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Thread: What if a single photon had been the result of the Big Bang?

  1. #1 What if a single photon had been the result of the Big Bang? 
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    Could it dissolve in space, dissipating its energy?

    Would the laws of thermo-dynamics prevail?


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  3. #2 Re: What if a single photon had been the result of the Big B 
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    Quote Originally Posted by himnextdoor
    Could it dissolve in space, dissipating its energy?

    Would the laws of thermo-dynamics prevail?
    Photons do not dissolve.

    Thermodynamics is the result of statistical behavior of large numbers of particles. One is not a large number.


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  4. #3 Re: What if a single photon had been the result of the Big B 
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by himnextdoor
    Could it dissolve in space, dissipating its energy?

    Would the laws of thermo-dynamics prevail?
    Photons do not dissolve.

    Thermodynamics is the result of statistical behavior of large numbers of particles. One is not a large number.
    Photons travel at the speed of light. Except when they are absorbed, right?

    What happens to a photon that has been absorbed by say, the nucleus of an iron atom; is it still travelling at the speed of light? It only carries one quanta of energy which cannot be divided so if a proton becomes energised, what happens to the energy-less shell that interacted with the nucleus in the first place?

    Are the insides of atoms littered with the dust of dead photons?
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  5. #4 Re: What if a single photon had been the result of the Big B 
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    Quote Originally Posted by himnextdoor
    Photons travel at the speed of light. Except when they are absorbed, right?
    Nope. Even when they are absorbed, they are still traveling at c. It just appears to us that they change speed, however, their speed is constant.
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  6. #5  
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    Would it be fair to say that:

    A photon to an electromagnetic wave is as a water molecule is to a wave travelling accross the surface of a pool.

    Similar in the sense that each molecule of water represents a 'packet of energy' at a single point on the wave that can be measured for vectorable components; velocity, direction, characteristics that could be plotted over time to show frequency, amplitude and, if we know the mass of a water-molecule, energy and momentum.

    Two aspects that appear to be missing from the water-molecule model are 'spin' and the constant 'c' for the speed of light but the water-molecule can be above or below the x-axis of the wave which gives us a plus and minus component that can be substituted for spin and the constant force of gravity, which is the same for all the water-molecules, could substitute for the speed of light.

    Is there anything in the photon model that can't be explained in the water-molecule model?

    I ask because I'd like to understand the implications of the differences between the two models as far as the topology and the dispersal of waves, divergence, is concerned in such a way that I can explain QM to a child and I want my analogies to be accurate.

    For instance, in the water model, there is space between the water molecules and one could not measure any properties at those points; the 'sensing' equipment needs to be in 'contact' with a water-molecule in order to generate any meaningful data. However, one might still obtain a reading that indicates an 'odd component' in the energy distribution of energy.

    Is this the case with photons? Are photons seperated by space and are they divergent; does the space between them increase?

    The topology of the surface of the water-wave is shaped by the molecules at the water-air boundary. If one were able to freeze the wave in time and represent its profile on an oscilloscope (roughly a sine-wave) and magnify the image to a molecular scale, the shape of the molecules would give the appearance of a high-frequency component superimposed on the low-frequency 'energy-wave'. This imposes a 'quantisation limit'; the size of the molecule limits the accuracy to which the shape of the wave, at any given point in time, can be plotted.

    Is this the case with photons? If photons have a shape then at some level of magnification, the profile of an electr-magnetic wave would be shaped by the photons, again, giving the impression that there is a high-frequency component superimposed on a low-frequency fundamental and again, imposes a 'quantisation limit'.

    Here's the thing though: If we measure the same water-wave at some point later in its journey, it has the same high-frequency component as it did before; the distance between the water-molecules that are carrying the wave's energy stays the same. This is in contrast with the photon model; the photons diverge over time which suggests that the surface topology of an electro-magnetic wave dissolves into points that become increasingly distant from each other.

    This leads naturally to the conclusion that if an elecro-magnetic wave detector were positioned at a sufficient distance from a radio transmitter, photons, from that source, could pass close by that detector and no electro-magnetic wave would be detected.

    Can this be so? Is it posible that an electro-magnetic wave moves through space gradually dissolving into increasingly distant points?

    I'm used to thinking of a light-wave as a membrane as opposed to buck-shot. 8)
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  7. #6  
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    (Have I seen this post before?)

    There are two types of wave that can be compared. Mechanical (water, sound, seismic etc) and electromagnetic (visible light, gamma, radio etc).

    A photon can have vectorable components (like and ), but you need to imagine them as being in dimensions other than those that we observe. The flaws in the photon model are QM anomalies like the dual-slit (w. observer) experiment.

    An interesting experiment to demonstrate mechanical/em waves is shining a strobe on a fast-vibrating, taught string at the same frequency.

    Does this help?
    Hardly have a phd in topics covered. I may be wrong - if this is the case, by all means, please do call me up on it.
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  8. #7  
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    Quote Originally Posted by himnextdoor
    Can this be so? Is it posible that an electro-magnetic wave moves through space gradually dissolving into increasingly distant points?

    I'm used to thinking of a light-wave as a membrane as opposed to buck-shot. 8)
    Yes. With enough distance relative to the intensity of the source, there would be a point where the mean distance between photons would be a foot/yard/meter/mile/etc. Browsing around I've seen an estimate (which I haven't double checked) says that a 100W bulb would put out about 10^20 photons per second.

    If you turned the light on for exactly 1 second, the density of the shell would be (on average) about photons per foot. Well, you wouldn't really expect to be able to see a 100W lightbulb from 1 light-second away.
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  9. #8 Re: What if a single photon had been the result of the Big B 
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    Quote Originally Posted by himnextdoor
    Quote Originally Posted by DrRocket
    Quote Originally Posted by himnextdoor
    Could it dissolve in space, dissipating its energy?

    Would the laws of thermo-dynamics prevail?
    Photons do not dissolve.

    Thermodynamics is the result of statistical behavior of large numbers of particles. One is not a large number.
    Photons travel at the speed of light. Except when they are absorbed, right?

    What happens to a photon that has been absorbed by say, the nucleus of an iron atom; is it still travelling at the speed of light? It only carries one quanta of energy which cannot be divided so if a proton becomes energised, what happens to the energy-less shell that interacted with the nucleus in the first place?

    Are the insides of atoms littered with the dust of dead photons?
    Let's stick with interactins between photons and electrons, which is the more common interaction and one that I understand fairly well. The situation in a nucleus involoves more than the electromagnetic force.

    A photon always travels at c. When a phtoon is absorved it ceases to exist.

    When a photon is absorved by an electron the electron moves to a higher energy state, an "excited state". It is then just an electron, not an electron and a photon. When it relaxes to a lower energy state it emits a photon of precisely the energy that is the difference between the excited state and the relaxed state. You then have an electron and photon.

    Recall the equation . It tells us that energy and matter are the same thing. An elementary particle is a blob of energy, and in interactions among elementary particles energy is consereved. So a photon absorved by an electron increases the mass'energy of the electron and a photon emitted carries away a bit of mass/energy. The photon always travels at c. It is never at rest. But elementary particles are created and destroyed in interactions all the time. But at each transformation energy and momentum are conserved.

    There is no such thing as the "dust of dead photons".
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  10. #9  
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    Voyager 1 is 12.6 light hours away and the power available to the whole craft is about 270 watts. We receive data from it all the time.

    How is it possible that the Deep Space Network is able to upload data from Voyager at all?

    How could one differentiate between a photon from Voyager and a photon from the Cosmic Background Radiation?
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  11. #10  
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    Two reasons. One is the CMB exists at a very specific band of energy (very, very low) and the transmissions from Voyager exist in a completely different band. Second is the the CMB is very diffuse and weak, whereas the transmissions from Voyager are much stronger and focused in our general direction.
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  12. #11  
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    Quote Originally Posted by DrRocket
    Let's stick with interactins between photons and electrons, which is the more common interaction and one that I understand fairly well. The situation in a nucleus involoves more than the electromagnetic force.

    A photon always travels at c. When a phtoon is absorved it ceases to exist.

    When a photon is absorved by an electron the electron moves to a higher energy state, an "excited state". It is then just an electron, not an electron and a photon. When it relaxes to a lower energy state it emits a photon of precisely the energy that is the difference between the excited state and the relaxed state. You then have an electron and photon.

    Recall the equation . It tells us that energy and matter are the same thing. An elementary particle is a blob of energy, and in interactions among elementary particles energy is consereved. So a photon absorved by an electron increases the mass'energy of the electron and a photon emitted carries away a bit of mass/energy. The photon always travels at c. It is never at rest. But elementary particles are created and destroyed in interactions all the time. But at each transformation energy and momentum are conserved.

    There is no such thing as the "dust of dead photons".
    Firstly, I'm happy to see you engage and secondly, I more or less completely agree with you as for as the principle of the energy-exchange mechanism operation is concerned; we disagree on the precise nature of the components that comprise matter/energy.

    It would be useful if we could find out where we agree and then we, or at least I, can consider the points within a common framework. This will help to avoid misunderstandings such as, to use navigation parlance, I call it 'the next left' where you call it 'the next right' and get bogged down because we fail to realise that I'm travelling north and you're going south; we're actually talking about the same turn.

    Would you accept this as an idealised model for a photon-emission surface:

    A number of electrons are arranged to form a flat sheet such that the entire space of the sheet is filled with electrons; that they are all side-by-side. I'm thinking of millions of identical ball-bearings packed tightly together in two dimensions. Funnily enough, the structure would appear to form a 'honeycomb.

    Behind this sheet is a source of energy with the same surface-area of the sheet; it is bombarding the back of the sheet with photons, an electro-magnetic force, at a constant wavelength. The arrangement is such that no photons from the energy source go past the edge of the 'electron-sheet' and, for the purpose of this discussion, none travel through the area between the electrons in the sheet either.

    And in front of the sheet, there is a photon detector.

    So, all the photons that are detected in front of the sheet are the result of absorbed photons from behind the sheet.

    Can we work with that?

    Is it the case that:

    1. In order for absorbtion/re-emission to occur, the photon has to make a physical contact with an electron?
    2. The incidental angle at which the photon interacts with the electron determines the wavelength of the emitted photon?
    3. One absorbtion causes one emission? (An electron can only emit one photon at a time)
    4. No detected photon can have a shorter wavelength than a photon from the source?
    5. All photons are the same size and travel at the velocity of light and represent the smallest unit of energy? and
    6. Photons do not interact with each other?

    I'm looking at a green leaf in white light and wondering why every single photon that doesn't represent green light is absorbed. At a very high magnification, the matter that makes up the leaf is composed largely of space; there are relatively large distances between the atomic nuclei and its associated electrons. Are photons too big to fit between the atomic particles that make up a leaf? How come most of the photons don't pass straight through the 'space' in the leaf?
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  13. #12  
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    1. Quantum uncertainty in position would make this a moot point. You're trying to imagine this as shooting at a bunch of baseballs with bullets, but it doesn't work that way.

    2. The incident angle has no effect, except for momentum, which is what causes the emitted photon to travel in the same direction.

    3. Generally true, but not always. As long as the energy adds up it's possible for the ratio to change. A single electron can absord two photons and emit one with twice the energy, or absorb 1 photon and emit 2 with (about) half the energy. It's rare, but known to happen.

    4. See #3.

    5. I'm not sure it makes sense to talk about the size of a photon, but all of them travel at c. Also, photons can have varying energies, so you'll have to be careful saying they're the smallest quantum of energy.

    6. AFAIK, photons only interact with each other very weakly. (Energy, including photons, gravitates, so two parallel photons in an otherwise empty universe would eventually come together.)

    Leaf: A large number of photons of all wavelengths do pass through the leaf. If you shine a red laser pointer at a leaf, you'll see a red dot on the other side. It looks green because less green light is absorbed than other visible frequencies.
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  14. #13  
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    Oh, and

    7. Any electron can emit a phot of any wavelength?

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  15. #14  
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    7: No. Electrons can only emit specific wavelengths. This discovery was the catalyst for all of quantum mechanics. Which wavelengths they emit depends on where on the atom they are.
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  16. #15  
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    Quote Originally Posted by himnextdoor
    Quote Originally Posted by DrRocket
    Let's stick with interactins between photons and electrons, which is the more common interaction and one that I understand fairly well. The situation in a nucleus involoves more than the electromagnetic force.

    A photon always travels at c. When a phtoon is absorved it ceases to exist.

    When a photon is absorved by an electron the electron moves to a higher energy state, an "excited state". It is then just an electron, not an electron and a photon. When it relaxes to a lower energy state it emits a photon of precisely the energy that is the difference between the excited state and the relaxed state. You then have an electron and photon.

    Recall the equation . It tells us that energy and matter are the same thing. An elementary particle is a blob of energy, and in interactions among elementary particles energy is consereved. So a photon absorved by an electron increases the mass'energy of the electron and a photon emitted carries away a bit of mass/energy. The photon always travels at c. It is never at rest. But elementary particles are created and destroyed in interactions all the time. But at each transformation energy and momentum are conserved.

    There is no such thing as the "dust of dead photons".
    Firstly, I'm happy to see you engage and secondly, I more or less completely agree with you as for as the principle of the energy-exchange mechanism operation is concerned; we disagree on the precise nature of the components that comprise matter/energy.

    It would be useful if we could find out where we agree and then we, or at least I, can consider the points within a common framework. This will help to avoid misunderstandings such as, to use navigation parlance, I call it 'the next left' where you call it 'the next right' and get bogged down because we fail to realise that I'm travelling north and you're going south; we're actually talking about the same turn.

    Would you accept this as an idealised model for a photon-emission surface:

    A number of electrons are arranged to form a flat sheet such that the entire space of the sheet is filled with electrons; that they are all side-by-side. I'm thinking of millions of identical ball-bearings packed tightly together in two dimensions. Funnily enough, the structure would appear to form a 'honeycomb.

    Behind this sheet is a source of energy with the same surface-area of the sheet; it is bombarding the back of the sheet with photons, an electro-magnetic force, at a constant wavelength. The arrangement is such that no photons from the energy source go past the edge of the 'electron-sheet' and, for the purpose of this discussion, none travel through the area between the electrons in the sheet either.

    And in front of the sheet, there is a photon detector.

    So, all the photons that are detected in front of the sheet are the result of absorbed photons from behind the sheet.

    Can we work with that?

    Is it the case that:

    1. In order for absorbtion/re-emission to occur, the photon has to make a physical contact with an electron?
    2. The incidental angle at which the photon interacts with the electron determines the wavelength of the emitted photon?
    3. One absorbtion causes one emission? (An electron can only emit one photon at a time)
    4. No detected photon can have a shorter wavelength than a photon from the source?
    5. All photons are the same size and travel at the velocity of light and represent the smallest unit of energy? and
    6. Photons do not interact with each other?

    I'm looking at a green leaf in white light and wondering why every single photon that doesn't represent green light is absorbed. At a very high magnification, the matter that makes up the leaf is composed largely of space; there are relatively large distances between the atomic nuclei and its associated electrons. Are photons too big to fit between the atomic particles that make up a leaf? How come most of the photons don't pass straight through the 'space' in the leaf?
    You are basically all wet and need to go study some quantum mechanics.

    1. Photons intereact with electrons, physical contact is not a meaningful idea, but if you mean by "physical contact" an interaction then that is simply a tautology.

    2. The incident angle has nothing to do with the wavelength of the emitted photon.

    3. One absorption usually results in one emissiosn but that is not necessarily the case. An electron can relax in a series of steps, with a photon emission at each step.

    4 If a an electron is in an excited state then it is possible for another photon to be absorved and a photon be emitted that is of higher energy than the absorved photon.

    5. In quantum electrodynamics photons are taken as point particles, which have no size whatever. The same is true of electrons. All photons travel at c. Photons can. in principle, have any amount of energy, dependent on the frequency of the photon. So, no a photon does not represent the smallest amount of energy.

    6. Photons ususally do not intereact with each other. However, for photons of sufficiently high energy photon-photon collisions are possible and the result can be a massive particle.

    So, not I do not accept your version of a "theory" which in fact is nothing but nonsense. I do not intend to "engage" you in any sort of debate. Your knowledge of physics is simply not up any debate. You don't have any idea what you are talking about and you need to go learn some basic quantum theory.
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  17. #16  
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    Quote Originally Posted by DrRocket
    So, not I do not accept your version of a "theory" which in fact is nothing but nonsense. I do not intend to "engage" you in any sort of debate. Your knowledge of physics is simply not up any debate. You don't have any idea what you are talking about and you need to go learn some basic quantum theory.
    What 'theory'? Photons are not my construct.
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