Gravitational Waves: Plane Wave Solutions and Polarization

Welcome back to our study of gravitational waves.

In the previous lesson, we accomplished the difficult but essential task of "fixing the gauge." We saw that the linearized Einstein equations simplify dramatically in a vacuum by choosing the Transverse-Traceless (TT) gauge. This choice eliminates the non-physical, coordinate-dependent parts of the metric perturbation, leaving us with a beautifully simple equation for the physical degrees of freedom:

□ h_{μν}^{T T} = 0

where $h_{μν}^{T T}$ is both transverse ( $h_{0 μ} = 0$ , $h_{i μ} k^{i} = 0$ ) and traceless ( $h_{μ}^{μ} = 0$ ).

Today, we will solve this equation and uncover the physical nature of its solutions. Your goal is to solve the linearized vacuum equations to find plane wave solutions and identify the two polarization states ( $h_{+}$ and $h_{\times}$ ). We will see precisely how these two modes of vibration correspond to distinct ways in which spacetime can stretch and squeeze.

1. Plane Wave Solutions

The equation $□ h_{μν}^{T T} = 0$ is the standard wave equation from relativistic physics. Its most fundamental solutions are plane waves. We will see that these solutions must travel at the speed of light.

Relativity 109d: Gravitational Waves - Transverse-Traceless Gauge (Plus and Cross Polarizations)

Let's begin by postulating a plane wave solution and exploring its fundamental properties. The following video from the 'eigenchris' channel provides a clear mathematical walkthrough of this first step.

Watch from 3:05 to 8:38. This segment will introduce the complex exponential form of a plane wave, (h_{\mu u} = \text{Re}[A_{\mu u} e^{i k_\sigma x^\sigma}]), and demonstrate a crucial consequence of the wave equation: the wave covector k_\mu must be null (k_\mu k^\mu = 0). This derivation formally proves that gravitational waves propagate at the speed of light.

As the video demonstrated, substituting the plane wave ansatz

h_{μν} (x) = Re [A_{μν} e^{i k_{σ} x^{σ}}]

into the wave equation $□ h_{μν} = η^{ρ σ} \partial_{ρ} \partial_{σ} h_{μν} = 0$ yields:

- η^{ρ σ} k_{ρ} k_{σ} A_{μν} e^{i k_{σ} x^{σ}} = 0

For a non-trivial wave ( $A_{μν} \neq = 0$ ), this requires the wave 4-vector $k^{μ} = (ω / c, k)$ to be null:

k_{μ} k^{μ} = η_{μν} k^{μ} k^{ν} = - (ω / c)^{2} + ∣ k ∣^{2} = 0 ⟹ ω = c ∣ k ∣

This confirms that the phase velocity of gravitational waves is the speed of light, $c$ .

2. The Polarization Tensor in the TT Gauge

In our last lesson, we found that for a wave propagating in the $z$ -direction, the TT gauge constraints reduce the 10 a priori independent components of a symmetric tensor $h_{μν}$ to just two. Let's re-derive this result now in the context of the constant amplitude tensor $A_{μν}$ .

6 Weak Fields and Gravitational Radiation

Applying the gauge conditions directly to the plane wave's amplitude tensor provides a concrete way to see how the physical degrees of freedom emerge. Sean Carroll's lecture notes offer an excellent, rigorous treatment of this process.

Please read from the subsection beginning 'A somewhat less simplistic application of the weak-field limit is to gravitational radiation' on page 148, down to equation (6.53) on page 151. Follow the argument closely: Start with the plane wave solution (eq. 6.30) and the null wavevector condition (eq. 6.32). See how the harmonic (Lorenz) gauge condition implies k_\mu C^{\mu\nu} = 0 (eq. 6.35) (Carroll uses C_{\mu\nu} for the amplitude tensor, which is our A_{\mu\nu}). Follow the use of residual gauge freedom to impose the transverse (C_{0\nu}=0) and traceless (C^\mu_\mu = 0) conditions. Finally, for a wave in the x^3 direction, observe how these constraints lead to the final matrix form in equation (6.53), revealing the two independent components.

Following Carroll's derivation (and our work from the last lesson), for a wave with $k^{μ} = (ω, 0, 0, ω)$ , the TT gauge conditions systematically eliminate components of the amplitude matrix $A_{μν}$ .

Transverse Condition ( $A_{0 μ} = 0$ ): All components in the first row and column are zero.
Lorenz Gauge Condition ( $k^{ν} A_{μν} = 0$ ): Combined with the transverse condition, this forces $A_{3 μ} = 0$ . All components in the last row and column are zero.
Traceless Condition ( $η^{μν} A_{μν} = 0$ ): The only remaining diagonal elements must sum to zero, so $A_{11} + A_{22} = 0$ , or $A_{22} = - A_{11}$ .
Symmetry: $A_{12} = A_{21}$ .

This leaves only two independent components, which we can call $A_{+}$ and $A_{\times}$ . The general amplitude tensor $A_{μν}$ in the TT gauge for a $z$ -propagating wave is:

A_{μν} = 0000 0 A_{11} A_{21} 0 0 A_{12} A_{22} 0 0000 = A_{+} \equiv e_{μν}^{(+)} 00000100 00 - 1 0 0000 + A_{\times} \equiv e_{μν}^{(\times)} 0000001001000000

The matrices $e_{μν}^{(+)}$ and $e_{μν}^{(\times)}$ are the polarization tensors. The full plane wave solution is a linear combination of these two fundamental modes:

h_{μν}^{T T} (x) = Re [(A_{+} e_{μν}^{(+)} + A_{\times} e_{μν}^{(\times)}) e^{i k_{σ} x^{σ}}]

We define $h_{+} (x) = Re [A_{+} e^{i k_{σ} x^{σ}}]$ and $h_{\times} (x) = Re [A_{\times} e^{i k_{σ} x^{σ}}]$ .

Polarization of Gravitational Waves — This image summarizes our findings. It shows the metric for a TT-gauge wave and provides the essential visualization of the plus (+) and cross (×) polarization states, which we will now derive.

3. Physical Effects and Geodesic Deviation

We have identified two mathematical modes of propagation. To understand their physical meaning, we must analyze their effect on matter. A single particle's worldline can be made straight in any coordinate system, so to see a real, physical effect, we must look at the relative motion of nearby particles. The tool for this is the geodesic deviation equation.

For a family of nearby, initially stationary test particles with 4-velocity $U^{μ} \approx (1, 0, 0, 0)$ and separation vector $S^{μ}$ , the equation is:

\frac{d ^{2} S ^{i}}{d t ^{2}} \approx R_{0 j 0}^{i} S^{j}

In the weak-field limit and TT gauge, the required component of the Riemann tensor simplifies to:

R_{0 j 0}^{i} = \frac{1}{2} \frac{\partial ^{2}}{\partial t ^{2}} h_{ij}^{T T} = \frac{1}{2} h ¨_{ij}^{T T}

So, the geodesic deviation equation becomes a beautifully simple expression relating the relative acceleration of particles directly to the second time derivative of the gravitational wave amplitude:

\frac{d ^{2} S ^{i}}{d t ^{2}} = \frac{1}{2} h ¨_{ij}^{T T} S^{j}

Let's now analyze the effect of each polarization separately on a ring of particles in the x-y plane.

6 Weak Fields and Gravitational Radiation

The final step is to plug our two polarization modes into the geodesic deviation equation and see what happens. The same section of Carroll's notes that we've been using provides a complete derivation of this physical effect.

Continue reading from the paragraph 'To get a feeling for the physical eﬀects...' on page 152 to the end of the section on page 154. This part explicitly: Calculates the relevant Riemann tensor components. Solves the geodesic deviation equation for each polarization mode. Describes and illustrates the resulting distortion of a ring of test particles for both '+' and '×' polarizations.

Plus (+) Polarization

For a pure plus-polarized wave, $h_{11}^{T T} = - h_{22}^{T T} = h_{+} (t, z)$ and $h_{12}^{T T} = 0$ . The geodesic deviation equations are:

\frac{d ^{2} S ^{1}}{d t ^{2}} = \frac{1}{2} h ¨_{+} S^{1} and \frac{d ^{2} S ^{2}}{d t ^{2}} = \frac{1}{2} h ¨_{22} S^{2} = - \frac{1}{2} h ¨_{+} S^{2}

Integrating twice with respect to time gives the change in separation:

δ S^{1} (t) \approx \frac{1}{2} h_{+} (t) S^{1} (0) and δ S^{2} (t) \approx - \frac{1}{2} h_{+} (t) S^{2} (0)

This means that as the wave passes, distances are stretched along one axis while being simultaneously compressed along the orthogonal axis. For a ring of particles in the x-y plane, this creates an oscillating pattern shaped like a plus sign (+).

Cross (×) Polarization

For a pure cross-polarized wave, $h_{11}^{T T} = h_{22}^{T T} = 0$ and $h_{12}^{T T} = h_{21}^{T T} = h_{\times} (t, z)$ . The equations are now:

\frac{d ^{2} S ^{1}}{d t ^{2}} = \frac{1}{2} h ¨_{\times} S^{2} and \frac{d ^{2} S ^{2}}{d t ^{2}} = \frac{1}{2} h ¨_{\times} S^{1}

Integrating these gives:

δ S^{1} (t) \approx \frac{1}{2} h_{\times} (t) S^{2} (0) and δ S^{2} (t) \approx \frac{1}{2} h_{\times} (t) S^{1} (0)

This pattern corresponds to stretching and squeezing along the diagonal axes $y = x$ and $y = - x$ . A ring of particles will be distorted into an oscillating pattern shaped like a cross (×).

Test your understanding!

Consider a gravitational wave that is purely "right-circularly polarized," defined by $A_{\times} = i A_{+}$ (where $A_{+}$ is real). The wave is then given by $h_{μν}^{T T} = Re [A_{+} (e_{μν}^{(+)} + i e_{μν}^{(\times)}) e^{iω (t - z)}]$ .

Qualitatively describe the motion of a particle initially on the positive x-axis ( $S^{μ} (0) = (0, L, 0, 0)$ ) as this wave passes. You don't need to solve the equations fully, but describe the particle's trajectory in the x-y plane.

Show answer

Let's write out the real parts of the metric components:

h_{+} (t, z) = Re [A_{+} e^{iω (t - z)}] = A_{+} cos (ω (t - z))

h_{\times} (t, z) = Re [i A_{+} e^{iω (t - z)}] = - A_{+} sin (ω (t - z))

The geodesic deviation equations for a particle at $S^{μ} (0) = (0, L, 0, 0)$ are:

δ S^{1} (t) \approx \frac{1}{2} h_{+} (t) S^{1} (0) + \frac{1}{2} h_{\times} (t) S^{2} (0) = \frac{1}{2} A_{+} cos (ω (t - z)) L

δ S^{2} (t) \approx \frac{1}{2} h_{\times} (t) S^{1} (0) - \frac{1}{2} h_{+} (t) S^{2} (0) = 0

This is incorrect. The equations are $δ S^{1} = \frac{1}{2} (h_{+} S^{1} + h_{\times} S^{2})$ and $δ S^{2} = \frac{1}{2} (h_{\times} S^{1} - h_{+} S^{2})$ . For a particle at $S = (L, 0, 0)$ :

S^{1} (t) = L + δ S^{1} \approx L (1 + \frac{1}{2} h_{+}) = L (1 + \frac{1}{2} A_{+} cos (ω t))

S^{2} (t) = 0 + δ S^{2} \approx \frac{1}{2} h_{\times} S^{1} (0) = \frac{L}{2} A_{+} (- sin (ω t))

The particle's position is approximately $(L (1 + \frac{1}{2} A_{+} cos (ω t)), - \frac{L}{2} A_{+} sin (ω t))$ . This is the equation of a small ellipse. The particle oscillates primarily along the x-axis, but also has a smaller oscillation along the y-axis that is 90 degrees out of phase. The combination of these motions means the particle traces out a small circle or ellipse in a clockwise (right-handed) direction around its initial position.

Conclusion

In this lesson, we have successfully solved the linearized vacuum Einstein equations and uncovered the physical nature of gravitational waves.

Key Takeaways:

The vacuum wave equation $□ h_{μν} = 0$ admits plane wave solutions of the form $h_{μν} = Re [A_{μν} e^{i k_{σ} x^{σ}}]$ .
The wave equation requires the wave 4-vector $k^{μ}$ to be null ( $k_{μ} k^{μ} = 0$ ), proving that gravitational waves travel at the speed of light.
Applying the Transverse-Traceless (TT) gauge constraints to the wave's amplitude tensor $A_{μν}$ reveals that there are only two independent, physical degrees of freedom.
These two degrees of freedom are the plus (+) and cross (×) polarizations, which correspond to two orthogonal patterns of spacetime distortion in the plane transverse to the wave's propagation.
The physical effect of these polarizations is to change the proper distance between test particles, causing a ring of particles to oscillate in characteristic '+' or '×' patterns.

Preview of Next Lesson:

We now understand what a gravitational wave is. The next questions are: How are they generated, and how much energy do they carry? In our next lesson, we will derive the famous quadrupole formula for the power radiated by a non-relativistic source, which is the cornerstone for calculating the emission of gravitational waves from astrophysical systems.

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