[Bored chemist]

It doesn't  mean, that I'm completely wrong here, but the subject is just far more complicated..

Yes it does.
You said this

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

which is completely wrong.

[Bored chemist]

How can the size of cavity be uncertain,?

Because of the uncertainty principle.

[CrazyScientist (OP)]

What frame are you talking about here?

The laboratory frame.
I have a box on teh bench with photons bouncing round it it.
I compress the box.
In doing so, I do work on the mirrors and they do work on the photons.
Where does that energy go?

Here's a hint; it has to go somewhere, and it can't be absorbed by the mirrors.

And where goes the momentum of waves, which get cancelled out due to their destructive interference?

https://physics.stackexchange.com/questions/23930/what-happens-to-the-energy-when-waves-perfectly-cancel-each-other

Yeah, good qustion... Somewhere? Nowhere? Even professional physicists aren't sure, what's going on in here...

Here's another hint.
It is the equivalent of heating a gas by compressing it.

Yeah - that's another of "those" question, where physicists dump on you couple megatons of information without giving any clear answers...

https://physics.stackexchange.com/questions/136408/what-happens-to-the-temperature-when-an-ideal-gas-is-compressed

You need to read about the Joule-Thompson effect in Wikipedia. In an ideal gas there is no temperature change upon compression or expansion. The only gases that come close to being ideal at room temperature are Helium, Hydrogen and Neon. They actually slightly cool on compression and heat on expansion at room temperature. This can be reversed at very low temperatures. Most non-ideal gases such as nitrogen, oxygen and carbon dioxide do heat on compression and cool on expansion. With carbon dioxide having the biggest temperature change for a given pressure change. This Joule-Thompson effect is due mostly to Van der Waal forces between the molecules.

[CrazyScientist (OP)]

Because of the uncertainty principle.

Cavity isn't made of photons - it's a fully defined physical object. You make a cavity at a specific size and you keep having it at this specific size

[Bored chemist]

And where goes the momentum of waves, which get cancelled out due to their destructive interference?

It's quite an interesting question- bright schoolkids usually ask a related one.
When you show them interference fringes on a screen, the clever pupils ask where the energy from the dark stripes has gone to.
And the answer is that it goes into the bright stripes.

It's the same with the momentum.

[Bored chemist]

Because of the uncertainty principle.

Cavity isn't made of photons - it's a fully defined physical object. You make a cavity at a specific size and you keep having it at this specific size

Are you really stupid, or just acting?

Do you realise that the "physical object" is made of atoms, that those atoms are particles and that the position of particles is uncertain because of the uncertainty principle?

You have so much to learn; you could start with a bit of humility.

[CrazyScientist (OP)]

It doesn't  mean, that I'm completely wrong here, but the subject is just far more complicated..

Yes it does.
You said this

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

which is completely wrong.

Yeah - I was wrong. You can get multiple wavelenghts in a given band superimposed in a single wavefunction. But this isn't exactly what I was trying to imply. What I had in mind, is the fact, that you can't overlap two EM fields at the same bandwidths in one volume of space and the intensity of radiation at a specific wavelenght can't grow to infinity - radiation reaches it's maximal intensity at a given frequency, when all the measured photons will be at that frequency

[Bored chemist]

You need to read about the Joule-Thompson effect in Wikipedia.

No, as usual, it's not me who needs to do some reading.
Eventually, you will realise that, like the FM radio and the bat and ball cases, you are not applying  the right science.

The JT effect applies when a gas expands, but does no work.

Here's the right physics.
https://www.grc.nasa.gov/www/k-12/airplane/compexp.html

As you see, it's all done with ideal gases.


https://www.tec-science.com/thermodynamics/thermodynamic-processes/why-does-pressure-and-temperature-increase-during-the-compression-of-a-gas/

[Bored chemist]

Yeah - I was wrong.

Repeatedly, in spite of being told what is right.

[Bored chemist]

that you can't overlap two EM fields at the same bandwidths in one volume of space

FFS!
Yes you can.
You can shine two beams of light through each other.
You can send two radio signals "through" eachother.
They can overlap; and they do, notably on my bedside table where there are two radios.


Seriously; stop posting, you are just wasting time and embarrassing yourself.
Go off to the Kahn academy or something and learn some science.

[CrazyScientist (OP)]

Because of the uncertainty principle.

Cavity isn't made of photons - it's a fully defined physical object. You make a cavity at a specific size and you keep having it at this specific size

Are you really stupid, or just acting?

Do you realise that the "physical object" is made of atoms, that those atoms are particles and that the position of particles is uncertain because of the uncertainty principle?

You have so much to learn; you could start with a bit of humility.

So I can't measure the lenght of each single hair on my head? Can't I take one hair, precisely measeure it's lenght, freeze it almost to absolute 0 and store it in some almost-perfectly sealed container - so anyone who will measure lenght of that hair during next thousand years, will get the same value, as I did? 

Can I then measure anything? How can we then know, if 10cm will be always equal to 100mm and not to any other lenght?

The answer is: YES - WE CAN MEASURE BOTH: THE SIZE OF CAVITY AND THE WAELENGHTS OF TRAPPED RADIATION

Uncertainty principle doesn't make you unable of a very precise measurement...

[Bored chemist]

Can I then measure anything?

You can't- unless you are prepared to accept some uncertainty.
That's what the uncertainty principle  tells you.

Seriously; stop posting, you are just wasting time and embarrassing yourself.
Go off to the Kahn academy or something and learn some science.

[Bored chemist]

Uncertainty principle doesn't make you unable of a very precise measurement.

It doesn't need to in this case; it only needs to prevent  you making a perfect measurement.

Strictly, what it does is tell you that a perfect measurement is meaningless.
It's not an issue of measuring things.
The diameter of the sphere does not exist to a greater precision than that permitted by the uncertainty principle.

From one side of the sphere's point of view, if the sphere behaved in the Newtonian way you seem to want it to (rather than the hundred year old QM way) the other side would be fixed in place- so its positional uncertainty would be zero, an it also would not be moving (the speed would be zero) - so the product of the uncertainties of the momentum and the position would be zero.
That's not allowed.

Considering how much time and space you have wasted telling me to look at new physics, you have a remarkably poor understanding of the stuff from a century ago.

[CrazyScientist (OP)]

that you can't overlap two EM fields at the same bandwidths in one volume of space

FFS!
Yes you can.
You can shine two beams of light through each other.
You can send two radio signals "through" eachother.
They can overlap; and they do, notably on my bedside table where there are two radios.


Seriously; stop posting, you are just wasting time and embarrassing yourself.
Go off to the Kahn academy or something and learn some science.

But without a gain medium, crossed beams won't result in the amplification of light
Yes -

You can send two radio signals "through" eachother.

And this is where you should now apply the uncertainty principle - you can find a photon at 2 wavelenghts in one volume of space, but you can't know it's exact position at each of those wavelenghts.. And each precise measurement of wavelenght and/or location of a photon, will make the second measured property of that photon unknown/uncertain

If you take 2 one-directional sources of white light at slightly different intensities and make them shine at us, so that both beams will intersect eachother on the way, amplitudes of both beams won't add up and won't result in the increase of light intensity in the intersecting area - instead white light emitted by the stronger source will be just as bright as before, while the white light at lower intensity will appear dark or even black if the differency of intensities gets high enough...

In shortcut you simply can't keep adding more and more light into a cavity, so long as it won't turn into a BH. Intensity of EM radiation doesn't work like density of matter...

[CrazyScientist (OP)]

Uncertainty principle doesn't make you unable of a very precise measurement.

It doesn't need to in this case; it only needs to prevent  you making a perfect measurement.

Strictly, what it does is tell you that a perfect measurement is meaningless.
It's not an issue of measuring things.
The diameter of the sphere does not exist to a greater precision than that permitted by the uncertainty principle.

From one side of the sphere's point of view, if the sphere behaved in the Newtonian way you seem to want it to (rather than the hundred year old QM way) the other side would be fixed in place- so its positional uncertainty would be zero, an it also would not be moving (the speed would be zero) - so the product of the uncertainties of the momentum and the position would be zero.
That's not allowed.

Considering how much time and space you have wasted telling me to look at new physics, you have a remarkably poor understanding of the stuff from a century ago.

Sure, sure... I just have no idea how any of this suppose to invalidate my claims about limited number of frequencies, which can form a standing wave in a given cavity and become amplified... This is exactly what is being observed in real-life and what the cavity QED is based on. But for some reason you keep insisting, that in case of my scenario, there's no need to speak about standing waves in a resonant cavity and that it has generally nothing to do with resonance of waves, while having a lot with baseball, cricket and black holes of light...

Here's how much all of this have to do with baaseball or cricket:

I'm sure that it would be quite easy to score a perfect hit on a ball that remains suspended in space - if that ball wouldn't be suspended at multiple locations at the same time...

It would be more like the ball would be standing on a stick in multiple positions on the field and there would be a specific propbability of hitting that ball with a baseball bat at each of those locations. I've heard that cricket has rather weird rules (I don't know them) - but this game would be for sure much more exotic...

TBC soon - it's time for me to sleep a bit...

[CrazyScientist (OP)]

Ah! One more thing - for standing EM waves in reesonant cavities, probability of detecting a photon close to the boundary of that cavity (mirror) has to be  zero - so in the baseball analogy, ball won't never get to the batsman, but will keep oscillting somewhere in the middle of the area between the thrower and the batsman... That would ne quite interesting indeed...

[Bored chemist]

I just have no idea

You are saying that only the wavelengths that "fit" in the cavity are allowed.
They "fit" if the cavity diameter  is an exact multiple of the wavelength.

But nether the wavelength nor the diameter is "exact" in the first place,
so your criterion makes no sense.

crossed beams won't result in the amplification of light

Nobody said it would.
But what you said was that they could not be there.

you can't overlap two EM fields at the same bandwidths in one volume of space

and, in reality you can.
So you are still wrong, and getting wronger.

If you take 2 one-directional sources of white light at slightly different intensities and make them shine at us, so that both beams will intersect eachother on the way, amplitudes of both beams won't add up and won't result in the increase of light intensity in the intersecting area - instead white light emitted by the stronger source will be just as bright as before, while the white light at lower intensity will appear dark or even black if the differency of intensities gets high enough...

This seems to be a pipe dream.
It's possible that there's a point in there somewhere, but it is buried so deep in rubbish that it's impossible to tell.

[Bored chemist]

Just keep reading this
"so photons, which are slowing down the mirror are giving up their own energy"
until you realise it is nonsense.

[CrazyScientist (OP)]

I just have no idea

You are saying that only the wavelengths that "fit" in the cavity are allowed.
They "fit" if the cavity diameter  is an exact multiple of the wavelength.

But nether the wavelength nor the diameter is "exact" in the first place,
so your criterion makes no sense.

Sure, there are limitations of our current technology, but still there's an equation, which allows you to calculate the EXACT number of wavelenghts, which will fit in the cavity.

crossed beams won't result in the amplification of light

Nobody said it would.

But in order to create a "Kugelblitz" you need to have a constant amplification of the emitted light

But what you said was that they could not be there.

you can't overlap two EM fields at the same bandwidths in one volume of space

and, in reality you can.
So you are still wrong, and getting wronger.

2 EM fields with similar bandwidth can SHARE the probability distribution in one volume of space. If they would overlap each other, their amplitudes would be added. Thing is, that when 2 (or more) EM waves at similar wavelenghts pass through the same space, probability of detecting a photon at one of those wavelenghts is being shared between interfering waves, according to their intensities of that probability

If you take 2 one-directional sources of white light at slightly different intensities and make them shine at us, so that both beams will intersect eachother on the way, amplitudes of both beams won't add up and won't result in the increase of light intensity in the intersecting area - instead white light emitted by the stronger source will be just as bright as before, while the white light at lower intensity will appear dark or even black if the differency of intensities gets high enough...

This seems to be a pipe dream.
It's possible that there's a point in there somewhere, but it is buried so deep in rubbish that it's impossible to tell.

But this is exactly, what happens. Just check out that movie or compare the brightness of your cell phone in a dark room and in the daylight. Screen will always appear to get darker in the sunlight...

Just keep reading this
"so photons, which are slowing down the mirror are giving up their own energy"
until you realise it is nonsense.

But they ARE giving up their energy...
Thing is, that in the frame of laboratory, this loss can be compensated due to Doppler shift of reflected light - but only if the mirror is moving fast enough and in opposite direction to incoming photons

https://physics.stackexchange.com/questions/539983/energy-conservation-in-reflection-of-light-from-a-perfect-mirror

[CrazyScientist (OP)]

I think, that when comes to optical cavities with a moving mirror, here's a pretty comprehensive info:
http://cpb.iphy.ac.cn/article/2018/1924/cpb_27_2_024204.html