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Quantized redshift anomaly
18 years 9 months ago #16998
by Michiel
Replied by Michiel on topic Reply from Michiel
I had a similar thought as in JMB's first point.
To create a spectrum consisting of one frequency you need a continuous sinewave which has always been running.
Anything else ( compare with amplitude-modulation of a radio transmitter ) will produce a bandwidth.
Strictly speaking, frequency is defined as 'how often something happens in a certain timespan'.
Of course, a single photon ,quantized as it is, happens only once. But how long does a photon last?
Maybe frequency is not the most fortunate term for a photon.
To create a spectrum consisting of one frequency you need a continuous sinewave which has always been running.
Anything else ( compare with amplitude-modulation of a radio transmitter ) will produce a bandwidth.
Strictly speaking, frequency is defined as 'how often something happens in a certain timespan'.
Of course, a single photon ,quantized as it is, happens only once. But how long does a photon last?
Maybe frequency is not the most fortunate term for a photon.
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18 years 9 months ago #14662
by Jim
Replied by Jim on topic Reply from
While following some of what is being kicked around here, it seems to me the focus is how the photon is understood. This has been a problem forever. As I see it the photon is the energy particle in both E=mc^2 and E=hf. Since both these statements are about "E" of energy then why not solve mc^2=hf? Would that perspective be any beter than debating weather the photon is a wave or particle or both wave and particle. In any case all photons have the same amount of mass and energy reguardless of what frequency they are observed at. It is a big error to use E=hf in such careless ways as is done by everyone-all you have to do is examine at the statement to see that is a describes a bundle of energy and not a photon.
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18 years 9 months ago #17226
by JMB
Replied by JMB on topic Reply from Jacques Moret-Bailly
For me, "photon" means "amount of energy corresponding to a transition between two stationary states of a system".
I cannot understand the photon as a particle:
- It has no known wave function (a lot of discussions : is the EM field its wave function while it has not the required properties, or is its wave function something else ?).
- It depends on many parameters.
- It is bound to an optical mode which is never defined.
- It is a source of many paradoxes and patches (EPR, radiation reaction...).
The purely classical theory, including Planck-Nernst zero point field, is always good, without any problem.
Kill the photon !
I cannot understand the photon as a particle:
- It has no known wave function (a lot of discussions : is the EM field its wave function while it has not the required properties, or is its wave function something else ?).
- It depends on many parameters.
- It is bound to an optical mode which is never defined.
- It is a source of many paradoxes and patches (EPR, radiation reaction...).
The purely classical theory, including Planck-Nernst zero point field, is always good, without any problem.
Kill the photon !
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18 years 9 months ago #14668
by JMB
Replied by JMB on topic Reply from Jacques Moret-Bailly
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
<br />Replying to JMB:
I mostly could not understand what you said. However, it seems relevant (not to mention simpler) to point out that "quantization" for light simply means that light consists of discrete waves instead of some sort of continuum. And each wave at a given frequency has a specific energy, so the total energy transferred by light to matter must come in integer multiples of the energy transferred from a single wave. -|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
It is the problem of writing Planck's relation with various conditions at the limits which lead to a space of the modes having continuous or not dimensions: you propose a discrete sum h= (Sigma)[e(f)/f], but in many cases it is an integral.
Interesting in maths, not in physics !
<br />Replying to JMB:
I mostly could not understand what you said. However, it seems relevant (not to mention simpler) to point out that "quantization" for light simply means that light consists of discrete waves instead of some sort of continuum. And each wave at a given frequency has a specific energy, so the total energy transferred by light to matter must come in integer multiples of the energy transferred from a single wave. -|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
It is the problem of writing Planck's relation with various conditions at the limits which lead to a space of the modes having continuous or not dimensions: you propose a discrete sum h= (Sigma)[e(f)/f], but in many cases it is an integral.
Interesting in maths, not in physics !
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18 years 9 months ago #17227
by Tommy
Replied by Tommy on topic Reply from Thomas Mandel
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">quote:
Originally posted by Tommy
What about as it makes its way THROUGH the atom?
Nope. Lightwaves are either propagating through vacuum or absorbed by matter.
quote:
Is it correct to say that a photon, a packet of waves, does not make the 13 billion year journey to here without hitting something on the way.
First, a "photon" is a hypothetical wave packet, a single entity, not multiple entities as in your "packet of waves". Each wave crest registers in detectors as another "photon".
And light mostly does travel the 13 billion lightyears to here without hitting anything. Photons that do hit hydrogen or other intergalactic atoms are absorbed, and show up here as spectral lines (missing photons). If photons did interact with oedinary matter, they would be slightly scattered, and images of distant galaxies would be fuzzy. But Hubble observations do not show this.
quote:
Is it correct to say that when a photon does hit something in the way, that something will emit a photon and the journey goes on.
The two possible fates are absorption or scattering. Most photons suffer neither, so they have not interacted with matter.
quote:
Light does what it is supposed to be doing, matter is where the energy loss is to be located.
Redshift (energy loss) mechanisms are many. The main mechanism in BB (Doppler) and that in MM (friction with gravitons) do not involve any interaction with ordinary matter. -|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
OK, methinks that a photon is a like a ball of light, but existing at the Planck scale. A plasma ball of light. That way I can see how it would be composed of a wave but act like a paricle in certains situations. It would be a wave that closes in around itself, compared to the ordinary electromagnetic field which spreads out as it moves.
However you look at it, light is energy doing something. What I want to know is where does this energy come from? I cannot accept that it is only as simple as what Maxwell's equations tell us, especially after I discovered Maxwell's equations are not complete, and are missing Maxwell's attempt to incorporate the Aether. I cannot accept that light,or all electromagnetics for that matter, is a perpetual motion machine. Especially when there is the Aether concept which would explain so much of what we call invisible stuff.
JMB thinks of the ZPF as some sort of field minimum, somehow still connected to the classical field itself. Once he said that when there is a measurable field, there is no ZPF. I am certain that Maxwell's treatment would bear this out. But I would like to know what the assumptions are.
Increasingly I find it better to think of it as the INSIDE of space and leave it undefined. To define is to create limits where it is apppropriate and where it is not. The INSIDE is not like that. It does not have any properties itself per se, rather it forms the source of properties.
So my assumptions is that electromagnetic energy is all forms is using energy. More so than simply trapping light inside a sperical mirror so to speak. It makes a lot of sense to assume that all processes involve a certain kind of "friction" if you will. And it makes just as much sense to assume that this used up energy is being supplied by a source. So it is not "nonsense" to assume that there is a source.
Originally posted by Tommy
What about as it makes its way THROUGH the atom?
Nope. Lightwaves are either propagating through vacuum or absorbed by matter.
quote:
Is it correct to say that a photon, a packet of waves, does not make the 13 billion year journey to here without hitting something on the way.
First, a "photon" is a hypothetical wave packet, a single entity, not multiple entities as in your "packet of waves". Each wave crest registers in detectors as another "photon".
And light mostly does travel the 13 billion lightyears to here without hitting anything. Photons that do hit hydrogen or other intergalactic atoms are absorbed, and show up here as spectral lines (missing photons). If photons did interact with oedinary matter, they would be slightly scattered, and images of distant galaxies would be fuzzy. But Hubble observations do not show this.
quote:
Is it correct to say that when a photon does hit something in the way, that something will emit a photon and the journey goes on.
The two possible fates are absorption or scattering. Most photons suffer neither, so they have not interacted with matter.
quote:
Light does what it is supposed to be doing, matter is where the energy loss is to be located.
Redshift (energy loss) mechanisms are many. The main mechanism in BB (Doppler) and that in MM (friction with gravitons) do not involve any interaction with ordinary matter. -|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
OK, methinks that a photon is a like a ball of light, but existing at the Planck scale. A plasma ball of light. That way I can see how it would be composed of a wave but act like a paricle in certains situations. It would be a wave that closes in around itself, compared to the ordinary electromagnetic field which spreads out as it moves.
However you look at it, light is energy doing something. What I want to know is where does this energy come from? I cannot accept that it is only as simple as what Maxwell's equations tell us, especially after I discovered Maxwell's equations are not complete, and are missing Maxwell's attempt to incorporate the Aether. I cannot accept that light,or all electromagnetics for that matter, is a perpetual motion machine. Especially when there is the Aether concept which would explain so much of what we call invisible stuff.
JMB thinks of the ZPF as some sort of field minimum, somehow still connected to the classical field itself. Once he said that when there is a measurable field, there is no ZPF. I am certain that Maxwell's treatment would bear this out. But I would like to know what the assumptions are.
Increasingly I find it better to think of it as the INSIDE of space and leave it undefined. To define is to create limits where it is apppropriate and where it is not. The INSIDE is not like that. It does not have any properties itself per se, rather it forms the source of properties.
So my assumptions is that electromagnetic energy is all forms is using energy. More so than simply trapping light inside a sperical mirror so to speak. It makes a lot of sense to assume that all processes involve a certain kind of "friction" if you will. And it makes just as much sense to assume that this used up energy is being supplied by a source. So it is not "nonsense" to assume that there is a source.
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18 years 9 months ago #14671
by Michiel
Replied by Michiel on topic Reply from Michiel
Tommy states that a photon could be a wave closing in around itself. That way we can consider the photon having an internal frequency, moving through its medium like a ringing bell. If for some reason the wave is not perfectly closed it could lose energy, resulting in redshift.
___
I don't want to make things too complicated, but here's another (hypothetical) thought:
In the Meta Model the light carrying medium consists of elysons. What would happen if one of these elysons ceases to exist at the moment a photon is passing by? I think the photon would lose some of its energy.
On the other hand if an elyson came into being while a photon is passing, the medium would get diluted, again resulting in energy loss.
___
I don't want to make things too complicated, but here's another (hypothetical) thought:
In the Meta Model the light carrying medium consists of elysons. What would happen if one of these elysons ceases to exist at the moment a photon is passing by? I think the photon would lose some of its energy.
On the other hand if an elyson came into being while a photon is passing, the medium would get diluted, again resulting in energy loss.
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