[CrazyScientist (OP)]

Of course! Because of the momentum conservation, when the ball is being hit by the bat, their relative velocity can be higher, than before their collision

OK, it's becoming increasingly clear that you are just posting nonsense, and you are doing it deliberately.

I did - and in terms of physics of baseball, it's a 100% valid answer. What else do you want?

I already said

from the point of view of the batsman?

Stop trolling, and answer the question.

Then obviously the answer is no... And...?
I still have no idea, what any of this has to do with cavity QED...

[Origin]

So I guess it means, that sunlight propagates as a buch of different photons, that occupy the same volume of space. This means, that it should be possible to simultaneously detect multiple photons at different frequencies, that are overlaping each other in a single volume of space...

However it seems, that such situation is hysically impossible..

Really, that seems impossible?  If only there were some sort of simple device that could separate the multiple wavelengths into separate wavelengths to see if it is possible, but I guess there just is no such futuristic device.

[Bored chemist]

Yes.... And each kind of EM radiation has a specific and measurable wavelenght... So...?

So... you say "yes", but you mean "no".

I mean exactly, what I say...

So you do understand that there's an uncertainty relation for energy which means that (in any finite time) the wavelength is not perfectly defined so you can't say whether or not it exactly fits into a cavity?

Yet you say that things can or can't happen depending on whether it fits or not.

[Bored chemist]

I still have no idea, what any of this has to do with cavity

Once again, we are back with the problem of your memory...

It's relevant to the idea of compressing a cavity with light in it until it becomes a blackhole.

You may, if prompted, remember this

If you compress a mirror box with light in then you do work against that photon pressure.
Where does that energy go?
I contend that it raises the frequencies of the photons in the box. What do you think happens to it?
Or are you really claiming that energy is not conserved?

And your absurd reply

Obviously into the atoms, that make the resonance cavity

To which I pointed out that , since the mirrors are perfectly reflective, you can't put energy into them that way.

And then you posted a diagram which you misunderstood and said  this

Nope. it's exactly opposite - momentum transfer during reflection of an EM wave increases it's wavelenght

Then you got shouty about a mirror's movement being relativistic- as if that was somehow magical- until I pointed out that a snail's movement is relativistic, it's just that the relativistic corrections are rather small.


But you still didn't seem to grasp the fact that bouncing a photon off a mirror that's coming towards you will mean that the photon comes back at you  with a higher energy, and the mirror gets slowed down a bit.
Fundamentally, you still have not answered the question.
What happens to the work done against photon pressure if you compress a box that has light in it?

Instead of answering, you cam up with this tosh

I guess, you slept during physics classes, when the Newton's laws of motion were discussed... If you wouldn't sleep, you would know, that to slow down an object, that moves at constant velocity, you need to use a force - so photons, which are slowing down the mirror are giving up their own energy... It's basic physics - aren't you ashamed to not know such things?

Now, since you framed your bilge in terms of high school physics, then, regardless of any weirdness due to quantum effects, it should work in terms of high school physics.
But you say that if a ball exerts a force on a moving  bat, the bat will gain energy (yet it slows down)  and the ball will lose energy (though its speed increases).

Do you really believe that?

Please don't waste time on energy losses and coefficients of restitution- this is an ideal, perfect mirror.
There are no losses.

[Bored chemist]

So I guess

It may be better if you stopped guessing and learned some science.

This means, that it should be possible to simultaneously detect multiple photons at different frequencies, that are overlaping each other in a single volume of space...

However it seems, that such situation is physically impossible...

You have just told me that I can't tune a radio to one frequency if someone else has their radio tuned to another frequency.

Do you want to think that through again?

Because creation of matter particles from radiation in the laboratory has nothing to do with the stable matter, that surrounds us

Obviously, electrons made in the lab are different- they have a little tag on them which says "made in England".
Or, possibly, you're talking nonsense again, and the two scenarios do have a lot of common ground.

Explain, what for you defines the mass/energy of EM radiation

I don't understand why you keep saying thing like "for you".
It's not me. You can argue the point whether, in this case it's de Broglie's equation  or Einstein's.
But it's not "me" that's defining anything.
You were the one trying to define new words- do you remember?

my idea of X-stationary photons

[Bored chemist]

I guess, it's too hard for you to understand, that visible light has a very short wavelenght, so the size of cavity matters mostly at the microscopic level.

Not only do I know that the wavelength is small, I know how to spell it.
Earlier on in lockdown, I got bored and I actually measured the wavelength of the green light from a frequency doubled Nd YAG laser using a plastic ruler from my local stationers shop.

Have you ever actually measured the wavelength?
I have also actually physically built a laser cavity, so I have some understanding of mode hopping and such.
Have you?

And the only actual size given for any cavity was a millimetre or so, which is vastly bigger than the wavelength of light.

You keep telling me it's terribly important that the waves fit in the cavity.
But you ignore the fact that my lightbulb proves you are wrong.
It's a broadband emitter.

[CrazyScientist (OP)]

Yes.... And each kind of EM radiation has a specific and measurable wavelenght... So...?

So... you say "yes", but you mean "no".

I mean exactly, what I say...

So you do understand that there's an uncertainty relation for energy which means that (in any finite time) the wavelength is not perfectly defined so you can't say whether or not it exactly fits into a cavity?

Yet you say that things can or can't happen depending on whether it fits or not.

But the wavelenght of EM radiation IS very well defined - we know for example the EXACT lenghts of EM waves for each color of visible spectrum - you even said, that you've measured one by yourself. Isn't that quite self-contradictory?

Besides you obviously have no idea, how much efforts are being put into the tuning of optical cavities, so the desired EM radiation can fit with it's wavelenght perfectly and create a standing wave...
https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-26-26-34965&id=403390

What did you think - that the make the microcavities with the right volumes just by accident?

I still have no idea, what any of this has to do with cavity

Once again, we are back with the problem of your memory...

It's relevant to the idea of compressing a cavity with light in it until it becomes a blackhole.

You may, if prompted, remember this

If you compress a mirror box with light in then you do work against that photon pressure.
Where does that energy go?
I contend that it raises the frequencies of the photons in the box. What do you think happens to it?
Or are you really claiming that energy is not conserved?

And your absurd reply

Obviously into the atoms, that make the resonance cavity

To which I pointed out that , since the mirrors are perfectly reflective, you can't put energy into them that way.

And then you posted a diagram which you misunderstood and said  this

Nope. it's exactly opposite - momentum transfer during reflection of an EM wave increases it's wavelenght

Then you got shouty about a mirror's movement being relativistic- as if that was somehow magical- until I pointed out that a snail's movement is relativistic, it's just that the relativistic corrections are rather small.


But you still didn't seem to grasp the fact that bouncing a photon off a mirror that's coming towards you will mean that the photon comes back at you  with a higher energy, and the mirror gets slowed down a bit.
Fundamentally, you still have not answered the question.
What happens to the work done against photon pressure if you compress a box that has light in it?

Instead of answering, you cam up with this tosh

I guess, you slept during physics classes, when the Newton's laws of motion were discussed... If you wouldn't sleep, you would know, that to slow down an object, that moves at constant velocity, you need to use a force - so photons, which are slowing down the mirror are giving up their own energy... It's basic physics - aren't you ashamed to not know such things?

Now, since you framed your bilge in terms of high school physics, then, regardless of any weirdness due to quantum effects, it should work in terms of high school physics.
But you say that if a ball exerts a force on a moving  bat, the bat will gain energy (yet it slows down)  and the ball will lose energy (though its speed increases).

Do you really believe that?

Please don't waste time on energy losses and coefficients of restitution- this is an ideal, perfect mirror.
There are no losses.

So you're still want to talk about this science-fiction, which is the "Kugelblitz"?

I guess, that you didn't even try to look what the present day science has to say about this fantasy... If  you would, you would probably know, that there's an entire branch of physics dedicated to deal with such scenarios, as the one, which you've proposed (EM radiation reflected by a moving mirror) - it's called cavity optomechanics
https://en.wikipedia.org/wiki/Cavity_optomechanics

There's quite a lot of information about this subject, but just so happens, that there isn't even a single source, where someone would mention ANYTHING about the possibility of creating a black hole made of radiation. Obviously modern-day scientists prefer sticking to experimentally proven reality, than to theoretical fantasies from the first half of XX century...

And it gets even better. It seems, that your idea of compressing an EM radiation, which is trapped in a cavity was already tested experimentally - and so happens, that the observed results were in 100% consistent with the model of cavity QED and not with the fairy tales about black holes made of light:
https://nige.wordpress.com/2009/08/25/casimir-force/

[CrazyScientist (OP)]

You have just told me that I can't tune a radio to one frequency if someone else has their radio tuned to another frequency.

Do you want to think that through again?

That's a completely invalid analogy. Instead try listening simultaneouly to 2 radio stations using one receiver with a single antenna... I might rethink my claims, when you'll manage to do it...

EDIT - you can use 2 receivers with one shared antenna, or even 2 antennas, which are wrapped around each other...

[CrazyScientist (OP)]

I don't understand why you keep saying thing like "for you".
It's not me. You can argue the point whether, in this case it's de Broglie's equation  or Einstein's.
But it's not "me" that's defining anything.
You were the one trying to define new words- do you remember?

So, what then defines the energy/mass of EM radiation - momentum or frequency of EM waves? It's a simple question with only 2 options.- please, try better this time...

[CrazyScientist (OP)]

And the only actual size given for any cavity was a millimetre or so, which is vastly bigger than the wavelength of light.

Well, sorry to inform you, that time keeps going on and science is constantly progressing... Modern-day nanocavities are capable of trapping a single photon...
https://pubs.rsc.org/en/content/articlelanding/2021/cc/d1cc01084k#!divAbstract

You keep telling me it's terribly important that the waves fit in the cavity.
But you ignore the fact that my lightbulb proves you are wrong.
It's a broadband emitter.

Tell this to people, who create modern day optical cavities

Obviously, electrons made in the lab are different- they have a little tag on them which says "made in England".
Or, possibly, you're talking nonsense again, and the two scenarios do have a lot of common ground.

Difference is, that electrons made in labs due to p-p scattering disappear out of existence couple nanoseconds after they were created...

[Bored chemist]

Well, sorry to inform you, that time keeps going on and science is constantly progressing... Modern-day nanocavities are capable of trapping a single photon...

I know.
For what it's worth, you can trap a single photon in a big cavity too.
But, since we were actually planning to use a bigger chamber your assertion, while true, is irrelevant, isn't it?

That's a completely invalid analogy. Instead try listening simultaneouly to 2 radio stations using one receiver with a single antenna... I might rethink my claims, when you'll manage to do it...

Using one receiver is moving the goalposts; I'm glad you dropped it.

EDIT - you can use 2 receivers with one shared antenna,

.
Using one antenna is easy enough; my house has a combined TV/ FM antenna both signals come down the same wire.

In principle, I  can do a bit better.
I can hook a long wire to the input of a modern oscilloscope and what I see is the sum of all the EM fields which the antenna picks up.
And then I can ask the 'scope to do an FFT on the signal and it will cheerfully "receive" and display all the radio stations etc in the area.
Would you like to buy me a new digital scope?

So, what then defines the energy/mass of EM radiation - momentum or frequency of EM waves? It's a simple question with only 2 options.- please, try better this time...

It's a simple question, but meaningless.
I can calculate the energy of a photon from it's momentum, or from its frequency.
So either of them define it.
Did you not realise that?

But the wavelenght of EM radiation IS very well defined - we know for example the EXACT lenghts of EM waves for each color of visible spectrum - you even said, that you've measured one by yourself. Isn't that quite self-contradictory?

Get back to us when you know what the uncertainty principle tells us about energy.
Until then, you don't have the knowledge to see why you must be wrong and there's no point carrying on the discussion.

Mind you, We won't expect much from the guy who can't face up to having said  that the point of a bat is to make the ball go slower.

Are you going to address this before, or after, learning what the uncertainty principle says about wavelengths?

But you still didn't seem to grasp the fact that bouncing a photon off a mirror that's coming towards you will mean that the photon comes back at you  with a higher energy, and the mirror gets slowed down a bit.
Fundamentally, you still have not answered the question.
What happens to the work done against photon pressure if you compress a box that has light in it?

Instead of answering, you cam up with this tosh

Quote from: CrazyScientist on 14/06/2021 17:11:04
I guess, you slept during physics classes, when the Newton's laws of motion were discussed... If you wouldn't sleep, you would know, that to slow down an object, that moves at constant velocity, you need to use a force - so photons, which are slowing down the mirror are giving up their own energy... It's basic physics - aren't you ashamed to not know such things?


Now, since you framed your bilge in terms of high school physics, then, regardless of any weirdness due to quantum effects, it should work in terms of high school physics.
But you say that if a ball exerts a force on a moving  bat, the bat will gain energy (yet it slows down)  and the ball will lose energy (though its speed increases).

Do you really believe that?

Please don't waste time on energy losses and coefficients of restitution- this is an ideal, perfect mirror.
There are no losses.

[CrazyScientist (OP)]

Sorry, that it took me so long - I had a pretty busy week...

Well, sorry to inform you, that time keeps going on and science is constantly progressing... Modern-day nanocavities are capable of trapping a single photon...

I know.
For what it's worth, you can trap a single photon in a big cavity too.
But, since we were actually planning to use a bigger chamber your assertion, while true, is irrelevant, isn't it?

Actually I didn't say anything about the size of cavity. In fact, what matters in this case, is the wavelenght of EM radiation. You can trap a single impulse of visible light in a cavity which is bigger, than the wavelenght of emitted radiation, but then it won't be amplified - as the highest amplitude of EM radiation (just like the sound) can be achieved only by the resonance of a standing wave.

On the other hand you can achieve exactly the same effect with a large cavity and EM radiation with bigger wavelenght. In the end, everything what matters IS the wavelenght and size of cavity...

Using one antenna is easy enough; my house has a combined TV/ FM antenna both signals come down the same wire.

Sure - only TV and FM use EM waves at different freuency bands...

FM radio uses frequency modulation, of course. The frequency band for FM radio is about 88 to 108 MHz. The information signal is music and voice which falls in the audio spectrum. The full audio spectrum ranges form 20 to 20,000 Hz, but FM radio limits the upper modulating frequency to 15 kHz.

TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz.

EM waves interfere with other EM waves at similar wavelenghts. Take 2 FM receivers and try to stick their antennas together, while listening to 2 different stations and you'll get static noise due to EM waves interference...

In principle, I  can do a bit better.
I can hook a long wire to the input of a modern oscilloscope and what I see is the sum of all the EM fields which the antenna picks up.
And then I can ask the 'scope to do an FFT on the signal and it will cheerfully "receive" and display all the radio stations etc in the area.
Would you like to buy me a new digital scope?

Of course. It's because all photons in an EM field are in superposition of all the avaliable wavelenghts - that's the potential aspect of the field. And then, when you try to receive one specific wavelenght in a given frequency band, photons placed around your antenna will collapse to that specific state - and this is the kinetic aspect of EM field.

What matters is, that photons will still remain in superposition at different frequency bands - so, each of them can make a FM wave and the visible light at the same time. That's exactly, what I described in the first post of this thread

So, what then defines the energy/mass of EM radiation - momentum or frequency of EM waves? It's a simple question with only 2 options.- please, try better this time...

It's a simple question, but meaningless.
I can calculate the energy of a photon from it's momentum, or from its frequency.
So either of them define it.
Did you not realise that?

It IS important, since there's a reversed corelation between those 2 properties - increase of frequency means decrease of momentum. It's pretty logical, that if we treat both of them as the same "mass/energy" of EM radiation, then the idea of "Kugelblitz" becomes impossible even in the theory...

TBC

[CrazyScientist (OP)]

But the wavelenght of EM radiation IS very well defined - we know for example the EXACT lenghts of EM waves for each color of visible spectrum - you even said, that you've measured one by yourself. Isn't that quite self-contradictory?

Get back to us when you know what the uncertainty principle tells us about energy.
Until then, you don't have the knowledge to see why you must be wrong and there's no point carrying on the discussion.

Mind you, We won't expect much from the guy who can't face up to having said  that the point of a bat is to make the ball go slower.

Are you going to address this before, or after, learning what the uncertainty principle says about wavelengths?

And...? We ARE capable to learn the exact wavelenght of measured radiation...

But you still didn't seem to grasp the fact that bouncing a photon off a mirror that's coming towards you will mean that the photon comes back at you  with a higher energy, and the mirror gets slowed down a bit.

Sure - but only if you and/or the mirror are moving towards eachother at a relative velocity, which is high enough, to compensate the energy loss due to transfer of momentum (what means pretty damn fast)

Fundamentally, you still have not answered the question.
What happens to the work done against photon pressure if you compress a box that has light in it?

It depends on:
1. Energy level of the trapped radiation
2. Velocity of compression

You would have to use a velocity of compression, which is in a perfect harmony with the frequency of trapped EM radiation - otherwise EM waves will cancel each other out due to destructive interference. And if you do, then depending on the trapped energy and in a cavity, which can't get smaller from the wavelenght of a single photon,
it should be possible to turn the trapped radiation in a Bose-Einstein condensate, or maybe even into a particle of matter.

And what will happen, if the cavity gets smaller than the wavelenght of trapped EM radiation? Well I guess, the energy will mostly cancel eachother out or in some part escape by quantum tunneling (I need to research it a bit).

Instead of answering, you cam up with this tosh

Quote from: CrazyScientist on 14/06/2021 17:11:04
I guess, you slept during physics classes, when the Newton's laws of motion were discussed... If you wouldn't sleep, you would know, that to slow down an object, that moves at constant velocity, you need to use a force - so photons, which are slowing down the mirror are giving up their own energy... It's basic physics - aren't you ashamed to not know such things?

It's because you insist to use the baseball analogy, which doesn't have anything to do with cavity QED, which SHOULD be used, to solve the presented scenario...

Now, since you framed your bilge in terms of high school physics, then, regardless of any weirdness due to quantum effects, it should work in terms of high school physics.
But you say that if a ball exerts a force on a moving  bat, the bat will gain energy (yet it slows down)  and the ball will lose energy (though its speed increases).

Do you really believe that?

Please don't waste time on energy losses and coefficients of restitution- this is an ideal, perfect mirror.
There are no losses.

Not only you still keep insisting to use this invalid analogy, but you also keep using a 3rd frame of some batsman guy, which can't be applied to interactions between the mirror (baseball bat) and the photon (ball). And what makes your baseball analogy completely wrong, is the fact, that photons suppose to move at a constant velocity of c, while in the rest frame of mirror, those photons are in fact causing acceleration opposite to the constant motion in your 3rd batsman frame...

[CrazyScientist (OP)]

Here's what I found:
https://www.researchgate.net/post/What_happens_to_a_photon_if_sent_into_a_cavity_whose_dimensions_are_not_fit_for_the_photons_wavelength

Perfect:

This is a really good question. Simple as it is stated, it turns out hard to answer, but is worth trying to. It took me and my colleagues a 21 pages paper Phys. Rev. D 91, 016005 (2015)  to address a simpler question.
The main issue is the radical change of boundary conditions the photon quantum undergoes when (instantly, or better in an insignificant amount of time) gets into the cavity.
Before entering into the cavity, the photon  is an elementary excitation of the em field in the whole space. It is defined in this way in QFT (I think you are thinking of this case, as a wave packet is only a superposition of elementary excitations), and it is described by a Fock state a(k) |o>. Notice that the a(k) and a(k)^+  operators are defined as those who annihilate or create an elementary excitation of momentum k - IN ALL SPACE R^3 - at an instant of time.
That is, these operators remove or add a normal mode of the wave equation in Minkowski space. Those modes are specified by the equation and the boundary conditions. Only by considering both together,  the allowed wave numbers emerge.
Beyond waves, in quantum physics -i.e. QFT - there are particles; photons in this case. To handle this, we need of a counting tool and the Fock treatment provides us with one, the number operator N(k) = a^+(k) a(k), which counts the number of elementary excitations of momentum k (wave number 1/k) present in the WHOLE SPACE R^3.
Coming back to your question; before entering into the cavity there is a photon of wave number 1/k as counted by N(k) (or of wave length λ according to your original wording).
The point is that photons inside the cavity are not created or annihilated by  the same operators as before, but by operators b(k_n)^+, b(k_n)  that add or remove one n-th normal mode of the cavity. Accordingly, photons inside the cavity will be counted by the operator b(k_n)^+b(k_n).
Finally, the same quantum state will have different expansions in terms of the R^3 normal modes than in terms of the cavity normal modes. In poor words, what was one free space photon outside may become a bunch of cavity photons inside.
Mathematically this is involved as both photon representations are unitarily inequivalent!
I hope this explanation will serve to add some light to your question. juan.leon@csic.es
I saw your interests in your Research Gate page

And here's the part, which is epecially important for me:
"The point is that photons inside the cavity are not created or annihilated by the same operators as before, but by operators b(k_n)^+, b(k_n)  that add or remove one n-th normal mode of the cavity. Accordingly, photons inside the cavity will be counted by the operator b(k_n)^+b(k_n)."

[CrazyScientist (OP)]

And this:
https://www.eenewseurope.com/news/light-trap-turns-photons-massive-quasiparticles

[Bored chemist]

Actually I didn't say anything about the size of cavity.

True, but I did.

I already made the point that the chamber size was arbitrary, and could be made "big enough"

Nobody was trying to make photons smaller than any limit  (apart from a rather arbitrary1mm which was chosen as a bit smaller than the radius of an earth mass BH)

Sure - only TV and FM use EM waves at different freuency bands...

Yes, that was my point.

You had said it was impossible to receive different frequencies, so I chose very different ones to make the point that you were wrong.

But my point about  hooking a spectrum analyser to an antenna overrides it anyway.
Your "reply" And then, when you try to receive one specific wavelenght in a given frequency band, photons placed around your antenna will collapse to that specific state " makes no sense.
The spectrum analysis works.
So it's receiving many channels at once.
It would be obvious from time to time when the AM radio stations broadcast a minute's silence- all the peaks would drop to zero simultaneously.
You can see the intensity of the broadcast at each frequency. So they are resolved; they are no longer in a superposition.

It proves that it's possible.

It's also common practice for blocks of flats to have a single aerial  feeding to all the flats.
You are trying to say that these don't exist.
https://www.screwfix.com/c/electrical-lighting/splitters/cat4740084

And that's silly.

It IS important,

It is important that either the momentum or the frequency of a photon will define its energy.
Which is why I said it, and it's the reason why your question (about was it one or the other) is silly.

Now, let's see if you understand the uncertainty principle.
You seem to think it only relates to momentum and position.
You don't have the sense to realise you are missing the important thing, even when I spell it out to you.

Get back to us when you know what the uncertainty principle tells us about energy.

You didn't even bother to look at the wiki page, did you?
If you had looked properly, you would have found this

"An informal, heuristic meaning of the principle is the following: A state that only exists for a short time cannot have a definite energy. To have a definite energy, the frequency of the state must be defined accurately, and this requires the state to hang around for many cycles, the reciprocal of the required accuracy. For example, in spectroscopy, excited states have a finite lifetime. By the time–energy uncertainty principle, they do not have a definite energy, and, each time they decay, the energy they release is slightly different. The average energy of the outgoing photon has a peak at the theoretical energy of the state, but the distribution has a finite width called the natural linewidth. Fast-decaying states have a broad linewidth, while slow-decaying states have a narrow linewidth"

And that tells you that an observation made in a finite time will only give you an uncertain assessment of the energy.
So the energy is (like momentum or position) always uncertain.
And that means the wavelength is always uncertain.

Before telling me that science has moved on in the 21st C, perhaps you should learn the stuff that the grown ups already knew in the 20th.

It hardly matters once you realise that the position of the wall of the cavity is uncertain.
If the cavity doesn't actually have a well defined size then it can't forbid photons fro being the wrong size.

the energy loss due to transfer of momentum

which is zero- because it's a perfect mirror.
So I don't have to do a lot to compensate for it.
However, in the latter stages of the compression when the thing is collapsing, on the way to become a BH, the mirrors are going to mover pretty fast anyway. So your bogus criterion gets mat by default.

Quote from: Bored chemist on 20/06/2021 15:06:01
Instead of answering, you cam up with this tosh

Quote from: CrazyScientist on 14/06/2021 17:11:04
I guess, you slept during physics classes, when the Newton's laws of motion were discussed... If you wouldn't sleep, you would know, that to slow down an object, that moves at constant velocity, you need to use a force - so photons, which are slowing down the mirror are giving up their own energy... It's basic physics - aren't you ashamed to not know such things?

It's because you insist to use the baseball analogy, which doesn't have anything to do with cavity QED, which SHOULD be used, to solve the presented scenario...

Quote from: Bored chemist on 20/06/2021 15:06:01
Now, since you framed your bilge in terms of high school physics, then, regardless of any weirdness due to quantum effects, it should work in terms of high school physics.
But you say that if a ball exerts a force on a moving  bat, the bat will gain energy (yet it slows down)  and the ball will lose energy (though its speed increases).

Do you really believe that?

Please don't waste time on energy losses and coefficients of restitution- this is an ideal, perfect mirror.
There are no losses.

Not only you still keep insisting to use this invalid analogy, but you also keep using a 3rd frame of some batsman guy, which can't be applied to interactions between the mirror (baseball bat) and the photon (ball). And what makes your baseball analogy completely wrong, is the fact, that photons suppose to move at a constant velocity of c, while in the rest frame of mirror, those photons are in fact causing acceleration opposite to the constant motion in your 3rd batsman frame...

So, are you still saying that the point of a cricket bat is to slow the ball down?

It's a simple yes or no.
You are the one who was going on about sleeping through  school science; not me.

Why did you get that school science wrong?

[Bored chemist]

And this:
https://www.eenewseurope.com/news/light-trap-turns-photons-massive-quasiparticles

If you put enough mass in a small enough space, it collapses.
That's still true if it's in the form of massive quasiparticles.

You seem to be arguing against yourself.

[CrazyScientist (OP)]

So I guess it means, that sunlight propagates as a buch of different photons, that occupy the same volume of space. This means, that it should be possible to simultaneously detect multiple photons at different frequencies, that are overlaping each other in a single volume of space...

However it seems, that such situation is hysically impossible..

Really, that seems impossible?  If only there were some sort of simple device that could separate the multiple wavelengths into separate wavelengths to see if it is possible, but I guess there just is no such futuristic device.

Thanks! Yes, I've made a stupid mistake here. It supposed to be :
So I guess it means, that sunlight propagates as a buch of different photons, that occupy the same volume of space. This means, that it should be possible to simultaneously detect multiple photons at similar frequencies, that are overlaping each other in a single volume of space..

[CrazyScientist (OP)]

You had said it was impossible to receive different frequencies, so I chose very different ones to make the point that you were wrong.

Yes. Sorry, I've made a mistake here. It should be "similar frequencies"

[CrazyScientist (OP)]

But my point about  hooking a spectrum analyser to an antenna overrides it anyway.
Your "reply" And then, when you try to receive one specific wavelenght in a given frequency band, photons placed around your antenna will collapse to that specific state " makes no sense.
The spectrum analysis works.
So it's receiving many channels at once.
It would be obvious from time to time when the AM radio stations broadcast a minute's silence- all the peaks would drop to zero simultaneously.
You can see the intensity of the broadcast at each frequency. So they are resolved; they are no longer in a superposition.

It proves that it's possible.

But you can't simultaneously receive multiple AM/FM stations at similar wavelenghts using one antenna - can you?

Intensity of an EM field is described as probability of detecting a photon at a given frequency. If in a potential EM field EM waves at similar wavelenghts are overlapping each other, then in the kinetic form of that field,each particular photon can collapse to just one of those states