4D d'Alembert Green's function for linearised metric

[ergospherical]

Q. Calculate the linearised metric of a spherically symmetric body $\epsilon M$ at the origin. The energy momentum tensor is $T_{ab} = \epsilon M \delta(\mathbf{r}) u_a u_b$. In the harmonic (de Donder) gauge $\square \bar{h}_{ab} = -16\pi G \epsilon^{-1} T_{ab}$ (proved in previous exercise) so
\begin{align*}
\partial_m \partial^m \bar{h}_{ab} =
\begin{cases}
-16\pi GM \delta(\mathbf{r}) & a = b = 0 \\
0 & \mathrm{otherwise}
\end{cases}
\end{align*}with $u^a = (1,\mathbf{0})$. What is the Green's function for Laplace d'Alembert in 4d?

 

[vanhees71]

Is it the Lapacian (Riemannian space) or rather the D'Alembertian (Lorentzian space) you are looking for?

 

[ergospherical]

Sorry, I mean the d'Alembertian $\square = \partial_m \partial^m = \partial_0^2 - \nabla^2$.

 

[vanhees71]

And it's with the Minkowski metric, right?

Then the most simple solution goes as follows: We look for a Green's function, defined by
$$(\partial_0^2-\Delta) G(x)=\delta^{(4)}(x). \qquad (*)$$
For $r^2=\vec{x}^2 \neq (x^0)^2$ the right-hand side is 0, and we can use the spherical symmetry to make the Ansatz
$$G(x^0,\vec{x})=g(x^0,r)$$,
where $r=|\vec{x}|$. Then the equation reads
$$\partial_0^2 g-\frac{1}{r} \partial_r^2 (r g)=0.$$
Thus $r g$ fulfills the 1D wave equation with the general solution
$$r g(x^0,r)=f_1(x^0-r) + f_2(x^0+r)$$
with arbitrary functions $f_1$ and $f_2$. Now making use of
$$\Delta \frac{1}{r} = -4 \pi \delta^{(3)}(\vec{x})$$
and plugging in this solution in (*) you get
$$4 \pi [f_1(x^0) + f_2(x^0)]=\delta(x_0),$$
i.e.,
$$f_1(x^0)=\frac{A}{4 \pi} \delta(x^0), \quad f_2(x^0)=\frac{1-A}{4 \pi} \delta(x^0)$$
with an arbitrary constant $A$. This implies that the most general Green's function for the d'Alembertian is
$$g(x^0,r)=\frac{1}{4 \pi r} [A \delta(x^0-r) + (1-A) \delta(x^0+r)].$$
Now usually you want the retarded propagator, for which $g(x^0,r)=0$ for $x^0<0$, for which you uniquely get $A=1$, i.e.,
$$g_{\text{ret}}(x^0,r)=\frac{\delta(x^0-r)}{4 \pi r}.$$
In manifestly covariant form you can express this as
$$g_{\text{ret}}(x^0,r)=\frac{\Theta(x^0)}{2 \pi} \delta(x \cdot x).$$

 

[Orodruin]

That said, it is just the regular Green’s function of the wave equation in three dimensions. It should be available in most textbooks on mathematical methods?

Also note that you don’t need it to solve your problem. It is clear that the problem has a stationary solution that is the Green’s function of the Laplace operator in three dimensions. The general time dependent solution can then be found by adding homogeneous solutions to that solution.