# The quality factor of cavity resonator

1. Dec 2, 2013

### Paul Black

Hello

i have to derive the quality factor Q

i must start with Q=2$\pi$$\frac{internal-stored-energy}{energy-loss-per-cycle}$

the quality factor should be in terms of λ so the last equation shall be Q=$\frac{4\pi L}{\lambda (1-R)}$

i searched in google but i always get the Q-factor in terms of bandwidth
i hope you can help me.
thank you very much

2. Dec 3, 2013

### f95toli

Is this homework?

If so this should be in the HW section, and you also need to show us some of your work,

I believe a good first step would be to think about what R means in the final equation....

3. Dec 3, 2013

### Paul Black

The factor ( R ) is the reflection coefficient of mirrors.

the step before the last one must be like this

Q=$\frac{4\pi L}{\lambda (1-R_{1}R_{2})}$

since one of the mirrors has a reflection coefficient of 100% and it is equal to 1 we get the last step.
i don't know what the ( internal stored energy ) and (energy loss per cycle ) refers to.
i cant write them in terms of $\lambda$
hope you can help me to solve this question.

4. Dec 3, 2013

### f95toli

It simply refers to the total energy stored in the EM field. Some of this energy is lost whenever the EM wave hits a wall and some of the energy "leaks out" (if the reflectivity is <1)

5. Dec 3, 2013

### Paul Black

thank you but i know what it means
i dont know how to write it mathematically ( in terms of R or L or lambda)
i found in google a new expression
Q=2$\pi$ $\frac{E}{\Delta E}$

but still dont know how to get the last expression

6. Dec 3, 2013

### nasu

ΔE is the energy lost per cycle.
Find how many reflections in one cycle. And how much energy (as a fraction of the initial energy) is lost at each reflection.

7. Dec 3, 2013

### Paul Black

i think there are two reflections each cycle. but i must assume that one of them ( the mirror )has a 100% reflection which is equal to one. therefore i got only R in the last equation. i still dont have any idea where to start. with which values shall i interpret the ( total internal energy E ) and the ( energy lost per cycle )?

8. Dec 3, 2013

### nasu

You don't need to find expressions for the total energy or the energy lost per cycle.
You need the fraction of energy lost for each reflection.
How do you define R? Isn't it related to this fraction of energy lost?

9. Dec 3, 2013

### Paul Black

sorry but i have no idea how to solve this question.
can you please give me a link for this theme? if you have one of course

10. Dec 3, 2013

### nasu

You mean about reflection coefficient? You don't have the meaning of it in your book or notes?

11. Dec 3, 2013

### sophiecentaur

Would the following help? It may be obvious but . . . .
It would be an idea to get to understand how the wavelength, length and the reflection coefficient come into the equation. A simple harmonic resonator has a Q factor which is governed by the resistance (loss) and you could say there is loss during each cycle of oscillation and only one cycle's worth of energy. In a cavity, there is only loss when the wave hits the ends(s) - at least to a first approximation. The cycles of the wave that are between the ends are not subject to loss. Hence, the proportion of energy lost is less per cycle. The loss is reduced by a factor corresponding to the number of wavelengths between the ends.
Or, putting it another way, with the same reflection coefficient, a wave of twice the frequency would have the same loss per cycle but twice as much energy in the cavity so the definition of Q would give double the value of Q.

12. Dec 4, 2013

### Paul Black

@nasu
i know what they mean ( reflection coefficient ) . i just wanted a pdf document where this type of resonator is discussed and maybe i can get the solution from such a document. i have searched a lot but i always get documents about electronic circuits which have resistance and Capacitor. i downloaded over 20 pdf documents but i just got the last equation or in terms of bandwidth but not how to derive it.

13. Dec 4, 2013

### Paul Black

i come to a solution but Im not sure if this is the right answer
i have attached the picture with the solution

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14. Dec 4, 2013

### f95toli

The derivation is trivial if you just combine the info you have been given above, especially the post by Sophiecentaur which tells you how lambda and L comes into it(hint: how many half-wavelenghts fit into the cavity?)

If you assume that the cavity contains an amount of energy equal to E, what is ΔE?
I.e how much energy leaks out of the cavity? Once you have an expression for ΔE you can plug that into your definition of Q above and you have your answer.

15. Dec 4, 2013

### Paul Black

sorry but i wrote it all in the attachment. delta E is is the energy loss per cycle. the energy difference which is the same as energy loss per cycle.
or maybe i didn't understand what you mean.
I'm not satisfied with my solution. I'm sure this is the wrong way to derive such an expression.

16. Dec 4, 2013

### sophiecentaur

There is just more energy in a long resonator, for the same amount of loss per cycle of oscillation. Forget the Maths derivation for a second and think of the Physics involved. Then return to the Maths.