  The Kurth solution is a particular non-isotropic steady state solution to the
gravitational Vlasov-Poisson system. It has the property that by means of a
suitable time-dependent transformation it can be turned into a family of
time-dependent solutions. Therefore, for a general steady state $Q(x,
v)=\tilde{Q}(e_Q, \beta)$, depending upon the particle energy $e_Q$ and
$\beta=\ell^2=|x\wedge v|^2$, the question arises if solutions $f$ could be
generated that are of the form \[ f(t)=\tilde{Q}\Big(e_Q(R(t), P(t), B(t)),
B(t)\Big) \] for suitable functions $R$, $P$ and $B$, all depending on $(t, r,
p_r, \beta)$ for $r=|x|$ and $p_r=\frac{x\cdot v}{|x|}$. We are going to show
that, under some mild assumptions, basically if $R$ and $P$ are independent of
$\beta$, and if $B=\beta$ is constant, then $Q$ already has to be the Kurth
solution.
  This paper is dedicated to the memory of Professor Robert Glassey.
