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?
  • $\begingroup$ Usually if the energy density is something like $E^* \cdot E/4$ means there was an average over the oscillation cycles. Does a cycle-averaged energy make sense in your context? $\endgroup$
    – ErickShock
    Jun 30, 2020 at 15:31
  • $\begingroup$ Yes, it may have been evaluated on a oscillation period. $\endgroup$
    – Kinka-Byo
    Jun 30, 2020 at 15:35

1 Answer 1


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).

  • $\begingroup$ Hello, I have another question which is related to this topic. We have seen that in time domain the average electrical energy density is u=1/2 <D,E>, while in frequency domain it is u=1/4 <D*,E>. Is this true also for the energy stored in a capacitor? I have always seen that u = 1/2 CV^2? Does it become u = 1/4 C V^2 (with V = peak voltage value)? $\endgroup$
    – Kinka-Byo
    Jul 6, 2020 at 16:24
  • 1
    $\begingroup$ @Kinka-Byo yes, some formulae also change. That's why some prefer to use the rms value of voltage/current. For the case of sinusoidal AC lines, the peak and rms values are related by $V_\text{rms} = V_\text{peak}/\sqrt 2$, and thus the formulae remains the same as in the static case if you agree to use rms values instead $\endgroup$
    – ErickShock
    Jul 7, 2020 at 15:52
  • $\begingroup$ perfect! Thank you very much $\endgroup$
    – Kinka-Byo
    Jul 7, 2020 at 15:58

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