Interference and conservation of energy in a resonator

[JerryY]

It is known that constructive interference in one place must be compensated for by destructive interference in another. Take a simple Fabry Perot resonator for example. The interference occurring at both sides of the first mirror (assuming one incident electric field) compensate each other out, leading to the familiar transmissivity/reflectivity curves.

My question is, what about the interference between the forward and backward propagating fields in the resonator itself? If we choose the resonator length such that this interference is constructive/destructive on average, there doesn't seem to be a place where opposite interference can occur to cancel it out.

 

[Twigg]

If you have a resonantor with infinite q-factor ringing down (well, not really "down" but you get my meaning), the energy inside the resonator does not change with time. The field inside is a standing wave, which changes from constructive to destructive interference every $\lambda / 2$ displacement along the axis. Which part troubles you?

 

[JerryY]

If you have a resonantor with infinite q-factor ringing down (well, not really "down" but you get my meaning), the energy inside the resonator does not change with time. The field inside is a standing wave, which changes from constructive to destructive interference every $\lambda / 2$ displacement along the axis. Which part troubles you?

Thanks for your reply, but I'm considering a resonator with mirror reflectivity < 1, i.e. finite Q-factor. I agree that the energy inside the resonator shouldn't change with time, but again due to the non-unity reflectivity, the two counter-propagating waves do not have the same magnitude, and the total field inside is not strictly a standing wave.
'which changes from constructive to destructive interference every $\lambda / 2$ displacement along the axis'
But what if the length of the resonator is not an integer multiple of $\lambda$? Then you'll have more constructive than destructive interference or vice versa inside the resonator.

 

[Twigg]

Then you'll have more constructive than destructive interference or vice versa inside the resonator

In this case, you get near-complete destructive interference, never constructive interference. What do you get when you average many sinusoidal with a random phase? It tends to 0.

 

[JerryY]

In this case, you get near-complete destructive interference, never constructive interference.

Yes, but destructive interference by itself doesn't conserve energy though, hence my initial question.

 

[BvU]

Yes, but destructive interference by itself doesn't conserve energy though, hence my initial question.

Oh ? In what way ?

$\$

 

[JerryY]

Oh ? In what way ?

$\$

Constructive interference in one place is compensated for by destructive interference in another for energy to be conserved. If my original question I picked a non-resonant Fabry Perot resonator. This leads to, on average, partial constructive or destructive interference within the resonator. And since this energy doesn't change in time, energy is apparently not conserved.

 

[BvU]

Where do you see more energy getting out of the cavity than what goes in ?

$\$

 

[pai535]

Thanks for your reply, but I'm considering a resonator with mirror reflectivity < 1, i.e. finite Q-factor. I agree that the energy inside the resonator shouldn't change with time, but again due to the non-unity reflectivity, the two counter-propagating waves do not have the same magnitude, and the total field inside is not strictly a standing wave.
'which changes from constructive to destructive interference every $\lambda / 2$ displacement along the axis'
But what if the length of the resonator is not an integer multiple of $\lambda$? Then you'll have more constructive than destructive interference or vice versa inside the resonator.

I think in your case if you don't consider absorption, only part of the light interferes and the residual light is transmitted through the mirror. If two waves with the same frequency but different amplitudes interfere, there will be a wave after interference even if the interference is destructive.

 

[JerryY]

Where do you see more energy getting out of the cavity than what goes in ?

$\$

Nowhere. My question concerns the intensity stored inside the resonator itself. Everywhere I look people take the absolute value squared of the two counter-propagating fields and claim that as the total intensity without taking the interference of the two fields into account. Why are they allowed to do so?

 

[Twigg]

Its an awkward way of phrasing it to say that constructive interference doesn't conserve energy, but I believe I understand what you are asking. When the cavity is off resonance, the field inside destructively interferes but the reflected beams constructively interfere. (More simply, the cavity off resonance looks like a single mirror.)

 

[JerryY]

Its an awkward way of phrasing it to say that constructive interference doesn't conserve energy, but I believe I understand what you are asking. When the cavity is off resonance, the field inside destructively interferes but the reflected beams constructively interfere. (More simply, the cavity off resonance looks like a single mirror.)

Sorry, but that still isn't what I'm asking. Take this diagram of a Fabry Perot I got from wikipedia.

It is well understood that the interference between E_refl,1 and E_back compensates for the interference between E_laun and E_RT. I'm asking about the interference between E_circ and E_bcirc. If this is partially constructive/destructive, how is energy still conserved?

 

[Twigg]

Having the incident beam makes things a bit messy. Let's instead imagine a ringdown scenario, where the incident beam $E_{inc}$ is on for a long time and then abruptly turned off. When $E_{inc}$ is turned off, $E_{refl,1}$ and $E_{laun}$ also go to 0. Now $E_{circ}$ and $E_{b-circ}$ are the only fields inside the cavity. Since this isn't steady state, energy is not conserved. In this case, any constructive interference between $E_{circ}$ and $E_{b-circ}$ constitutes the stored energy of the resonator, and that energy will be dissipated by the outgoing beams $E_{back}$ and $E_{trans}$.

If we had an ideal, perfect resonator where $r_1 = r_2 = 1$, then $E_{back}$ and $E_{trans}$ would be 0 and energy would be conserved. The only non-zero fields would be $E_{circ}$ and $E_{b-circ}$. In this case, there could be constructive interference inside the resonator and NO beams outside the resonator. This really hones in on the point: constructive interference inside the resonator translates to energy stored, but not energy gained. It doesn't need to be balanced by any other beams. It's the beams that go into or out from the cavity that have to be energy-balanced.

Edit: fixed a typo. It said $r_1=r_2=0$ before.

 

[JerryY]

Thanks for the explanation. I have one more question. Physically, the absolute value squared of Ecirc and Ebcirc is proportional to the number of photons propagating in the forward and backward directions respectively. So logically speaking summing these two up gives you the number of photons inside the resonator. This however disregards the interference between the two fields, so do we still that into account when calculating the total intensity inside the resonator?

 

[Twigg]

so do we still that into account when calculating the total intensity inside the resonator?

Yes. The photon number is proportional to $|E_{circ} + E_{bcirc}|^2$, NOT $|E_{circ}|^2 + |E_{bcirc}|^2$. The former accounts for interference and is the accurate formula for the intensity of the circulating fields, the latter has no interference and is meaningless.

 

[JerryY]

Yes. The photon number is proportional to $|E_{circ} + E_{bcirc}|^2$, NOT $|E_{circ}|^2 + |E_{bcirc}|^2$. The former accounts for interference and is the accurate formula for the intensity of the circulating fields, the latter has no interference and is meaningless.

I can see how $|E_{circ}|^2$ ($|E_{bcirc}|^2$) corresponds to the photons traveling in the forward (backward) direction, but what does the interference term physically correspond to though?

 

[vanhees71]

Forget photons completely! All you need is the electromagnetic field, Maxwell's equations and the conservation laws, here particularly "Poynting's theorem".

Before you start learning relativistic QFT, which is the only way to understand what photons are, you have learn classical electromagnetics!

 

[JerryY]

Unfortunately, the course I'm currently studying deals with entirely semi-classical derivations, e.g. the Schawlow Townes linewidth is derived without considering quantum fluctuations, and I believe there is a semi-classical explanation to relate the interference with the photon number.

 

[Twigg]

If you inject one photon's worth of $E_{circ}$ into a resonator with perfectly reflecting mirrors, that photon's reflection $E_{bcirc}$ will interfere with itself and form a standing wave. If you need a reason for this, it's because there's nothing at all in the resonator that would project the photon's state in position space so it will act like a wave and interfere with itself. Same logic as the two-slit experiment when you don't detect which slit the photon passes through.

The end result of this is that the photons populate standing waves, not forward and backward propagating traveling waves. If you were to stick some kind of direction-filtering optic into the resonator (a Faraday isolator, maybe?), you would be able to project the standing wave into either a forward or backward propagating wave. But until you introduce such an optic, the photon will occupy a standing wave state.

Because of this, asking "how many photons are forward/backward propagating?" is like asking "how many electrons are in a spin up/spin down state?" when the electrons are in the superposition state $\frac{1}{\sqrt{2}} (|\uparrow \rangle + |\downarrow \rangle )$.

Does that help at all? Sorry if my explanation isn't clear, running on little sleep.

 

[sophiecentaur]

So logically speaking summing these two up gives you the number of photons inside the resonator. This however disregards the interference between the two fields, so do we still that into account when calculating the total intensity inside the resonator?

I would not be happy with including photons and waves in the same sentence and expecting to get sense out of it. Photons will not help us here.
Waves travel back and forth. At each end there will be some transmission and some reflection. The total energy in the cavity will take time to build up until a steady state condition is arrived at (Power in = power out). All things being equal, the total amount of energy in the cavity will be the same, with or without resonance. At or even near resonance, the energy will be mostly in the vicinity of the maxes. Any system will have a finite Q and Q will be determined by the ratio of the peak energy and the energy admitted (passed through) per cycle.

 

[sophiecentaur]

All things being equal,

The power admitted to the input will be affected by the Impedance presented by the cavity and that will depend (I think) on the frequency and the natural resonance curve. This is often (nearly always) ignored when discussing a transmission line resonance. The line acts as a transformer between the source impedance and the load impedance so I think that would imply the matching will be different at different frequencies / line lengths.

 

[JerryY]

I would not be happy with including photons and waves in the same sentence and expecting to get sense out of it. Photons will not help us here.

Perhaps not, but the photon number is the quantity I'm interested in, for use in further calculations. Regarding the rest of your comment, yes I've now understood how energy is conserved in such a system with the help of previous comments and yours. But the question still remains: do I take interference into account when calculating the total intensity inside a resonator?

 

[JerryY]

The end result of this is that the photons populate standing waves, not forward and backward propagating traveling waves.

I'm not sure I agree. Firstly, due to the non-unity reflectivity of the mirrors, we don't actually get a standing wave inside the cavity. Second, if photons populate standing waves then nothing would couple in and out of the cavity, and we know that photons travel with speed c₀/n where n is the refractive index. They don't just sit still.

 

[vanhees71]

Perhaps not, but the photon number is the quantity I'm interested in, for use in further calculations. Regarding the rest of your comment, yes I've now understood how energy is conserved in such a system with the help of previous comments and yours. But the question still remains: do I take interference into account when calculating the total intensity inside a resonator?

You solve the Maxwell equations with the boundary conditions for the field. For that you can first look for a complete set of eigenmodes, i.e., with time dependence $\exp(-\mathrm{i} \omega t)$. The general solution is then given by superposition of these eigenmodes. In this sense you "take interference into account", i.e., you use the superposition principle which is valid for linear differential equations as the Maxwell equations are.

 

[sophiecentaur]

You solve the Maxwell equations with the boundary conditions for the field. For that you can first look for a complete set of eigenmodes, i.e., with time dependence $\exp(-\mathrm{i} \omega t)$. The general solution is then given by superposition of these eigenmodes. In this sense you "take interference into account", i.e., you use the superposition principle which is valid for linear differential equations as the Maxwell equations are.

I'd say that is the best possible reason for not wanting to work this out 'with photons'. Perhaps the best way to bring in the photon idea is to relate the probability of a photon / charged system interaction (i.e. probability of a photon being detected) to the intensity along the tube. We have no idea (don't care) 'where the photons are', within the cavity because their extent cannot be known.
The 'photon count' will be proportional to the intensity as it varies along the cavity. The 'photon throughput' will be the output power divided by hf.
I don't think it's unfashionable to put it in the above way; you don't know where or when a photon is until it interacts with a massive particle and photon/photon interactions don't take place so the peaks and troughs can just be regarded as a wave phenomenon.

So logically speaking summing these two up gives you the number of photons inside the resonator.

Yes - that's effectively what I wrote above. Interference affects the probability at any point in the resonator and we have no idea what those photons are actually doing or where they are.
It must be possible to write all that in terms of photons and probabilities but, as has already been mentioned, you should get totally familiar with EM waves first and then (tadaaa!) you will get the same answers for this sort of situation when you manage to do it all with photons.

 

[sophiecentaur]

Firstly, due to the non-unity reflectivity of the mirrors, we don't actually get a standing wave inside the cavity.

I just read this again. You are implying a standing wave involves no power in or out of a cavity. Not true (for the normal definition of a standing wave). When there are any losses (internal or leakage out) the standing wave will have a 'standing wave ratio' (look it up) where there is incomplete cancellation (filled up nulls) and incomplete addition (failure to reach full peak amplitude). For total reflection / no losses, the SWR will be infinite.

 

[vanhees71]

The calculation for "photons", i.e., the quantized electromagnetic field is exactly the same. The only difference is that the fields are operator valued and the coefficients in the mode expansion are annihilation and creation operators for the corresponding mode.

 

[tech99]

For a cavity resonator, the mystery of where does the destructive interference energy go is easily explained by comparison with an LC circuit. The energy is alternately stored in magnetic and electric fields. These fields are always at 90 degrees phase and are in different locations. So when cancellation happens with one, the energy is hiding in the other. If we use a cavity that is not resonant, then in the same way as an LC circuit, it presents a reactance to the outside world, so that it will store energy for half a cycle and give it back the next. If we use a resonant cavity which has losses, or maybe a load on the far end, then the generator, instead of seeing zero or infinite resistance, will see a loss resistance, and once having filled the resonant store with energy, will supply the loss resistance with its entire output by sending a traveling wave to it.

 

[sophiecentaur]

For a cavity resonator, the mystery of where does the destructive interference energy go is easily explained by comparison with an LC circuit.

Something that tends to be ignored when trying for a qualitative explanation of this stuff is the net energy flow in or out. In any resonator there is always loss involved so there will never be perfect nulls, either whilst resonance is building up or in the equilibrium state. The only 'relevant' energy flow is (I would say) from the source to the load and the SWR can never be infinite.