Gravitational Waves ========================== In the weak field regime of sourceless Einstein's equation (:math:T^{\mu\nu}=0), the equation for metric with perturbations is reduced to a wave equation, .. math:: \left( - \frac{\partial^2}{ \partial t^2 } + \nabla^2 \right) \bar h^{\alpha\beta} = 0, where :math:\bar h^{\alpha\beta} is the trace-reversed perturbation of the metric on top of Minkowski metric background, i.e., .. math:: \bar h^{\alpha\beta} = h^{\alpha\beta} - \frac{1}{2} \eta^{\alpha\beta} h, where :math:h^{\alpha\beta} = g^{\alpha\beta} - \eta^{\alpha\beta} and :math:h is the trace of metric perturbation :math:h^{\alpha\beta}. .. admonition:: Trace Reverse :class: toggle The tensor :math:\bar h^{\alpha\beta} is called trace reverse of :math:h^{\alpha\beta} for its trace is :math:-h. Gauge --------------------- To solve the equation we introduce a solution of the form :math:\hat h^{\alpha\beta} = A^{\alpha\beta}e^{i k_\mu x^\mu }, which simiplifies the equation .. math:: \eta^{\mu\nu} k_{\mu}k_\nu \bar h^{\alpha\beta} = 0. To solve the amplitude :math:A^{\alpha} we need constraints on it. We can derive that gravitational waves are always null, that is :math:k^\mu k_\mu=0. Some of the conditions requires a gauge transformation. In any case, we have the second gauge condition as .. math:: A_{\alpha\beta} U^{\beta} = 0, which specifies that :math:A_{\alpha\beta} is orthogonal to the vector we chose :math:U^{\beta}. A practical choice of :math:U^\beta is a four velocity. This removes another **four degrees of freedom**. For illustration purpose, we choose :math:U^{\beta} \to ( 1, 0, 0, 0 ) since it's a null vector. The degrees of freedom removed can be visualized as the first rwo and column. The second one we can think of is a transverse condition, .. math:: A_{\alpha\beta} k^\beta = 0, which removes **another three degrees of freedom**. This specifies that the wave is transverse, i.e., :math:A_{\alpha\beta} can not have elements that is in the direction of four wavevector. We specify a wavevector :math:k^\beta \to (\omega, 0, 0, \omega ), which leads to the removal of the remaining elements of the fourth row and column. The matrix we have now becomes .. math:: A_{\alpha\beta} \to \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & A_{xx} & A_{xy} & 0 \\ 0 & A_{yx} & A_{yy} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}. The last gauge condition is traceless condition :math:A^\alpha_\alpha = 0 which also requires the gauge transformation. This condition fixes the phase relations between different spatial directions, that is $A_{xx} = e^{i\pi} A_{yy} = - A_{yy}$. This conditions insists that the two directions of distance oscillations should be quadrupole-like, i.e., contracts in one direction (say x) while extend in the other direction (say y). .. admonition:: Slicing :class: note The first two conditions are basically specifying slicings of spacetime. .. admonition:: Physical Significance of Transverse-traceless Gauge :class: important Transverse-traceless gauge is the very gauge that determines a coordinate system that a test particle is stationary in terms of coordinates. To show this we assume that we have a test particle being stationary initially, i.e., :math:U^\alpha\vert_{\tau=0} \to (1,0,0,0)^{\mathrm T}. The particle should travel on geodesics, .. math:: \frac{d}{d\tau} U^\alpha + \Gamma^\alpha_{ \mu\nu } U^\mu U^\nu =0, which leads to .. math:: \frac{ d }{ d\tau } U^\alpha \vert_{\tau = 0} = - \Gamma^\alpha_{00} = 0. The four acceleration is 0 for the test particle. No motion would be detected within the coordinate system. The same is true for a particle moving in :math:z direction. However, the conclusion doesn't hold for other motions. p