Is the energy flow due to a an alternating current constant or cyclic
In the time domain, it is certainly not constant, it is cyclic and, if the load is reactive, alternating.
Consider the case of an AC current source driving a resistor. The power delivered to the resistor is:
$$p_R(t) = i^2(t)R = I^2_{max}\cos^2(\omega t) R = \dfrac{I^2_{max}R}{2}[1 + \cos(2\omega t)]$$
So, for a purely resistive load, the power cycles between
$$0 \leq p(t) \leq I^2_{max}R $$
Now, consider replacing the resistor with with a purely reactive load, e.g., an inductor. Then:
$$p_L(t) = v_L(t) i(t) = L \dfrac{di(t)}{dt}i(t) = -\omega L \sin(\omega t) cos (\omega t) = -\dfrac{\omega L I^2_{max}}{2}\sin(2 \omega t) $$
Note that the power associated with the inductor alternates between positive and negative, i.e., the inductor alternately absorbs and delivers power.
For a purely reactive load, energy "sloshes" back and forth between the source and load.
For a complex load, there is a combination of the above; a non-zero net power delivered to the resistive part and an alternating component associated with the reactive part.
The above can analyzed in the phasor domain too but, I think, it is especially transparent in the time domain.
To answer the question, it's called energy harvesting because in most systems the rate of collection (input power) is very different from the rate of use (output power).
Most such systems run intermittently, so a key parameter is how much energy the the application consumes each time it runs (output power times run time). This then determines how often the application can run, because it takes a certain amount of time to collect at least that much energy (input power times collection time) in the energy storage device (capacitor or battery).
Best Answer
We can divide the losses in FET logic (all processors are made in FET logic) in to categories:
The reason is simple: Because they use FETs, the transistors don't need any current to flow through their gate to control the output. Therefore, the transistors use no current at all – aside from leakage.
When switching, however, the charge in the gate capacitor of a FET has to be changed – which means a current needs to flow. Since resistances are non-zero, with P=V·I and Ohm's law, it follows that P = I²·R.
To switch faster, i.e. to have a higher clock frequency, you need to have a higher current flowing in (simple: a gate capacitor exposed to a higher voltage charges faster, like every other capacitor; current is amount of charge per time) or out the gate capacitors every clock cycle. Therefore, I is (at least!) proportional to f, I = µ·f.
Therefore, P = I²·R = µ²·f²·R.
µ and R are material/structural constants of your semiconductor technology (this is a bit simplifying, but it doesn't really matter whether losses are purely ohmic or also have higher potences of the voltages involved here).
Therefore, P is proportional to the square of frequency, at the very least.
That's why the Pentium IV generation of processors, designed to clock incredibly fast, was extremely power-hungry.