Conformal Killing fields on Schwarzschild

Question

I am trying to understand which are the conformal Killing Fields on the Schwarzschild spacetime. I say that $X$ is a conformal Killing field on $S$ ($S$ is Schwarzschild) if there exists a function $f: S \to \mathbb{R}$ such that \begin{equation} \mathcal{L}_X g = fg, \end{equation} where $g$ is the Schwarzschild metric, and $\mathcal{L}$ is the Lie derivative.

I know that the time translation, $\partial_t$, and the rotations, $\Omega_{ij}$ are Killing fields, therefore conformal Killing fields, with $f$ constant and equal to $0$.

In the Minkowski spacetime, for example, parametrized by $(x^0, x^1, x^2, x^3)$, I know that the ``dilation'' field, i.e. the field $$ \sum_{i=0}^3 x^\lambda \partial_{x^\lambda} $$ is a conformal Killing field, with $f=2$. I would like to understand if there is an analogous in Schwarzschild.

Answer 1

In the Schwarzschild geometry, the Schwarzschild radius breaks naive dilation symmetry. In the simple case of a radial dilation $r \to \lambda r$, the geometry is only preserved by $R_S \to \lambda R_S$. So, it naively seems like it would be difficult to find a working dilation, even just a radial dilation.

I went to some effort (as an exercise for myself) to find a working dilation, and failed. What I have found is that the vector field

$$ X = t \partial_t + r \sqrt{1-\frac{R_S}{r}} \, \partial_r $$

which approaches $0$ as $r \to R_S$ and $r \partial_r$ as $r \to \infty$ is almost a conformal Killing field. However, the lie derivative of the metric is

$$ \mathcal{L}_X g = 2 \sqrt{1-\frac{R_S}{r}} \left( \begin{array}{cccc} -\sqrt{1-\frac{R_S}{r}}- \frac{R_S}{2r} & & & \\ & \left( 1 - \frac{R_S}{r} \right)^{-1} & & \\ & & r^2 & \\ & & & r^2 \sin^2 \theta \end{array} \right) $$

So, the tt component of the metric spoils everything. I spend a small amount of effort trying to modify this vector field (adding it a timelike component, adding in explicit time dependence, etc.), but so far nothing has worked.