Lenses

There you were reasonable. However, at the same time that there might be ways to say that multiple frequencies exist in that packet, all the definitions that I have seen, until this weekend, have implied that each photon has its own specific wavelength. This wavelength is derived from its energy. I was taught a long time ago that the "frequency" was derived by dividing the speed of light by that specific wavelength, so you get a single frequency by convention if not by physical fact.

 

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MetaKron, your post 61 shows that still do not understand what either I or James are telling you. Again here it is:

For there to be one precise wavelength or "pure frequency" the energy packet of the photon must be infinitely long. No photon is; however, some are very nearly pure frequencies / have precise wave length. Look up some and you will find some wavelengths given to 8 or even 10 significant figures. These nearly pure frequency photons are the ones I told you were "meters long."

As the energy and wavelenghts become less well defined, the photons are shorter, but I do not think any are less than 10cm long. Certainly none even slightly resemble your little balls one wavelength in diameter. If they were only 1 cm long, then they would have spectrographic film images that were a blur, not anything one could call a "spectral line."

You really need to learn the rudiments of Fourier Theory to undersatnd. If you re-read my posts many times, perhaps that would help you also to understand.

 

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Give a reference.

Also, see #61.

 

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You can quibble about the "frequency" but there is an established convention that photons are discrete packets of energy of a certain physical size, a wavelength determined by the amount of energy contained in each packet, and "frequency" is determined by wavelength. Billy started this quibble over my talking about the physical size of a photon. His information totally contradicts the established definition of a photon, its wavelength, and so on, and he insists that I am the idiot for going with the established definitions.

I've seen bullies in real life pretend to be even stupider than they were just to bother me. That's what is happening here.

Billy, I doubt if you can see reason, but I have references, I have an established convention, and I was talking about the estimated size of a given photon using the conventions that you will find in most physics textbooks. Your objections are completely specious and serve no useful purpose.

 

I will be ignoring MetaKrons post in this thread now, but did make an addition to post 58 for James R's benefit mainly. (Readers have been more than adequately warned that MetaKron is ill informed about the nature of photons and does not even understand correctly the things he simply quotes. Mainly his false ideas stem for lack of understanding what the literature is stating when saying things like: "The photon is quantized packet of energy." - He has falsely concluded from this that the photon is a tiny ball one wave length in diameter. This error has caused him to make several other errors.)


Unless you are quite knowledgable you would not have noticed what some, not well informed about modern Fourier Theory, might think is a subtle error my prior posts or at least implicitly suggested. (I do not actually state any error, even if you are not up on the modern developments to Fourier theory - I was careful not to.)

There is not even an implict error, if one knows that Fourier Theory has been extended since Fourier died to include many basis set in addition to the original sin and cos basis set.

Specifically when speaking of "Fourier Theory" I was including basis sets like Gabor invented about 60 years ago.

See:

http://en.wikipedia.org/wiki/Gabor_transform

where their application to photons is explicitly mentioned (via the first paragraph link to "time frequency analysis", but that is still more general even than the above link so will be even more difficult for the people I am posting here to help).

 

People used to slash your tires a lot, didn't they, Billy?

 

Anyway, as I was saying, the typical visible photon has a diameter more than a thousand times that of an atom. It is a large, diffuse object.

 

I consider this statement to be an outright lie and the material about Fourier theory to be a red herring. The idea of the photon being something like a tiny ball one wavelength in diameter is basic to physics and Billy has offered no evidence to the contrary, or even any evidence that makes sense.

I know what James is talking about. Neither I nor Billy has a freaking clue what Billy is talking about. Billy isn't even being coherent. He's not even wrong.

 

I agree 99.9% with this. Your 0.1% error is a minor quibble known as the Gibb's effect, if memory serves me correctly.

Namely for a continuous function F(x) defined in the range x = a to x = b and with F(a) = A & F(b) = A also, the Fourier transform will exactly reproduce F(x);
BUT if F(b) = B; and B is not equal to A, then the Fourier series (or integral) will NOT "exactly reproduce the function." As this A not equal B is commonly the case, I am going to count this as the fourth error of yours I have caught. {You make so few, I cannot let even this one slip buy uncounted.}

I will designate the values reproduced by the Fourier series (or integral) with a prime: I.e. F '(x). The value of F '(a) = A + (B-A)/2 and not F(a) Likewise F '(b) is B + (A-B)/2 not F(b). {I think I have the signs correct. - It takes the "mid point" values at ALL the discontinuous gaps.} This is because F '(x) is not really reproducing the continuous F(x) defined only in the above stated range. F '(x) is reproducing a DISCONTINUOUS, function which has an infinite range, which I will call F "(x) which is identical to F(x) in the range: a < x < b. (Note that if F(x) itself has a discontinuity within the range <a to b> F '(x) will take the "mid point" value there also. Thus when F(x) has a step within the range, your statement "will exactly reproduce the function." is FALSE, even if A = B at the range limits.)

Nowhere does F "(x) have multiple values. At x = a & b it has either no derivatives or a "left" and "right" derivatives, which are equal only by chance [or by design of the function F(x) to have a "local extremium" (Max. or Min.)at the limits of the range. (They need not be the same.)] The "no derivatives" option is not the one normally consider to be the case - it gets down to exactly what your definition of a derivative is. The usually definition is via a limiting process and this produces the "Left" and "Right" derivatives.

Now back to the physics and my posts:

The reason one really does need to use the Gabor functions (or something like them) to represent a photon is that the Fourier series (or integral) representation makes not one, but an infinite number of them. With Gabor functions, you can avoid this error instead of ignore it (focus your attention only one). Admittedly every reasonable person would ignore all but one, but in my posts I did IMPLICITELY imply that one could represent and analyze for frequency contents etc. a photon with the Fourier series (or in this case actually the Fourier integral is required - but in my posts I always try to avoid speaking about integrals as most of the people I try to help would not get much if I were that accurate.)

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*If you had had "Inside" instead of "In" you would have been "home free" with no error for continuous functions. (But still in error if there are discontinuities in the range.) It’s your choice: You can claim it was "just a typo" or admit to making a slight error by not being careful enough.

PS. Do not give up - keep reading my post in this thread. There may be a real error in my posts. :shrug: I make a lot more than you do.

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You can pick any post of MetaKron's and easily find errors there.

 

Why don't you two boys play in your own sandbox for a while. I have better things to do.

 

It's hard to do that when all that you do is try to feed me the little treats that kitties leave in sandboxes.