A Strange Question On Cosmic Microwave Background Radiation.

More Articles

About Ladislav

Born in Prague (CZ), studied physics. Started with algol programming on GEAR-1 in Rez of the shell model of nuclei in 1966. Moved to Bergen, Norway...

View Ladislav's Profile

Ladislav Kocbach

I just ask. If I look at a glowing something with an antenna, I will not get any signal, I think. This is, I believe, because the radiation comes as photons, more or less, because it is a result of thermal motion. Each of the myriads of electrons in the glowing material moves independently and each emits the energy as a photon. The antennas can not see photons. They can see only electromagnetic waves.

To get electromagnetic waves, I must have an ordered motion, some oscillating current of some kind. In the optical region of frequencies we have lasers which also produce coherent radiation, i.e. something looking more like a wave than as a stream of individual photons, as R. J. Glauber got a 2005 Nobel prize for explaining in the sixties. The laser light I cannot detect by antenna, because that does not work. But many atomic processes due to lasers confirm this picture, the intense pulses are not just streams of photons. They are coherent. This is what I work with.

Having this is mind, I then wonder. How is this with the cosmic microwave background

( http://en.wikipedia.org/wiki/Cosmic_microwave_background_radiation )? It is detected as waves ( http://en.wikipedia.org/wiki/Radio_telescope ), not as photons. Definitely. Those are antennas detecting it. So what was lasing in the Big Bang or just after it? Can somebody explain this? Why is the cosmic microwave background a coherent radiation? Or - what is wrong with my question?

Comments

Ladislav -- the cosmic microwave background is black body radiation. So nothing is 'lasing' or 'masing'. You seem to make a distinction between photons and EM waves, with the latter associated with 'lasing', and the former with incoherent radiation. That is incorrect, EM waves are nothing more than the large scale manifestation (the classical limit) of photons. So an antenna does detect (the aggregate effect of) photons. Radio wavelength photons, that is.

Our understand of black body radiation is only a century old. So it is not that simple. Energy quantization of the EM field plays a dominant role, and so does the assumption of thermal equilibrium of the EM field (the photons) and the object that radiates. See here for a summary of the derivation.

Johannes Koelman | 02/15/12 | 08:11 AM

Dear Johannes, there is clearly a big difference between incoherent photons and an observable coherent wave, like the radiowave we use to distribute advertisements and music on the AM radio (the FM is already a bit mixed up, but still a wave). Thinking about photons when talking about AM radio or the RF signal in magnetic resonance imaging would make you not pass your exam with me. I am state authorized to not to let you pass. So please, go and read the Glauber papers and lots of similar literature. Wikipedia got it probably equally wrong as you did.
But you are excused. My question is one of the uncomfortable questions, which immediately triggers answers of the type you present.

In short, your answer is not accepted, tou do not pass. But thank you for trying!

Ladislav Kocbach | 02/15/12 | 10:20 AM

Thinking about photons when talking about AM radio or the RF signal in magnetic resonance imaging would make you not pass your exam with me. I am state authorized to not to let you pass.

Oh - that is how science works in your state. Which state is that? One where all the random stuff can never lead to biology and so god must have added some coherence? ;-)
Hate to side with the lazy armchair philosopher on anything, but BB = BB (Big Bang = Black Body).

Sascha Vongehr | 02/15/12 | 20:12 PM

On the rare occasions that Johannes and Sascha agree, there is no hope in hell for any alternative view. So, I suggest the author of this piece to humbly hand back to his state his examination rights for physics. Seriously, how can anyone maintain thermal EM radiation does not exist? Ever seen an incandescent light? Yes, that is the same radiation as CMB.

MaxP (not verified) | 02/15/12 | 21:19 PM

This is a bit longer, but I do not think it can pass either. Glauber's coherent states explain to me that when in a situation looking as a wave, I must have precisely that - coherent states. I can not have a bunch of photons of roughly the same frequencies or a whole spectrum of frequencies - a situation which I would have in a "black body". In any thermal situation I simply can not have coherent waves. I could be heating it by RF for example to bring it to the wanted temperature, but when I have thermalized it, there is no coherence left.
Think about your microwave oven - you have standing waves of some 12cm wavelength, but your soup will be in a nice black-body manner emit infrared corresponding to some 360 K with no coherence whatsoever, you can rely on that. Waves in, incoherent photons out.

As I (perhaps wrongly, but not on a physical side, maybe on the technical side) understand it, the background microwave should be of the blackbody, i.e. thermal type. Thus incoherent thermal noise at about hundred times longer wavelengths spectrum than your soup.

But it might be so that the telescopes work as some types of resonators which simply absorb a given frequency component from any noise - so the answer to my strange question would explain how the radio telescopes work. I do not know, but would like to know.

What I know is that many of my fellow physicists have issues with understanding photons and "classical" or rather "macroscopic" fields.

In your answer, dear Derek, I do not get it why to heat some antennas.

My point was that we could not see a thermal noise consisting of incoherent radiation at all by an antenna. It seems to me that the fact that we can detect it by an antenna type arrangement excludes its thermal origin as we understand it. It is a strange question, but I hope for some good answer.

Ladislav Kocbach | 02/15/12 | 10:48 AM

Ladislav:

Perhaps you need to read more about the actual Discovery of cosmic microwave background radiation. (That's a Wikipedia source, but there are more fundamental sources that go into further detail.) The principle issue was (and still is) with noise, not with normal "wave-like" signals.

I would hope that you would recognize that from a Quantum Mechanical perspective (especially Quantum Field Theory), electromagnetic "waves"/radiation are made up of photons (the electromagnetic quanta) whether one is talking about coherent or incoherent cases—whether one can detect "wave-like" behavior, or not.

The famous R. J. Glauber papers are focussed entirely upon "coherent radiation". So they are not to be confused with any argument as to whether incoherent "states" are "waves", or "photons", or something all-together different.

Come on, Ladislav, I'm sure you can "get it".

David

David Halliday | 02/15/12 | 12:21 PM

Thank you David for the link, yes I should have seen that before. This quote from there might have answered my strange question: "When Penzias and Wilson reduced their data they found a low, steady, mysterious noise that persisted in their receiver. This residual noise was 100 times more intense than they had expected". So yes, noise, no waves. But now I would like to know more anyway. Because I checked the energy flux and it seems very small, but then can wonder. So first part of your comment is definitely close to answering my "strange question", but opening for others.
But the rest is very interesting. No, I m not so sure that I would agree to that all electromagnetic radiation is "made of photons". It seems to be the opposite: the photons are energy quanta of the electromagnetic radiation, which might not quite be the same. Yes, we have in the textbooks, but do we need to believe it?

Look at this: It is in the textbooks that electric field is caused by exchange of photons (maybe virtual in the better ones). Do you believe that? That field of MegaVolt/meter in some commercial van de Graafs caused by photons? I do not think that we have a good theory for that. It has been somehow postponed till later.

You wrote:

The famous R. J. Glauber papers are focussed entirely upon "coherent radiation". So they are not to be confused with any argument as to whether incoherent "states" are "waves", or "photons", or something all-together different.

Well, when we have coherent radiation, it looks more or less as a wave. There is a strong electric field present, oscillating with the given frequency. It is not useful to speak about "photons", because their number is not known. For the totally incoherent radiation, the number of photons "is known" whatever that means. Among other things it means that the electric field is "not known". This is a simple account found in more up to date textbooks. But then we can have all those accidental coherence and fluctuations in the "real life". It has to do with the origin of the radiation. If we assume that it originates from transitions between discrete states, and without much stimulated emission, it should be in the form of quanta. If there can be a lot of stimulated emission, one could have a lot of coherence and thus a lot of wave-likeness. I guess it would depend on both the energy density and the matter density, but just now I have not the answers. But then the radiation density depends only on the temperature, i.e. it determines the temperature in thermal equilibrium.

To make it short: Your link probably answered my strange question: it is noise, no "waves". Now the question is even stranger: Is it noisy enough? And if it is not, can it say something about the state of the matter which got decoupled from the radiation at the "time of last scattering"?

Ladislav Kocbach | 02/15/12 | 14:01 PM

Ladislav:

It often seems like most good answers lead to many more questions. ;)

Note, of course, that when I wrote (emphasis added) "electromagnetic 'waves'/radiation are made up of photons (the electromagnetic quanta) whether one is talking about coherent or incoherent cases—whether one can detect 'wave-like' behavior, or not" I did preface this with "from a Quantum Mechanical perspective (especially Quantum Field Theory)". In other words, I did not try to assert that this is the one true and only perspective/explanation. (The latter would be unscientific, actually.)

You state:

Look at this: It is in the textbooks that electric field is caused by exchange of photons (maybe virtual in the better ones). Do you believe that? That field of MegaVolt/meter in some commercial van de Graafs caused by photons? I do not think that we have a good theory for that. It has been somehow postponed till later.

Well, haven't you heard of Quantum Electrodynamics (QED)? It is an application of the more general field of Quantum Field Theory (QFT, that I mentioned before). This theory works quite well for the above mentioned "issue". In fact, it works far better than classical electromagnetism in handling fine structures and energies that we can measure in the electromagnetic interactions of atoms. (It is arguably the most precisely tested theory mankind has.)

As another point of fact, before Planck derived his black-body radiation formula, he, and others, had already tried to use the continuum electrodynamics in their calculations, and failed miserably! Then, perhaps almost entirely as a means for approximate computation, and handling the statistics properly, Planck decided to work with "discrete" pieces of electromagnetic energy—almost certainly expecting to take the limit as the discretization "size" went to zero (a continuum).

However, since there was never any prior suggestion that electromagnetic fields came in any finite sized "packets", it must have been quite a surprise to him to find that the closest match was obtained with some finite sized discretization: Given by Planck's constant, significantly before anyone thought of quantum anything. (Einstein's own idea of using discrete pieces of light, in his explanation of the photoelectric effect was inspired almost exclusively by Planck's black-body work.)

As for your assertion:

Well, when we have coherent radiation, it looks more or less as a wave. There is a strong electric field present, oscillating with the given frequency. It is not useful to speak about "photons", because their number is not known. ...

This is quite incorrect. Having coherent radiation is the easiest for determining how many photons are involved—especially if the radiation is monochromatic. However, you are correct if you want to try and determine when each photon is emitted, for instance. (Good old uncertainty relationship. Of course, this relationship holds just as true [with somewhat different constants] for any continuum wave phenomenon.)

You go on with:

... If we assume that it originates from transitions between discrete states, and without much stimulated emission, it should be in the form of quanta. If there can be a lot of stimulated emission, one could have a lot of coherence and thus a lot of wave-likeness. ...

Uhm. Are you unfamiliar with the underlying quantum mechanical nature of "stimulated emission"? Hint, it's very much about discrete "photons", and "transitions between discrete states". It even relies upon the statistical nature of particles with integer spin (bosons).

As to whether the Cosmic Microwave Background Radiation (CMBR) is "noisy enough". You tell me: Which form of radiation tends to form broad spectral emissions; coherent, or incoherent processes? Which spectrum will be more continuous; noise, or music?

David

David Halliday | 02/19/12 | 16:26 PM

That field of MegaVolt/meter in some commercial van de Graafs caused by photons? I do not think that we have a good theory for that.

~ "That amazing human body with its gazillions of intricate cells all working together and provided with a conscious soul all caused by molecules randomly bumping around? I do not think that we have a good theory for that."

Sascha Vongehr | 02/15/12 | 21:56 PM

Ladislav, would you consider the light we see from the sky coherent?

EM fields coherent or not are all photons, we just call long wave photons radio waves, and much shorter waves photons ir,visible light,uv, x-rays and finally gamma rays.
Once you get into the waves lengths we call light, electrons are big enough(long enough, have enough freedom of movement I'm not sure which) to act as an antenna. Conversely conductive metals of the proper length(antenna) act as if they're a big electron and capture EM fields.
Like wise gamma rays are so short they almost all come from nuclei.

—

Never is a long time.

Mi Cro | 02/15/12 | 13:25 PM

Look above, please, answer to David. There I try to explain the difference between wave and stream of "photons". Ideal laser pulse looks as electromagnetic field, as a wave. Wave can be understod only as a coherent state of the quantized field, if we need to quantize - and we do. Radiation coming from transition between discrete states - and in most cases all states of microsystems are "discrete", and if it is not result of stimulated emission, looks like quanta. Each different from any other. There is no measurable field. Each of the quanta can be absorbed only by another microsystem - because the energy is so tiny. And then it becomes heat.
Only a wave can generate ordered response. That is the origin of my "strange question". Now I have got a partial answer - see above, David, - showing that the detected signal is a "noise". Now i only wonder whether it is noisy enough and more. So My first strange question is now slightly modified, but still not completely answered.
Unfortunately, your answer can not be accepted - you see it does not include the "classical" field states mentioned in this answer. I refered to them as Glauber States, but in fact Schrödinger invented them already in 1927, by pure luck.

Ladislav Kocbach | 02/15/12 | 14:19 PM

May be it is a solution to look at the problem from the point of view of low dose rates. Shot noise is a term that short radio wave amateurs call the signal at very low signal levels. It is a feature that characterizes Poisson processes. Instead of out of a wave, the highly amplified signal consists out of a series of spikes. Only at higher dose rates a wave like signal appears. So the quanta that constitute the wave have a probability density distribution that has the shape of a wave. You can see similar effects, when you look at the images that are produced at low dose rates. For example a night vision device that works under starlight conditions. Or see http://heasarc.nasa.gov/docs/rosat/gallery/misc_moon2.html .

So even antenna's show quantum detection behavior at sufficiently low dose rates.

—

If you think, think twice

Hans van Leunen | 02/15/12 | 13:51 PM

Interesting, I should look into this!

Ladislav Kocbach | 02/15/12 | 14:16 PM

This is exactly that. It's only at long exposures does the shot noise become a picture.

And if you zoom in, you'll see there's still noise.

But you also didn't answer my question about blue sky, and really in the same venue how photodiodes detect light, it's definitely quantum interactions (hence the shot noise in astrophotography).

—

Never is a long time.

Mi Cro | 02/15/12 | 14:52 PM

Ah yes, but in the case of photography you are dealing with quanta of an electron volt or two so there's plenty of energy for just one or two to be registered as a dot. So a picture made of noise is to be expected.

I am surprised to hear from Hans that short-wave radio quanta can be discerned above thermal noise in any system. The energy of a 30 MHz quantum would be 120neV corresponding to a temperature of about 1.5 mK. Hmmm...!!!???

Derek Potter | 02/15/12 | 15:19 PM

The amplifier may be guilty of raising extra noise, but in the noise often the modulated music or voice can be discerned.
The human visual system and audio senses seem to be optimized to discern information in noisy input. All vertebrates also enjoy the same visual setup. http://vixra.org/abs/1101.0065

—

If you think, think twice

Hans van Leunen | 02/15/12 | 15:37 PM

Derek Potter | 02/15/12 | 22:12 PM

Does this mean that photons radiating from an antenna powered by a radio transmitter are coherent? It would have to be that way otherwise a receiving antenna would not pick up any signal because the photons would all have random phases and would thus cancel out producing no signal.

David_S (not verified) | 02/16/12 | 01:24 AM

The detected signal was a random sample from a stochastic distribution that had a modulated wave form. Wave functions are probabitity amplitude distributions and EM-waves have in fact a similar structure. It is best understood in a quaternionic format. In that case the problem turns into a streaming problem. In the case of EM-waves, the streaming items are detectable quanta.
See: http://www.crypts-of-physics.eu/ConciseHilbertBookModel.pdf or http://www.crypts-of-physics.eu/Hilbert_Book_Model.pdf .

—

If you think, think twice

Hans van Leunen | 02/16/12 | 03:02 AM

i might reply to your question but:
1. "you not pass your exam with me"
2.i'm getting "comment waiting approval..."
3."don't argue with a fool,he will get smart and you only lose time-what for?"-a good friend of mine said that.
1+2+3=agravation.see you next time.

mobber (not verified) | 02/29/12 | 19:34 PM

ok.If you look at a glowing something with an antenna, you will get NOISE.in this arrangement the antenna "sees" the un-related photons,a thermal bath.if the antenna looks at a transmitter you will get coherent-in-time voltage at the terminals,an electric tension witch is far more predictable that noise-pure sinus or am/fm is incredibly simpler than noise-you can say that an electromagnetic wave is seen.coherence is a statistical relationship between individual photons and this coherence is TRANSLATED by the antenna into electrical current's properties.suppose a radio telescope is watching a pulsar:if you put the amplified "output" of the biggest antenna onto an oscilloscope you will see only noise.because there is still a relation,a coherence,a periodicity,the electrical signal can be processed,by periodic averaging,fourier transform and the result displayed nicely like a hart on EKG. in quantum optics,where in an experiment you have to account for everything you know of, even the vacuum must be taken into account,it's not just nothing,zero.example:you have a beam splitter at 45 degrees;a ray can pass or get deflected at 90 degrees.in this cross 3 sides are occupied by the beam ,the 4-th side by the void/vacuum/dark space-this must be described and taken in consideration.

mobber (not verified) | 03/01/12 | 18:35 PM

Comments

Add a comment

Know Science And Want To Write?

What's Happening

Books By Writers Here

Related Articles on Science 2.0

Comments

Add a comment

Know Science And Want To Write?

What's Happening

Books By Writers Here