[CrazyScientist (OP)]

You have an "interesting" definition of a perfect mirror; it absorbs energy from the light it reflects.
Would you like to try again with a better definition?

I just pasted here, what other people anwsered to a similar problem.
But on the other hand, in my previous posts I didn't include the fact, that even if the light is in 100% reflected from a perfect mirror it still DOES induce mechanical pressure on the reflective surface. In fact, the more energy is being reflected, the stronger is the pressure of a reflected EM wave - energy is still being transferred into the reflective surface, despite the complete lack of energy absorption. So, in the end the entire system will any way reach a constant energetic eequilibrium after some time...

[Bored chemist]

In fact, the more energy is being reflected, the stronger is the pressure of a reflected EM wave - energy is still being transferred into the reflective surface, despite the complete lack of energy absorption.

Not quite.
The force which the photons exert does not necessarily mean there is  transfer of energy. Energy is transferred when the force moves through a distance
In order to be perfectly reflective the walls have to be infinitely massive (this causes other problems).

However, there's another way to do it.
You can imagine a nearly massless mirror.
When a photon hits it, it will move and take some energy from the photon. But that means that, when another photon hits it on the other side, it will add energy to that photon.
Overall, the sum of the energies will be conserved The wavelengths of the photons will be "scrambled" and will settle down to a black-body distribution.
The energy (on average) imparted to the light, rigid mirror will be Boltzmann's constant times 3 times the temperature. (That's the same energy as would be carried by an electron or proton at that temperature.)
As more photons are added, the temperature will rise. If you do that slowly, you can easily calculate the energy of the mirror.

However, in the end, one of two things will still happen; the sphere will break or it will collapse into a BH.

[CrazyScientist (OP)]

In fact, the more energy is being reflected, the stronger is the pressure of a reflected EM wave - energy is still being transferred into the reflective surface, despite the complete lack of energy absorption.

Not quite.
The force which the photons exert does not necessarily mean there is  transfer of energy. Energy is transferred when the force moves through a distance
In order to be perfectly reflective the walls have to be infinitely massive (this causes other problems).

But as long as the pressure is being applied to the reflective surface, there is some transfer of energy between photons and the mirror. Question is, if this process is sufficient enough, to maintain a constant energy level in the system...

However, there's another way to do it.
You can imagine54 a nearly massless mirror.
When a photon hits it, it will move and take some energy from the photon. But that means that, when another photon hits it on the other side, it will add energy to that photon.
Overall, the sum of the energies will be conserved The wavelengths of the photons will be "scrambled" and will settle down to a black-body distribution.
The energy (on average) imparted to the light, rigid mirror will be Boltzmann's constant times 3 times the temperature. (That's the same energy as would be carried by an electron or proton at that temperature.)
As more photons are added, the temperature will rise. If you do that slowly, you can easily calculate the energy of the mirror.

Generally, I have now a problem with the pressure induced by light on the reflective surface and I have now to make a deeper research of this subject, to get some answers.

My question is, if EM waves reflected from a perfect mirror keep their original wavelenght?

If so, then after traveling to the other side of the sphere, they will induce the same amount of pressure during the second reflection (and all other reflections, which will then follow). This would mean, that the kinetic pressure of light is a 100% free energy, that is being added to the system from nowhere.

The only possible solution, I can think of, is that frequency of photons get lower with every reflection, despite the lack of energy absorption...

And if not, then I really don't know the solution - maybe you can help?

However, in the end, one of two things will still happen; the sphere will break or it will collapse into a BH

I'm sure, that if it would be the case, someone would already use this process, to create some kind of weapon of mass destruction, or just to make a cheap source of heat - a real-life mirror can reflect up to 99% of energy, so it would be enough,  to create a BH using an EM radiation with high enough frequency.

Besides you seem to ignore the fact, that photon-photon interactions were achieved using the laws of QM and it is now experimentally proven, that photons don't behave in the way, which is predicted by GR. So, it appears that light collabsing into BH is nothing more, than a completely unproven theoretical fantasy...

[Bored chemist]

But as long as the pressure is being applied to the reflective surface, there is some transfer of energy between photons and the mirror.

Only if it moves.

And, if it moves then, in due course, it transfers energy back to other photons.

I'm sure, that if it would be the case, someone already use this process, to create some kind of weapon of mass destruction, or just to make a cheap source of heat.

Do you understand that the requirements- both the perfect mirror and the

  ideal optical diode to let light in but keep it inside

do not, and can not exist?

[Bored chemist]

My question is, if EM waves reflected from a perfect mirror keep their original wavelenght?

They must, by definition.
Because a change in wavelength would mean a change in energy and that is inconsistent with the definition of a "perfect" mirror.

[CrazyScientist (OP)]

But as long as the pressure is being applied to the reflective surface, there is some transfer of energy between photons and the mirror.

Only if it moves.

And, if it moves then, in due course, it transfers energy back to other photons.

I'm sure, that if it would be the case, someone already use this process, to create some kind of weapon of mass destruction, or just to make a cheap source of heat.

Do you understand that the requirements- both the perfect mirror and the

  ideal optical diode to let light in but keep it inside

do not, and can not exist?

Sure. I've included this fact in the latest edit of my previous post

[CrazyScientist (OP)]

My question is, if EM waves reflected from a perfect mirror keep their original wavelenght?

They must, by definition.
Because a change in wavelength would mean a change in energy and that is inconsistent with the definition of a "perfect" mirror.

And you are peffectly aware, that this makes each single photon an almost ideal source of a constant and infinite kinetic energy?

Since real life mirrors can use up to 99% of that energy to produce a kinetic energy. All you have to do, is to keep each photon "alive" as long as it is possible.

[Origin]

And you are peffectly aware, that this makes each single photon an almost ideal source of a constant and infinite kinetic energy?

Why would you think that?

[Colin2B]

And you are peffectly aware, that this makes each single photon an almost ideal source of a constant and infinite kinetic energy?

he’s aware of the constant, but not the infinite. By definition the energy available is hf

All you have to do, is to keep each photon "alive" as long as it is possible.

To do that you have to keep it from being absorbed by an atom or kicking an electron out of its potential well.

Sorry, I keep meaning to respond to your posts, but haven’t had time to read beyond the 1st. I’ll put some thoughts on it.

[Bored chemist]

And you are peffectly aware, that this makes each single photon an almost ideal source of a constant and infinite kinetic energy?

Since real life mirrors can use up to 99% of that energy to produce a kinetic energy. All you have to do, is to keep each photon "alive" as long as it is possible.

That is equivalent to saying that  a bag of pasta can  provide energy forever- until you eat it.

[CrazyScientist (OP)]

So this is what I found:

I guess this settles the issue of photon momentum transfer during reflection.

Sorry to prove you being wrong. Well you can't be always correct, can you? 

[CrazyScientist (OP)]

Your problem is, that you think that light emmision inside a spherical shell, is like inflating an empty balloon - and this is totally wrong.

Think about this scenario as about some source of a buzzing sound, which is submerged in a free sphere of water. Even if you will constantly add energy to the system by emission of sound, frequency of vibration won't go beyond a certain treshold

[Bored chemist]

Your problem is, that you think that light emmision inside a spherical shell, is like inflating an empty balloon - and this is totally wrong.

Your problem is claiming that you know what I think; you are totally wrong.

I guess this settles the issue of photon momentum transfer during reflection.

It was never in dispute; you have shown that I am right.

[Bored chemist]

Think about this scenario as about some source of a buzzing sound

Why think about a really bad analogy when we can think about the actual science?

[CrazyScientist (OP)]

Think about this scenario as about some source of a buzzing sound

Why think about a really bad analogy when we can think about the actual science?

Because it properly demonstrates the most possible outcome of my scenario.

Well at least it doesn't turn into a black hole made of pure sound - what should happen if we apply here your way of thinking...

[CrazyScientist (OP)]

Think about this scenario as about some source of a buzzing sound

Why think about a really bad analogy when we can think about the actual science?

So you don't consider quantum physics as an actual science? What is then actual science for you? Black holes made of pure light?

[Bored chemist]

So you don't consider quantum physics as an actual science?

You seem to have got me mixed up with you.
I think that science works.
You think that we should pretend that light is the same as sound.

[Bored chemist]

what should happen if we apply here your way of thinking

What would happen is that we wouldn't waste time thinking about sound- which is very different from light.

[Bored chemist]

The reason why sound isn't the same as light (among other things) is that you can't even get close to being able to make this claim.

EM waves which propagate in vacuum don't interact with each other -

Sound waves clearly do interact- especially at high intensities.

[CrazyScientist (OP)]

I guess this settles the issue of photon momentum transfer during reflection.

It was never in dispute; you have shown that I am right.

Pretty much not - pressure of photons on the reflective surface is the result of momentum transfer - and it results in change of frequency for the reflected photon.

You claimed that photons will always maintain the same frequency, despite applying pressure on reflective matter.

Maybe you don't actually understand the things, you're trying to discuss