# Propagating degrees of freedom of gravitational field

I'm following Carroll's Spacetime and Geometry, chapter on Linearized Gravity, pag. 282. He splits up the metric perturbation in scalar, vector and tensor components, writes the Einstein tensor with respect to these components and plugs into the Einstein field equation. He claims that the vector part $w_i$ and the two scalars ($\Phi$ being a component of the metric by itself, $\Psi$ being the trace of the tensor part) are not "propagating degrees of freedom", because they appear in equations like $$\text{space derivatives (field of interest)} = f \text{ (other fields and their derivatives of any kind)}$$

For example ($\nabla^2 = \delta^{ij}\partial_i \partial_j$):

$$\nabla^2\Psi=k T_{00} -\frac{1}{2}\partial_i\partial_j s^{ij}$$

This is an equation for $\Psi$ with no time derivatives; if we know what $T_{00}$ and $s_{ij}$ are doing at any time, we can determine what $\Psi$ must be (up to boundary conditions at spatial infinity). Thus, $\Psi$ is not by itself a propagating degree of freedom; it is determined by the energy-momentum tensor and the gravitational strain $s_{ij}$.

I do not understand this. What does it mean to be a propagating degree of freedom? What would be the difference if we had time derivatives acting on $\Psi$?

The equation you wrote is a Poisson-like equation. To put it in a - hopefully - more familiar way, let's rewrite it as: $$\nabla^2 \phi = \left( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2} \right)\phi= -\rho(\vec{r},t),$$ where for simplicity I wrote the Cartesian expression of the Laplacian operator $\nabla^2$. The solution to this equation, for well posed initial value conditions, can be found via the Green's function method and gives rise to the well known Newton-like potential: $$\phi(\vec{r},t)=\frac{1}{4\pi}\int\frac{\rho(\vec{r}',t)}{|\vec{r}-\vec{r}'|}d^3\vec{r}'.$$ This solution is said to be a static potential because the functional dependence is only on the spatial variable $\vec{r}$. That is, its dependence is directly related to that of the source $\rho(\vec{r},t)$. For this reason the solution doesn't evolve in time by itself but only through variations in $\rho(\vec{r},t)$. Notice also that the potential $\phi$ is identically zero if $\rho=0$.
This is not the case for the strain potential $s_{ij}$, whose field equation is an inhomogeneous wave-equation, which is a partial differential equation involving both $\vec{r}$ and $t$, and whose solutions exist even in the absence of sources (i.e. in the homogeneous case: $\square f=\big(\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\big) f = 0$) and are, fore example, plane waves. In general the solution to this equation depend on $(\omega t\pm \vec{k}\cdot\vec{r})$ and this gives rise to their characteristic oscillatory propagating behaviour.