Electron in a microwave cavity

[thaiqi]

Hello, imagine there is a microwave cavity and the standing wave inside it oscillates, and then place one electron at the wave node point . Will the electron stay at rest? And what if there is some perturbation? Any idea? Thanks.

 

[sophiecentaur]

This depends on the electron actually 'standing still enough for an effect to be seen. It would need to be in an electrically neutral medium so the answer is not trivial (i.e. I don't know the real answer)

As a problem in Classical Physics, you have a charge in the presence of a varying electric field, I guess you could say that the electron will experience a varying force. There is also a varying magnetic field so this would modify the electron's path due to the Lorenz Force. Electrons in a metal would satisfy the need for neutrality but the displacement of an electron (which can be thought of as travelling at only millimetres per second on a metal) would mean very little motion in an alternating field of 2GHz.

This same effect is demonstrated in the effect of low and medium frequency radio waves on charged particles in the ionosphere (a very low density medium, of course). The Appleton Hartree formula (1930' research) describes this sort of effect and 'explains' the resulting refractive index. The theory is good enough to predict (+/- quite a lot) the likely path of radio wave transmissions when they are reflected over the horizon (which was a surprise to earlier workers). The ionosphere is electrically neutral - a plasma, generated by radiation from the Sun.

 

[thaiqi]

Basically I want to learn how the electron inside the cavity will exchange energy with the field.

 

[sophiecentaur]

Basically I want to learn how the electron inside the cavity will exchange energy with the field.

The potential energy is essentially due to the 'electron in the field'. There can be kinetic and potential energy and the share can vary. This is like a gravitational orbit where there is both forms of energy and the proportions can change in an elliptical orbit.

 

[sophiecentaur]

Basically I want to learn how the electron inside the cavity will exchange energy with the field.

There has to be both an electron and a field for potential energy. Assume a field already exists somewhere then putting the electron in that place (starting from an infinite distance away) will involve work and that's how much PE is in that system.
Alternatively, you can put an electron somewhere and ramp up the field from zero and the same energy is there. Sorry if that's bloomin' obvious but it needs to be said.

There is some discussion of how this relates to photons and quanta but energy quanta when doing quasi-static changes involves infinitesimally low energy quanta / photons. Both situations are the same.

 

[tech99]

For a resonant cavity, the magnetic and electric fields are each in the form of a standing wave. This is a wave that stands still. If the cavity has no heat losses, then no flow of overall energy is taking place and it is just an energy store. The two standing waves are displaced from each other so that a node of one coincides with an antinode of the other. The two waves are also separated by 90 degrees in time - they are in time quadrature. If the electron is at a zero, or node, of the electric wave, then it will not experience any force and will not move. If it is moved away from an electric node, then it will experience force from both fields, and as they are in quadrature, I believe the electron moves in an ellipse. I am looking for an example of where this happens. The ionosphere is not quite the same as we have a the Earth's steady field as well.

 

[sophiecentaur]

I believe the electron moves in an ellipse.

Agreed. The fields are in time quadrature and the H deflection is proportional to the product of velocity and H field.

The ionosphere is not quite the same as we have a the Earth's steady field as well.

A radio signal has E and H fields but they are co-phased, unlike the quadrature situation in a cavity. There are differences - like the Earth's H field is not related to the RF signal fields. My point was that free electrons follow paths according to the applied RF field.

The question in the OP is about behaviour in a node and either side of it. There would be no net field in the direction of the flow of the two travelling waves (except for a small field tilt due to losses and finite power transfer). So I'd imagine the electron would be 'fairly' stable in its position.