Editing Multipactor effect (section)

==Frequency-gap product in two-surface multipactor==
The conditions under which multipactor will occur in two surface multipactor can be described by a quantity called the frequency-gap product. Consider a two surface setup with the following definitions:
:<math> d </math>, distance or gap between the surfaces 
:<math> \omega </math>, angular frequency of the RF field
:<math> V_0 </math>, peak plate-to-plate RF voltage
:<math> E_0 </math>, peak electric field between the surfaces, equal to <math>V_0</math>/<math>d</math>.

The RF voltage varies sinusoidally. Consider the time at which the voltage at electrode A passes through 0 and starts to become negative. Assuming that there is at least 1 free electron near A, that electron will begin to accelerate to the right toward electrode B. It will continue to accelerate and reach a maximum velocity half a cycle later, just as the voltage at electrode B begins to become negative. If the electron(s) from electrode A strike electrode B at this time and produce additional free electrons, these new free electrons will begin to accelerate toward electrode A. The process may then repeat causing multipactor. We now find the relationship between the plate spacing, RF frequency, and RF voltage that causes the strongest multipactor resonance.

Consider a point in time at which electrons have just collided with electrode A at position -d/2. The electric field is at zero and is beginning to point to the left so that the newly freed electrons are accelerated toward the right. Newton's equation of motion of the free electrons is
:<math> a(t)=\frac{F(t)}{m} </math>
:<math> \ddot{x}(t) = \frac{qE_0}{m}~\sin(\omega t) </math>

The solution to this differential equation is
:<math> x(t) = -\frac{qE_0}{m \omega^2}\sin(\omega t) + \frac{qE_0}{m \omega}t - \frac{d}{2} </math>
where it is assumed that when the electrons initially leave the electrode they have zero velocity. We know that resonance happens if the electrons arrive at the rightmost electrode after one half of the period of the RF field, <math> t_{\frac{1}{2}}=\frac{\pi}{\omega} </math>. Plugging this into our solution for <math>x(t)</math> we get
:<math> x(t_{\frac{1}{2}}) = -\frac{qE_0}{m \omega^2}\sin(\omega t_{\frac{1}{2}}) + \frac{qE_0}{m \omega} t_{\frac{1}{2}} - \frac{d}{2} </math>
:<math> \frac{d}{2} = -\frac{qE_0}{m \omega^2}\sin(\omega \frac{\pi}{\omega}) + \frac{qE_0}{m \omega} \frac{\pi}{\omega} - \frac{d}{2} </math>

Rearranging and using the frequency <math>f</math> instead of the angular frequency gives
:<math> fd = \frac{1}{2\sqrt\pi}\sqrt \frac{qV_0}{m}</math>.

The product <math> fd </math> is called the frequency-gap product. Keep in mind that this equation is a criterion for greatest amount of resonance, but multipactor can still occur when this equation is not satisfied.