Magnetic and Electric energy for AC fields In "Applied Frequency Domain Electromagnetics" (here the page) there are these two equations for the computation of the magnetic and electric energies stored in a certain volume $V_0$:

*

*$$W_{e}= \frac{1}{4} \cdot \int_{V_0} D^* \cdot E \,\,\,dV$$

*$$W_{h}= \frac{1}{4} \cdot \int_{V_0} B^* \cdot H \,\,\,dV$$
Where E,D,H,B are phasors.
I do not understand:

*

*why there is "1/4" instead of "1/2";

*the meaning of "stored energy" in case of purely sinusoidal fields. I think that stored energy is the integral between 0 and infinite of instant power as function of time, but I do not know what the result represents in this case: should it be 0 because of zero mean value of fields? Or infinite because of their infinite duration in time?

 A: As I mentioned, looks like the author averaged over the oscillation cycle. This is useful when the instantaneous energy is not important. The form of the cycle-average he uses, $E^* E/2$, appears as a 'trick' when dealing with complex oscillating fields. Since the energy is quadratic in the fields (and thus nonlinear) one should use:
$$
u = \frac 1 2 \langle\text{Re}(D)\cdot\text{Re}(E)\rangle
$$
for the cycle averaged energy density because only the real part is physically measurable. When both fields evolve with the time factor $e^{-i\omega t}$, the cycle average is easy to calculate:
$$
\langle\text{Re}(D)\cdot\text{Re}(E)\rangle = \frac 1 T \int_T \frac 1 4 (DE + D^* E^* + D^* E + DE^*) \, dt
$$
The terms $DE$ and $(DE)^*$ oscillate with twice the frequency while $D^* E$ and $DE^*$ don't oscillate at all, thus only the last two terms survive the cycle average. The result is
$$
\langle\text{Re}(D)\cdot\text{Re}(E)\rangle = \frac{ D^* E + DE^*}{4} = \frac 1 2 \text{Re}(D^* E)
$$
and that's where the extra $1/2$ comes from.
This stored energy is the average energy the fields carry around time $t$. If the fields are purely oscillatory, the average is the same for all times. But if there's absorption or energy pumping into the fields the cycle-average increases over time. Integrating the instantaneous power will give you the instantaneous net energy in your interval. For instance, integrating from $t = 0$ to $\infty$ may give you zero net energy because on average the energy remains in the field (the same amount is traded back and forth, so the average change is zero).
