= Solution
For nonrelativistic particles the <energy density> is dominated by rest energy, while their isotropic kinetic <pressure> is approximately $p_m=n m\langle v^2\rangle/3$. Therefore
$$
w_m=\frac{p_m}{\rho_mc^2}
\simeq\frac{\langle v^2\rangle}{3c^2}\ll1,
\qquad
\boxed{w_m\simeq0.}
$$
The matter <pressure> is negligible in this approximation, including in the gravitational combination $\rho_m+3p_m/c^2$. For isotropic <radiation in cosmology>, $E=pc$ implies one third of the <energy density> in each diagonal spatial stress:
$$
p_r=\frac13\rho_rc^2,\qquad \boxed{w_r=\frac13.}
$$
Because the two fluids do not exchange energy or momentum, they are <separately conserved cosmological fluids>: $\nabla_\alpha T_i^{\alpha\beta}=0$ for each, rather than merely conservation of their sum. In a comoving volume proportional to $a^3$, the <first law of thermodynamics> gives
$$
d(\rho_i c^2a^3)=-p_i\,d(a^3).
$$
Equivalently this follows by projecting the individual <stress-energy conservation> equation along the common comoving <four-velocity>. Differentiating and dividing by $\rho_i c^2a^3$ yields
$$
\boxed{
\frac{\dot\rho_i}{\rho_i}
+3(1+w_i)\frac{\dot a}{a}=0.
}
$$
For constant <equation-of-state parameter> $w_i$,
$$
\rho_i\propto a^{-3(1+w_i)},\qquad
\boxed{\rho_m\propto a^{-3},\quad \rho_r\propto a^{-4}.}
$$
The extra power for <radiation in cosmology> represents the <cosmological redshift> of photon energies in addition to number dilution. Differentiating the total <Friedmann equation> gives only the total continuity equation; the separate equations also use the stated absence of interactions between components.
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