r(a)=\frac{c}{H_{0}}\int _{a}^{1}\frac{da^{\prime }}{\sqrt{\Omega _{m,0}a^{\prime }+\Omega _{\Lambda ,0}a^{\prime 4}}}$$ $$d_{A}(a)=a\cdot \frac{c}{H_{0}}\int _{a}^{1}\frac{da^{\prime }}{\sqrt{\Omega _{m,0}a^{\prime }+\Omega _{\Lambda ,0}a^{\prime 4}}}$$ $${d}_{L}(a)\mathbf{=}\frac{\mathbf{c}}{\mathbf{aH}_{\mathbf{0}}}\int _{\mathbf{a}}^{\mathbf{1}}\frac{\mathbf{da}^{\prime }}{\sqrt{\mathbf{\Omega }_{\mathbf{m,0}}\mathbf{a}^{\prime }\mathbf{+\Omega }_{\mathbf{\Lambda ,0}}\mathbf{a}^{\prime \mathbf{4}}}}$$ ## 4. Angular Diameter Distance in Terms of Redshift Substituting the co-moving distance: $$d_A(z) = \frac{c}{H_0(1+z)} \int_0^z \frac{dz'}{\sqrt{\Omega_{m,0}(1+z')^3 + \Omega_{r,0}(1+z')^4 + \Omega_{\Lambda,0}}}$$ Make sure to plot the logarithm of the distances in units of h^{−1} Mpc versus z, and plot the age on a linear scale in units of h^{−1} Gyr for redshifts in the interval z ∈ [0, 10] ## 5. Angular Diameter Distance in Terms of Scale Factor Using the relationship $1+z = 1/a$ and $dz = -da/a^2$, we can transform to scale factor space: $$d_A(a) = \frac{c}{H_0} \cdot a \int_a^1 \frac{da'}{a'^2 \sqrt{\Omega_{m,0} a'^{-3} + \Omega_{r,0} a'^{-4} + \Omega_{\Lambda,0}}}$$ Simplifying: $$d_A(a) = \frac{c}{H_0} \cdot a \int_a^1 \frac{da'}{\sqrt{\Omega_{m,0} a' + \Omega_{r,0} + \Omega_{\Lambda,0} a'^4}}$$