Elementary problems concerning electromagnetic spectrum

1. Aug 12, 2012

prane

All electromangtic radiation comes under one continuous spectrum.

http://en.wikipedia.org/wiki/File:Electromagnetic-Spectrum.png

Different wavelengths of this radiation has different properties which is very evident in real life. However the whole concept remains extremely mysterious to me.

I am familiar with the wave-particle duality theory for light and I know for some purposes it makes more sense to describe light as a wave rather than a particle. With respect to the electromagnetic spectrum it makes more sense to think of light as a 'wave' with a certain wavelength. I know light is neither a wave nor a particle it is just LIGHT going along doing its usual thing it just happens that thinking of it as a particle sometimes and a wave othertimes helps explain certain macroscopic properties

However, what about other types of radiation? For example, what constitutes a radio wave? Is it a photon with a longer wavelength? Surely then it is no longer a photon? Also what happens at boundaries of the electromagnetic spectrum. Radio waves become microwaves somewhere around 10^(-1)m in wavelength but what is about these types of radiation which are at this unclear boundaries?

2. Aug 12, 2012

PeetPb

Well photons are transmitters of electromagnetic force so the radio waves are photones as well. Only difference is that human eyes cannot detect them because of their wavelength/frequency. Wikipedia says about the gamma radiation (which is about at 10^(-11)meters) that it has no lower limit on it's wavelength. So I guess everything under 10^(-11)m is considered gamma radiation.

3. Aug 12, 2012

mikeph

The boundaries are arbitrary, man-made divisions of a continuous spectrum. A photon is a photon, whatever its frequency.

4. Aug 12, 2012

sophiecentaur

Photons don't have 'a wavelength', as such - it's not the way to look at this. The photon may be looked upon as a particle but don't try to associate a size / extent with the photon. It's the wave that has the wavelength.

There is no essential difference between photons for different frequencies of EM - it's just the fact that they interact with different charge systems (i.e. electrons and protons in a nucleus, charges in a wire, nucleons within a nucleus etc) according to the energy involved. Once you start dealing with really low frequencies ('radio') the energy of individual photons is so small that they are pretty much impossible to identify because the noise in any detection system is too high to be able to spot individuals. The changeover is round about Infra Red frequencies, as far as I know.

5. Aug 12, 2012

davenn

no its not, radio waves cover a huge amount of the E-M spectrum .... eg from kHz to TeraHerz we just call them "radio waves" as a way to differentiate them from say infra red radiation, visible light, X-rays etc
in the radio spectrum we have sections named like ( as an approximate guide)
ELF --- Extra Low Frequency below 10 kHz
VLF --- Very Low Frequency ~ 10 kHz to 100kHz
LF ---- Low Frequency ~ 100 kHz to 1MHz
HF ---- High Frequency 1MHz to 30 MHz
VHF -- Very High Frequency 30MHz to 300 MHz
UHF -- Ultra High Frequency 300 MHz to ~ 1000 MHz
SHF -- Super High Frequency 1000 MHz to 30 GHz
EHF -- Extra High Frequency 30 GHz to 100 GHz

They are ALL part of the radio part of the E-M spectrum

radio waves DONT become microwaves, this is a common misunderstanding often seen in various forums .... microwaves ARE radio waves, to put it more clearly, the microwave radio frequencies start ~ 1000 MHz ( 30cm wavelength) and go upwards in frequency from there, they are still called microwaves right to the infrared part of the spectrum

cheers
Dave

6. Aug 13, 2012

prane

So you're saying all EM radiation can be thought of as radiowaves? The energies of the waves are then responsible for the different observable effects due to the wavelength being of a size comparable to the thing it is interacting with. Are you also saying that the particle responsible for this phenomenon is always a photon just with varying energies?

7. Aug 13, 2012

sophiecentaur

Yep.
[edit: it might be better to relate the energy of the photon to the energy in the system rather than the "size".]

Last edited: Aug 13, 2012
8. Aug 16, 2012

curiousatlarg

My understanding is that all of the above are EM, and all are photons, and the photons travel at the speed of light. The energy of a photon is inversely propotional to the wavelngth. The waves per meter per second for example would show the energy required to make that many ocillations per unit time. The more ocillations, the more energy required (and the more enery is being transmitted by the photon), the fewer ocillations, the less energy. The photons react vastly differently with different materials. One thing in common is that they travel through a vacuum. Other aspects include the refraction, reflection, and absorption. Yes it is mysterious! I am just beginnin to explore it myself, and having lot of fun.

9. Aug 17, 2012

sophiecentaur

The problem with that sort of model is that it seems to 'go the wrong way'. Relating the energy to the number of waves per metre would seem to imply that, for a given flux of energy, you would need fewer, rather than more photons for lower frequency EM waves. Why not go along with the accepted description? It works well enough.

10. Aug 17, 2012

curiousatlarg

Huh? No it wouldn't. The smaller the wavelength, the greater the energy of the photon. Overall flux can be raised by increasing the intensity of the photon stream, or by increasing the energy of the photons. A UV photon of say 200 nm has a greater energy than an IR photon, and much more than photons at longer wavelengths such as microwaves etc... But, if you want to present the "accepted view" I don't mind a bit. I am here to increase my understanding. The wave represents the magenetic and electric field right? The area between the path of the photon and the wave surface represents the energy if I am correct. The area is greater with more waves. Amplitude is variable regardless of frequency right? So then for all wavelengths, energy flux will increase proportional to frequency. Please correct me if I am wrong.

11. Aug 17, 2012

sophiecentaur

What does that mean? There is no "area". You are describing the EM wave as if it is some sort of surface wave with a spatial displacement which increases as the amplitude increases. The E and H fields are not spatial displacements - they can be represented by vectors but that's only a way of looking at it. Also, you make a huge assumption that the photon takes any particular path at all. The photon only 'acquires' a location at either end of its journey (at emitter or detector).
Yes, more photons corresponds to more energy flux but what does "more waves" mean? There is just one wave front for one frequency and you can't assume that one 'individual' photon is just one wave on its own. I often find it helps to think of coherent radio waves, rather than the non-coherent waves that come from most light sources whereas I often read, here, of laser light being somehow special. But is it different from what comes from a simple CW radio source?

12. Aug 17, 2012

curiousatlarg

OK, as for the waves, I refer to a representative graph of light waves. However, are not high frequencies also high energy ( E = hf)? As for the path, couldn't the flux be measured anywhere on the path? If so, then, for practical purposes, the photon follows a "path", although, having it appear and reappear is fine. Amplitude is the number of energy packets is it not? The electric field can be measured in two dimensions can't it? It could be measured with an antenna.

13. Aug 18, 2012

sophiecentaur

For high frequency EM, the Photons have higher energy. That is true.
What "practical purpose" does the 'path' of a photon serve in the argument? As soon as you commit to where the photon is on its path, QM tells you that the wave function collapses and it ceases to be a part of the wave. (Just read the millions of words which are written about the two slits etc. etc..)
You are using an unfortunate combination of wave and particle in your description.
The description that you gave was suggesting that, somehow, there is a 'wiggly line', traced by the photon. That is not a valid description. When you say that the electric field can be measured in two dimensions, (in fact it's three) that doesn't say anything about motion or change of spatial position; it just concerns the direction in which the E and H fields are pointing. The 'lengths' of the field vectors do not represent positions in geometrical space; they represent the amplitude at a particular position (not the same thing at all).

I don't understand how the comment about the antenna is relevant to this.

14. Aug 19, 2012

curiousatlarg

That is useful. I am still trying to visualize the E an H fields. I am working on reading the "millions of words".

15. Aug 19, 2012

sophiecentaur

Well - at least some of them haha.

There is a great temptation to hop from waves to photons and back again and it is very risky within one situation. Quite frankly, the wave treatment is the best for most circumstances that you meet in everyday life (other people may jump on me for that statement but I am talking generalities).
There is nothing 'wrong' with sticking to wave theory when discussing propagation, diffraction, RF circuits etc.. You will not be 'criticised' for ignoring photons when apppropriate. Remember, both approaches are valid and it is only when quantisation becomes relevant that you need to consider photons. It is definitely 'third year work' to try to explain the two slits using QM and it isn't necessary for predicting accurately the results. It is easy to fall over when trying to discuss what happens to photons on their journey from A to B (that is actually a vary naive thing to try to do) because they can be said only to exist at either end and to be 'anywhere and at no particular time' in between.
Try to avoid the 'little bullets' model.