Define $p_X(x)\equiv p(x_1,x_2,x_3,x_4,X_1,X_2,X_3,X_4)$.
Let $x_1=t,x_2=r,x_3=\theta,x_4=\phi,X_1=T,X_2=R,X_3=\Theta,X_4=\Phi$.
Define $\int d^4x\equiv r^2\sin\phi \int_{-\infty}^{\infty}dt\int_{0}^{\infty}dr\int_{0}^{2\pi}d\theta\int_{0}^{\pi}d\phi$.
Find $p_X(x)$ that satisfies all the following constraints: \begin{equation} \int d^4x\frac{1}{p_X(x)}\left(\frac{\partial p_X(x)}{\partial T}\right)^2=-\left(1-\frac{1}{R}\right) \end{equation} \begin{equation} \int d^4x\frac{1}{p_X(x)}\left(\frac{\partial p_X(x)}{\partial R}\right)^2=\left(1-\frac{1}{R}\right)^{-1} \end{equation} \begin{equation} \int d^4x\frac{1}{p_X(x)}\left(\frac{1}{R}\frac{\partial p_X(x)}{\partial \Theta}\right)^2=R^2 \end{equation} \begin{equation} \int d^4x\frac{1}{p_X(x)}\left(\frac{1}{R\sin(\Theta)}\frac{\partial p_X(x)}{\partial \Phi}\right)^2=R^2\sin^2\Theta \end{equation} \begin{equation} \int d^4x\frac{1}{p_X(x)}\frac{\partial p_X(x)}{\partial X_\mu}\frac{\partial p_X(x)}{\partial X_\nu}=0\quad\quad\text{if }\mu\neq\nu \end{equation}
Aside:
For the cartesian case ($\int d^4x\equiv \int_{-\infty}^{\infty} dx_1dx_2dx_3dx_4$),
\begin{equation} \int d^4x\frac{1}{p_X(x)}\frac{\partial p_X(x)}{\partial X_\mu}\frac{\partial p_X(x)}{\partial X_\nu}=1\quad\quad\text{if }\mu=\nu \end{equation} \begin{equation} \int d^4x\frac{1}{p_X(x)}\frac{\partial p_X(x)}{\partial X_\mu}\frac{\partial p_X(x)}{\partial X_\nu}=0\quad\quad\text{if }\mu\neq\nu \end{equation} a known solution is, \begin{equation} p_X(x)=(2\pi)^{-3/2}e^{-\frac12|\boldsymbol{x}-\boldsymbol{X}|^2} \end{equation}
Any help, including partial solutions, suggestions for possible lines of attack, helpful theorems, would be appreciated :)