\) (when \(H^2 = \Lambda/3\)). This agrees with the known result. ✓

Medium

M-1. Newtonian Limit of Schwarzschild Geodesics

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Solution Strategy

Substitute the Christoffel symbols from the formula collection into the \(r\)-component geodesic equation and take the low-velocity, weak-field limit.

Detailed Calculation

Substitution of Christoffel symbols:

Assuming the equatorial plane \(\theta = \pi/2\) (without loss of generality):

\[ \frac{d^2 r}{d\tau^2} + \frac{M}{r^2}\left(1-\frac{2M}{r}\right)\left(\frac{dt}{d\tau}\right)^2 - \frac{M}{r^2}\left(1-\frac{2M}{r}\right)^{-1}\left(\frac{dr}{d\tau}\right)^2 - (r-2M)\left(\frac{d\varphi}{d\tau}\right)^2 = 0 \]

Here the \(\Gamma^r{}_{\theta\theta}(d\theta/d\tau)^2\) term is omitted since \(d\theta/d\tau = 0\) on the equatorial plane.

Low-velocity, weak-field limit:

Conditions: - \(dr/d\tau \approx 0\), \(d\varphi/d\tau \approx 0\) (low velocity) - \(r \gg 2M\) (weak field) so \(1 - 2M/r \approx 1\) - \(dt/d\tau \approx 1\) (time dilation is negligible) - \(d\tau \approx dt\)

All terms except the second are negligible:

\[ \frac{d^2 r}{d\tau^2} + \frac{M}{r^2}\cdot 1\cdot 1^2 \approx 0 \]

Using \(d\tau \approx dt\):

\[ \frac{d^2 r}{dt^2} \approx -\frac{M}{r^2} \]

Final Answer

\[ \boxed{\frac{d^2 r}{dt^2} \approx -\frac{M}{r^2}} \]

This is precisely the equation of motion from Newton's law of universal gravitation \(F = -GMm/r^2\) (with \(G = 1\)).

Verification

Dimensional analysis: \([M/r^2] = \text{(length)}/\text{(length)}^2 = 1/\text{(length)}\). In geometric units (\(G = c = 1\)), \(M\) has dimensions of length, so \(M/r^2\) has dimensions of \(1/\text{length}\) = dimensions of acceleration. ✓

M-2. FRW and Friedmann Equations - Conservation Law

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Solution Strategy

Use the \((\hat{t},\hat{t})\) and \((\hat{r},\hat{r})\) components of the Einstein equation \(G_{\hat{\mu}\hat{\nu}} = 8\pi T_{\hat{\mu}\hat{\nu}}\).

Detailed Calculation

First Friedmann Equation:

Substituting \(G_{\hat{t}\hat{t}} = 3(\dot{a}^2 + k)/a^2\) and \(T_{\hat{t}\hat{t}} = \rho\) into \(G_{\hat{t}\hat{t}} = 8\pi T_{\hat{t}\hat{t}}\):

\[ 3\frac{\dot{a}^2 + k}{a^2} = 8\pi\rho \]
\[ \boxed{H^2 \equiv \left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi\rho}{3} - \frac{k}{a^2}} \]

Second Friedmann Equation (Acceleration Equation):

Substituting \(G_{\hat{r}\hat{r}} = -2\ddot{a}/a - (\dot{a}^2 + k)/a^2\) and \(T_{\hat{r}\hat{r}} = p\) into \(G_{\hat{r}\hat{r}} = 8\pi T_{\hat{r}\hat{r}}\):

\[ -\frac{2\ddot{a}}{a} - \frac{\dot{a}^2 + k}{a^2} = 8\pi p \]

Substituting \((\dot{a}^2 + k)/a^2 = 8\pi\rho/3\) from the first Friedmann equation:

\[ -\frac{2\ddot{a}}{a} - \frac{8\pi\rho}{3} = 8\pi p \]
\[ \frac{\ddot{a}}{a} = -\frac{4\pi}{3}(\rho + 3p) \]
\[ \boxed{\frac{\ddot{a}}{a} = -\frac{4\pi}{3}(\rho + 3p)} \]

Derivation of the Continuity Equation:

Taking the time derivative of the first Friedmann equation:

\[ \frac{d}{dt}\left(\dot{a}^2 + k\right) = \frac{d}{dt}\left(\frac{8\pi\rho}{3}a^2\right) \]
\[ 2\dot{a}\ddot{a} = \frac{8\pi}{3}\left(\dot{\rho}\,a^2 + 2\rho\,a\dot{a}\right) \]

Dividing both sides by \(2a\dot{a}\) (assuming \(\dot{a} \neq 0\)):

\[ \frac{\ddot{a}}{a} = \frac{8\pi}{3}\left(\frac{\dot{\rho}}{2}\frac{a}{\dot{a}} + \rho\right) = \frac{4\pi}{3}\frac{\dot{\rho}\,a}{\dot{a}} + \frac{8\pi\rho}{3} \]

Substituting the acceleration equation \(\ddot{a}/a = -\frac{4\pi}{3}(\rho + 3p)\) into the left-hand side:

\[ -\frac{4\pi}{3}(\rho + 3p) = \frac{4\pi}{3}\frac{\dot{\rho}\,a}{\dot{a}} + \frac{8\pi\rho}{3} \]
\[ -\frac{4\pi}{3}\rho - 4\pi p = \frac{4\pi}{3}\frac{\dot{\rho}\,a}{\dot{a}} + \frac{8\pi\rho}{3} \]
\[ -4\pi p - \frac{4\pi}{3}\rho - \frac{8\pi\rho}{3} = \frac{4\pi}{3}\frac{\dot{\rho}\,a}{\dot{a}} \]
\[ -4\pi p - 4\pi\rho = \frac{4\pi}{3}\frac{\dot{\rho}\,a}{\dot{a}} \]
\[ -3(\rho + p) = \frac{\dot{\rho}\,a}{\dot{a}} \]
\[ \dot{\rho} = -3\frac{\dot{a}}{a}(\rho + p) \]

Final Answer

\[ \boxed{\dot{\rho} + 3\frac{\dot{a}}{a}(\rho + p) = 0} \]

Verification

For dust (\(p = 0\)): \(\dot{\rho}/\rho = -3\dot{a}/a\) → \(\rho \propto a^{-3}\). Since volume is proportional to \(a^3\), this is consistent with mass conservation \(\rho \propto 1/\text{volume}\). ✓

For radiation (\(p = \rho/3\)): \(\dot{\rho}/\rho = -4\dot{a}/a\) → \(\rho \propto a^{-4}\). The density decreases by an extra factor due to energy loss from redshift, consistent with the known result. ✓

M-3. Kretschmann Scalar for Schwarzschild

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Solution Strategy

Enumerate the independent Riemann tensor components in an orthonormal basis, and use symmetries to count the number of times each component contributes to \(K = R_{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}}R^{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}}\).

Detailed Calculation

Independent non-zero components (from reference tables):

Component Value
\(R_{\hat{r}\hat{t}\hat{r}\hat{t}}\) \(-2M/r^3\)
\(R_{\hat{\theta}\hat{t}\hat{\theta}\hat{t}}\) \(M/r^3\)
\(R_{\hat{\varphi}\hat{t}\hat{\varphi}\hat{t}}\) \(M/r^3\)
\(R_{\hat{r}\hat{\theta}\hat{r}\hat{\theta}}\) \(-M/r^3\)
\(R_{\hat{r}\hat{\varphi}\hat{r}\hat{\varphi}}\) \(-M/r^3\)
\(R_{\hat{\theta}\hat{\varphi}\hat{\theta}\hat{\varphi}}\) \(2M/r^3\)

Raising indices:

In an orthonormal basis, indices are raised with \(\eta^{\hat{\alpha}\hat{\beta}} = \text{diag}(-1,+1,+1,+1)\).

\[ R^{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}} = \eta^{\hat{\alpha}\hat{\mu}}\eta^{\hat{\beta}\hat{\nu}}\eta^{\hat{\gamma}\hat{\rho}}\eta^{\hat{\delta}\hat{\sigma}}R_{\hat{\mu}\hat{\nu}\hat{\rho}\hat{\sigma}} \]

For each independent component, we check the product of the four \(\eta\) factors. Due to the antisymmetry of the Riemann tensor, each index pair \((ab)\) and \((cd)\) contains either 0 or 1 time index. Therefore the total number of time indices is either 0 or 2, and the product of \(\eta\) factors is \((-1)^0 = 1\) or \((-1)^2 = 1\).

For example, for \(R_{\hat{r}\hat{t}\hat{r}\hat{t}}\): $$ R^{\hat{r}\hat{t}\hat{r}\hat{t}} = \eta^{\hat{r}\hat{r}}\eta^{\hat{t}\hat{t}}\eta^{\hat{r}\hat{r}}\eta^{\hat{t}\hat{t}}R_{\hat{r}\hat{t}\hat{r}\hat{t}} = (+1)(-1)(+1)(-1)R_{\hat{r}\hat{t}\hat{r}\hat{t}} = R_{\hat{r}\hat{t}\hat{r}\hat{t}} $$

For \(R_{\hat{\theta}\hat{t}\hat{\theta}\hat{t}}\): $$ R^{\hat{\theta}\hat{t}\hat{\theta}\hat{t}} = \eta^{\hat{\theta}\hat{\theta}}\eta^{\hat{t}\hat{t}}\eta^{\hat{\theta}\hat{\theta}}\eta^{\hat{t}\hat{t}}R_{\hat{\theta}\hat{t}\hat{\theta}\hat{t}} = (+1)(-1)(+1)(-1)R_{\hat{\theta}\hat{t}\hat{\theta}\hat{t}} = R_{\hat{\theta}\hat{t}\hat{\theta}\hat{t}} $$

For \(R_{\hat{r}\hat{\theta}\hat{r}\hat{\theta}}\): $$ R^{\hat{r}\hat{\theta}\hat{r}\hat{\theta}} = (+1)(+1)(+1)(+1)R_{\hat{r}\hat{\theta}\hat{r}\hat{\theta}} = R_{\hat{r}\hat{\theta}\hat{r}\hat{\theta}} $$

Therefore, for all independent components, \(R^{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}} = R_{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}}\) holds.

Counting contributions from each independent component:

Using the Riemann tensor symmetries \(R_{abcd} = -R_{bacd} = -R_{abdc} = R_{cdab}\), we count the non-zero components arising from one independent component \(R_{abcd}\) (with \(a \neq b\), \(c \neq d\)).

All six independent components have the form \(R_{\hat{A}\hat{B}\hat{A}\hat{B}}\) (where the index pair \((ab) = (cd)\)). In this case:

  • \((AB, AB)\): the original component
  • \((BA, AB)\): antisymmetry in the first pair → \(-R_{ABAB}\)
  • \((AB, BA)\): antisymmetry in the second pair → \(-R_{ABAB}\)
  • \((BA, BA)\): antisymmetry in both pairs → \(+R_{ABAB}\)

This gives 4 permutations. The pair exchange \(R_{ABAB} = R_{ABAB}\) is trivial and does not generate new components.

Therefore, each independent component contributes 4 times to the sum for \(K\), with each contribution being \((R_{ABAB})^2\) (since \(R^{ABAB} = R_{ABAB}\)).

Computing \(K\):

\[ K = 4\left[\left(-\frac{2M}{r^3}\right)^2 + \left(\frac{M}{r^3}\right)^2 + \left(\frac{M}{r^3}\right)^2 + \left(-\frac{M}{r^3}\right)^2 + \left(-\frac{M}{r^3}\right)^2 + \left(\frac{2M}{r^3}\right)^2\right] \]
\[ = 4\left[\frac{4M^2}{r^6} + \frac{M^2}{r^6} + \frac{M^2}{r^6} + \frac{M^2}{r^6} + \frac{M^2}{r^6} + \frac{4M^2}{r^6}\right] \]
\[ = 4\cdot\frac{(4 + 1 + 1 + 1 + 1 + 4)M^2}{r^6} = 4\cdot\frac{12M^2}{r^6} = \frac{48M^2}{r^6} \]

Final Answer

\[ \boxed{K = R_{\alpha\beta\gamma\delta}\,R^{\alpha\beta\gamma\delta} = \frac{48M^2}{r^6}} \]

Verification

  • \(K \to 0\) as \(r \to \infty\): consistent with asymptotic flatness. ✓
  • At \(r = 2M\), \(K = 48M^2/(2M)^6 = 48/(64M^4) = 3/(4M^4)\): a finite value. This confirms it is a coordinate singularity, not a physical singularity. ✓
  • \(K \to \infty\) as \(r \to 0\): a true singularity. ✓
  • Dimensional analysis: \([M^2/r^6] = \text{(length)}^2/\text{(length)}^6 = 1/\text{(length)}^4\). Since this is the square of the curvature tensor, \(1/\text{(length)}^4\) is correct. ✓

M-4. Derivation of the Mass Function

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Solution Strategy

Substitute \(e^{-\lambda} = 1 - 2m(r)/r\) into the expression for \(G_{\hat{t}\hat{t}}\) and simplify.

Detailed Calculation

Derivative of \(e^{-\lambda}\):

\[ \frac{d}{dr}e^{-\lambda} = -\lambda'\,e^{-\lambda} \]

On the other hand,

\[ \frac{d}{dr}\left(1 - \frac{2m}{r}\right) = -\frac{2m'r - 2m}{r^2} = \frac{2m - 2m'r}{r^2} \]

Therefore:

\[ -\lambda'\,e^{-\lambda} = \frac{2m - 2m'r}{r^2} \]
\[ \lambda'\,e^{-\lambda} = \frac{2m'r - 2m}{r^2} \]

Substitution into \(G_{\hat{t}\hat{t}}\):

\[ G_{\hat{t}\hat{t}} = \frac{1}{r^2}\,e^{-\lambda}\left(\lambda' r - 1 + e^{\lambda}\right) \]

Computing \(\lambda' r\,e^{-\lambda}\):

\[ \lambda' r\,e^{-\lambda} = r\cdot\frac{2m'r - 2m}{r^2} = \frac{2m'r - 2m}{r} \]

Since \(e^{-\lambda}\cdot e^{\lambda} = 1\), we rearrange the expression for \(G_{\hat{t}\hat{t}}\):

\[ G_{\hat{t}\hat{t}} = \frac{1}{r^2}\left[\lambda' r\,e^{-\lambda} - e^{-\lambda} + 1\right] \]

Substituting each term:

\[ = \frac{1}{r^2}\left[\frac{2m'r - 2m}{r} - \left(1 - \frac{2m}{r}\right) + 1\right] \]
\[ = \frac{1}{r^2}\left[\frac{2m'r - 2m}{r} - 1 + \frac{2m}{r} + 1\right] \]
\[ = \frac{1}{r^2}\cdot\frac{2m'r - 2m + 2m}{r} = \frac{1}{r^2}\cdot\frac{2m'r}{r} = \frac{2m'}{r^2} \]

Applying the Einstein equation:

\[ G_{\hat{t}\hat{t}} = 8\pi\rho \quad \Longrightarrow \quad \frac{2m'}{r^2} = 8\pi\rho \]
\[ m' = \frac{dm}{dr} = 4\pi r^2\rho \]

Final Answer

\[ \boxed{\frac{dm}{dr} = 4\pi r^2 \rho} \]

Physical interpretation: This equation states that the mass increment of a spherical shell at radius \(r\) is \(dm = \rho \cdot 4\pi r^2\,dr\) (energy density × coordinate volume element of the shell). That is, \(m(r)\) is the gravitational mass (Misner-Sharp mass) contained within radius \(r\), and represents the general relativistic version of the Newtonian mass integral formula

\[ m(r) = \int_0^r 4\pi r'^2\,\rho(r')\,dr' \]

However, in general relativity, \(\rho\) is the local energy density, and the effect of gravitational binding energy is implicitly included through \(e^{\lambda}\) (the proper volume element is \(4\pi r^2 e^{\lambda/2}dr\), which differs from the coordinate volume element \(4\pi r^2 dr\)).

Verification

In the exterior vacuum region (\(\rho = 0\)), we have \(dm/dr = 0\) → \(m = \text{const} = M\). This gives \(e^{-\lambda} = 1 - 2M/r\), which reduces to the Schwarzschild solution. ✓

Advanced

A-1. Schwarzschild Circular Orbits and Tidal Forces

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(a) Derivation of Kepler's Third Law

Solution strategy: Substitute the conditions for a circular orbit in the equatorial plane (\(r = \text{const}\), \(\theta = \pi/2\)) into the \(r\)-component of the geodesic equation.

Detailed calculation:

Circular orbit conditions: \(dr/d\tau = 0\), \(d^2r/d\tau^2 = 0\), \(\theta = \pi/2\), \(d\theta/d\tau = 0\).

The \(r\)-component of the geodesic equation:

\[ 0 + \Gamma^r{}_{tt}\left(\frac{dt}{d\tau}\right)^2 + 0 + 0 + \Gamma^r{}_{\varphi\varphi}\left(\frac{d\varphi}{d\tau}\right)^2 = 0 \]

From the formula reference (at \(\theta = \pi/2\)):

\[ \Gamma^r{}_{tt} = \frac{M}{r^2}\left(1 - \frac{2M}{r}\right), \qquad \Gamma^r{}_{\varphi\varphi} = -(r - 2M)\sin^2\theta = -(r - 2M) \]

Substituting:

\[ \frac{M}{r^2}\left(1 - \frac{2M}{r}\right)\left(\frac{dt}{d\tau}\right)^2 - (r - 2M)\left(\frac{d\varphi}{d\tau}\right)^2 = 0 \]

Using \((1 - 2M/r) = (r - 2M)/r\):

\[ \frac{M}{r^2}\cdot\frac{r - 2M}{r}\left(\frac{dt}{d\tau}\right)^2 = (r - 2M)\left(\frac{d\varphi}{d\tau}\right)^2 \]

Since \(r > 2M\) implies \(r - 2M \neq 0\), we divide both sides by \((r - 2M)\):

\[ \frac{M}{r^3}\left(\frac{dt}{d\tau}\right)^2 = \left(\frac{d\varphi}{d\tau}\right)^2 \]

Using the angular velocity \(\Omega = d\varphi/dt = (d\varphi/d\tau)/(dt/d\tau)\):

\[ \Omega^2 = \frac{(d\varphi/d\tau)^2}{(dt/d\tau)^2} = \frac{M}{r^3} \]
\[ \boxed{\Omega^2 = \frac{M}{r^3}} \]

This has the same form as Kepler's third law in Newtonian mechanics, \(\omega^2 = GM/r^3\) (with \(G = 1\)). In general relativity, the same relation holds for the coordinate angular velocity.

(b) Evaluation of Tidal Forces

Solution strategy: Use the geodesic deviation equation.

\[ \frac{D^2\xi^{\hat{\alpha}}}{d\tau^2} = -R^{\hat{\alpha}}{}_{\hat{\beta}\hat{\gamma}\hat{\delta}}\,u^{\hat{\beta}}\,u^{\hat{\gamma}}\,\xi^{\hat{\delta}} \]

The 4-velocity of an observer on a circular orbit is approximately \(u^{\hat{\beta}} \approx (u^{\hat{t}}, 0, 0, u^{\hat{\varphi}})\) in an orthonormal basis, but for \(r \gg M\) or when examining the leading term of the tidal force, we can approximate \(u^{\hat{\beta}} \approx (1, 0, 0, 0)\). Here we evaluate the tidal force for a static observer (\(u^{\hat{\beta}} = (1, 0, 0, 0)\)) in general.

Detailed calculation:

The radial relative acceleration for a radial deviation \(\xi^{\hat{\delta}} = (0, \delta r, 0, 0)\):

\[ \frac{D^2\xi^{\hat{r}}}{d\tau^2} = -R^{\hat{r}}{}_{\hat{\beta}\hat{\gamma}\hat{\delta}}\,u^{\hat{\beta}}\,u^{\hat{\gamma}}\,\xi^{\hat{\delta}} \]

Setting \(u^{\hat{\beta}} = \delta^{\hat{\beta}}_{\hat{t}}\) and \(\xi^{\hat{\delta}} = \delta^{\hat{\delta}}_{\hat{r}}\,\delta r\):

\[ \frac{D^2\xi^{\hat{r}}}{d\tau^2} = -R^{\hat{r}}{}_{\hat{t}\hat{r}\hat{t}}\,\delta r \]

Raising the index:

\[ R^{\hat{r}}{}_{\hat{t}\hat{r}\hat{t}} = \eta^{\hat{r}\hat{r}}R_{\hat{r}\hat{t}\hat{r}\hat{t}} = (+1)\cdot\left(-\frac{2M}{r^3}\right) = -\frac{2M}{r^3} \]

Therefore:

\[ \frac{D^2\xi^{\hat{r}}}{d\tau^2} = -\left(-\frac{2M}{r^3}\right)\delta r = \frac{2M}{r^3}\,\delta r \]

The magnitude of the relative acceleration:

\[ \boxed{\left|\frac{D^2\xi^{\hat{r}}}{d\tau^2}\right| = \frac{2M}{r^3}\,\delta r} \]

This has the same form as the Newtonian tidal force \(2GM\,\delta r/r^3\).

(c) Ratio of Tidal Forces at the ISCO and Event Horizon

Since the tidal force is \(\propto M/r^3\):

\[ \frac{(\text{tidal force})_{r=6M}}{(\text{tidal force})_{r=2M}} = \frac{(2M)^3}{(6M)^3} = \frac{8M^3}{216M^3} = \frac{1}{27} \]
\[ \boxed{\frac{(\text{tidal force})_{\text{ISCO}}}{(\text{tidal force})_{\text{horizon}}} = \frac{1}{27}} \]

Verification

  • (a) At \(r = 6M\): \(\Omega^2 = M/(216M^3) = 1/(216M^2)\), \(\Omega = 1/(6\sqrt{6}\,M)\). A reasonable value. ✓
  • (b) Dimensions: \([M/r^3] = \text{(length)}/\text{(length)}^3 = 1/\text{(length)}^2\). Multiplying by \(\delta r\) gives \(1/\text{(length)}\) = dimensions of acceleration. ✓
  • (c) \(6^3 = 216\), \(2^3 = 8\), \(8/216 = 1/27\). ✓

A-2. de Sitter and the Cosmological Constant

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(a) Derivation of the Modified Friedmann Equations

Calculation details:

The Einstein equation is \(G_{\hat{\mu}\hat{\nu}} + \Lambda\,\eta_{\hat{\mu}\hat{\nu}} = 8\pi T_{\hat{\mu}\hat{\nu}}\).

\((\hat{t},\hat{t})\) component:

\[ G_{\hat{t}\hat{t}} + \Lambda\,\eta_{\hat{t}\hat{t}} = 8\pi T_{\hat{t}\hat{t}} \]
\[ 3\frac{\dot{a}^2 + k}{a^2} + \Lambda(-1) = 8\pi\rho \]
\[ 3\frac{\dot{a}^2 + k}{a^2} = 8\pi\rho + \Lambda \]
\[ \boxed{H^2 = \frac{8\pi\rho}{3} + \frac{\Lambda}{3} - \frac{k}{a^2}} \]

\((\hat{r},\hat{r})\) component:

\[ G_{\hat{r}\hat{r}} + \Lambda\,\eta_{\hat{r}\hat{r}} = 8\pi T_{\hat{r}\hat{r}} \]
\[ -\frac{2\ddot{a}}{a} - \frac{\dot{a}^2 + k}{a^2} + \Lambda(+1) = 8\pi p \]

Substituting \((\dot{a}^2 + k)/a^2 = \frac{8\pi\rho}{3} + \frac{\Lambda}{3}\) from the first Friedmann equation:

\[ -\frac{2\ddot{a}}{a} - \frac{8\pi\rho}{3} - \frac{\Lambda}{3} + \Lambda = 8\pi p \]
\[ -\frac{2\ddot{a}}{a} = 8\pi p + \frac{8\pi\rho}{3} - \frac{2\Lambda}{3} \]
\[ \frac{\ddot{a}}{a} = -\frac{4\pi}{3}(\rho + 3p) + \frac{\Lambda}{3} \]
\[ \boxed{\frac{\ddot{a}}{a} = -\frac{4\pi}{3}(\rho + 3p) + \frac{\Lambda}{3}} \]

(b) de Sitter Spacetime

Conditions: \(\rho = p = 0\), \(k = 0\).

First Friedmann equation:

\[ H^2 = \frac{\Lambda}{3} \]

Since \(H = \dot{a}/a = \text{const}\):

\[ \frac{\dot{a}}{a} = \sqrt{\frac{\Lambda}{3}} \equiv H = \text{const} \]

Solution of this differential equation:

\[ a(t) = a_0\,\exp\left(\sqrt{\frac{\Lambda}{3}}\,t\right) \propto e^{Ht} \]

Verification (acceleration equation):

\[ \frac{\ddot{a}}{a} = H^2 = \frac{\Lambda}{3} \quad \checkmark \]
\[ \boxed{a(t) \propto e^{Ht}, \qquad H = \sqrt{\frac{\Lambda}{3}}} \]

(c) Verification that the Riemann Tensor of de Sitter Spacetime Takes the Maximally Symmetric Space Form

Strategy: Compute the Ricci tensor and scalar curvature of de Sitter spacetime, and compare with the general form of the Riemann tensor for a maximally symmetric space. Additionally, directly compute the FRW Riemann tensor components for confirmation.

Calculation of the scalar curvature:

From the results of D10:

\[ R = 6\left(\frac{\ddot{a}}{a} + \frac{\dot{a}^2 + k}{a^2}\right) \]

For de Sitter spacetime (\(k = 0\), \(\ddot{a}/a = H^2\), \(\dot{a}^2/a^2 = H^2\)):

\[ R = 6(H^2 + H^2) = 12H^2 = 12\cdot\frac{\Lambda}{3} = 4\Lambda \]

Calculation of the Ricci tensor:

From the definition of the Einstein tensor \(G_{\hat{\mu}\hat{\nu}} = R_{\hat{\mu}\hat{\nu}} - \frac{1}{2}\eta_{\hat{\mu}\hat{\nu}}R\):

\[ R_{\hat{\mu}\hat{\nu}} = G_{\hat{\mu}\hat{\nu}} + \frac{1}{2}\eta_{\hat{\mu}\hat{\nu}}R \]

In de Sitter spacetime, the Einstein equation gives \(G_{\hat{\mu}\hat{\nu}} = -\Lambda\,\eta_{\hat{\mu}\hat{\nu}}\) (since \(T_{\hat{\mu}\hat{\nu}} = 0\)):

\[ R_{\hat{\mu}\hat{\nu}} = -\Lambda\,\eta_{\hat{\mu}\hat{\nu}} + \frac{1}{2}\eta_{\hat{\mu}\hat{\nu}}\cdot 4\Lambda = -\Lambda\,\eta_{\hat{\mu}\hat{\nu}} + 2\Lambda\,\eta_{\hat{\mu}\hat{\nu}} = \Lambda\,\eta_{\hat{\mu}\hat{\nu}} \]

That is, de Sitter spacetime is an Einstein space (\(R_{\hat{\mu}\hat{\nu}} = \frac{R}{n}\eta_{\hat{\mu}\hat{\nu}}\), \(n = 4\)):

\[ R_{\hat{\mu}\hat{\nu}} = \Lambda\,\eta_{\hat{\mu}\hat{\nu}} = \frac{R}{4}\eta_{\hat{\mu}\hat{\nu}} \]

Riemann tensor of a maximally symmetric space:

For an \(n\)-dimensional maximally symmetric space:

\[ R_{\alpha\beta\gamma\delta} = \frac{R}{n(n-1)}(g_{\alpha\gamma}g_{\beta\delta} - g_{\alpha\delta}g_{\beta\gamma}) \]

Substituting \(n = 4\), \(R = 4\Lambda\):

\[ R_{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}} = \frac{4\Lambda}{4\cdot 3}(\eta_{\hat{\alpha}\hat{\gamma}}\eta_{\hat{\beta}\hat{\delta}} - \eta_{\hat{\alpha}\hat{\delta}}\eta_{\hat{\beta}\hat{\gamma}}) = \frac{\Lambda}{3}(\eta_{\hat{\alpha}\hat{\gamma}}\eta_{\hat{\beta}\hat{\delta}} - \eta_{\hat{\alpha}\hat{\delta}}\eta_{\hat{\beta}\hat{\gamma}}) \]

Direct verification from the FRW Riemann tensor:

We compute the FRW Riemann tensor components for de Sitter spacetime (\(k = 0\), \(\dot{a}/a = H\), \(\ddot{a}/a = H^2\), \(H^2 = \Lambda/3\)).

The Riemann tensor components in the orthonormal basis for FRW spacetime have, in general, two types of independent components (from FRW symmetry):

$$R_{\hat{t}\hat{i}\hat{t}\hat{j}} = -\frac{\ddot{a}}{a}\,\delta_{ij} $$ $$R_{\hat{i}\hat{j}\hat{k}\hat{l}} = \frac{\dot{a}^2 + k}{a^2}\,(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk}) $$ where \(\hat{i}, \hat{j}, \hat{k}, \hat{l}\) are spatial orthonormal basis indices (\(1, 2, 3\)).

Substituting de Sitter spacetime values:

For de Sitter spacetime with \(k = 0\), \(\ddot{a}/a = H^2\), \(\dot{a}^2/a^2 = H^2\), \(H^2 = \Lambda/3\):

\[R_{\hat{t}\hat{i}\hat{t}\hat{j}} = -H^2\,\delta_{ij} = -\frac{\Lambda}{3}\,\delta_{ij} $$ $$R_{\hat{i}\hat{j}\hat{k}\hat{l}} = H^2\,(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk}) = \frac{\Lambda}{3}\,(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk}) \]

Comparison with the maximally symmetric space form:

We expand each component of the maximally symmetric Riemann tensor \(R_{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}} = \frac{\Lambda}{3}(\eta_{\hat{\alpha}\hat{\gamma}}\eta_{\hat{\beta}\hat{\delta}} - \eta_{\hat{\alpha}\hat{\delta}}\eta_{\hat{\beta}\hat{\gamma}})\).

\((\hat{t},\hat{i},\hat{t},\hat{j})\) component:

$$\frac{\Lambda}{3}(\eta_{\hat{t}\hat{t}}\eta_{\hat{i}\hat{j}} - \eta_{\hat{t}\hat{j}}\eta_{\hat{i}\hat{t}}) = \frac{\Lambda}{3}((-1)\delta_{ij} - 0) = -\frac{\Lambda}{3}\,\delta_{ij} $$ This agrees with \(R_{\hat{t}\hat{i}\hat{t}\hat{j}} = -\frac{\Lambda}{3}\,\delta_{ij}\) computed above. ✓

\((\hat{i},\hat{j},\hat{k},\hat{l})\) component (space-space):

$$\frac{\Lambda}{3}(\eta_{\hat{i}\hat{k}}\eta_{\hat{j}\hat{l}} - \eta_{\hat{i}\hat{l}}\eta_{\hat{j}\hat{k}}) = \frac{\Lambda}{3}(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk}) $$ This agrees with \(R_{\hat{i}\hat{j}\hat{k}\hat{l}} = \frac{\Lambda}{3}(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk})\) computed above. ✓

\((\hat{t},\hat{i},\hat{j},\hat{k})\) component (one temporal + three spatial):

$$\frac{\Lambda}{3}(\eta_{\hat{t}\hat{j}}\eta_{\hat{i}\hat{k}} - \eta_{\hat{t}\hat{k}}\eta_{\hat{i}\hat{j}}) = \frac{\Lambda}{3}(0 - 0) = 0 $$ This component also vanishes from FRW symmetry. ✓

Conclusion:

It has been confirmed by direct computation of the FRW Riemann tensor components that all components of the de Sitter spacetime Riemann tensor agree with the general form for a maximally symmetric space:

\[\boxed{R_{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}} = \frac{\Lambda}{3}\left(\eta_{\hat{\alpha}\hat{\gamma}}\eta_{\hat{\beta}\hat{\delta}} - \eta_{\hat{\alpha}\hat{\delta}}\eta_{\hat{\beta}\hat{\gamma}}\right)} \]

This means that de Sitter spacetime is a 4-dimensional maximally symmetric space (with the maximum number of Killing vectors: \(4\times 5/2 = 10\)). De Sitter spacetime is a Lorentzian manifold with positive constant curvature, and can be embedded as a one-sheeted hyperboloid \(-T^2 + X_1^2 + X_2^2 + X_3^2 + X_4^2 = 3/\Lambda\) in 5-dimensional Minkowski space.

Verification

  1. Consistency of the Ricci tensor: Computing the Ricci tensor from the maximally symmetric Riemann tensor: \(R_{\hat{\mu}\hat{\nu}} = \eta^{\hat{\alpha}\hat{\beta}}R_{\hat{\alpha}\hat{\mu}\hat{\beta}\hat{\nu}} = \frac{\Lambda}{3}\eta^{\hat{\alpha}\hat{\beta}}(\eta_{\hat{\alpha}\hat{\beta}}\eta_{\hat{\mu}\hat{\nu}} - \eta_{\hat{\alpha}\hat{\nu}}\eta_{\hat{\mu}\hat{\beta}}) = \frac{\Lambda}{3}(4\eta_{\hat{\mu}\hat{\nu}} - \eta_{\hat{\mu}\hat{\nu}}) = \Lambda\,\eta_{\hat{\mu}\hat{\nu}}\). This agrees with the earlier result. ✓

  2. Consistency of the scalar curvature: \(R = \eta^{\hat{\mu}\hat{\nu}}R_{\hat{\mu}\hat{\nu}} = \Lambda\,\eta^{\hat{\mu}\hat{\nu}}\eta_{\hat{\mu}\hat{\nu}} = \Lambda \cdot 4 = 4\Lambda\). This agrees with the earlier result. ✓

  3. Schwarzschild limit: As \(\Lambda \to 0\), \(R_{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}} \to 0\) (Minkowski spacetime). De Sitter spacetime is an "empty universe" with \(\Lambda > 0\), and reduces to flat spacetime when \(\Lambda = 0\). ✓