1. Introduction
The gauge issue is a long-standing problem in general relativity. Related to gravitational waves (GWs), the gauge issue ever confused people including Einstein for almost half a century [1]. For isolated GW sources which correspond to asymptotically flat spacetime, Bondi–Metzner–Sachs theory [2, 3] solves the gauge problem elegantly and gives a beautiful description of GWs. In the cases of asymptotically non-flat spacetimes, including the cosmological situations, metric perturbation with special gauge choice is used to describe GWs [4, 5].
Since asymptotically flat spacetime can be approximated as a Minkowski perturbation, the transverse-traceless (TT) gauge condition becomes a convenient tool to describe GWs. Many textbooks use the TT gauge condition as the basic theory to present GWs [6–8]. The TT gauge condition provides a fundamental picture of GWs to many researchers.
It is well known that the TT gauge condition is a special harmonic gauge. It is the vacuum Einstein equation that aids us to pick out the special one from harmonic gauges. Although many textbooks use plane GWs as examples to illustrate the choice procedure of the TT gauge condition, the plane wave is not a necessity, which we will clarify later in the current paper. In contrast, the vacuum condition must be satisfied to guarantee the existence of the TT gauge condition. Our analysis later will make this fact apparent.
In real situations, we need to consider that GWs propagates in matter [9, 10]. This is because that interstellar medium is ubiquitous in the Universe although it is thin. And we need to consider the interaction between GWs and matter including the Weber bar GW detectors [11], GW-excited lunar vibrations [12, 13] and others. In the existing literature, people neglect the matter completely when they describe GWs in these situations. Intuitively we can believe this treatment is reasonable and quantitatively accurate enough. But it is still interesting to ask the mathematical principle behind. In fact Prof. Yuan-Zhong Zhang ever discussed this problem with one of us several times.
In the current paper we aim to discuss the concept of GWS when matter is presented. More specifically we will discuss the generalization of the TT gauge condition to situations when matter is presented. In the next two sections we describe the mathematics needed by our analysis. Although the mathematical principle has been scattered in the existing literature, the description provided here is more systematic and comprehensive. Based on this mathematical tool, we briefly review the decomposition of the metric perturbation of Minkowsky in section 4 . Besides the introduction of the well known gauge invariant quantities, we design three gauge sensitive variables which are useful to discuss the gauge conditions later. After that we give a clearer analysis of harmonic gauge conditions from a new viewpoint. Within this framework we discuss the TT gauge condition choice in section 6 . Along with that a natural extension of the usual TT gauge condition to situations when matters are present is proposed. Alternatively we analyze the CZ gauge condition proposed by other authors in section 7 . Intuitively people may think the CZ gauge condition is similar but different to harmonic gauge condition. Our analysis indicates that the CZ gauge is more similar to the usual TT gauge condition. The similarity is that harmonic gauge is a family of gauges while the usual TT gauge and the CZ gauge are unique. Our analysis shows that the CZ gauge in general is not harmonic. But the CZ gauge condition returns to the TT gauge condition when the spacetime is a vacuum. In this sense the CZ gauge condition can also be looked as a generalization of TT gauge condition to situations when matters are present. We summarize the paper with some discussions about the generalized TT gauge condition in the last section.
The geometric units with c = G = 1 are used throughout the paper.
2. Decomposition of a vector field
For any vector field $\vec{v}$ with asymptotic condition $\vec{v}(\vec{r}\to \infty )=0$ we have the following decomposition
$\begin{eqnarray}\vec{v}={\rm{\nabla }}\phi +{\rm{\nabla }}\times \vec{u},\end{eqnarray}$
where the scalar field φ is determined by $\begin{eqnarray}{{\rm{\nabla }}}^{2}\phi ={\rm{\nabla }}\cdot \vec{v}.\end{eqnarray}$
According to the partial differential equation theory for the Poisson equation, there is an unique solution for φ to the above equation. As the solution of the above mentioned Poisson equation, asymptotically we have $\phi (\vec{r}\to \infty )=0$.Then the decomposition condition equation (1 ) leads us to
$\begin{eqnarray}{\rm{\nabla }}\times \vec{u}=\vec{v}-{\rm{\nabla }}\phi ,\end{eqnarray}$
$\begin{eqnarray}{\rm{\nabla }}\times ({\rm{\nabla }}\times \vec{u})={\rm{\nabla }}\times \vec{v}-{\rm{\nabla }}\times ({\rm{\nabla }}\phi )={\rm{\nabla }}\times \vec{v},\end{eqnarray}$
$\begin{eqnarray}{\rm{\nabla }}({\rm{\nabla }}\cdot \vec{u})-{{\rm{\nabla }}}^{2}\vec{u}={\rm{\nabla }}\times \vec{v}.\end{eqnarray}$
We assume ${\rm{\nabla }}\cdot \vec{u}=0$, thus the last equation becomes $\begin{eqnarray}{{\rm{\nabla }}}^{2}\vec{u}=-{\rm{\nabla }}\times \vec{v}.\end{eqnarray}$
According to the partial differential equation theory for the Poisson equation again, there is an unique solution for $\vec{u}$ to the above equation. And asymptotically we have $\vec{u}(\vec{r}\to \infty )=0$.Based on equation (6 ) we have6 ).
$\begin{eqnarray}{\rm{\nabla }}\cdot {{\rm{\nabla }}}^{2}\vec{u}=-{\rm{\nabla }}\cdot ({\rm{\nabla }}\times \vec{v})=0\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}({\rm{\nabla }}\cdot \vec{u})=0.\end{eqnarray}$
Using the partial differential equation theory for the Laplace equation, we confirm $\begin{eqnarray}{\rm{\nabla }}\cdot \vec{u}=0,\end{eqnarray}$
which satisfies the requirement to obtain equation (In other words, any vector $\vec{v}$ with asymptotic condition $\vec{v}(\vec{r}\to \infty )=0$ can be uniquely decomposed to a transverse part ${\rm{\nabla }}\times \vec{u}$, where $\vec{u}$ is determined by equation (6 ) satisfying ${\rm{\nabla }}\cdot \vec{u}=0$, and longitudinal part ∇φ, where φ is determined by equation (2 ) .
The decomposition equation (1 ) admits an interesting property which is related to harmonic gauge condition involved in the GW theory. In order to illustrate this property we explain a theorem first. Considering a Poisson equation
$\begin{eqnarray}{{\rm{\nabla }}}^{2}u=s,\end{eqnarray}$
if the source term s is an harmonic function which means $\square$s = 0, then $\begin{eqnarray}\square {{\rm{\nabla }}}^{2}u=\square s=0,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}\square u=0.\end{eqnarray}$
Based on asymptotic boundary condition, the above equation means $\square$u = 0. This is to say the solution of a Poisson equation with a harmonic source is a harmonic function.The above mentioned theorem together with equations (2 ) and (6 ) tell us that if the original vector field $\vec{v}$ is harmonic $\square \vec{v}=0$, then the decomposition components φ and $\vec{u}$ are also harmonic satisfying $\square$φ = 0 and $\square \vec{u}=0$.
Actually the aforementioned property can even be stronger as follows. Assuming that φ and $\vec{u}$ are decomposition components of a given field $\vec{v}$, then $\square$φ and $\square \vec{u}$ are decomposition components of the vector field $\square \vec{v}$.
3. Decomposition of a tensor field
Closely following the trick of the above decomposition of a vector field, for any tensor field hij with asymptotic condition ${h}_{{ij}}(\vec{r}\to \infty )=0$ we have the following decomposition.
$\begin{eqnarray}{h}_{{ij}}=\displaystyle \frac{1}{3}H{\delta }_{{ij}}+\left[{\partial }_{i}{\partial }_{j}-\displaystyle \frac{1}{3}{\delta }_{{ij}}{{\rm{\nabla }}}^{2}\right]\lambda +{\partial }_{i}{\epsilon }_{j}+{\partial }_{j}{\epsilon }_{i}+{h}_{{ij}}^{\mathrm{TT}}\end{eqnarray}$
where the scalar fields H and λ are determined by $\begin{eqnarray}H={\delta }^{{ij}}{h}_{{ij}},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}\tau =\displaystyle \frac{3}{2}{\partial }^{j}{\partial }^{i}\left({h}_{{ij}}-\displaystyle \frac{1}{3}H{\delta }_{{ij}}\right)=\displaystyle \frac{3}{2}{\partial }^{j}{\partial }^{i}{h}_{{ij}}-\displaystyle \frac{1}{2}{{\rm{\nabla }}}^{2}H,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}\lambda =\tau .\end{eqnarray}$
According to the partial differential equation theory for the Poisson equation, there are unique solutions for τ and λ to the above equations. Asymptotically we have $H(\vec{r}\to \infty )=0$, $\tau (\vec{r}\to \infty )=0$ and $\lambda (\vec{r}\to \infty )=0$.The transverse-traceless notation TT in equation (13 ) means13 ) leads us to21 ) becomes23 ) leads to
$\begin{eqnarray}{\partial }^{i}{h}_{{ij}}^{\mathrm{TT}}=0,\end{eqnarray}$
$\begin{eqnarray}{\delta }^{{ij}}{h}_{{ij}}^{\mathrm{TT}}=0.\end{eqnarray}$
Consequently, equation ( $\begin{eqnarray}{\partial }_{i}{\epsilon }_{j}+{\partial }_{j}{\epsilon }_{i}={h}_{{ij}}-\displaystyle \frac{1}{3}H{\delta }_{{ij}}-\left[{\partial }_{i}{\partial }_{j}-\displaystyle \frac{1}{3}{\delta }_{{ij}}{{\rm{\nabla }}}^{2}\right]\lambda -{h}_{{ij}}^{\mathrm{TT}},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}{\epsilon }_{j}+{\partial }_{j}{\partial }^{i}{\epsilon }_{i}={\partial }^{i}{h}_{{ij}}-\displaystyle \frac{1}{3}{\partial }_{j}H-\left[{{\rm{\nabla }}}^{2}{\partial }_{j}-\displaystyle \frac{1}{3}{\partial }_{j}{{\rm{\nabla }}}^{2}\right]\lambda ,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}{\epsilon }_{j}+{\partial }_{j}{\partial }^{i}{\epsilon }_{i}={\partial }^{i}{h}_{{ij}}-\displaystyle \frac{1}{3}{\partial }_{j}H-\displaystyle \frac{2}{3}{\partial }_{j}\tau .\end{eqnarray}$
In addition, we require εi to be divergence free $\begin{eqnarray}{\partial }^{i}{\epsilon }_{i}=0.\end{eqnarray}$
Consequently, equation ( $\begin{eqnarray}{{\rm{\nabla }}}^{2}{\epsilon }_{j}={\partial }^{i}{h}_{{ij}}-\displaystyle \frac{1}{3}{\partial }_{j}H-\displaystyle \frac{2}{3}{\partial }_{j}\tau .\end{eqnarray}$
According to the partial differential equation theory for the Poisson equation, there is a unique solution for εj to the above equation. In the mean time equation ( $\begin{eqnarray}{\partial }^{i}{{\rm{\nabla }}}^{2}{\epsilon }_{i}={\partial }^{j}{\partial }^{i}{h}_{{ij}}-\displaystyle \frac{1}{3}{{\rm{\nabla }}}^{2}H-\displaystyle \frac{2}{3}{{\rm{\nabla }}}^{2}\tau ,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}{\partial }^{i}{\epsilon }_{i}=0.\end{eqnarray}$
Using the partial differential equation theory for the Laplace equation, we know solution εi is automatically divergence free satisfying the assumption we made before. This equivalently means the tensor ∂iεj is traceless.Based on the determined H, λ and εi, we can determine ${h}_{{ij}}^{\mathrm{TT}}$ as
$\begin{eqnarray}{h}_{{ij}}^{\mathrm{TT}}\equiv {h}_{{ij}}-\displaystyle \frac{1}{3}H{\delta }_{{ij}}-\left[{\partial }_{i}{\partial }_{j}-\displaystyle \frac{1}{3}{\delta }_{{ij}}{{\rm{\nabla }}}^{2}\right]\lambda -2{\partial }_{\left(i\right.}{\epsilon }_{\left.j\right)}.\end{eqnarray}$
When hij is symmetric, ${h}_{{ij}}^{\mathrm{TT}}$ is symmetric, transverse and traceless [4, 14, 15].Specifically for a planar GW14 ) gives us15 ) we get16 ) we get23 ) we get26 ) we get
$\begin{eqnarray}{h}_{{ij}}={A}_{{ij}}{{\rm{e}}}^{I\vec{n}\cdot \vec{x}},\end{eqnarray}$
thus equation ( $\begin{eqnarray}H={{A}{\rm{e}}}^{I\vec{n}\cdot \vec{x}},\end{eqnarray}$
$\begin{eqnarray}A\equiv {\delta }^{{ij}}{A}_{{ij}}.\end{eqnarray}$
Here we have used I to denote $\sqrt{-1}$. Plugging the above relations into equation ( $\begin{eqnarray}{{\rm{\nabla }}}^{2}\tau =-\displaystyle \frac{3}{2}{n}^{i}{n}^{j}{h}_{{ij}}+\displaystyle \frac{1}{2}{n}^{2}H,\end{eqnarray}$
$\begin{eqnarray}\tau =\displaystyle \frac{3}{2{n}^{2}}{n}^{i}{n}^{j}{h}_{{ij}}-\displaystyle \frac{1}{2}H.\end{eqnarray}$
Continually plugging the above relations into equation ( $\begin{eqnarray}\lambda =-\displaystyle \frac{3}{2{n}^{4}}{n}^{i}{n}^{j}{h}_{{ij}}+\displaystyle \frac{1}{2{n}^{2}}H.\end{eqnarray}$
Plugging the above relations into equation ( $\begin{eqnarray}{{\rm{\nabla }}}^{2}{\epsilon }_{j}={{In}}^{i}{h}_{{ij}}-{{In}}_{j}\displaystyle \frac{1}{{n}^{2}}{n}^{k}{n}^{l}{h}_{{kl}},\end{eqnarray}$
$\begin{eqnarray}{\epsilon }_{j}=-I\displaystyle \frac{{n}^{i}}{{n}^{2}}{h}_{{ij}}+I\displaystyle \frac{{n}_{j}}{{n}^{4}}{n}^{k}{n}^{l}{h}_{{kl}}.\end{eqnarray}$
Collecting these results into equation ( $\begin{eqnarray}\begin{array}{rcl}{h}_{{ij}}^{\mathrm{TT}} & \equiv & {h}_{{ij}}-{n}_{i}{n}^{k}{h}_{{kj}}-{n}_{j}{n}^{l}{h}_{{il}}\\ & & -\displaystyle \frac{1}{2}\left(H{\delta }_{{ij}}-{{Hn}}_{i}{n}_{j}-{\delta }_{{ij}}{n}^{k}{n}^{l}{h}_{{kl}}-{n}_{i}{n}_{j}{n}^{k}{n}^{l}{h}_{{kl}}\right).\end{array}\end{eqnarray}$
This is to say ${h}_{{ij}}^{\mathrm{TT}}$ can also be alternatively expressed as [11] $\begin{eqnarray}{h}_{{ij}}^{\mathrm{TT}}={{\rm{\Lambda }}}_{{ij}}{}^{{kl}}{h}_{{kl}},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Lambda }}}_{{ij}}{}^{{kl}}\equiv {P}_{i}{}^{k}{P}_{j}{}^{l}-\displaystyle \frac{1}{2}{P}_{{ij}}{P}^{{kl}},\end{eqnarray}$
where the projector operator Pik is defined as $\begin{eqnarray}{P}_{i}{}^{k}{\tilde{h}}_{{kj}}(\vec{n})\equiv {\tilde{h}}_{{ij}}(\vec{n})-{n}_{i}{n}^{k}{\tilde{h}}_{{kj}}(\vec{n})\end{eqnarray}$
with respect to the Fourier component ${\tilde{h}}_{{kj}}(\vec{n})$ $\begin{eqnarray}{h}_{{ij}}\equiv \int {\tilde{h}}_{{ij}}(\vec{n}){e}^{I\vec{n}\cdot \vec{x}}{d}^{3}\vec{n}.\end{eqnarray}$
4. Decomposition of four dimensional rank 2 tensor
Given a Minkowski perturbation hμν satisfying ${h}_{\mu \nu }(\vec{r}\to \infty )\to 0$, we can decompose its components with the techniques shown above as follows [4]3 we have absorbed the factor 2 into εi here.
$\begin{eqnarray}{h}_{{tt}}=2\phi ,\end{eqnarray}$
$\begin{eqnarray}{h}_{{ti}}={\beta }_{i}+{\partial }_{i}\gamma ,\end{eqnarray}$
$\begin{eqnarray}{h}_{{ij}}={h}_{{ij}}^{\mathrm{TT}}+\displaystyle \frac{1}{3}H{\delta }_{{ij}}+{\partial }_{\left(i\right.}{\epsilon }_{\left.j\right)}+\left({\partial }_{i}{\partial }_{j}-\displaystyle \frac{1}{3}{\delta }_{{ij}}{{\rm{\nabla }}}^{2}\right)\lambda .\end{eqnarray}$
Different to section Similarly we have decomposition for infinitely small gauge transformation ξμ satisfying ${\xi }_{\mu }(\vec{r}\to \infty )\to 0$
$\begin{eqnarray}{\xi }_{t}=A,\end{eqnarray}$
$\begin{eqnarray}{\xi }_{i}={B}_{i}+{\partial }_{i}C.\end{eqnarray}$
Similar to the property of vector decomposition shown in the previous section, if hμν is harmonic, then all the decomposition components are harmonic. If the infinitely small gauge transformation ξμ is harmonic, the components A, Bi and C are also harmonic.
Under such gauge transformation ξμ, the aforementioned decomposition components of hμν will change as
$\begin{eqnarray}\phi \to \phi -\dot{A},\end{eqnarray}$
$\begin{eqnarray}{\beta }_{i}\to {\beta }_{i}-{\dot{B}}_{i},\end{eqnarray}$
$\begin{eqnarray}\gamma \to \gamma -A-\dot{C},\end{eqnarray}$
$\begin{eqnarray}H\to H-2{{\rm{\nabla }}}^{2}C,\end{eqnarray}$
$\begin{eqnarray}\lambda \to \lambda -2C,\end{eqnarray}$
$\begin{eqnarray}{\epsilon }_{i}\to {\epsilon }_{i}-2{B}_{i},\end{eqnarray}$
$\begin{eqnarray}{h}_{{ij}}^{\mathrm{TT}}\to {h}_{{ij}}^{\mathrm{TT}}.\end{eqnarray}$
Based on the above transformation rules for the decomposition components, we can easily construct gauge invariant quantities [4] $\begin{eqnarray}{\rm{\Phi }}\equiv -\phi +\dot{\gamma }-\displaystyle \frac{1}{2}\ddot{\lambda },\end{eqnarray}$
$\begin{eqnarray}{\rm{\Theta }}\equiv \displaystyle \frac{1}{3}\left(H-{{\rm{\nabla }}}^{2}\lambda \right),\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Xi }}}_{i}\equiv {\beta }_{i}-\displaystyle \frac{1}{2}{\dot{\epsilon }}_{i},\end{eqnarray}$
$\begin{eqnarray}{h}_{{ij}}^{\mathrm{TT}}.\end{eqnarray}$
Straightforward calculation shows that the linearized Einstein tensor can be expressed with the above gauge invariant quantities as [4] $\begin{eqnarray}{G}_{{tt}}=-{{\rm{\nabla }}}^{2}{\rm{\Theta }},\end{eqnarray}$
$\begin{eqnarray}{G}_{{ti}}=-\displaystyle \frac{1}{2}{{\rm{\nabla }}}^{2}{{\rm{\Xi }}}_{i}-{\partial }_{i}\dot{{\rm{\Theta }}},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{G}_{{ij}} & = & -\displaystyle \frac{1}{2}\square {h}_{{ij}}^{\mathrm{TT}}+{\delta }_{{ij}}\left[\displaystyle \frac{2}{3}{{\rm{\nabla }}}^{2}({\rm{\Phi }}+\displaystyle \frac{1}{2}{\rm{\Theta }})-\ddot{{\rm{\Theta }}}\right]-{\partial }_{\left(i\right.}{\dot{{\rm{\Xi }}}}_{\left.j\right)}\\ & & -\left({\partial }_{i}{\partial }_{j}-\displaystyle \frac{1}{3}{\delta }_{{ij}}{{\rm{\nabla }}}^{2}\right)({\rm{\Phi }}+\displaystyle \frac{1}{2}{\rm{\Theta }}).\end{array}\end{eqnarray}$
It can be seen more that the above form is also the decomposition form of the Einstein tensor.
We decompose the stress-energy tensor Tμν also as
$\begin{eqnarray}{T}_{{tt}}=\rho ,\end{eqnarray}$
$\begin{eqnarray}{T}_{{ti}}={S}_{i}+{\partial }_{i}S,\end{eqnarray}$
$\begin{eqnarray}{T}_{{ij}}={T}_{{ij}}^{\mathrm{TT}}+P{\delta }_{{ij}}+{\partial }_{\left(i\right.}{\sigma }_{\left.j\right)}+\left({\partial }_{i}{\partial }_{j}-\displaystyle \frac{1}{3}{\delta }_{{ij}}{{\rm{\nabla }}}^{2}\right)\sigma .\end{eqnarray}$
One interesting point is that Tμν is gauge invariant up to high order approximation. This is because Tμν is a first order small quantity in our cases and the gauge transformation matrix of infinitely small gauge transformation is the sum of a identity matrix and a first order small quantity.Based on the decomposition form of Gμν and Tμν, the Einstein equation Gμν = 8πTμν can be expressed as
$\begin{eqnarray}{{\rm{\nabla }}}^{2}{\rm{\Theta }}=-8\pi \rho ,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}{{\rm{\Xi }}}_{i}=-16\pi {S}_{i},\end{eqnarray}$
$\begin{eqnarray}\dot{{\rm{\Theta }}}=-8\pi S,\end{eqnarray}$
$\begin{eqnarray}\square {h}_{{ij}}^{\mathrm{TT}}=-16\pi {T}_{{ij}}^{\mathrm{TT}},\end{eqnarray}$
$\begin{eqnarray}\displaystyle \frac{2}{3}{{\rm{\nabla }}}^{2}({\rm{\Phi }}+\displaystyle \frac{1}{2}{\rm{\Theta }})-\ddot{{\rm{\Theta }}}=8\pi P,\end{eqnarray}$
$\begin{eqnarray}{\dot{{\rm{\Xi }}}}_{i}=-8\pi {\sigma }_{i},\end{eqnarray}$
$\begin{eqnarray}{\rm{\Phi }}+\displaystyle \frac{1}{2}{\rm{\Theta }}=-8\pi \sigma .\end{eqnarray}$
Combining equations (62 ), (64 ) and (66 ) we get
$\begin{eqnarray}{{\rm{\nabla }}}^{2}{\rm{\Phi }}=4\pi (\rho +3P-3\dot{S}).\end{eqnarray}$
Regarding Tμν the conservation law ∇μTμν = 0 leads us to [4]54 ). The last line of the above equation about Σi is due to equation (67 ). Under the gauge transformation defined by ξμ, these gauge sensitive variables will change as77 ) we have54 ) we have
$\begin{eqnarray}\dot{\rho }-{{\rm{\nabla }}}^{2}S=0,\end{eqnarray}$
$\begin{eqnarray}\displaystyle \frac{3}{2}P-\displaystyle \frac{3}{2}\dot{S}+{{\rm{\nabla }}}^{2}\sigma =0,\end{eqnarray}$
$\begin{eqnarray}2{\dot{S}}_{i}-{{\rm{\nabla }}}^{2}{\sigma }_{i}=0.\end{eqnarray}$
These three relations guide us to introduce three gauge sensitive variables $\begin{eqnarray}{\rm{\Pi }}\equiv {{\rm{\nabla }}}^{2}\gamma -\dot{\phi }-\displaystyle \frac{1}{2}\dot{H},\end{eqnarray}$
$\begin{eqnarray}{\rm{\Lambda }}\equiv -\displaystyle \frac{1}{4}H+\displaystyle \frac{3}{2}\phi +{{\rm{\nabla }}}^{2}\lambda -\displaystyle \frac{3}{2}\dot{\gamma },\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Sigma }}}_{i}\equiv {\dot{\beta }}_{i}-\displaystyle \frac{1}{2}{{\rm{\nabla }}}^{2}{\epsilon }_{i}\end{eqnarray}$
$\begin{eqnarray}={\dot{{\rm{\Xi }}}}_{i}+\displaystyle \frac{1}{2}{\ddot{\epsilon }}_{i}-\displaystyle \frac{1}{2}{{\rm{\nabla }}}^{2}{\epsilon }_{i}\end{eqnarray}$
$\begin{eqnarray}={\dot{{\rm{\Xi }}}}_{i}-\displaystyle \frac{1}{2}\square {\epsilon }_{i}\end{eqnarray}$
$\begin{eqnarray}=-8\pi {\sigma }_{i}-\displaystyle \frac{1}{2}\square {\epsilon }_{i}.\end{eqnarray}$
The second line of the above equation about Σi is due to equation ( $\begin{eqnarray}{\rm{\Pi }}\to {\rm{\Pi }}-\square A,\end{eqnarray}$
$\begin{eqnarray}{\rm{\Lambda }}\to {\rm{\Lambda }}-\displaystyle \frac{3}{2}\square C,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Sigma }}}_{i}\to {{\rm{\Sigma }}}_{i}+\square {B}_{i}.\end{eqnarray}$
Given Ξi and Σi we can construct εi and βi as follows. Based on equation ( $\begin{eqnarray}\square {\epsilon }_{i}=2({\dot{{\rm{\Xi }}}}_{i}-{{\rm{\Sigma }}}_{i}).\end{eqnarray}$
Up to initial data of εi and ${\dot{\epsilon }}_{i}$ we can get εi. Then based on equation ( $\begin{eqnarray}{\beta }_{i}={{\rm{\Xi }}}_{i}+\displaystyle \frac{1}{2}{\dot{\epsilon }}_{i}.\end{eqnarray}$
Given Θ, Φ and Λ we can construct λ through $\begin{eqnarray}\square \lambda ={\rm{\Theta }}+2{\rm{\Phi }}+\displaystyle \frac{4}{3}{\rm{\Lambda }}.\end{eqnarray}$
Then λ is determined up to initial data of λ and $\dot{\lambda }$. Based on the determined λ and the given Θ we can determine H as $\begin{eqnarray}H=3{\rm{\Theta }}+{{\rm{\nabla }}}^{2}\lambda .\end{eqnarray}$
Based on the determined λ and H and the given Π and Φ we can construct γ through $\begin{eqnarray}\square \gamma ={\rm{\Pi }}-\dot{{\rm{\Phi }}}+\displaystyle \frac{1}{2}\dot{H}-\displaystyle \frac{1}{2}\dddot{\lambda }.\end{eqnarray}$
Then γ is determined up to initial data of γ and $\dot{\gamma }$. Finally we can determine φ through $\begin{eqnarray}\phi =\dot{\gamma }-\displaystyle \frac{1}{2}\ddot{\lambda }-{\rm{\Phi }}.\end{eqnarray}$
5. Harmonic gauge conditions
The linearized Einstein equation under the harmonic gauge condition reads as40 )–(42 ) we have92 ) and (93 ) can be expressed with these gauge sensitive variables as99 ) tells us Σi = 0. Consequently equation (102 ) becomes69 ), (62 ) and (63 ), the perturbation metric components can be determined up to initial data of εi, ${\dot{\epsilon }}_{i}$, λ, $\dot{\lambda }$, γ and $\dot{\gamma }$ under harmonic gauge condition. Alternatively we can check equations (94 )–(100 ) which can determine the solutions up to initial data. But we need to notice that the initial data for ${h}_{{ij}}^{\mathrm{TT}}$ correspond to initial GW content. The initial data for εi, λ, and γ correspond to the gauge freedom (propagating freedom) within the harmonic gauge conditions. While the initial data for φ, H and βi should satisfy constrain equations (105 )–(107 ) given by the harmonic gauge conditions.
$\begin{eqnarray}\square {\bar{h}}_{\mu \nu }=-16\pi {T}_{\mu \nu },\end{eqnarray}$
here ${\bar{h}}_{\mu \nu }\equiv {h}_{\mu \nu }-\tfrac{1}{2}h{\eta }_{\mu \nu }$ means the corresponding trace-reversed metric perturbation of hμν. The relation between the four dimensional trace h and three dimensional trace H is h = H − 2φ. Corresponding to equations ( $\begin{eqnarray}{\bar{h}}_{{tt}}=\phi +\displaystyle \frac{1}{2}H,\end{eqnarray}$
$\begin{eqnarray}{\bar{h}}_{{ti}}={\beta }_{i}+{\partial }_{i}\gamma ,\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{\bar{h}}_{{ij}} & = & {h}_{{ij}}^{\mathrm{TT}}+\displaystyle \frac{1}{3}(3\phi -\displaystyle \frac{1}{2}H){\delta }_{{ij}}+{\partial }_{\left(i\right.}{\epsilon }_{\left.j\right)}\\ & & +\left({\partial }_{i}{\partial }_{j}-\displaystyle \frac{1}{3}{\delta }_{{ij}}{{\rm{\nabla }}}^{2}\right)\lambda .\end{array}\end{eqnarray}$
The harmonic gauge condition ${\eta }^{\mu \sigma }{\partial }_{\mu }{\bar{h}}_{\sigma \nu }=0$ can be expressed as $\begin{eqnarray}{{\rm{\nabla }}}^{2}\gamma -\dot{\phi }-\displaystyle \frac{1}{2}\dot{H}=0,\end{eqnarray}$
$\begin{eqnarray}\displaystyle \frac{2}{3}{{\rm{\nabla }}}^{2}{\partial }_{i}\lambda +\displaystyle \frac{1}{2}{{\rm{\nabla }}}^{2}{\epsilon }_{i}+{\partial }_{i}\phi -\displaystyle \frac{1}{6}{\partial }_{i}H-{\dot{\beta }}_{i}-{\partial }_{i}\dot{\gamma }=0.\end{eqnarray}$
The linearized Einstein equation under the harmonic gauge condition becomes $\begin{eqnarray}\square \phi =-4\pi (\rho +3P),\end{eqnarray}$
$\begin{eqnarray}\square {\beta }_{i}=-16\pi {S}_{i},\end{eqnarray}$
$\begin{eqnarray}\square \gamma =-16\pi S,\end{eqnarray}$
$\begin{eqnarray}\square H=24\pi (P-\rho ),\end{eqnarray}$
$\begin{eqnarray}\square \lambda =-16\pi \sigma ,\end{eqnarray}$
$\begin{eqnarray}\square {\epsilon }_{i}=-16\pi {\sigma }_{i},\end{eqnarray}$
$\begin{eqnarray}\square {h}_{{ij}}^{\mathrm{TT}}=-16\pi {T}_{{ij}}^{\mathrm{TT}}.\end{eqnarray}$
Interestingly the harmonic gauge condition equations ( $\begin{eqnarray}{\rm{\Pi }}=0,\end{eqnarray}$
$\begin{eqnarray}\displaystyle \frac{2}{3}{\partial }_{i}{\rm{\Lambda }}-{{\rm{\Sigma }}}_{i}=0.\end{eqnarray}$
And more within harmonic gauge conditions, equation ( $\begin{eqnarray}{\partial }_{i}{\rm{\Lambda }}=0.\end{eqnarray}$
Because Λ goes to zero when $\vec{r}\to \infty $, the solution of the above equation is $\begin{eqnarray}{\rm{\Lambda }}=0.\end{eqnarray}$
In another word, harmonic gauge condition is equivalent to $\begin{eqnarray}{\rm{\Pi }}=0,\end{eqnarray}$
$\begin{eqnarray}{\rm{\Lambda }}=0,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Sigma }}}_{i}=0.\end{eqnarray}$
Together with the gauge invariant variables Φ, Θ, Ξi and ${h}_{{ij}}^{\mathrm{TT}}$ determined by equations (6. TT gauge condition
Equations (94 )–(100 ) tell us that all the decomposition components φ, βi, γ, H, λ, εi and ${h}_{{ij}}^{\mathrm{TT}}$ are harmonic functions for the vacuum case. Equations (62 ), (63 ) and (69 ) tell us that Θ = Ξi = Φ = 0 for the vacuum case. This fact results in relations108 )–(110 ) guarantee that H, βi and φ become zero automatically. Consequently hμν has only ${h}_{{ij}}^{\mathrm{TT}}$ left. This new gauge condition is nothing but the TT gauge condition.
$\begin{eqnarray}H={{\rm{\nabla }}}^{2}\lambda ,\end{eqnarray}$
$\begin{eqnarray}{\beta }_{i}=\displaystyle \frac{1}{2}{\dot{\epsilon }}_{i},\end{eqnarray}$
$\begin{eqnarray}\phi =\dot{\gamma }-\displaystyle \frac{1}{2}\ddot{\lambda },\end{eqnarray}$
for any harmonic gauge condition if only Tμν = 0. Starting from any harmonic gauge condition, we can choose $\begin{eqnarray}C=\displaystyle \frac{1}{2}\lambda ,\end{eqnarray}$
$\begin{eqnarray}{B}_{i}=\displaystyle \frac{1}{2}{\epsilon }_{i},\end{eqnarray}$
$\begin{eqnarray}A=\gamma -\displaystyle \frac{1}{2}\dot{\lambda },\end{eqnarray}$
to form ξμ and perform a gauge transformation. Since λ, εi and γ are harmonic functions, this gauge transformation will result in a harmonic gauge. At the mean time λ, εi and γ become zero in the new gauge condition. Equations (When matter is presented, equations (94 )–(100 ) tell us that all the decomposition components φ, βi, γ, H, λ, εi and ${h}_{{ij}}^{\mathrm{TT}}$ cannot be zero, and cannot even be harmonic functions. So the usual TT gauge condition hμν equals ${h}_{{ij}}^{\mathrm{TT}}$ does not exist in general. But we can still use the gauge transformation equations (111 )–(113 ) to set initial data of εi, λ, and γ to zero. Then the initial data for φ, H and βi are determined by matter through
$\begin{eqnarray}\phi =-{\rm{\Phi }},\end{eqnarray}$
$\begin{eqnarray}H=3{\rm{\Theta }},\end{eqnarray}$
$\begin{eqnarray}{\beta }_{i}={{\rm{\Xi }}}_{i}.\end{eqnarray}$
So when matter is presented, we can call the above initial data choice as generalized TT gauge condition.7. CZ gauge condition
In [15, 16] CZ gauge condition is proposed119 ) we get119 ) becomes
$\begin{eqnarray}{\partial }^{i}{h}_{i\mu }-\displaystyle \frac{1}{2}{\partial }_{\mu }{h}^{i}{}_{i}=0.\end{eqnarray}$
With the decomposition components of perturbation metric, the above condition can be equivalently expressed as $\begin{eqnarray}{{\rm{\nabla }}}^{2}\gamma =\displaystyle \frac{1}{2}\dot{H},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}{\epsilon }_{i}=\displaystyle \frac{1}{3}{\partial }_{i}H-\displaystyle \frac{4}{3}{{\rm{\nabla }}}^{2}{\partial }_{i}\lambda .\end{eqnarray}$
Take divergence of equation ( $\begin{eqnarray}0=\displaystyle \frac{1}{3}{{\rm{\nabla }}}^{2}H-\displaystyle \frac{4}{3}{{\rm{\nabla }}}^{2}{{\rm{\nabla }}}^{2}\lambda ,\end{eqnarray}$
$\begin{eqnarray}H=4{{\rm{\nabla }}}^{2}\lambda .\end{eqnarray}$
Consequently equation ( $\begin{eqnarray}{{\rm{\nabla }}}^{2}{\epsilon }_{i}=0,\end{eqnarray}$
$\begin{eqnarray}{\epsilon }_{i}=0.\end{eqnarray}$
Under the CZ gauge condition, the linearized Einstein equations reduce to [15]117 ), we have130 ) and (131 ) gives us129 ) becomes
$\begin{eqnarray}{{\rm{\nabla }}}^{2}{h}_{0\mu }=-16\pi {S}_{0\mu },\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}({\partial }_{i}{h}^{i}{}_{j})=-16\pi {\partial }_{j}{T}_{00},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}({h}^{i}{}_{i})=-32\pi {T}_{00},\end{eqnarray}$
$\begin{eqnarray}\square {\hat{h}}_{{ij}}=-16\pi {\hat{S}}_{{ij}},\end{eqnarray}$
where ${S}_{\mu \nu }\equiv {T}_{\mu \nu }-\tfrac{1}{2}{\eta }_{\mu \nu }T$ is the trace-reversed stress-energy tensor with T the trace of Tμν. The quantities ${\hat{S}}_{\mu \nu }$ and ${\hat{h}}_{\mu \nu }$ are defined by the Poisson equations $\begin{eqnarray}{{\rm{\nabla }}}^{2}({S}_{\mu \nu }-{\hat{S}}_{\mu \nu })={\partial }_{\mu }{\partial }_{i}{S}^{i}{}_{\nu }+{\partial }_{\nu }{\partial }_{i}{S}^{i}{}_{\mu }-{\partial }_{\mu }{\partial }_{\nu }{S}^{i}{}_{i},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}({h}_{\mu \nu }-{\hat{h}}_{\mu \nu })={\partial }_{\mu }{\partial }_{k}{h}^{k}{}_{\nu }+{\partial }_{\nu }{\partial }_{k}{h}^{k}{}_{\mu }-{\partial }_{\mu }{\partial }_{\nu }{h}^{k}{}_{k}.\end{eqnarray}$
Due to equation ( $\begin{eqnarray}{\partial }_{\nu }{\partial }^{i}{h}_{i\mu }-\displaystyle \frac{1}{2}{\partial }_{\nu }{\partial }_{\mu }{h}^{i}{}_{i}=0.\end{eqnarray}$
Equivalently we have $\begin{eqnarray}{\partial }_{\mu }{\partial }^{i}{h}_{i\nu }-\displaystyle \frac{1}{2}{\partial }_{\mu }{\partial }_{\nu }{h}^{i}{}_{i}=0.\end{eqnarray}$
The summation of equations ( $\begin{eqnarray}{\partial }_{\mu }{\partial }^{i}{h}_{i\nu }+{\partial }_{\nu }{\partial }^{i}{h}_{i\mu }-{\partial }_{\mu }{\partial }_{\nu }{h}^{i}{}_{i}=0.\end{eqnarray}$
Then equation ( $\begin{eqnarray}{{\rm{\nabla }}}^{2}({h}_{\mu \nu }-{\hat{h}}_{\mu \nu })=0,\end{eqnarray}$
which means $\begin{eqnarray}{\hat{h}}_{\mu \nu }={h}_{\mu \nu },\end{eqnarray}$
under the CZ gauge condition.Equation (124 ) can be written as
$\begin{eqnarray}{{\rm{\nabla }}}^{2}\phi =-4\pi (\rho +3P),\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}{\beta }_{i}=-16\pi {S}_{i},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}^{2}\gamma =-16\pi S,\end{eqnarray}$
which determines φ, βi and γ completely.Equation (126 ) can be written as121 ) determined λ completely.
$\begin{eqnarray}{{\rm{\nabla }}}^{2}H=-32\pi \rho ,\end{eqnarray}$
which determines H completely. Based on the determined H, equation (Comparing equations (135 )–(138 ) and equations (62 )–(69 ) we have relations127 ). The initial data of ${h}_{{ij}}^{\mathrm{TT}}$ is determined by the initial GW content. That is to say the CZ gauge is determined completely by matter content. There is no more gauge freedom (non-propagating freedom) in the CZ gauge condition.
$\begin{eqnarray}{{\rm{\nabla }}}^{2}(\phi +{\rm{\Phi }})=\displaystyle \frac{3}{2}\ddot{{\rm{\Theta }}},\end{eqnarray}$
$\begin{eqnarray}{\beta }_{i}={{\rm{\Xi }}}_{i},\end{eqnarray}$
$\begin{eqnarray}H=4{\rm{\Theta }}.\end{eqnarray}$
The left ${h}_{{ij}}^{\mathrm{TT}}$ is controlled by equation (In the vacuum case, equations (135 )–(138 ) and (121 ) determine φ = βi = γ = H = λ = εi = 0 which correspond to the usual TT gauge. So the CZ gauge condition can be viewed as another generalization of the TT gauge condition.
It is interesting to ask if the CZ gauge condition is identical to the generalized TT gauge condition defined in the previous section. In order to answer this question we can investigate the behavior of the gauge sensitive variables Π, Λ and Σi under the CZ gauge condition. Due to equations (118 ), (121 ) and (123 ) we have135 ) and (137 ) tell us135 ) and (136 ), we can see Π, Λ and Σi cannot be zero in general when matter is presented, which means the CZ gauge is harmonic if and only if it is a vacuum. This is to say that the CZ gauge condition is different to the generalized TT gauge condition defined in the previous section when matter is presented.
$\begin{eqnarray}{\rm{\Pi }}=-\dot{\phi },\end{eqnarray}$
$\begin{eqnarray}{\rm{\Lambda }}=\displaystyle \frac{3}{2}(\phi -\dot{\gamma }),\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Sigma }}}_{i}={\dot{\beta }}_{i}.\end{eqnarray}$
equations ( $\begin{eqnarray}{{\rm{\nabla }}}^{2}(\phi -\dot{\gamma })=-4\pi (\rho +3P-4\dot{S}).\end{eqnarray}$
From the above equation and equations (8. Conclusion and discussion
Along with metric perturbation, the TT gauge condition provides a clear physical picture for GWs. It is well known the TT gauge condition is not valid any more when matter is present. In the current paper we use scalar-vector-tensor decomposition to analyze harmonic gauge conditions systematically and in depth. Based on our analysis it becomes clear why the TT gauge condition cannot be satisfied when matter is present. Accordingly, we generalize the usual TT condition within the harmonic gauge condition framework.
Alternative to the harmonic gauge condition, we also analyzed the CZ gauge condition. We find out that the CZ gauge condition is different to the harmonic in general. The CZ gauge is harmonic when and only when the spacetime is a vacuum.
Mathematically, the harmonic gauge conditions are controlled by the wave equation while the CZ gauge conditions are controlled by the Poisson equation. Since the boundary conditions for the equations are always given by physical conditions, the CZ gauge condition is uniquely determined while the harmonic gauge conditions can be different to each other up to the initial condition for the equation. When the spacetime is a vacuum, the uniquely determined CZ gauge condition is nothing but the TT gauge condition. Consequently the CZ gauge condition can be looked at as the generalized TT gauge condition when matter is present.
Within the TT gauge condition, the perturbation metric components except ${h}_{{ij}}^{\mathrm{TT}}$ vanishes. This is why people are familiar with the gGW description with ${h}_{{ij}}^{\mathrm{TT}}$.
When matters are present, the gauge invariant quantity ${h}_{{ij}}^{\mathrm{TT}}$ is controlled by a wave equation. The propagating property makes ${h}_{{ij}}^{\mathrm{TT}}\sim \tfrac{1}{r}$ near null infnity. In contrast, the rest gauge invariant quantities Φ, Θ and Ξi are all controlled by Poisson equations. The multiple expansion of the source results in
$\begin{eqnarray*}\begin{array}{l}{\rm{\Theta }}=\displaystyle \frac{2M}{r}+O\Space{0ex}{2.5ex}{0ex}(\displaystyle \frac{1}{{r}^{2}}\Space{0ex}{2.5ex}{0ex}),\\ {\rm{\Phi }}=-\displaystyle \frac{{M}}{r}+O\Space{0ex}{2.5ex}{0ex}(\displaystyle \frac{1}{{r}^{2}}\Space{0ex}{2.5ex}{0ex}),\\ {\rm{\Xi }}=\displaystyle \frac{4{P}_{i}}{r}+O\Space{0ex}{2.5ex}{0ex}(\displaystyle \frac{1}{{r}^{2}}\Space{0ex}{2.5ex}{0ex}),\end{array}\end{eqnarray*}$
where M ≡ ∫$\rho$d3x is the total mass of the source and Pi ≡ ∫Sid3x is the total linear momentum of the source. Due to the conservation of mass and linear momentum, if we just keep accuracy as $\displaystyle \frac{1}{r}$ order near null infnity, we can say there is only ${h}_{{ij}}^{\mathrm{TT}}$ admits wave behavior. Once again it is consistent to our familiar picture that ${h}_{{ij}}^{\mathrm{TT}}$ describes gravitational waves.If we use the generalized TT gauge within harmonic gauge conditions when matter is present, all the perturbation metric components are controlled by wave equations. Consequently, all of these components behave as $\sim \tfrac{1}{r}$ near null infinity.
Differently, if we use the CZ gauge when matter is present, the perturbation metric component ${h}_{{ij}}^{\mathrm{TT}}$ is controlled by a wave equation and other components are controlled by Poisson equations. Consequently if we just keep accuracy as $\tfrac{1}{r}$ order near null infinity, the metric behaves as ${h}_{\mu \nu }={h}_{{ij}}^{\mathrm{TT}}\sim \tfrac{1}{r}$.