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Consider a tensor field $h_{\mu \nu}(t,\vec{x})$ which obeys the wave equation, given by (in units where $c = G = 1$):

$$ \Box h_{\mu \nu} = a \text{ } \bar{T}_{\mu \nu} \tag{1}$$ where $a$ is some constant. The solution to this can be given by: $$h_{\mu \nu} = b\int d^3x^{\prime} \frac{\bar{T}_{\mu \nu}(\vec{x^{\prime}},t_R )}{|{\vec{x} - \vec{x}^{\prime}}|} \tag{2}$$

where the retarded time $t_R = t - |\vec{x} - \vec{x}^{\prime}|$ and $b$ is some other constant. Here, Greek indices run over 4 indices $\mu = 0,1,2,3$ with 0 referring to the 'time coordinate.' But I have marked spatial 3-vectors using an arrow. My question is the following.

If I consider Minkowski background, I am able to show (by simply operating the $\Box = \partial_t^2 - \nabla^2$ on $h_{\mu \nu}$) that (2) solves (1). However, I want to solve (1) in some curved background. My confusion is that I do not know how the PDE (1) can differentiate between flat background in curvilinear coordinates, and true curvature via a curved background. Further, I can expand the operator $$\Box = \frac{1}{\sqrt{-g}}\partial_{\mu}\left(\sqrt{-g}g^{\mu \nu} \partial_\nu\right)$$ for both flat space (in any coordinates) and curved space. So I am not sure how to modify (2) in order to obtain a solution for (1) in a curved background.

EDIT: For some context, this is the motivation behind this question. I want to consider linearized gravity around curved space time. So my full metric $\bar{g}_{\mu \nu} = g_{\mu \nu} + h_{\mu \nu}$. Normally one takes $g_{\mu \nu} = \eta_{\mu \nu}$. In this case, the wave equation (1) is obtained using the Fierz Pauli action (equation 5.8 here, for example). But if I start with a curved $g_{\mu \nu}$ my plan to obtain the wave equation is to modify the FP action via minimal coupling, change the partial derivatives to covariant derivatives, and I find that I still obtain an equation of the same form as (1) but now the $\Box$ contains information about the background geometry. How would I use the solution in (2) to to construct a solution for (1) in the curved space case?

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2 Answers 2

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We have $\square=g^{\mu\nu}\nabla_\mu\nabla_\nu$ (or with a minus sign depending on convention) and because of the covariant derivative, the expansion depends on if it is applied to a scalar or tensor field. I don't know where yours is from as when using Christoffel symbols, $\sqrt{-g}$ only arises in the contraction $\Gamma_{\mu\sigma}^\sigma=\frac{\partial_\mu\sqrt{-g}}{\sqrt{-g}}$, which can not appear as the contravariant index in the Christoffel symbol is always contracted with a covariant index from a different tensor. For a scalar field, we have: \begin{equation} \square\psi =g^{\mu\nu}\nabla_\mu\nabla_\nu\psi =g^{\mu\nu}\nabla_\mu\partial_\nu\psi =g^{\mu\nu}(\partial_\mu\partial_\nu-\underbrace{\Gamma_{\mu\nu}^\sigma\partial_\sigma}_{=\Gamma^\sigma})\psi \end{equation} and $\Gamma^\sigma$ vanishes when using the deDonder gauge. Since it transforms as $\Gamma^\sigma=\frac{\partial x^\sigma}{\partial{x'}^\rho}{\Gamma'}^\rho-\square'x^\sigma$, solving the partial differential equation $\square'x^\sigma=\frac{\partial x^\sigma}{\partial{x'}^\rho}{\Gamma'}^\rho$ will give you a suitable coordinate system $x$.

For a tensor field, the expansion will get much more complicated: \begin{equation} \square h_{\kappa\lambda} =g^{\mu\nu}\nabla_\mu\nabla_\nu h_{\kappa\lambda} =g^{\mu\nu}\nabla_\mu\left( \partial_\nu h_{\kappa\lambda} -\Gamma_{\nu\kappa}^\sigma h_{\sigma\lambda} -\Gamma_{\nu\lambda}^\sigma h_{\kappa\sigma}\right) =\ldots \end{equation} and therefore there is no simple integral to solve it.

But it looks to me like you are considering a small pertubation $h_{\mu\nu}$ on a flat spacetime $\eta_{\mu\nu}$ and therefore a metric $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$. If you consider a nonflat background metric and therefore a metric of the form $\widetilde{g}_{\mu\nu}=g_{\mu\nu}+h_{\mu\nu}$, then the wave equation gets a ton of additional terms, which also makes it harder to solve. Those terms arise, when you express the Christoffel symbol, Ricci tensor and Eisntein tensor as a sum or all terms without pertubation and all those with (but to make it simpler, only the ones in first order are considered). You can then use this, to seperate the field equations into two equations for both kinds of curvature.

How this works and which equations arise is pretty difficult. I have recently written a short summary here together with a link to a document, that explains each step in detail.

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  • $\begingroup$ If I want to look at a small perturbation $h_{\mu \nu}$ against some non-flat metric $g_{\mu\nu}$ only up to first order, then can't I start with the Fierz-Pauli action, and replace partial derivatives with covariant derivates? Then, the equations of motion I get from that action (after suitably imposing de Donder gauge) should be the perturbative version of the EFE right? That is how I was able to derive (1) from my question, even when the background was curved. $\endgroup$
    – newtothis
    Commented Jun 28, 2022 at 17:54
  • $\begingroup$ Unfortnutly, I don't know enough about the Fierz-Pauli action to answer that question. But I know you can also derive equation (1) by considering a metric of the form $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$. I have the code written down, if you want me to add it to my answer. $\endgroup$ Commented Jun 28, 2022 at 18:01
  • $\begingroup$ Considering $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$ and then writing down the EFE to first order to obtain (1) (in de Donder gauge), can equivalently be obtained by considering an action that is second order in $h_{\mu \nu}$ (since up to first order, that would simply be zero upon imposing the gauge), and the action which gives (1) is simply the Fierz Pauli gauge. This was my understanding of how (1) is derived. $\endgroup$
    – newtothis
    Commented Jun 28, 2022 at 18:06
  • $\begingroup$ @newtothis You are right that you can replace partial derivatives with covariant derivatives to first order in $h$ -- so you can get the form of the equation of motion (at least formally) -- but that doesn't change Samuel's point that there's no way in general to write a closed form integral expression for the Green's function of that operator. $\endgroup$
    – Andrew
    Commented Jul 3, 2022 at 19:04
  • $\begingroup$ @newtothis Since you put up a bounty, can you tell me, what you think is still missing from my answer, maybe I can add something to it? I have all the calculations how to calculate gravitational waves in first order pertubation written down. $\endgroup$ Commented Jul 3, 2022 at 20:35
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Note, that OP's equation (2) could be written as $$ h_{\mu \nu}(x) = b\int d^4x^{\prime}G_\text{ret}(x,x') \bar{T}_{\mu \nu}(x^{\prime}), $$ where the retarded Green's function $G_\text{ret}$ is: $$ G_\text{ret}(x,x^\prime)=2\theta(t-t')\delta[(x^\mu-x'^μ)(x_\mu-x'_μ)]=\frac{\delta(t-t'-|\mathbf{x}-\mathbf{x^\prime}|)}{|\mathbf{x}-\mathbf{x^\prime}|},$$ with $\theta(t)$ and $\delta(t)$ being the Heaviside step and Dirac δ functions.

In Minkowski spacetime we have the following properties that greatly simplify work with Green's functions of the wave equation:

  • The support of retarded Green's function is the past lightcone of the point $x$, so we can replace integration over the whole 4D spacetime with integration over 3D space evaluated at retarded time.

  • Parallel transport of tensors is trivial in Minkowski spacetime, so the fact that $T_{μν}(x^\prime)$ and $h_{μν}(x)$ are tensors at different points does not complicate the Green's function structure.

In curved spacetime we also can define the Green's function of the linearized PDE for the metric perturbation $h_{\mu\nu}$, but

  • Generically Green's functions for wave equations in curved spacetime are nonzero also inside the lightcone, so the integration must be over the 4D domain of the causal past of the point $x$. In other words, gravitational perturbations propagate with all the velocities smaller or equal to the speed of light.

  • Green's functions for tensor PDEs are bitensors, tensorial functions of two points of spacetime. Notationally those object would have two types of indices, greek ($α,β$ …) corresponding to the point $x$ and primed greek ($α^\prime,β^\prime$ …) corresponding to second spacetime point ($x^\prime$).

Linearized equations for metric perturbations around the curved background look simpler for trace-reversed potentials $γ_{αβ}$ defined via $$ {\gamma _{\alpha \beta}} = {h_{\alpha \beta}} - {1 \over 2}\left({{g^{\gamma \delta}}{h_{\gamma \delta}}} \right){g_{\alpha \beta}},$$ with imposed Lorentz gauge condition $$ \gamma _{\;\;\;;\beta}^{\alpha \beta} = 0.$$ Indices of $γ$ are raised and lowered and covariant derivatives evaluated via the unperturbed metric $g$.

Those potentials satisfy the following wave equation: $${\square \gamma ^{\alpha \beta}} + 2R_{\gamma \;\delta}^{\;\alpha \;\beta}{\gamma ^{\gamma \delta}} = - 16\pi {\bar{T}^{\alpha \beta}},$$ where $□ = g^{αβ} ∇_α ∇_ β $ is the wave operator and $R_{αβγδ}$ is the Riemann curvature tensor.

The solution of this wave equation written in terms of retarded Green's function ${G_{\text{ret}\;\gamma^ \prime \delta ^\prime}^{\alpha \beta}} (x,x^\prime)$ is: $$ {\gamma ^{\alpha \beta}}(x) = 4\int {G_{\text{ret}\;\gamma^ \prime \delta ^\prime}^{\alpha \beta}} (x,x^\prime){\bar{T}^{\gamma ^\prime \delta^\prime}}(x^\prime)\sqrt {- g^\prime} \,{d^4}x^\prime , $$

The equations above were take from section 16.1 (with minor notation adjustments) of the following review paper:

  • Poisson, E., Pound, A., & Vega, I. (2011). The motion of point particles in curved spacetime. Living Reviews in Relativity, 14(1), 7, doi:10.12942/lrr-2011-7.

In addition, this paper contains general theory of bitensors and theory of Green's function in curved spacetime with references to original works.

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