# B Frequency in refraction

1. Oct 4, 2016

### Biker

I would like to know why the frequency of a wave doesn't change after a refraction.

I have seen some arguments about Energy conservation that in order to conserve energy "Assuming no reflection" the same amount of waves must be coming out in the same time.

But That raises a question, Why cant it just change frequency and it change the amplitude as for example lower the frequency, increase the amplitude to have the same "ratio of transfering energy"

Do I have to take it as an observed fact?

Still new to the waves theory, I would just like to learn it to its grounds
Note: Highschool physics

2. Oct 4, 2016

### A.T.

I think the assumption is that wave-fronts aren't created or made disappear. Which is a stronger assumption than Energy conservation.

I guess this would require some means to buffer energy temporally, which is assumed not to be happening here.

3. Oct 4, 2016

### Biker

Sorry not getting what you are implying here, Could you explain a bit more please?

4. Oct 4, 2016

### Staff: Mentor

The energy of a photon is proportional to its frequency. There is no independent amplitude. Therefore, to conserve energy with photons means conserving the frequency of each photon.

5. Oct 4, 2016

### Biker

Does this apply to all mechanical waves? because I am specifically asking for those.
In my book, It says energy is proportional to the square of amplitude. It doesn't depend on frequency.

6. Oct 4, 2016

### Staff: Mentor

Ay, sorry. I thought your original question was about refraction of light.

7. Oct 4, 2016

### Biker

It is fine :).
All I found on the internet was about light.

Just wondering if it is has an explanation or just an observed fact for mechanical waves
Hopefully someone will clear this out.

8. Oct 4, 2016

### DarkBabylon

Then the book is not about quantum mechanics or the introduction to it. As far as Electromagnetic waves are concerned, it is indeed the case that the total SUM of all photons detected is the energy of the wave, which would be simpler to just detect the electric or magnetic wave and deal with it from there, no need to sum 'gazillions' of particles individually if to use a technical term, you already do that by measuring the fields.

As for why in a reflection the frequency does not change, we must also remember that their speed doesn't change as it depends on universal constants, thus to satisfy c=λf, it must mean that the wavelength also does not change. From here we can work our way forward using Einstein's equation for Energy and momentum relationship in Spacial relativity. We also need to use the finding from Black Body Radiation that the energy is coming in packets, essentially there is a minimum amount of electric field fluctuation which is associated with a photon, as you must introduce in the equations that light is both a wave and a particle.
Math warning:
Using the following equations:
E2=(mc2)2+(pc)2 , m-mass, E-energy, p-momentum, c-speed of light
E=hf, for photons (and remember, photons have momentum), h-planck's constant, f-frequency of the photon
c=λf, λ-wavelength of the photon
And conservation of momentum you will be able to arrive at the situation that should a free electron be hit by a photon, not only does the electron get deflected, a photon of a different wavelength must be emitted, and thus satisfying elastic collisions, there really is no other way to get rid of the excess energy other than heat (which is emitted by photons).
You will also get this relationship with some hard work and probably another equation I may have left out:
λ'-λ=h[1-cos(θ)]/(mec)
where λ' is the new wavelength, me is the mass of the electron, and θ is the angle at which the photon is deflected.
Without a re-emission you'd get only one term to change without any other term to change thus meaning the speed of light, planck constant or the mass of the electron change at classical limits, and using Einstein's equation we arrive at mc=0.
So electrons should move and photons change wavelength depending on angle of re-emission, right?

Except in material we don't find free electrons unless it is a conductor(which is an interesting topic by itself but not necessary for us here). Thing is, the electrons according to the Bohr Model are bound, and can change their kinetic energy (and the picture now gets ridiculous as now not every photon can be received), so the electron can now 'eat' the photon entirely and get more energy, now no photon must be re-emitted in order to conserve both momentum and energy. Except electrons will lose energy spontaneously and emit new photons which sum up to the energy of the original one, many times the same frequency as well, and also most of them will adhere to the law of reflection on the macro scale, because nature works like that, the minimum.

TL;DR
Assuming the reflector remains stationary, which for electrons bound in an atom is true, and walls of material for mechanical waves (slight fib, but won't go deeper), the reflection will adhere to the angle of reflection is the same as the angle of attack.

NOTE: There might be a problem with my second paragraph as it seems like I am missing 1 or 2 equations to get to the final equation of free electron hit by a photon.

Last edited: Oct 4, 2016
9. Oct 4, 2016

### Biker

I really appreciate the contribute and anyone who is asking about why light has the same frequency will definitely get benefits from this.

I was talking about mechanical waves, You just went through a lot of concept ( relativity, photons, electrons) which I know basic concepts about only.

Anyway, I might have found an explanation for it. It is just that a mechanical waves is made by the molecules of the medium is dragging up and down each other (transverse wave) so when it comes to the medium, the last particle of the first medium is attached to the particle in the medium so as that first particle moves the 2nd does and so on. This might be a good explanation for why the frequency is the same.

10. Oct 4, 2016

### DarkBabylon

Ah I am sorry then, my mistake. But it is a similar principle, all waves have momentum, so do objects, momentum conservation and energy conservation will do the trick for you, plus the hint that the object doesn't move.

11. Oct 4, 2016

### Drakkith

Staff Emeritus
That seems like a good explanation to me. I was just going to say something like that.

12. Oct 5, 2016

### sophiecentaur

I was just going to write something like that.
Dealing with the particles on either side of the interface ( wave travelling from left to right), the force acting on the 'right hand particle will be due to the position of the left hand particle. The right hand particle has to follow that force and stay in phase. It cannot 'do its own thing'; there has to be continuity across the boundary.
The reason for the confusion about this is to do with Quantum Mechanics, the energy of photons and the speed of light. It is 'well known' that the wavelength of light changes as it passes between two different media. People wrongly take that to mean that the frequency changes because the wave speed is fλ. But the wave speed is only c in a vacuum; the wavelength changes within any medium.
Not a problem if you think about this in terms of classical waves. The energy of the photons is the same wherever you detect them and that can give some confusion if you associate the slowing up at an interface with reducing the flux of energy. Fact is that there will be a transition time, when the light is turned on, giving a delay in the appearance of the first photons to emerge from the other side of the glass. Thereafter, the rate of photons passing is the same before and after the glass (and through it). Take a motorway with a slow section. Once things have settled down, the same number of vehicles leave the end per hour as the number entering the other end.

Last edited: Oct 5, 2016