c + a + b)},$$ where $f(t)$ and $g(\psi)$ are arbitrary functions of their arguments. The effective energy density $\rho^{(eff)} \equiv T_{0}^{0}$ and pressure $p^{(eff)} \equiv - T_{1}^{1} = - T_{2}^{2} = - T_{3}^{3}$ are given by $$\label{density and pressure, perfect fluid}
\rho^{(eff)} = p^{(eff)}, \;\;\;8\pi G \rho^{(eff)} = \frac{c a^2}{(a + b)^2 F^{2a^2/(a + b)}}.$$

Ultra-relativistic matter and radiation
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It is well-known that in the case of radiation as well as for ultra-relativistic matter (i.e., particles with finite rest mass moving close to the speed of light) the trace of the EMT vanishes identically. From (\[EMT\]) we find that $T = T_{0}^{0} + T_{1}^{1} + T_{2}^{2} + T_{3}^{3} = 0$ requires $$\label{ultra-relativistic matter}
(a + b)\left(\frac{\ddot{F}}{\dot{F}} - \frac{\ddot{V}}{\dot{V}}\right) + 2 c \left(\frac{\dot{F}}{F}\right) = 0.$$ This equation is the analogue of (\[perfect fluid case\]). Following the same procedure as above we find $F = \left[f(t) + g(\psi)\right]^{(a + b)/(2 c + a + b)}$. Therefore, the solution for radiation-like matter resembles that of perfect fluid in the sense that the effective stresses $p_{x}^{(eff)} \equiv - T_{1}^{1}$, $p_{y}^{(eff)} \equiv - T_{2}^{2}$, $p_{z}^{(eff)} \equiv - T_{3}^{3}$ are proportional to the energy density, viz., $$\label{effective stresses, radiation-like solution}
p_{x}^{(eff)} = n_{x}\rho^{(eff)}, \;\;\;p_{y}^{(eff)} 