[Bored chemist]

Obviously into the atoms, that make the resonance cavity

So, because the energy is absorbed by the walls, it isn't reflected.
So it isn't a perfectly reflective wall- as specified.
So your "explanation" only makes sense when you move the goal posts.

Of course you are fully aware, that the the mass of the trapping box has billions times greater relativistic mass, than all the photons trapped in that box - so even your expactactions are fundamentally flawed

Well, yes and no.
I'm the one who pointed out that a strictly perfect mirror needs to have infinite mass.

But I also pointed out that, if you add enough photons the box collapses- regardless of the original mass of the box.
The box was always pretty much doomed anyway- it was going to be right next to a BH.

[Bored chemist]

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

Imagine doing that experiment, as shown in that diagram and filming it.
Then imagine watching the film backwards.
An electron heading towards the source of radiation (on the right of the picture) strikes the photon and , like hitting a ball with a bat, the photon is sent off towards the top left at a higher energy than it started with (and the electron is slowed down).

That's the scenario we are talking about.
The box wall is moving towards the photon and when it hits it, the energy of the photon is raised.

It's fairly closely analogous to compressing a gas and getting an increase in temperature.

[Bored chemist]

And what exactly does it have to do with EM radiation trapped inside a cavity with decreasing volume?

It is the experiment where someone actually did bounce photons off a fast moving mirror and show that the photons' energy was raised.
It experimentally demonstrates that your idea-

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

is wrong if the mirror moves towards the source.

The picture you show is the case  where the photons cause the electron to move.
If you hang a bat on a length of rope and throw a ball at it, the ball will transfer momentum to the bat and will bounce off more slowly than it started out.
But that's not how it would work if you were deliberately squashing the box.

Here:
for example your claim is that:

It still isn't "my" claim.
It's Kirchhoff's.
And it's a direct consequence of the conservation of energy.

Things that absorb will also emit.

[Bored chemist]

"If the antenna is inside a reflecting cavity, however, its behavior changes

Yes.
We know.
I already pointed that out.
Given that I wrote

This is still true if the excited state has been perturbed by a cavity or, indeed, anything else.

the question is now, did you not read that, or not understand it?

[Bored chemist]

The suppression of spontaneous emission at an optical frequency requires much smaller cavities. In 1986 one of us (Haroche), along with other physicists at Yale University, made a micron-wide structure by stacking two optically flat mirrors separated by extremely thin metallic spacers. The workers sent atoms through this passage, thereby preventing them from radiating for as long as 13 times the normal excited-state lifetime. Researchers at the University of Rome used similar micron-wide gaps to inhibit emission by excited dye molecules."

Well, I have to admit this is rather surprising.

I never thought that the research  work I did as a student would ever actually be of any use to me.

As is often the case, I was pretty much copying a piece of research done earlier by someone else.
And I actually did the experiment of putting a fluorescent molecule (well, lots of them) near a mirror- i.e. in a "one sided" cavity.
And I measured their fluorescence half life as a function of distance from the mirror.
Here are the results I got.


Part II shrunk.jpg (39.28 kB . 330x440 - viewed 4928 times)

So,yes, I have known for over 30 years, and from personal experience that this is the case.
(My structures were even smaller than the ones your reference refers to: the smallest was just 60 nm wide.- not bad going for 1988 technology)

I'm curious: what was your research field in the late 80s?

[CrazyScientist (OP)]

Obviously into the atoms, that make the resonance cavity

So, because the energy is absorbed by the walls, it isn't reflected.
So it isn't a perfectly reflective wall- as specified.
So your "explanation" only makes sense when you move the goal posts.

Problem is, that when an EM wave hits a surface, it can interact with it only in two ways - the energy can be absorbed or it can be reflected. However even if the surface is 100% reflective, so the radiation isn't absorbed and turned into thermal energy, momentum transfer turns part of the energy carried by EM waves into kinetic energy of atoms that make the surface.

So, in real-life there's no way out of this situation, as energy of EM radiation will be lost, even if the surface is 100% reflective (and yes - 100% reflective surfaces actually DO exist). Besides, even if we'd make the cavity out of a black hole, the outcome won't change anyway - as phhotons can't gain energy due to reflection, so they will simply keep the same energy level all the time

Of course you are fully aware, that the the mass of the trapping box has billions times greater relativistic mass, than all the photons trapped in that box - so even your expactactions are fundamentally flawed

Well, yes and no.
I'm the one who pointed out that a strictly perfect mirror needs to have infinite mass.

But I also pointed out that, if you add enough photons the box collapses- regardless of the original mass of the box.
The box was always pretty much doomed anyway- it was going to be right next to a BH.

And this turns out to be incorrect, as each volume of space has a specific "capacity" of probability distribution, so intensity of EM radiation in that volume can't exceed a specific level. When that capacity is fully used by a constant emission of EM radiation at a constant frequency, further emission is prohibited

[CrazyScientist (OP)]

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

Imagine doing that experiment, as shown in that diagram and filming it.
Then imagine watching the film backwards.
An electron heading towards the source of radiation (on the right of the picture) strikes the photon and , like hitting a ball with a bat, the photon is sent off towards the top left at a higher energy than it started with (and the electron is slowed down).

That's the scenario we are talking about.
The box wall is moving towards the photon and when it hits it, the energy of the photon is raised.

It's fairly closely analogous to compressing a gas and getting an increase in temperature.

The only way, in which photons can gain energy due to reflection, is to accelerate the mirror to relativistic velocity (what was done in the paper you presented) - and then energy used to accelerate the mirror is transferred to the reflected photon, allowing conservation of energy in the system. Any other scenario can't lead to the increase of energy level in the system - it would violate everything what we know about physics...

And of cource you didn't mention about the speed at which the cavity is shrinking, so obviously you didn't take it into account and your scenario won't lead to the increase of energy level in the cavity

[CrazyScientist (OP)]

And what exactly does it have to do with EM radiation trapped inside a cavity with decreasing volume?

It is the experiment where someone actually did bounce photons off a fast moving mirror and show that the photons' energy was raised.
It experimentally demonstrates that your idea-

Yes - but:

1. It wasn't made in a cavity
2. mirror was moving at relativistic velocity

So, this scenario has nothing to do with the one, which I presented in the first post of this thread... Sorry...

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

is wrong if the mirror moves towards the source.

The picture you show is the case  where the photons cause the electron to move.
If you hang a bat on a length of rope and throw a ball at it, the ball will transfer momentum to the bat and will bounce off more slowly than it started out.
But that's not how it would work if you were deliberately squashing the box.

You know, that even 10yo kids know from physics and chemistry classes, that picturing photons as tiny balls/marbles moving through space, is completely incorrect

Here:
for example your claim is that:

It still isn't "my" claim.
It's Kirchhoff's.
And it's a direct consequence of the conservation of energy.

Things that absorb will also emit.

So, obviously this statement can't be applied to a scenario where EM radiation is emitted inside a cavity

[CrazyScientist (OP)]

"If the antenna is inside a reflecting cavity, however, its behavior changes

Yes.
We know.
I already pointed that out.
Given that I wrote

This is still true if the excited state has been perturbed by a cavity or, indeed, anything else.

the question is now, did you not read that, or not understand it?

You just proved yourself wrong with that statement. Each system will reach an energetic equilibrium, as long as the input energy remains constant

[Bored chemist]

Problem is, that when an EM wave hits a surface, it can interact with it only in two ways - the energy can be absorbed or it can be reflected. However even if the surface is 100% reflective, so the radiation isn't absorbed and turned into thermal energy, momentum transfer turns part of the energy carried by EM waves into kinetic energy of atoms that make the surface.

No.
As I have pointed out a few times, and posted a link to the experimental verification , there are three things that can happen.
If a photon moving left to right bounces off a mirror that is moving right to left then the photon gains energy (at the expanse of the mirror) and it leaves with a shorter wavelength than it starts.

You need to stop ignoring that fact.

So, in real-life there's no way out of this situation,

Yes there is. I pointed it out earlier in the thread.

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.

But you ignored it.

(and yes - 100% reflective surfaces actually DO exist)

No, they do not. (Unless they are infinitely massive).
Because, if a photon hits them, it transfers momentum to the mirror and (usually) loses energy.
They are "perfect" in the sense that the number of photos bouncing off s the same as the number that hit, but they are not perfect in the sense of no energy transfer happening.

And this turns out to be incorrect,

Proof by repeated assertion isn't going to work here.
Particular not when you are doing the asserting, since it took you at least three goes to recognise Kirchhoff's law.

[ edited to add that I was mistaken; CrazyScientist still hasn't recognised it. Maybe he will learn this time]

[Bored chemist]

Yes - but:

1. It wasn't made in a cavity
2. mirror was moving at relativistic velocity

So, this scenario has nothing to do with the one, which I presented in the first post of this thread... Sorry...

There's no way the photon can tell if there's another mirror moving away behind it, so there's no way that the second mirror in a cavity could make a difference.

So your first objection is absurd.

Your second objection is absurd because, from the point of view of a  photon, all mirrors are relativistic. They never hit them slowly.

You know, that even 10yo kids know from physics and chemistry classes, that picturing photons as tiny balls/marbles moving through space, is completely incorrect

The man who was previously trying to pretend that sound waves act the same was as light waves just broke the irony record.

It's not good enough to say that you don't like the analogy, because what I was relying on wasn't the analog, but the experimental fact that reflection from a moving mirror will shorten the wavelength if the mirror is traveling towards the source.

So, obviously this statement can't be applied to a scenario where EM radiation is emitted inside a cavity

The conservation of energy can be applied universally.

Each system will reach an energetic equilibrium, as long as the input energy remains constant

Are you muddling power an energy here?

If you keep putting more net energy into a system, it does not reach equilibrium.

[CrazyScientist (OP)]

I'm curious: what was your research field in the late 80s?

Around this time I was researching the ability to walk and speak

[CrazyScientist (OP)]

Problem is, that when an EM wave hits a surface, it can interact with it only in two ways - the energy can be absorbed or it can be reflected. However even if the surface is 100% reflective, so the radiation isn't absorbed and turned into thermal energy, momentum transfer turns part of the energy carried by EM waves into kinetic energy of atoms that make the surface.

No.
As I have pointed out a few times, and posted a link to the experimental verification , there are three things that can happen.
If a photon moving left to right bounces off a mirror that is moving right to left then the photon gains energy (at the expanse of the mirror) and it leaves with a shorter wavelength than it starts.

You need to stop ignoring that fact.

1. MIRROR HAS TO MOVE AT RELATIVISTIC VELOCITY
2. IT DOESN'T WORK FOR A STANDING WAVE IN A CAVITY
3. IT HAS NOTHING TO DO WITH THE SCENARIO, WHICH I PROPOSED IN THE FIRST POST

So, in real-life there's no way out of this situation,

Yes there is. I pointed it out earlier in the thread.

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.

But you ignored it.

1. It would require the existence of EM radiation at the same frequency as the one emitted inside the cavity beyond that cavity - and my scenario doesn't include that factor

2. It still wouldn't increase the energy level inside the cavity

(and yes - 100% reflective surfaces actually DO exist)

No, they do not. (Unless they are infinitely massive).
Because, if a photon hits them, it transfers momentum to the mirror and (usually) loses energy.
They are "perfect" in the sense that the number of photos bouncing off s the same as the number that hit, but they are not perfect in the sense of no energy transfer happening.

Yes. But still even if a 100% perfect mirror would exist, it still wouldn't increase the energy level inside cavity

And this turns out to be incorrect,

Proof by repeated assertion isn't going to work here.
Particular not when you are doing the asserting, since it took you at least three goes to recognise Kirchhoff's law.

[ edited to add that I was mistaken; CrazyScientist still hasn't recognised it. Maybe he will learn this time]

I don't care. What matters, is the fact, that it was YOU, who tried to apply it to the proposed scenario - and it didn't work well.

But it's great, that you are capable to recognize your own mistakes - many people have serious problems with that...

[CrazyScientist (OP)]

Yes - but:

1. It wasn't made in a cavity
2. mirror was moving at relativistic velocity

So, this scenario has nothing to do with the one, which I presented in the first post of this thread... Sorry...

There's no way the photon can tell if there's another mirror moving away behind it, so there's no way that the second mirror in a cavity could make a difference.

So your first objection is absurd.

And this is where you're wrong again...
Photons DO "know" somehow, if they create a standing wave or not... Research it...

Your second objection is absurd because, from the point of view of a  photon, all mirrors are relativistic. They never hit them slowly.

Then why the frequency of EM waves doesn't increase, when they are reflected by a stationarry mirror or one, that doesn't move fast enough?

You know, that even 10yo kids know from physics and chemistry classes, that picturing photons as tiny balls/marbles moving through space, is completely incorrect

The man who was previously trying to pretend that sound waves act the same was as light waves just broke the irony record.

Yes - and my analogy of a buzzing speaker submerged in a sphere of water IS in 100% valid...

Learn about the research of PHONONS

It's not good enough to say that you don't like the analogy, because what I was relying on wasn't the analog, but the experimental fact that reflection from a moving mirror will shorten the wavelength if the mirror is traveling towards the source.

Sure - only it doesn't work in cavities...

So, obviously this statement can't be applied to a scenario where EM radiation is emitted inside a cavity

The conservation of energy can be applied universally.

Each system will reach an energetic equilibrium, as long as the input energy remains constant

Are you muddling power an energy here?

If you keep putting more net energy into a system, it does not reach equilibrium.

I never said, that the system is powered by a source, that is increasing it's power output...
If the source of radiation is powered by a battery with a specific and constant power output, then the system WILL reach equilibrium - despite you saying otherwise

[CrazyScientist (OP)]

It's also interesting, how perfectly you managed to completely ignore this:

Ok, so let's now assume, that the battery is actually placed somewhere outside of the spherical cavity and is powering up a source of heat (thermal radiation), allowing it to maintain a constant temprature of 1000°C. Can you explain the mechanism, which in this case could possibly lead to creation of a BH out of the thermal radiation, which is constantly emitted by the source of heat inside the cavity?

Tell me, how you will solve that. I would love to hear some cool fairytale for good night

[Bored chemist]

MIRROR HAS TO MOVE AT RELATIVISTIC VELOCITY

A snail moves at a relativistic velocity. But the relativistic correction is small.

IT DOESN'T WORK FOR A STANDING WAVE IN A CAVITY

It's not meaningful to talk about a standing wave in a cavity which is shrinking or moving WRT the source.

IT HAS NOTHING TO DO WITH THE SCENARIO, WHICH I PROPOSED IN THE FIRST POST

The scenario you proposed was dull.
All it did was flatten a battery and warm up some stuff in a fancy box.

If there was enough stored energy in the sphere to create a BH the size of that sphere (or bigger) then it would collapse anyway.

If there was not enough mass to do so then it wouldn't collapse anyway unless you did something more complicated than turn on a light.

So the situation can only be interesting if  you talk about the other scenario which you mentioned.
The one with a valve that lets light in, or (roughly equivalently) where the wires feeding the lamp lead outside the sphere.

Learn about the research of PHONONS

I did that a bit before the time you should have been learning about where to use CAPITAL LETTERS, and when not to,

Sure - only it doesn't work in cavities...

Except, obviously, if the cavity is big then the "far end" is distant enough that its existence is unknown to the interactions at this end  because the information about it's presence would only arrive at the speed of light.
In that case, the photons don't  know about it.
They set out from the source, get hit by a moving mirror, their wavelength is shorted (and you can demonstrate that classically if you get bored) and they make their way back the way they came in a shorter interval than any information about the far end of the cavity could reach them.

You still seem to ignore the fact that time is reversible. If a photon hitting a stationary atoms sets it moving then (in an ideal system) that same photon could be sent back and bring it to a halt by bouncing off it again.

And you still seem to ignore the fact that , if a mirror hits a photon, momentum is transferred to the mirror.
If the mirror was originally moving towards the photon, then the mirror will be slowed down by the momentum transfer.
If the mirror is slowed down then it loses energy.
And the conservation laws say that energy has to go somewhere.

The only possibility is that it goes into the photon, and raises its energy.

Where else?

I never said, that the system is powered by a source, that is increasing it's power output...

Nobody said you did.


If you have a mirrored sphere with a "light valve" in it, and you shine a 1 Watt laser into it then the energy in the sphere rises by 1 J/s.
And that means the mass rises by 1/C^2 Kg/ sec.

It won't reach an equilibrium- because there's always more energy being added.

If the sphere had a "leak" then yes, there would be a point where energy left at the same rate that it entered.
But, unless you are moving the goalposts to make the system boring, there is no leak because the mirrors are prefect.
There is (as I explained twice) a way round the infinite mass requirement.
If the mirror is allowed to move slightly then it will thermalise the photons in the sphere. You can keep putting more and more mass (as energy) into it until it collapses.

You may recall that I said that earlier. In doing so, I made it clear that I know that photons aren't little balls (which wouldn't thermalise)- but yet you somehow thought I did.

Yes - and my analogy of a buzzing speaker submerged in a sphere of water IS in 100% valid...

No. Because the compressibility is non linear at high sound levels. Essentially, in air it's possible to get a pressure peak of 2 atmospheres, but you can't get a trough of minus two.
However because, as you say, there's not much interaction between photons, there's no such (relevant) limitation on light energy until there are so many that gravity takes an effect.

[Bored chemist]

It's also interesting, how perfectly you managed to completely ignore this:

Ok, so let's now assume, that the battery is actually placed somewhere outside of the spherical cavity and is powering up a source of heat (thermal radiation), allowing it to maintain a constant temprature of 1000°C. Can you explain the mechanism, which in this case could possibly lead to creation of a BH out of the thermal radiation, which is constantly emitted by the source of heat inside the cavity?

Tell me, how you will solve that. I would love to hear some cool fairytale for good night

It's not much of a story.
This bit

allowing it to maintain a constant temprature of 1000°C.

implies that there's some system which measures the temperature of the sensor and adjusts the power in order to maintain 1000C - a thermostat.
The hot body is enclosed by a perfect mirror. (I'm assuming it's in a vacuum too and is thus thermally insulated).
Power is fed to it and that heats it.
As it reaches 1000C the sensors of the thermostat throttle back the power from the battery.
Once the body is at 1000 C the power fed to it drops to zero.
The body stays hot because there's nowhere for the thermal energy to go.
You can take the battery away now, if you like.

I have a different question or two.
If you remove the thermostat and just let the heater deliver 1 watt of heat to the inside of the sphere (from which none can escape) , and if the heater is magic so it won't burn out,
(1) what happens to the mass of the sphere as a function of time and
(2) what happens win the end?

[CrazyScientist (OP)]

It's also interesting, how perfectly you managed to completely ignore this:

Ok, so let's now assume, that the battery is actually placed somewhere outside of the spherical cavity and is powering up a source of heat (thermal radiation), allowing it to maintain a constant temprature of 1000°C. Can you explain the mechanism, which in this case could possibly lead to creation of a BH out of the thermal radiation, which is constantly emitted by the source of heat inside the cavity?

Tell me, how you will solve that. I would love to hear some cool fairytale for good night

It's not much of a story.
This bit

allowing it to maintain a constant temprature of 1000°C.

implies that there's some system which measures the temperature of the sensor and adjusts the power in order to maintain 1000C - a thermostat.
The hot body is enclosed by a perfect mirror. (I'm assuming it's in a vacuum too and is thus thermally insulated).
Power is fed to it and that heats it.
As it reaches 1000C the sensors of the thermostat throttle back the power from the battery.
Once the body is at 1000 C the power fed to it drops to zero.
The body stays hot because there's nowhere for the thermal energy to go.
You can take the battery away now, if you like.

Then no creation of black hole will occur?

If so, then what is the difference between a source of light and a source of heat, so that one is able to create a black hole and the other isn't?

My answer is: NONE. Scenario, which I presented can't lead to creation of a BH, no matter what kind of EM radiation will be trapped inside the cavity

You can take the battery away now, if you like.

Only then some part of the energy inside the cavity will get absorbed back by the source...

[EDIT] - I see no sense in responding to your previous post, as the outcome of my scenario, which you've presented here, pretty much invalidates everything, what you said before and proves most of my claims from this thread - constant emission of EM radiation inside a reflective cavity, will ALWAYS lead to energetic equilibrium of the system... If you don't agree, you'll have to prove that thermal radiation is somehow completely different from other types of EM radiation and only because of that, this particular scenario (thermal radiation) will have a different outcome.

[Bored chemist]

I see no sense in responding to your previous post,

There are many things you do not see; principally because you are starting from mistaken beliefs.
Neither repetition of a falsehood, nor putting it in CAPITALS, makes it true.
 

Only then some part of the energy inside the cavity will get absorbed back by the source...

Do you know that flames absorb light?
It's used a lot in chemical analysis.
https://en.wikipedia.org/wiki/Atomic_absorption_spectroscopy

So your idea that a hot body won't absorb until after you stop supplying power to it is simply wrong.
A fact observed in many labs around the world every day.
(And, of course, it's still one of the documented laws of physics).

But the big problem is still your refusal to accept the conservation of energy.
You really need to revisit your view on this.

And you still seem to ignore the fact that , if a mirror hits a photon, momentum is transferred to the mirror.
If the mirror was originally moving towards the photon, then the mirror will be slowed down by the momentum transfer.
If the mirror is slowed down then it loses energy.
And the conservation laws say that energy has to go somewhere.

The only possibility is that it goes into the photon, and raises its energy.

Where else?

[Bored chemist]

If you don't agree, you'll have to prove that thermal radiation is somehow completely different from other types of EM radiation and only because of that, this particular scenario (thermal radiation) will have a different outcome.

The different outcome arises from your decision to turn off the heater when it reaches 1000C.
That's why I asked about the case where that does not happen.
I guess you realise that, if the energy kept going in then you would eventually get a black hole , but don't want to admit it.
That would certainly explain your refusal to discuss it.