What is retardation effect? I have in my book a passage which states:
"We also remark that the dipole approximation (obtained by setting $\exp(i \textbf{k} \cdot \textbf{r}) = 1$ in the matrix element Eq. #) yields the leading term inthe differential cross section, such that the retardation effects of order $v_f/c$ are neglected. As seen from Eq. #, these retardation effects become more important if the wave number $k$ (and hence frequency $v$) of the photon increases, as expected from our discussion of the dipole approximation in Section #"
So basically, what is a retardation effect ? In the excerpts I have from the book, it says nothing about it, and I can't find much about on Google. 
 A: when $v/c$ is no longer very small, the characteristic radiation wavelength becomes comparable to the characteristic length of the radiating system, so that the charge configuration would change significantly over the course of one radiation period. If such is the case, then you cannot use the great simplification of dropping the integration variable dependent term ($\vec{x}'\cdot\hat{x}/c$ ) in the dipole term of the definition of the retarded potentials at far away distances 
\begin{align*}
\phi(\vec{x}) &= \frac{1}{|\vec{x}|}\int d^3 x' \rho_{t - |\vec{x}|/c -\vec{x}'\cdot\hat{x}/c}  + \mathcal{O}\left(|\vec{x}|^{-2}\right)\\
\vec{A}(\vec{x}) &= \frac{1}{c|\vec{x}|}\int d^3 x' \vec{j}_{t - |\vec{x}|/c -\vec{x}'\cdot\hat{x}/c} + \mathcal{O}\left(|\vec{x}|^{-2}\right)\\
\end{align*}
Which would have otherwise lead to the very simple expression for dipole radiation intensity 
$$
dI = \frac{1}{4\pi c^3} \left( \ddot{\vec{d}}\times\hat{x} \right) ^2 d\Omega 
$$
Where instead now you get 
$$
dI = \frac{e^2}{4\pi c^3} \left( \frac{2(\hat{x}\cdot \vec{a})(\hat{x}\cdot\vec{v})}{c\left(1 - \frac{\vec{v}\cdot\hat{x}}{c}\right)^5} + \frac{\vec{a}^2}{\left(1 - \frac{\vec{v}\cdot\hat{x}}{c}\right)^4} - \frac{(1-v^2/c^2)(\hat{x}\cdot\vec{a})^2}{\left(1 - \frac{\vec{v}\cdot\hat{x}}{c}\right)^6} \right)  d\Omega 
$$
Whose non relativistic limit ($c\rightarrow\infty$) readily reproduces the former result (remember that $(\vec{a}\times\hat{x})^2 = \vec{a}^2\hat{x}^2 - (\vec{a}\cdot\hat{x})^2$. You can also expand order by order in $v/c$ 
