[Graybeard]

Originally Posted by N0B0DY

I'm unsure about Greg's correlation between AM and FM. Does it mean that a change in amplitude will produce a change in frequency?

Also, were you referring to the spatial measurement of the wavelength instead of the Hz, HBD?

Originally Posted by humanbydefault

Imagine that light was a bunch of horses ready to start the race. The fastest horse (the longer wavelength of the components of the given light) will brake from the rest and will go ahead in the race. Obviously the larger wavelength of all [component of what we know as light] will be the INFRARED.

Let me know even if you disagree totally with my thoughts... especially so.

HUMANBYDEFAULT

I am trying to understand what your saying .....


I can't come to terms with the above statement ... I have tried.

Different colours are different wavelengths. If I made a two pole generator with a North and a South pole then one revolution in one second would chart a complete Sinewave.

This ONE wavelength would be 300,000 kilometres long.

If I made a four pole generator then one revolution would chart two Sinewaves, each 150,000 kilometres long.

Both of these examples would take the same time to travel 300,000 kilometres ... they would take one second.

Shortening of lengthing the frequency or amplitude does not change the speed ???

Could you explain why you don't see this as correct ??

I also have other problems with your theory ... in particular those Neutralino mentioned. 'Inter-atomic distances' and the 'different size of atoms' is not how QM explains the interactions of particle/waves.

cool bananas ... greg

[humanbydefault]

Originally Posted by N0B0DY

I'm unsure about Greg's correlation between AM and FM. Does it mean that a change in amplitude will produce a change in frequency?

Also, were you referring to the spatial measurement of the wavelength instead of the Hz, HBD?

Fisrt question:

In radio communication technology information can be send in AM or FM [modulation of the signal].
Let's say that you want to send a group of 1 and 0 in the following way... 110100011.
If we are referring to Amplitude Modulated signals, then each 1 will coincide with a wave or group of waves followed one by the other. the 0 will be the ABSENCE of wave(s) being sent.

In the case of Frequency Modulated signals>>> and this is where my interpretation of modulation oppose the orthodox one<<< each 1 will be "imprinted" in a carrier as a given number of HARMONICS. It is important to see the difference between HARMONICS along a CARRIER WAVE and the actual CARRIER. The CARRIER does not 'carry' any sort of information from point A to POINT B. It is the de-modulation of those HARMONICS previously imprinted in the 'main wave' that carries the information between two points.

If you look for the way FM is represented graphically, you'll see a continuous wave along the CARRIER... inside the CARRIER WAVE. The difference is that HARMONICS are key important in the actual symmetry of the CARRIER WAVE [It's actual existence].

In the complete CYCLE of the CARRIER WAVE [4 four frames I used to explain it] the same exact number of HARMONICS must be repeated four times inside one CYCLE. If this condition is violated by any reason the wave cease to exist.

In the following site [of my theory] I have represented >very easy-to-catch< the way those 4 four frames are divided in time [the SHIFT between the magnetic and the electric fields].

http://www.theory-of-bonding-harmonics.com/graphics.htm

What it means is that the North pole of the magnetic filed of the CARRIER WAVE and the South Pole of the same wave are DIVIDED IN TIME BY 2 (TWO) EXACT FRAMES.
But the most important point here is that the SHIFT and the TIME the SHIFT is produced IS ALWAYS THE SAME IN ALL E-M WAVES >>> NO MATTER THE FREQUENCY OR WAVELENGTH OF THE ACTUAL WAVE<<< This is where I disagree upon the orthodox views.

Allow me now to show you how HARMONICS CAN A DO EXIST IN CARRIER WAVES in nature all the time CONSERVING THE SYMMETRY NEEDED FOR THE EXISTENCE OF THOSE WAVES.

Go to:

http://www.theory-of-bonding-harmonics.com/Fig.7A-D.htm

As you can easuly see, I've represented an atomic orbit of elements of group 7. Each FRAME [very important to understand it] EACH FRAME -I repeat- HOLDS [CARRIES OR CONTAIN... As you prefer to call it] the exact same NUMBER OF HARMONICS IN EACH FRAME. >>>Which also means during the SHIFT-PERIODS of TIME.
-------------------------------------------------------------------------------------------

THE ANSWER to your second question is now easier to understand, I suppose...

IF you accept and understand [now] that NO MATTER THE WAVELENGTH of a given CARRIER WAVE the TIME involved in its MOMENTUM along the eather is always the same. Then as long as you consider any type of MODULATION [infrared, ultraviolet> number if imprinted harmonics along the carrier] the speed [of the carrier] will coincide with what we know as "c" or speed of light. That will be the case of the light coming from the sun to Earth.

NOW..It comes the tricky part:

What would happen when the amplitude of those CARRIERS are increased?

The higher the amplitude of the CARRIER WAVE the more chances are to "imprint" more HARMONICS along the wave... got it? The modulation -as long as it does not breaks the symmetry needed- will increase in its capacity to hold more HARMONICS.

If the same CARRIER HAD A LONGER WAVELENGTH... WE WOULD ALSO BE CAPABLE OF PUTTING MORE HARMONICS ALONG THE WAVE... right? >>> keeping the symmetry intact of course.

That's why I'm saying that "c" is not always a constant when it comes to define the speed of light or any E-M WAVE for that matter.

The speed of the MODULATION [INFORMATION=COLORS AND FREQUENCIES INSIDE THE CARRIER] will always depend on the CARRIER.

Therefore... THE LONGER THE CARRIER (wavelength) THE HIGHER THE ACTUAL "CAPACITY" OF THE WAVE TO "TRANSPORT" harmonics from A to B and at the same time THE MORE SPACE IT WILL COVER IN THE SAME EXACT TIME. When I refer to TIME I'm referring to "real time".

Now! Could a CARRIER's WAVELENGTH be equal to the same frequency of HARMONICS?
YES! But then the TIME involved in the SHIFT from frames one to four will coincide with the rest of E-M waves in nature.

My point is very simple: If what I said is somehow troubbling for you to catch... try to see this>>>

1. MODULATION OF HARMONICS inside a CARRIER WAVE is possible due to the fact that SYMMETRY is never violated along the entire wavelength [4 frames of existence].

2. The modulation of HARMONICS -along the CARRIER- will always contain the exact same number of HARMONICS during the lapse of 4 frames of one cycle... one complete wave. A change in the 'modulation' [number of harmonics "transported" by the CARRIER WAVE] INSIDE the SHIFT or before the 4 frames are complete will violate SYMMETRY and the actual modulation will cease to exist.

3. Once any type of MODULATION [the actual information] is converted in CARRIERS themsleves, they will ADOPT the same conditions of any CARRIER>>> SINCHRONIZING ITS SPEED WITH THE REST OF E-M IN THE UNIVERSE.

I hope this explanation will serve for the others too...

HBD

And thanks for your participation too!

[N0B0DY]

My pleasure, HBD. You give a clear explanation, and I am going to study your site to see if I can find some answers to questions I have myself. I'm interested in analog-digital conversions and differences, and how much information is lost in the digital signals.

You answered the question that frequency changes wavelength, wavelength doesn't change frequency and momentum, and judging from your initial explanation I gather amplitude has nothing to do with frequency.

Just one thing I don't understand so far, why the four frames can't be further divided?

[RascalPuff]

It may be that I"m missing something in the interesting and reasonable explanation for why we receive a red shift from distant light sources that are receding, whereas, what I don't understand and request an explanation for, is, in the case of approaching light sources that exhibit a blue shift.

Given the offered proposal as I understand it, because of the succession of the 'colors' (frequencies), ROYGBIV, the red shift prevails because of the succession of frequencies; yet, the blue end of the visible spectrum precedes red, in the case of an approaching light source, i.e., there is a 'blue shift' also, corresponding to the estimated speed of approaching light sources, which does seem to give credence to the premise that the red shift is a valid way of determining that the spatial universe is expanding...

Best regards,
- RP

[Graybeard]

Originally Posted by RascalPuff

...... what I don't understand and request an explanation for, is, in the case of approaching light sources that exhibit a blue shift. .......

Given the offered proposal as I understand it, because of the succession of the 'colors' (frequencies), ROYGBIV, the red shift prevails because of the succession of frequencies; yet, the blue end of the visible spectrum precedes red, in the case of an approaching light source, i.e., there is a 'blue shift' also, corresponding to the estimated speed of approaching light sources, which does seem to give credence to the premise that the red shift is a valid way of determining that the spatial universe is expanding...

Best regards,
- RP

Rascal .... are you requesting an explanation of what the term 'redshift' means. I don't know what you mean by the 'succession' of frequencies?

I am having difficulties with HBD's explanation, its over my head. But if you label the spectrum 'ROYGBIV' then redshift just means the entire spectrum has gone down in frequency by an 'octave', blue means up an 'octave'.

The amount of the spectrum shift, (which we know has a fixed frequency band and cannot change) determines the velocity of 'recession' or 'procession'. Particular measurements are taken of the absorption lines which can be focussed in high contrast.

The Ca II K line is a strong spectral line associated with once-ionized Calcium. It has a wavelength of 393.4 nm (billionths of a meter, in the blue part of the spectrum). If this absorption line was to be measured 393.5 nm then we would know that we are viewing a 'doppler shift' towards the lower frequency (red) end of the spectrum. The amount the wavelength has increased (slang word = stretched) is used to calculate the velocity of recession.

We know this must be true as CA II has a fixed frequency and cannot change its properties. (There are numerous other factors involved here such as temp, density and quantity of CA II K in the emitting star. But Science, not me, is satisfied in the way they account for these factors.

So .....
(-ROYGBIV) would mean a doppler shift towards the higher, or blue end.
(ROYGBIV) would mean both us (observers) and the emitting star are stationary in relation to each other.
(+ROYGBIV) would mean a doppler shift towards the lower, or red end.

Hope it helps a bit.

cool bananas ... greg

[bottomlander]

Originally Posted by N0B0DY

Does it mean that a change in amplitude will produce a change in frequency?

With reference to my poor physics knowledge, amplitude is a y-variable or called an depend variable. Frequency/wavelength is a x-variable or called independent variable.

For an EM wave in vacuum without red/blue shift, the curve of amplitude is always sinuous.
That means, usually, amplitude is decided by x-valves (position in a sine curve). Maximum amplitude is thus one forth of a wave length.

Mathematically speaking, from the max amplitude, we can calculate back the wavelength. But seems unable to say that amplitude determines/affects its wavelength.

The square of an amplitude is correlated to its intensity. In a quantum wave, it is the probability of being found locating there.
For the emission of radio em waves, it seems that the higher the amplitude, the more powerful, the more photons sending out. But the energy of each photon is still solely depends on its frequency, not amplitude.
(Amplitude tells us the group numbers of photons per unit volume).

Correct me if I am wrong. I no longer remember my high school physics well.

Best Regards. Bottomlander

[RascalPuff]

Originally Posted by Graybeard

Rascal .... are you requesting an explanation of what the term 'redshift' means. I don't know what you mean by the 'succession' of frequencies?

I am having difficulties with HBD's explanation, its over my head. But if you label the spectrum 'ROYGBIV' then redshift just means the entire spectrum has gone down in frequency by an 'octave', blue means up an 'octave'.

The amount of the spectrum shift, (which we know has a fixed frequency band and cannot change) determines the velocity of 'recession' or 'procession'. Particular measurements are taken of the absorption lines which can be focussed in high contrast.

The Ca II K line is a strong spectral line associated with once-ionized Calcium. It has a wavelength of 393.4 nm (billionths of a meter, in the blue part of the spectrum). If this absorption line was to be measured 393.5 nm then we would know that we are viewing a 'doppler shift' towards the lower frequency (red) end of the spectrum. The amount the wavelength has increased (slang word = stretched) is used to calculate the velocity of recession.

We know this must be true as CA II has a fixed frequency and cannot change its properties. (There are numerous other factors involved here such as temp, density and quantity of CA II K in the emitting star. But Science, not me, is satisfied in the way they account for these factors.

So .....
(-ROYGBIV) would mean a doppler shift towards the higher, or blue end.
(ROYGBIV) would mean both us (observers) and the emitting star are stationary in relation to each other.
(+ROYGBIV) would mean a doppler shift towards the lower, or red end.

Hope it helps a bit.

cool bananas ... greg

Hey Greg, thanks, I learned a lot about the details of spectro analysis from your post, but I'm not sure I understand HBD's intention.

HBD, are you saying that the spectral red shift is not an indicator that the spatial universe is expanding - that it's the dominant paradigm from light sources in a non-spatially expanding universe? Because red precedes all the other color frequencies of the visible spectrum?

Best regards,
- RP

[N0B0DY]

"For the emission of radio em waves, it seems that the higher the amplitude, the more powerful, the more photons sending out. But the energy of each photon is still solely depends on its frequency, not amplitude. (Amplitude tells us the group numbers of photons per unit volume)."

Spoken like a true university scholar, bottomlander. I think there may have been a typo on Greg's part. I'm not sure about your reference to maximum amplitude being limited to one fourth the wavelength.

RP,

Perhaps the inverse explanation would apply to the blueshift, then explaining contractive measures.

[neutralino]

Originally Posted by N0B0DY

You answered the question that frequency changes wavelength,

Correct.

wavelength doesn't change frequency and momentum,

This is not correct. The frequency of an electromagnetic wave is proportional to the wavelength. Thus, if a change in the frequency affects the wavelength, then the converse is necessarily true. Further one can show, depending on ones definition of momentum of an electromagnetic wave, I shall use p=E/c that, since the energy of a wave (more specifically, then energy of a photon) is proportional to the frequency, then the momentum will also depend upon the wavelength.

and judging from your initial explanation I gather amplitude has nothing to do with frequency.

This is true, and easier to see in the particle picture; the amplitude can be thought of as the number of photons with a certain energy/frequency (I think Greg mentioned this above). [/size]

[N0B0DY]

I get the bottom part of your response, neutralino, but if the frequency of oscillations changes the wavelength, how would wave inteference change the frequency? I thought that in variable media, wavelength changes and frequency remains the same.

Also, I've heard that amplitude effects velocity, is this right?