Exercise Solution Guide

This concise guide outlines the essential steps to solve the GR (General Relativity) exercises.

Thank for all the contibutors in the telegram group (year 2024/2025):

• Jean-Pierre
• Belli Luigi
• Fulterz
• Margherita
• Sergio Lucci
• NikoG
• Sergio
• Llewyn Merrill
• Franceschino

1. Metric Tensor Transformation

1. Identify the Original Metric

   • Start with the metric tensor $g_{\mu\nu}$ in the initial coordinates.
   • In Cartesian coordinates $(x, y)$:
     • $g_{\mu\nu} = \left[ 1 \quad 0 \atop 0 \quad 1 \right]$

2. Define the Transformation

   • Express the new coordinates in terms of the original ones.
   • $u = \frac{x + y}{2}, \quad v = \frac{x - y}{2}.$

3. Compute Partial Derivatives

   • Calculate $\frac{\partial x^\mu}{\partial x^{\mu'}}$ for all relevant indices:
     • $\frac{\partial x}{\partial u} = 1, \quad \frac{\partial x}{\partial v} = 1, \quad \frac{\partial y}{\partial u} = 1, \quad \frac{\partial y}{\partial v} = -1.$

4. Apply the Transformation Rule

   • Use the tensor transformation rule:
     • $g_{\mu'\nu'} = \frac{\partial x^\mu}{\partial x^{\mu'}} , \frac{\partial x^\nu}{\partial x^{\nu'}} , g_{\mu\nu}.$

5. Simplify the Results

   • Determine the new metric tensor components:
     • $g_{\mu'\nu'} = \left[ 2 \quad 0 \atop 0 \quad 2 \right]$

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2. Computing Christoffel Symbols

1. Count the Independent Symbols

   • In 2D, there are $2 \times 3 = 6$ independent Christoffel symbols considering symmetry.

2. Identify Non-Vanishing Symbols

   • Condition 1: Indices in partial derivatives must align with non-zero components of the metric.
   • Condition 2: Only one partial derivative of the metric is non-zero at a time.

3. Apply the Connection Formula

   • Compute using: $\Gamma^\mu_{\nu\lambda} = \frac{1}{2} , g^{\mu\rho} \bigl( \partial_\nu g_{\lambda\rho} + \partial_\lambda g_{\nu\rho} - \partial_\rho g_{\nu\lambda} \bigr).$

4. List the Non-Zero Symbols

   • In polar coordinates:

     $\Gamma^r_{\phi\phi} = -r, \quad \Gamma^\phi_{r\phi} = \Gamma^\phi_{\phi r} = \frac{1}{r}.$

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3. Christoffel Symbols in a 2D Spacetime

1. Identify the Metric

   • $ds^2 = -(1+x)^2 dt^2 + dx^2$
   • $g_{\mu\nu} = \left[ -(1+x)^2 \ \quad 0\ \atop 0 \quad \ \ \ \ \ 1 \right]$

2. Compute Non-Vanishing Derivatives

   • $\partial_x g_{tt} = -2(1+x)$

3. Calculate Christoffel Symbols

   • $\Gamma^t_{tx} = \Gamma^t_{xt} = \frac{1}{2} g^{tt} \partial_x g_{tt} = \frac{1}{1+x}$
   • $\Gamma^x_{tt} = \frac{1}{2} g^{xx} \partial_x g_{tt} = 1+x$

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4. Christoffel Symbols with Coordinate-Dependent Metric

1. Identify the Metric

   • $ds^2 = (1+x^2)dx^2 + (1+y^2)dy^2$
   • $g_{\mu\nu} = \left[ 1+x^2 \quad 0 \atop 0 \quad 1+y^2 \right]$

2. Compute Non-Vanishing Derivatives

   • $\partial_x g_{xx} = 2x$
   • $\partial_y g_{yy} = 2y$

3. Calculate Christoffel Symbols

   • $\Gamma^x_{xx} = \frac{1}{2} g^{xx} \partial_x g_{xx} = \frac{x}{1+x^2}$
   • $\Gamma^x_{yy} = 0$

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5. Gravitational Time Dilation

1. Metric Approximation:

   • $ds^2 = -\left(1+2\Phi\right)dt^2 + \left(1-2\Phi\right)dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\phi^2$
   • $\Phi = -\frac{GM_E}{r}$

2. Scenario:

   • Two clocks: one at $r = R_E$ and another at $r = R_E + h$.

3. Proper Time Calculation:

   • $d\tau = \sqrt{1 + 2\Phi(r)} dt$
   • Clock 1: $\tau_1 = \sqrt{1 - \frac{2GM_E}{R_E}} \times t$
   • Clock 2: $\tau_2 = \sqrt{1 - \frac{2GM_E}{R_E + h}} \times t$

4. Proper Time Ratio:

   • $\frac{\tau_2}{\tau_1} = \frac{\sqrt{1 - \frac{2GM_E}{R_E + h}}}{\sqrt{1 - \frac{2GM_E}{R_E}}}$

5. Approximation for $h \ll R_E$:

   • Using binomial expansion:
   • $\frac{\tau_2}{\tau_1} \approx \frac {\left(\frac{R_E}{GM_E} - 1 + \frac{h}{R_E}\right)} {\left(\frac{R_E}{GM_E} - 1\right)} $
   • Where $\frac{GM_E}{R_E} \approx 6.957 \times 10^{-10}$ (calculated using SI units and $c=1$).

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6. Geodesics on a Sphere

1. Identify the Metric

   • $ds^2 = d\theta^2 + \sin^2\theta , d\phi^2$

2. Identify Non-Vanishing Christoffel Symbols

   • $\Gamma^\theta_{\phi\phi} = -\sin\theta \cos\theta$
   • $\Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \cot\theta$

3. Write Down the Geodesic Equations

   • $\frac{d^2 \theta}{d\lambda^2} - \sin\theta \cos\theta \left(\frac{d\phi}{d\lambda}\right)^2 = 0$
   • $\frac{d^2 \phi}{d\lambda^2} + 2 \cot\theta \frac{d\theta}{d\lambda} \frac{d\phi}{d\lambda} = 0$

4. Lines of Constant Longitude ($\phi = \text{const.}$)

   • Substitute $\frac{d\phi}{d\lambda} = 0$ and $\frac{d^2\phi}{d\lambda^2} = 0$ into the geodesic equations.
   • The equations are satisfied, confirming that lines of constant longitude are geodesics.

5. Lines of Constant Latitude ($\theta = \text{const.}$)

   • Substitute $\frac{d\theta}{d\lambda} = 0$ and $\frac{d^2\theta}{d\lambda^2} = 0$ into the geodesic equations.
   • The first equation reduces to $\sin\theta \cos\theta \left(\frac{d\phi}{d\lambda}\right)^2 = 0$.
   • This is satisfied if $\theta = \frac{\pi}{2}$ (equator) or if $\frac{d\phi}{d\lambda} = 0$ (meridian).
   • The second equation is satisfied, indicating that $\phi$ changes linearly with $\lambda$.
   • Therefore, the only line of constant latitude that is a geodesic is the equator.

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7. Covariant Derivatives on a Sphere

1. Identify the Metric and Coordinates

   • $ds^2 = d\theta^2 + \sin^2\theta , d\phi^2$
   • $x^\mu = (\theta, \phi)$

2. Recall Non-Vanishing Christoffel Symbols

   • $\Gamma^\theta_{\phi\phi} = -\sin\theta \cos\theta$
   • $\Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \cot\theta$

3. Compute Covariant Derivative of a Contravariant Vector

   • $\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu\lambda} V^\lambda$
   • $\nabla_\theta V^\theta = \partial_\theta V^\theta$
   • $\nabla_\theta V^\phi = \partial_\theta V^\phi + \cot\theta V^\phi$
   • $\nabla_\phi V^\theta = \partial_\phi V^\theta - \sin\theta \cos\theta V^\phi$
   • $\nabla_\phi V^\phi = \partial_\phi V^\phi + \cot\theta V^\theta$

4. Compute Covariant Derivative of a Covariant Vector

   • $\nabla_\mu V_\nu = \partial_\mu V_\nu - \Gamma^\lambda_{\mu\nu} V_\lambda$
   • $\nabla_\theta V_\theta = \partial_\theta V_\theta$
   • $\nabla_\theta V_\phi = \partial_\theta V_\phi - \cot\theta V_\phi$
   • $\nabla_\phi V_\theta = \partial_\phi V_\theta - \cot\theta V_\phi$
   • $\nabla_\phi V_\phi = \partial_\phi V_\phi + \sin\theta \cos\theta V_\theta$

5. Show $\nabla_\mu (V^\nu V_\nu) = \partial_\mu (V^\nu V_\nu)$

   • Use the product rule: $\nabla_\mu (V^\nu V_\nu) = V^\nu \nabla_\mu V_\nu + V_\nu \nabla_\mu V^\nu$
   • Substitute the expressions for $\nabla_\mu V^\nu$ and $\nabla_\mu V_\nu$.
   • Show that the terms involving Christoffel symbols cancel out after renaming dummy indices.
   • The result follows: $\nabla_\mu (V^\nu V_\nu) = \partial_\mu (V^\nu V_\nu)$.

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8. Commutator of Covariant Derivatives on a (2,0) Tensor

1. General Commutator Formula

   For a $(2,0)$ tensor $W^{\mu\nu}$, $[\nabla_\alpha, \nabla_\beta]\ W^{\mu\nu} \ =\ R^\mu_{\ \ \sigma\alpha\beta},W^{\sigma\nu} \ +\ R^\nu_{\ \ \sigma\alpha\beta},W^{\mu\sigma}.$

2. Contraction and Conditions

   After contracting suitable indices (so that we look at $[\nabla_{\mu}, \nabla_{\nu}],W^{\mu\nu}$), one obtains $[\nabla_{\mu}, \nabla_{\nu}],W^{\mu\nu} \ =\ R_{\sigma\nu},\bigl(W^{\sigma\nu} \ -\ W^{\nu\sigma}\bigr).$ This expression vanishes under either of these conditions:

   • $W^{\mu\nu}$ is symmetric, i.e.\ $W^{\mu\nu} = W^{\nu\mu}$.
   • The spacetime is Ricci-flat, $R_{\mu\nu} = 0$.

3. Different approach

   Note: also a different approach is possible, if we assume zero torsion, then the commutator of covariant derivatives is always zero. (see box)

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9. Photon Geodesics in 2D Spacetime

1. Metric and Null Condition

   • The 2D metric is: $ds^2 = -\Bigl(1 + \frac{x^2}{\ell^2}\Bigr),dt^2 \ +\ dx^2.$
   • A photon satisfies $ds^2 = 0$ (null geodesic).

2. Conserved Quantity

   • The metric does not depend on $t$, so $p_t = g_{t\alpha} , p^\alpha$ is conserved.
   • Define $p_t = -,E$, leading to $\Bigl(1 + \frac{x^2}{\ell^2}\Bigr),\frac{dt}{d\lambda} = E.$

3. Solve for (\frac{dx}{d\lambda})

   • From the null condition: $0 = -\Bigl(1 + \frac{x^2}{\ell^2}\Bigr),\Bigl(\frac{dt}{d\lambda}\Bigr)^2 \ +\ \Bigl(\frac{dx}{d\lambda}\Bigr)^2.$
   • Substitute $\tfrac{dt}{d\lambda} = \tfrac{E}{1 + x^2/\ell^2}$ to get $\Bigl(\frac{dx}{d\lambda}\Bigr)^2 = \frac{E^2}{,1 + \frac{x^2}{\ell^2},}.$
   • Therefore, $\frac{dx}{d\lambda} = \pm\ \frac{E}{\sqrt{,1 + \frac{x^2}{\ell^2},}}.$

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10. Christoffel Symbols in Schwarzschild Coordinates

1. Schwarzschild Metric

   • $ds^2 = -\left(1 - \frac{2GM}{r}\right)dt^2 + \left(1 - \frac{2GM}{r}\right)^{-1}dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\phi^2.$

2. Compute (\Gamma^r_{rr})

   • Recall
   • $\Gamma^r_{rr} = \frac{1}{2}g^{rr}\partial_r g_{rr}, \quad g_{rr} = \left(1 - \frac{2GM}{r}\right)^{-1}, \quad g^{rr} = \left(1 - \frac{2GM}{r}\right).$
   • One finds
   • $\Gamma^r_{rr} = -\frac{GM}{r^2}\left(1 - \frac{2GM}{r}\right)^{-1}.$

3. Compute (\Gamma^\phi_{r\phi})

   • Recall
   • $\Gamma^\phi_{r\phi} = \frac{1}{2}g^{\phi\phi}\partial_r g_{\phi\phi},\quad g_{\phi\phi} = r^2\sin^2\theta, \quad g^{\phi\phi} = \frac{1}{r^2\sin^2\theta}.$
   • Hence
   • $\Gamma^\phi_{r\phi} = \frac{1}{r}.$

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11. Stable Circular Orbits in Schwarzschild Geometry

1. Schwarzschild Metric and Conserved Quantities

   • The Schwarzschild metric is: $ds^2 = - \left(1 - \frac{2GM}{r}\right)dt^2 + \left(1 - \frac{2GM}{r}\right)^{-1}dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2.$
   • Conserved quantities for geodesic motion are: $E = \left(1 - \frac{2GM}{r}\right) \frac{dt}{d\tau}$ (energy per unit mass) $L = r^2 \frac{d\phi}{d\tau}$ (angular momentum per unit mass)

2. Normalization Condition and Effective Potential

   • For a massive particle, the four-velocity normalization is: $g_{\mu\nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} = -1$
   • Restricting motion to the equatorial plane ($\theta = \frac{\pi}{2}$), the normalization condition becomes: $-\left(1 - \frac{2GM}{r}\right)\left(\frac{dt}{d\tau}\right)^2 + \left(1 - \frac{2GM}{r}\right)^{-1} \left(\frac{dr}{d\tau}\right)^2 + r^2 \left(\frac{d\phi}{d\tau}\right)^2 = -1$
   • Substituting $E$ and $L$ and rearranging, we get: $\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 + V_{\mathrm{eff}}(r) = \frac{E^2}{2}$
   • The effective potential $V_{\mathrm{eff}}(r)$ is: $V_{\mathrm{eff}}(r) = -\frac{GM}{r} + \frac{L^2}{2r^2} - \frac{GML^2}{r^3} + \frac{1}{2} = \left(1-\frac{2GM}{r}\right)\left(\frac{L^2}{2r^2}+\frac{1}{2}\right)$

3. Circular Orbits

   • For circular orbits, $\frac{dr}{d\tau} = 0$, implying an extremum of the effective potential: $\frac{dV_{\mathrm{eff}}}{dr} = 0$
   • This condition leads to the equation: $\frac{GM}{r^2} - \frac{L^2}{r^3} + \frac{3 G M L^2}{r^4} = 0$
   • Multiplying by $r^4$ gives a quadratic equation in $r$: $GMr^2 - L^2r + 3GML^2 = 0$

4. Smallest Stable Radius (ISCO)

   • The quadratic equation has solutions: $r = \frac{L^2 \pm \sqrt{L^4 - 12 G^2 M^2 L^2}}{2 G M}$
   • Real solutions exist only if $L^2 \ge 12 G^2 M^2$.
   • At the threshold $L^2 = 12 G^2 M^2$, the two solutions coincide, giving the innermost stable circular orbit ( ISCO): $r_{\text{ISCO}} = 6GM$

12. Stable Circular Orbits in Schwarzschild Geometry (Continued)

1. Effective Potential

   • The effective potential $V_{\mathrm{eff}}(r)$ is: $V_{\mathrm{eff}}(r) = -\frac{GM}{r} + \frac{L^2}{2r^2} - \frac{GML^2}{r^3} + \frac{1}{2} = \left(1-\frac{2GM}{r}\right)\left(\frac{L^2}{2r^2}+\frac{1}{2}\right)$

2. Circular Orbits

   • For circular orbits, $\frac{dr}{d\tau} = 0$, implying an extremum of the effective potential: $\frac{dV_{\mathrm{eff}}}{dr} = 0$
   • This condition leads to the equation: $\frac{GM}{r^2} - \frac{L^2}{r^3} + \frac{3 G M L^2}{r^4} = 0$
   • Multiplying by $r^4$ gives a quadratic equation in $r$: $GMr^2 - L^2r + 3GML^2 = 0$

3. Smallest Stable Radius (ISCO)

   • The quadratic equation has solutions: $r = \frac{L^2 \pm \sqrt{L^4 - 12 G^2 M^2 L^2}}{2 G M}$
   • Real solutions exist only if $L^2 \ge 12 G^2 M^2$.
   • At the threshold $L^2 = 12 G^2 M^2$, the two solutions coincide, giving the innermost stable circular orbit ( ISCO): $r_{\text{ISCO}} = 6GM$

4. Stable Orbit in the (L \to \infty) Limit

   • For large (L), the stable orbit radius is approximately: $r \approx \frac{L^2}{GM}$
   • This matches the Newtonian result in the same limit.

13. Circular Orbits for Massless Objects in Schwarzschild Geometry

1. Schwarzschild Metric and Conserved Quantities

   • The Schwarzschild metric is: $ds^2 = - \left(1 - \frac{2GM}{r}\right)dt^2 + \left(1 - \frac{2GM}{r}\right)^{-1}dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2.$
   • Conserved quantities for geodesic motion are: $E = \left(1 - \frac{2GM}{r}\right) \frac{dt}{d\lambda}$ (energy) $L = r^2 \frac{d\phi}{d\lambda}$ (angular momentum)

2. Normalization Condition and Effective Potential

   • For a massless particle, the four-momentum normalization is: $g_{\mu\nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} = 0$
   • Restricting motion to the equatorial plane ($\theta = \frac{\pi}{2}$), the normalization condition becomes: $-\left(1 - \frac{2GM}{r}\right)\left(\frac{dt}{d\lambda}\right)^2 + \left(1 - \frac{2GM}{r}\right)^{-1} \left(\frac{dr}{d\lambda}\right)^2 + r^2 \left(\frac{d\phi}{d\lambda}\right)^2 = 0$
   • Substituting $E$ and $L$ and rearranging, we get: $\frac{1}{2}\left(\frac{dr}{d\lambda}\right)^2 + V_{\mathrm{eff}}(r) = \frac{E^2}{2}$
   • The effective potential $V_{\mathrm{eff}}(r)$ is: $V_{\mathrm{eff}}(r) = \frac{L^2}{2r^2}\left(1-\frac{2GM}{r}\right)$

3. Circular Orbits

   • For circular orbits, $\frac{dr}{d\lambda} = 0$, implying an extremum of the effective potential: $\frac{dV_{\mathrm{eff}}}{dr} = 0$
   • Setting $GM=1$ and $L=10$, this condition leads to the equation: $-\frac{100}{r^3} + \frac{300}{r^4} = 0$, that is solved for $r = 3GM$.

• Thus, the radius of the circular orbit for a massless particle is: $r = 3GM$

4. Stability of the Circular Orbit
   • To check stability, we compute the second derivative of the effective potential: $\frac{d^2 V_{\text{eff}}}{dr^2} = \frac{3L^2}{r^4} - \frac{12GML^2}{r^5}$
   • Evaluating at $r = 3GM$, we get: $\frac{d^2 V_{\text{eff}}}{dr^2}\Big|_{r=3GM} = -\frac{L^2}{81(GM)^4} < 0$
   • Since the second derivative is negative, the circular orbit at $r = 3GM$ is unstable.

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14. Minimum Energy for Photon to Reach Singularity in Schwarzschild Geometry

1. Schwarzschild Metric and Conserved Quantities

   • The Schwarzschild metric is: $ds^2 = - \left(1 - \frac{2GM}{r}\right)dt^2 + \left(1 - \frac{2GM}{r}\right)^{-1}dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2.$
   • Conserved quantities for geodesic motion are: $E = \left(1 - \frac{2GM}{r}\right) \frac{dt}{d\lambda}$ (energy) $L = r^2 \frac{d\phi}{d\lambda}$ (angular momentum)

2. Normalization Condition and Effective Potential

   • For a massless particle (photon), the four-momentum normalization is: $g_{\mu\nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} = 0$
   • Restricting motion to the equatorial plane ($\theta = \frac{\pi}{2}$), the normalization condition becomes: $-\left(1 - \frac{2GM}{r}\right)\left(\frac{dt}{d\lambda}\right)^2 + \left(1 - \frac{2GM}{r}\right)^{-1} \left(\frac{dr}{d\lambda}\right)^2 + r^2 \left(\frac{d\phi}{d\lambda}\right)^2 = 0$
   • Substituting $E$ and $L$ and rearranging, we get: $\frac{1}{2}\left(\frac{dr}{d\lambda}\right)^2 + V_{\mathrm{eff}}(r) = \frac{E^2}{2}$
   • The effective potential $V_{\mathrm{eff}}(r)$ is: $V_{\mathrm{eff}}(r) = \frac{L^2}{2r^2}\left(1-\frac{2GM}{r}\right)$

3. Conditions for Reaching the Singularity

   • Setting $GM=1$ and $L=10$, the effective potential becomes: $V_{\mathrm{eff}}(r) = \frac{50}{r^2} - \frac{100}{r^3}$
   • To reach the singularity at $r=0$, the photon's energy must be greater than or equal to the maximum value of the effective potential.
   • The maximum of $V_{\mathrm{eff}}$ occurs at $r_{\text{max}} = 3GM = 3$ (in units where $GM=1$).
   • The value of the effective potential at this maximum is: $V_{\mathrm{eff}}(3) = \frac{50}{27}$

4. Minimum Energy

   • The photon can reach the singularity if $\frac{E^2}{2} \ge V_{\mathrm{eff}}(3)$.
   • Therefore, the minimum energy required is: $E_{\min} = \sqrt{2 V_{\mathrm{eff}}(3)} = \frac{10}{3\sqrt{3}}$

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15. Gauge Transformations in Linearized Gravity

1. Definition of $V_\mu$

   • Define the vector $V_\mu$ as:
     • $V_\mu = \partial_\nu h^\nu_\mu - \frac{1}{2} \partial_\mu h^\nu_\nu$
   • $h^\nu_\mu = \eta^{\nu\alpha}h_{\mu\alpha}$ and $h^\nu_\nu = \eta^{\nu\alpha}h_{\nu\alpha}$ is the trace of $h_{\mu\nu}$.

2. Infinitesimal Gauge Transformation

   • Consider a small coordinate transformation:
     • $x^\mu \to x^\mu + \xi^\mu$
   • The metric perturbation $h_{\mu\nu}$ transforms as:
     • $h_{\mu\nu} \to h_{\mu\nu} + \partial_\mu \xi_\nu + \partial_\nu \xi_\mu$
   • $\xi_\mu = \eta_{\mu\alpha} \xi^\alpha$

3. Transformation of $V_\mu$

   • Under the gauge transformation, $V_\mu$ transforms as:
     • $V_\mu \to V_\mu + \Box \xi_\mu$
   • $\Box = \partial^\alpha \partial_\alpha$ is the d'Alembert operator.

4. Existence of Lorenz Gauge

   • It is always possible to find a gauge where $V_\mu = 0$ (Lorenz gauge).
   • This is achieved by choosing $\xi_\mu$ such that:
     • $\Box \xi_\mu = -V_\mu$
   • The existence of a solution to this inhomogeneous wave equation guarantees that the Lorenz gauge can always be reached.

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16. Gravitational Wave Polarization and Gauge Transformations

1. Plane Wave in TT Gauge

   • Consider a plane wave propagating in the positive z-direction:
     • $h_{\mu\nu}^{TT} = Re(C_{\mu\nu} e^{ik_\sigma x^\sigma})$
   • $k^\mu = (\omega, 0, 0, \omega)$ for a wave propagating along the positive z-axis.

2. Polarization Tensor $C_{\mu\nu}$ for '+' Polarization

   • $C_{\mu\nu}$ is symmetric: $C_{\mu\nu} = C_{\nu\mu}$.
   • Transversality: $k^\mu C_{\mu\nu} = 0 \implies C_{0\nu} = -C_{3\nu}$
   • Tracelessness: $C^\mu_\mu = 0 \implies C_{00} = C_{33}, C_{11} = -C_{22}$.
   • For '+' polarization in standard TT gauge: $C_{11} = -C_{22} = C_+$ and $C_{12} = C_{21} = 0$, $C_{0\nu} = C_{3\nu} = 0$.
   • Explicitly:
     • $C_{00} = 0$
     • $C_{01} = 0$
     • $C_{02} = 0$
     • $C_{03} = 0$
     • $C_{10} = 0$
     • $C_{11} = C_+$
     • $C_{12} = 0$
     • $C_{13} = 0$
     • $C_{20} = 0$
     • $C_{21} = 0$
     • $C_{22} = -C_+$
     • $C_{23} = 0$
     • $C_{30} = 0$
     • $C_{31} = 0$
     • $C_{32} = 0$
     • $C_{33} = 0$

3. Gauge Transformation

   • Consider a gauge transformation:
     • $\xi^\mu = (\alpha(x^\mu), \beta(x^\mu), 0, 0)$
   • The metric perturbation transforms as:
     • $h_{\mu\nu} \to h_{\mu\nu} + \partial_\mu \xi_\nu + \partial_\nu \xi_\mu$
   • Compute $\partial_\mu \xi_\nu + \partial_\nu \xi_\mu$, where $\xi_\nu = (-\alpha, \beta, 0, 0)$, by explicitly calculating all the partial derivatives.
   • The transformed metric perturbation is:
     • $h'_{00} = -2\partial_0 \alpha$
     • $h'_{01} = \partial_0 \beta - \partial_1 \alpha$
     • $h'_{02} = -\partial_2 \alpha$
     • $h'_{03} = -\partial_3 \alpha$
     • $h'_{10} = \partial_0 \beta - \partial_1 \alpha$
     • $h'_{11} = C^{+} \cos(\omega(t-z)) + 2\partial_1 \beta$
     • $h'_{12} = \partial_2 \beta$
     • $h'_{13} = \partial_3 \beta$
     • $h'_{20} = -\partial_2 \alpha$
     • $h'_{21} = \partial_2 \beta$
     • $h'_{22} = -C^{+} \cos(\omega(t-z))$
     • $h'_{23} = 0$
     • $h'_{30} = -\partial_3 \alpha$
     • $h'_{31} = \partial_3 \beta$
     • $h'_{32} = 0$
     • $h'_{33} = 0$

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17. Gravitational Wave Polarization and Coordinate Transformations

1. Relation between $\omega$ and $k$

   • We start with a wave vector $k^\mu = (\omega,,0,,k,,k)$.
   • Since gravitational waves propagate at the speed of light in vacuum, we must have $k_\mu,k^\mu = 0$.
   • Using the Minkowski metric $\eta_{\mu\nu} = \mathrm{diag}(-1,,+1,,+1,,+1)$, we find $$ k_\mu = (-\omega,,0,,k,,k). $$
   • The null condition $-\omega^2 + k^2 + k^2 = 0$ implies $\omega^2 = 2,k^2$, hence $\omega = \sqrt{2},k$.

2. Polarization Tensor in the Rotated Frame

   • To describe the wave propagating along $(0,,k,,k)$, define new coordinates $(t',,x',,y',,z')$ by:
     • $x' = x$,
     • $z' = \tfrac{1}{\sqrt{2}},(y + z)$ (the propagation axis),
     • $y' = \tfrac{1}{\sqrt{2}},(y - z)$ (so that ${x',,y',,z'}$ is a right-handed system).
   • In the primed frame (TT gauge) for a wave propagating along $z'$, the non-zero components of the amplitude tensor for the plus and cross polarizations are:
     • $C^{'}{x'x'} = h'{+}$,
     • $C^{'}{y'y'} = -,h'{+}$,
     • $C^{'}{x'y'} = C^{'}{y'x'} = h'_{\times}$,
     • all other $C^{'}_{\mu'\nu'} = 0$.

3. Coordinate Transformation and Final $C_{\mu\nu}$

   • The old coordinates $(t,,x,,y,,z)$ are related to the new ones via the transformation matrix $\Lambda^{\mu'}_{;\mu} = \tfrac{\partial x^{\mu'}}{\partial x^{\mu}}$.
   • After applying this transformation back to the original system, the non-zero components of $C_{\mu\nu}$ become:
     • $C_{xx} = h'_{+}$,
     • $C_{xy} = \tfrac{1}{\sqrt{2}},h'_{\times}$,
     • $C_{xz} = -,\tfrac{1}{\sqrt{2}},h'_{\times}$,
     • $C_{yy} = -,\tfrac{1}{2},h'_{+}$,
     • $C_{yz} = \tfrac{1}{2},h'_{+}$,
     • $C_{zz} = -,\tfrac{1}{2},h'_{+}$,
     • $C_{t\mu} = 0$ for all $\mu$.
   • These results ensure:
     • Transversality: $k^\mu,C_{\mu\nu} = 0$, which means $C_{t\mu} = 0$ here.
     • Tracelessness: $C_{xx} + C_{yy} + C_{zz} = 0$ in the spatial part.
     • Clear separation of polarizations: the off-diagonal parts $(xy,;xz)$ depend only on $h'{\times}$, and the diagonal parts $(xx;yy;zz)$ depend only on $h^{'}+$.

By examining this procedure, one sees explicitly how the $45^\circ$ rotation in the $(y,,z)$ plane isolates the usual plus and cross polarizations in the original $(t,,x,,y,,z)$ coordinates.

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18. Deriving the Acceleration Equation in Cosmology

1. Start with the Friedmann and Fluid Equations

• Friedmann:
  $H^2 = \frac{8\pi G}{3}\rho - \frac{\kappa}{a^2}$
• Fluid (continuity):
  $\dot{\rho} + 3H(\rho + p) = 0$

2. Differentiate the Friedmann Equation

$2H!\left(\frac{\ddot{a}}{a} - H^2\right) = \frac{8\pi G}{3},\dot{\rho} + 2,\frac{\kappa}{a^2},H$

3. Substitute the Fluid Equation

Use $\dot{\rho} = -3H(\rho + p)$: $\frac{\ddot{a}}{a} - H^2 = -\frac{4\pi G}{3}\bigl[3H(\rho+p)\bigr] + \frac{\kappa}{a^2}$

4. Rearrange and Insert $H^2$ from Friedmann

$\frac{\ddot{a}}{a} = \frac{8\pi G}{3}\rho - \frac{\kappa}{a^2} - 4\pi G,H(\rho+p) + \frac{\kappa}{a^2}$

The $\frac{\kappa}{a^2}$ terms cancel out, so $\frac{\ddot{a}}{a} = \frac{8\pi G}{3}\rho - 4\pi G,H(\rho + p)$

5. Final Simplification

Substitute $H = \frac{\dot{a}}{a}$ and use $\dot{\rho} = -3H(\rho + p)$: $\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\bigl(\rho + 3p\bigr)$

Hence, the acceleration equation is: $\boxed{\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\bigl(\rho + 3p\bigr).}$

19. Evolution of a Flat Universe with a Specific Equation of State

1. Energy Density as a Function of Scale Factor

   • Given the equation of state $p = \frac{2}{3}\rho$, we use the fluid equation:
     • $\dot{\rho} + 3\frac{\dot{a}}{a}(\rho + p) = 0$
     • $\frac{d\rho}{\rho} = -5 \frac{da}{a}$
   • Integrating, we obtain:
     • $\rho(a) = \rho_0 a^{-5}$ (since $a_0 = 1$)

2. Scale Factor as a Function of Time

   • Substituting $\rho(a)$ into the Friedmann equation for a flat universe ($\kappa = 0$):
     • $\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3} \rho_0 a^{-5}$
     • $\dot{a} = \sqrt{\frac{8\pi G \rho_0}{3}} a^{-3/2}$
   • Integrating, we find:
     • $a(t) = \left(\frac{25}{4}\frac{8\pi G \rho_0}{3}\right)^{1/5} t^{2/5} = \left(\frac{50\pi G \rho_0}{3}\right)^{1/5} t^{2/5}$
   • Using the condition $a(t_0) = 1$, we get:
     • $a(t) = \left(\frac{t}{t_0}\right)^{2/5}$ with $t_0 = \left(\frac{3}{50\pi G \rho_0}\right)^{1/2}$

3. Hubble Parameter as a Function of Time

   • Using $H(t) = \frac{\dot{a}(t)}{a(t)}$ and the expression for $a(t)$:
     • $H(t) = \frac{2}{5t}$

4. Energy Density as a Function of Time

   • Substituting $a(t)$ into $\rho(a) = \rho_0 a^{-5}$:
     • $\rho(t) = \rho_0 \left(\frac{3}{50\pi G \rho_0}\right) t^{-2} = \frac{3}{50\pi G t^2}$

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20. Evolution of a Flat Universe with Non-Relativistic Matter and Cosmological Constant

1. Derivation of (\Omega_M(a))

   • Given a spatially flat universe ($\kappa=0$) with non-relativistic matter and a cosmological constant $\Lambda$.
   • Friedmann equation: $H^2 = \frac{8\pi G}{3}(\rho_M + \rho_\Lambda) = \frac{8\pi G}{3}\rho_M + \frac{\Lambda}{3}$
   • $\rho_M(a) = \rho_0 a^{-3}$ (since $a_0 = 1$)
   • $\rho_\Lambda = \frac{\Lambda}{8\pi G}$
   • Critical density: $\rho_{\text{cr}}(t) = \frac{3H^2(t)}{8\pi G}$
   • Current critical density: $\rho_{\text{cr},0} = \frac{3H_0^2}{8\pi G}$
   • Current matter density parameter: $\Omega_0 = \frac{\rho_0}{\rho_{\text{cr},0}} = 0.3$
   • At present time: $H_0^2 = \frac{8\pi G}{3}\rho_0 + \frac{\Lambda}{3}$
   • Dividing the Friedmann equation by $H^2$: $1 = \frac{\rho_M}{\rho_{\text{cr}}} + \frac{\Lambda}{3H^2}$
   • Density parameter for matter: $\Omega_M(a) = \frac{\rho_M(a)}{\rho_{\text{cr}}(a)} = \frac{8\pi G \rho_0}{3H^2(a)a^3} = \Omega_0 \frac{H_0^2}{H^2(a)a^3}$
   • Expressing $\frac{\Lambda}{3}$ in terms of $H_0$ and $\Omega_0$: $\frac{\Lambda}{3} = H_0^2(1-\Omega_0)$
   • Substituting into $H^2(a)$: $H^2(a) = H_0^2(\Omega_0 a^{-3} + 1 - \Omega_0)$
   • Finally: $\Omega_M(a) = \frac{\Omega_0 a^{-3}}{\Omega_0 a^{-3} + 1 - \Omega_0}$

2. Scale Factor at the Onset of Acceleration

   • Acceleration equation: $\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho_M - 2\rho_\Lambda)$
   • Acceleration begins when $\ddot{a} > 0$, i.e., $\rho_M < 2\rho_\Lambda$
   • Using $\rho_M(a) = \rho_0 a^{-3}$ and $\rho_\Lambda = \frac{\rho_0}{\Omega_0}(1-\Omega_0)$:
     • $\rho_0 a^{-3} < 2 \frac{\rho_0}{\Omega_0}(1-\Omega_0)$
     • $a^3 > \frac{\Omega_0}{2(1-\Omega_0)}$
   • Scale factor at the onset of acceleration: $a_\Lambda = \left(\frac{\Omega_0}{2(1-\Omega_0)}\right)^{1/3}$
   • With $\Omega_0 = 0.3$: $a_\Lambda = \left(\frac{0.3}{2(1-0.3)}\right)^{1/3} \approx 0.6$

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