In order to use the advantages of rescaling, I want to show that a rescaled solution of the Vlasov equation is again a solution. The setting is a gravitational Vlasov-Poisson system, which is a system of partial differential equations, meaning that $f$ is a distribution function depending on $x, v \in \mathbb{R}^3$, $t \in \mathbb{R}_{\geq 0}$ and we have the potential given by
$$U_f = \frac{1}{4 \pi} \iint \frac{f(x,v,t)}{|x - y|} dv dy.$$
To show: If $f(x,v,t)$ solves the Vlasov equation, i.e. $$\partial_t f + v \cdot \nabla_x f - \nabla_x U \cdot \nabla_v f = 0,$$, then for every $\alpha, \beta \in \mathbb{R}$, $\frac{\beta}{\alpha} f \left( \frac{x}{\alpha}, \beta v, \frac{t}{\alpha \beta} \right)$ is also a solution to the Vlasov equation.
My attempt: Let $g = \frac{\beta}{\alpha} f \left( \frac{x}{\alpha}, \beta v, \frac{t}{\alpha \beta} \right)$. Then
$$U_g = \frac{\alpha^4}{\beta^2} U_f$$
and
$$\partial_t g + v \cdot \nabla_x g - \nabla_x U_g \cdot \nabla_v g = \frac{1}{\alpha \beta} \frac{\beta}{\alpha^2} Df(y, w, s) + \frac{1}{\alpha} \frac{\beta}{\alpha^2} v \cdot \nabla_x f(y, w, s) - \frac{\alpha^4}{\beta^2} \nabla_x U_{f(y, w, s)} \cdot \beta \nabla_v f (y, w, s)$$ $$= \frac{1}{\alpha^3} Df(y, w, s) + \frac{1}{\alpha^3} w \cdot \nabla_x f(y, w, s) - \frac{\alpha^4}{\beta} \nabla_x U_{f(y, w, s)} \cdot \nabla_v f (y, w, s),$$
where $y = \frac{x}{\alpha}, w =\beta v, s = \frac{t}{\alpha \beta}.$
Do you spot my mistake or could you help me, doing it correctly?
Thank you in advance!