(\alpha + \beta)T_{2}^{2} -  (\alpha + \gamma)T_{3}^{3} &=&  (\gamma - \beta)T_{0}^{0}.\end{aligned}$$

Let us notice some particular cases: (i) If two of the parameters are equal to each other (axial symmetry), say $\alpha = \beta$, then $T_{1}^{1} =  T_{2}^{2}$; (ii) If $T_{1}^{1} =  T_{2}^{2}$ but $\alpha \neq \beta$, then $T_{1}^{1} =  T_{2}^{2} = T_{3}^{3} = - T_{0}^{0}$; (iii) If $\alpha = - \beta$, then $T_{3}^{3} = -  T_{0}^{0}$; (iv) In the case of isotropic expansion $(\alpha = \beta = \gamma)$ then $T_{1}^{1} =  T_{2}^{2} = T_{3}^{3}$ (but not necessarily $T_{1}^{1} =  T_{2}^{2} = T_{3}^{3} = - T_{0}^{0}$).

Perfect Fluid
-------------

Let us consider the case where the effective EMT behaves like a perfect fluid. From (\[EMT\]) we find that $T_{1}^{1} =  T_{2}^{2} = T_{3}^{3}$ requires $$\label{perfect fluid case}
(a + b)\left(\frac{\ddot{F}}{\dot{F}} - \frac{\ddot{V}}{\dot{V}}\right) + 4 c \left(\frac{\dot{F}}{F}\right) = 0,$$ which implies $\dot{V} \propto \dot{F} F^{4 c/(a + b)}$. Substituting this into (\[diagonal 5D metric\]) and using $(\partial \dot{V}/\partial \psi) = (\partial {V'}/\partial t)$ we find that $F$ must satisfy the equation[^6] $$\label{equation for F, perfect fluid}
(a + b)F \dot{F}' + 4 c \dot{F}F' = 0,$$ from which we get $$\label{F for perfect fluid}
F = \left[f(t) + g(\psi)\right]^{(a + b)/(4 