- Sun 27 January 2019
- Physics
- #green-function, #laplacian
In \(d\)-dimensional euclidean space, the laplacian operator is
\begin{equation*}
\partial^{2} \equiv \delta^{ab} \partial_{a} \partial_{b}
\end{equation*}
The laplacian operator is the trace of the \(\partial_{a} \partial_{b}\) operator. The Green function for the laplacian operator is such that
\begin{equation*}
\partial^{2} G = - A \delta(x)
\end{equation*}
Here \(A\) is a constant and \(\delta(x)\) is a \(d\)-dimensional Dirac delta function.
In this post I will evaluate the following matrix:
\begin{equation*}
D_{ab} \equiv \partial_{a} \partial_{b} G
\end{equation*}
Note that \(D_{ab}\) is a symmetric matrix. The trace of \(D_{ab}\) must give the Dirac delta:
\begin{equation*}
\delta^{ab} D_{ab} = -A \delta(x)
\end{equation*}
So maybe \(D_{ab}\) can be split into a traceless and non-traceless parts:
\begin{equation*}
D_{ab} = -\frac{A}{d}\delta_{ab}\delta(x) + T_{ab}
\end{equation*}
Here \(T_{ab}\) is symmetric and traceless. Maybe it can be written as
\begin{equation*}
T_{ab} = F(x) \left(\delta_{ab} - d\frac{x_{a}x_{b}}{x^{2}}\right)
\end{equation*}
Now, dimensional analysis suggest
\begin{equation*}
F(x) = B\left(\frac{1}{x^{2}}\right)^{d/2}
\end{equation*}
Here, \(B\) is a constant. Thus, apparently, you would have
\begin{equation*}
D_{ab} = B\left(\frac{1}{x^{2}}\right)^{d/2} \left(\delta_{ab} - d\frac{x_{a}x_{b}}{x^{2}}\right) - \frac{A}{d}\delta_{ab}\delta(x)
\end{equation*}
The last step is to find the values of \(A\) and \(B\).