@ Billy T - I can hardly be accused of picking on the oldies (for I might be older than you) but I found your comments once again a hard to take.
Will CO2 absorb photon in all directions?
- Thread starter Robittybob1
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You seem to think that the bond is made by an electron that belongs to one of the bonded pair of atoms. There really is no classically correct POV, but often one does speak of these several valence electrons, the bonding electrons, as a cloud (Their "orbitals" deform with stretching). I.e. The photon´s energy is inititially stored in the stretching of these bonds as the bonded atoms vibrate - in quantized level of energy.
Here clearly you are thinking that IR absorption by a molecule is like electronic excitation where only one of the electrons is excited from a lower energy level in an atom to a higher level.
That is not at all correct, even in a classical POV. There is no particular electron that is excited - the cloud of many electrons participating in the bond is stretched by the vibratory motion of the atoms they are bonding is a much better POV than that some one electron absorbed the IR photon´s energy. I was just "prooving" what you asked me to. - I.e. it is not just one electron but the whole molecule the IR photon excites. (Although if the IR photon was not very strong, the vibration energy probably will just remain as an excited state of one of the bonds when there are several other bonds, expecially if one of the bonded atoms is just the proton of hydrogen and it has only that one bond.)
Also note that not until post 220, did you ever mention the photons energy being stored in a bond vibration - before that it was always stored intitially by ONE electron, yet in post 220 you then claim "the energy entered via the one bond. That is clearly what it says in my opinion." I.e repeat what I had just taught to correct your false POV, as if you had always held that more correct POV.
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I sense we are not actually saying things that are drastically different. Maybe a slight different model of the electron is used in our thoughts, but its still essentially the same.
Now the picture I have at the moment is that every photon passing close enough to an electron's domain may excite it to some extent. I.e. if it is too powerful ionization will occur. If it happened to be just the right wavelength the electron could absorb the photon, and go up a number of energy levels, 1-X levels, where X is the number of levels available before the electron is ionized.
Intermediary level wavelength photons are passed over but the electron's wave pattern is non-the-less affected for a moment, but the energy level did not match the atom.
At lower levels the electron is excited and the wave pattern sets up vibrations in the rest of the molecule from the epicenter (stretch bending depending on the position of the electron's domain).
If the vibration matches the atom the energy and momentum are taken from the photon and it is also absorbed, like the energy in an out of balance washing machine shaking the house.
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I tend to agree with this statement made on Physicsforums, quote:"Since photons have momentum, conservation of momentum will result in a change in velocity of the object that absorbed the photon (the entire atom would have to move, not the just the electron). That means a change in kinetic energy of the object in addition to the electron moving to a higher orbital" [end quote] or mode of vibration.
now previously we had looked at how much this momentum would imply a change in velocity. But that speed increase is over and above the speed of an already "fast" moving molecule which even in the cool Earth's atmosphere could be on average 505 m/s (still did not add on the planet's rotational velocity).
But earlier in this thread we had calculated the the momentum in the photon was only capable of this: "So does that mean after the photon has attacted itself to the Carbon dioxide there will be an increase in velocity in the direction of the momentum of either 7.10 or 2.02 x 10^ -14 meters/sec."
Lets for ease of calculation say 5 X 10^-14 meters/sec velocity increase.
So we could see how much kinetic energy would be drained from the photon to give the CO2 molecule such a small increase in velocity. (Vf^2 -Vi^2 = 5.05E-11)
Kinetic Energy (final) - Kinetic Energy (initial) = 0.5*mass*5.05E-11
Mass of CO2 in kg = 7.31 * 10^-26
Delta KE = 1.84578E-36 Joules but that compared to the average energy in the photons absorbed 9.99E-31 J means that still only a fraction of the energy goes into satisfying the momentum requirement, in fact on average just 1.84723E-06 of it (like 2/one-millionth part of it).
But I still want to stress that since the photon has no mass this can only be satified with an increase in velocity rather than a reduction, so it only happens if the molecule and the photon have a "like" motion vector to begin with.
I wonder if the 4-momentum equations can be used to show this.
now previously we had looked at how much this momentum would imply a change in velocity. But that speed increase is over and above the speed of an already "fast" moving molecule which even in the cool Earth's atmosphere could be on average 505 m/s (still did not add on the planet's rotational velocity).
But earlier in this thread we had calculated the the momentum in the photon was only capable of this: "So does that mean after the photon has attacted itself to the Carbon dioxide there will be an increase in velocity in the direction of the momentum of either 7.10 or 2.02 x 10^ -14 meters/sec."
Lets for ease of calculation say 5 X 10^-14 meters/sec velocity increase.
So we could see how much kinetic energy would be drained from the photon to give the CO2 molecule such a small increase in velocity. (Vf^2 -Vi^2 = 5.05E-11)
Kinetic Energy (final) - Kinetic Energy (initial) = 0.5*mass*5.05E-11
Mass of CO2 in kg = 7.31 * 10^-26
Delta KE = 1.84578E-36 Joules but that compared to the average energy in the photons absorbed 9.99E-31 J means that still only a fraction of the energy goes into satisfying the momentum requirement, in fact on average just 1.84723E-06 of it (like 2/one-millionth part of it).
But I still want to stress that since the photon has no mass this can only be satified with an increase in velocity rather than a reduction, so it only happens if the molecule and the photon have a "like" motion vector to begin with.
I wonder if the 4-momentum equations can be used to show this.
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While it is logical to assume that there is some change in momentum for the atom as a whole, at least a portion of the photon's momentum contributes to a change in the energy state of the electron and the total mass of the atom. Momentum can be conserved without an equivalent change in the atom's kinetic energy.
The whole momentum does not have to be accounted for as adding a kinetic motion to the atom. Increased mass adds momentum even without a change in velocity etc.
When increased mass adds momentum without adding velocity - the only thing that springs to mind is when two objects going at the same velocity link up and combine their momentum then considered one object instead of two.
Can you think of a situation that would serve as the example of what you propose?
Would that mass be in the nucleus or in the electron? And would that mass disappear if the motion was stopped? This whole "relativistic mass" is an odd one, for as I see it there is a case where the so called relativistic mass of photon adds to the rest mass of the electron once absorbed and it seems to remain once the electron is made to be at rest. Which is different to the "relativistic mass" from the kinetic energy added in particle accelerators.
I for one have not got the whole topic understood at this stage.
@Only Me - the amount of momentum/energy tied up with the velocity change was only 2 parts per million, so you might be right, for there still is another 999998 parts of the energy still to be accounted for. Can all of the additional energy just be in the vibration? It was hard to imagine for we are talking of movements so small you do wonder if it is possible. Converting some of the photon's energy to small amount of mass would be a quick and easy way to hide the energy. But I don't have any scientific way of confirming this.
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If the vibration was extreme nearly to the point where the bonds between the C=O are nearly to the point of breaking then I think this amount of energy could be contained in these vibrations. I had previously used the equations estimating the spring strength but the number I calculated had no relevance as I had nothing to compare it to.
That is much better understanding than you had.
Only significant error in it is the complete lack of appreciation that the IR wavelength is typically at least 1000 times larger than all inorganic molecules (polymers considered as their monomers)
Thus EVERY electron of the molecule is in the same strength Electric filed of the photon´s EM wave. No reason why only one is accelerated - they all feel the very same force, all have the same mass thus ALL experience the same accelerations. - I.e. you have not yet fully abandoned your false ideas about ONE electron ONLY interacting with the Photon´s electric field. You cannot carry over to IR absorption by molecules the ideas you had about visisble & UV absorption by atoms as you have in every post, including this last one.
In crude classical terms: The WHOLE molecule is shaken by the E-field first one way for half a cycle of the EM wave then the other way and there may be a million or more of these alternating directions of shaking before the photon´s energy is stored in the bond(s) as the atoms of the molecule oscillate back and forth about their unexcited spatial separation points.
You should fully believe in that as it is correct. I even stated it in my prior post discussing atomic hydrogen you had complained I was ignoring your question about. Focus on the now bold part below:
I add now, to drive home hard the point that almost all of the solar radiation CANNOT be absorbed, by noting that for atomic hydrogen there are only 4 very narrow wavelenghts* in the ENTIRE continusous visible sprectrum of the sun that H can absorb. For the dominate gases of the earth´s atmosphere, there are NONE.
* They are the first 4 lines of the Balmer series. I.e.excited state n=2 going to even more excited states upper states with n = 3,4,5,& 6. Note also that even these 4 absorptions are very improbable as the hydrogen must already be in excited state n = 2 when the photon arrives. The N =2 to n= 1 transition occures almost immediately (life time of upper state before radiative decay less than a ms). It is the weakest line of the Lyman series and has the shortest excited state life time of all. - This is because all selections rules are obeyed AND the n=2 & n=1 orbitals have considerable spatial over lap, to speak classically. I will make a wild guess that less than 1 in a 100 trillion non-uv solar photons can be absorbed by hydrogen (or O2 or N2 also) - not much to build your original theory on (or even its replacement about DIFFERENTIAL absorption in the limbs)!
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I have already noted that Jupitor´s high winds are not solar driven, but due to the still cooling of Jupitor from the graviationaly energy released as it formed (and still some as it contracts). You wanted to know about Venus. I don´t know the details but am nearly sure the basic story as to why it has high (400 mph) winds high up, were we can measure them is due to:
(1) Venus is is much closer to the sun than earth so every square meter of cross section has much greater solar energy incident upon it.
(2) Venus has and opaque atmosphere. Thus essentially 100% of this stronger solar input is absorbed. Earth has many white clouds that reflect solar energy so we don´t get all of the lesser amount the sun is sending to Earth.
(3) As Venus is opaque, it act like a black body radiator, and a quite hot temperature one.
(4) Thus there is great radiative cooling occuring on the side turned away from the sun - significantly lower temrperature there after a few hour of being in the dark with this large radiative loss of energy.
(5) In contrast on the sunlight side there is great heating of this black body absorber in the very intense sunlight.
(6) Points (4) & (5) make for atmospheric temprature gradients much greater than on Earth (even contrasting the equator with the poles) I would guess at least 500 degrees F temperature differentials.
(7) Large atmospheric temperature gradients ALWAYS drive very strong winds.
Frankly I am surprised some winds of Venus are not faster than 400 mph. When that cooled off during the night time gas mass first comes back into the sunlight the max thermal gradient I would guess exceeds 100 F / mile! i.e. Stroms like you can not image, if this is true. - Seeing this "reheating region" of Venus from its "top side looking down" from Earth is not easy as the line of sight would need to be close to the sun, if not thru the sun.
SUMMARY: There is no need for your theories, which contain several violations of well established physic, to "explain" the high winds of either Venus or Jupitor. Well accepted physics does quite a good job of that. although a few minor details may still need computer modeling to get complete agreement.
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This is what I call your first false theory of why planets have strong winds. Not only is the whole idea false but even you have the postulated direction of the effect WRONG (making prograde rotation as you later call it) as you were ignorant of the blue shift giving more photons with UV enough to be absorbed (ionizing) any atom or molecule you wish to consider. (Also blue shift gives more that can disassociate O2 & N2 molecules) The "push" on the atmosphere you were speaking of above thus tends to make retrograde winds.
You have done a lot of "back peddling" in your later posts, I.e. claiming you were speaking of molecular absorption in bonds all along, not by one single electron in the molecule, and that you never suggested IR absorption by GHG were the cause of the high winds, etc. - When I first called that your "first false theory" you said I had misunderstood the OP, but I was referring to this clear statement in post 26 and many later repeats of the idea that took me many posts to get you to abandon (replace with your "its the differential absorption in the two limbs (one approaching the sun & other receding from it) that causes the high winds.
If I have been critical of you personnally (don´t think I have been) it would be because this back peddling to claim you never said what is easy to show you did; That is DISHONEST.
(As well as foolish since quotes of your posts with those statements can be displayed as I did above.)
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Look - you make the actions of a photon a bit different than is usual. Ionisation and excitation seem to be actions directed to one electron only. As soon as some of the energy of a photon is absorbed is the photon the same as it was before? I cannot yet accept that you have the right concept as yet.
Your idea that as the photon passes the transfer is intensified means as the photon passes it loses energy over time. In your view is a photon just the one wavelength or a series of waves? Above you seem to say the photon is million or more of these? Is there science supporting such a view?
Light has both wave and particle properties. I would prefer to think that these energy transfers are involving the particle nature of light yet you favour the wave.
When increased mass adds momentum without adding velocity - the only thing that springs to mind is when two objects going at the same velocity link up and combine their momentum then considered one object instead of two.
Can you think of a situation that would serve as the example of what you propose?
Would that mass be in the nucleus or in the electron? And would that mass disappear if the motion was stopped? This whole "relativistic mass" is an odd one, for as I see it there is a case where the so called relativistic mass of photon adds to the rest mass of the electron once absorbed and it seems to remain once the electron is made to be at rest. Which is different to the "relativistic mass" from the kinetic energy added in particle accelerators.
I for one have not got the whole topic understood at this stage.
@Only Me - the amount of momentum/energy tied up with the velocity change was only 2 parts per million, so you might be right, for there still is another 999998 parts of the energy still to be accounted for. Can all of the additional energy just be in the vibration? It was hard to imagine for we are talking of movements so small you do wonder if it is possible. Converting some of the photon's energy to small amount of mass would be a quick and easy way to hide the energy. But I don't have any scientific way of confirming this.
The only real point I was making is that, assuming I interpret the equation E = mc^2 properly, an absorbed photon increases the mass of an atom which would increase its momentum, even where the kinetic energy (velocity) of the atom as a whole cannot account for all of the photon's momentum.
Light has a lot of energy and not much momentum, so it is not hard to account for its momentum. If light could just be absorbed as mass all light would disappear in a flash.
James R agreed that the mass of the electron is increased when a photon absorbs to it, but only a few photons can be added till the electron is ionized, and at the same time the electron is trying to go back to where it was, so light has these momentary interactions with atoms.
With the GHG the photon does not convert to mass as the net effect of the absorptions is heating which is a reflection of the kinetic energy not mass.
So I agree with the bending and stretching ideas, but I think of it as a punch to an electron but Billy likens it to a building resonance.
No I have a very standard view that 99% of Ph.D.s in physics have. (Although many of them have never stopped to consider the lenght of a photon with well defined energy.)
After the photon´s energy is absorbed by interaction with matter, it exist no more. It is an all or nothing process (but in a scattering process like Compton scattering a photon of lesser energy exists as there has been momentum and energy tranferred to the scatter. Whether or not you want to consider the weaker photon the same or a new one is your choice of how you want to speak. Never during the statering was there no photon - Scattering is not an absorption process with re-emission for most Ph.D.s so I tend to think of the lower energy photon post scattering as a modification in energy and direction of travel ot the original photon.)
What you don´t understand is that it is nonsense to speak as you do of the photon being absorbed at some point in time. There is a fundamental uncertainity about when this occurs (delta T). If the energy of the photon is very well defined (a long photon with millions of cycles of E-field alternation) then the uncertainity in the energy (delta E) is very small. Nature, not man´s measurement limitations sets a minimum value for the product (delta T)(delta E) Thus the (Delta T) is relatively large. It is nonsense to speak of the photon being absorbed at some unique time - best you can say is that during (delta T) the photon becomes energy in the eletron atom system.
Thus when the energy is absorbed, if you want to think classically and incorrectly ignoring the fact that (delta T) is relative large for a long photon is uncertain - could be when the first cycle of the EM wave arrives at the absorber or when the last leaves. There really is no defined time of the absorption.
Definitely not my view as that is nonsense. It make no more sense than to ask at what age do unicorns get their horn? (Clearly were not with horn when born or pre-natally.) I have tried to show you, by point out that the typical wave length is 1000 or more times the size of the molecule so that there is more than one electron in the molecule in the same E-field of the current cycle of the photon ONLY to try to get you to stop speaking of IR absorption by a molecule as being by ONE electron of the molecule - at best localization it is by one bonding cloud between two atoms of the molecule.
Yes I have made than very clear in the post telling how I measured some Sodium D photons to be 30 cm long and that the green line of the Norther lights photon is at least 20 meters long. Considering how short (but long compared to a molecule) the wave lenght are, yes, typically many milions of cycles in a photon. For details of how I measured a ~30cm length for some shorter than average photons see:
http://www.sciforums.com/showpost.php?p=2918932&postcount=178
No a photon is neither. It is a photon, with characteristic humans can not understand. At time in some experiments it behaves more like a particle (EG in the photo- eletric effect) and in other experiments (EG the two slit interferomenter) more like a wave.
In the case of absorption both characteristic are present: the energy is extended over its length, like a wave, yet when it is gone, it is all gone with the absorber particle recoiling as if hit by a particle, etc.
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Light has a lot of energy and not much momentum, so it is not hard to account for its momentum. If light could just be absorbed as mass all light would disappear in a flash.
James R agreed that the mass of the electron is increased when a photon absorbs to it, but only a few photons can be added till the electron is ionized, and at the same time the electron is trying to go back to where it was, so light has these momentary interactions with atoms.
With the GHG the photon does not convert to mass as the net effect of the absorptions is heating which is a reflection of the kinetic energy not mass.
So I agree with the bending and stretching ideas, but I think of it as a punch to an electron but Billy likens it to a building resonance.
What generally happens is that a photon is absorbed, raising the energy state of an electron which then emits a photon as it falls back to a lower energy state.
If an atom whose electrons are already at their highest enegry states is subjected to an interaction with a sufficiently energetic photon, it may then result in ionization or the ejection of an electron.
For photons which would normally be involved in absorbtion/emission with a particular atom, ionizing events should be rare when compared to the general absorbtion emission cycle.
Ionization can occur through other interactions also.
rearranging that equation M = E/C^2
Now the higher frequency IR photon that CO2 absorbed 2349 Hz and frequency times Plack's Constant gives photon energy
2349 Hz * 6.63E-34 = 1.56E-30Joules
dividing that by C^2 you are going to get a mass increase (1/(9*10^16)* 1.56E-30) kg.
The mass increase is small and could account for the momentum and the energy but not the physics of the GHG using the energy as heat not mass.
This is all true as far as I know, but you have not mentioned the photon's momentum. The photon is not just energy, it has two other types of momentum as well. Generally I have only worried about the linear momentum.
And that is where my theory differs for I say these absorptions only happen when there is a similar linear momentum vector between atom/molecule and photon. So the momentum always speeds up the interacted particle, never slowing it down.
It is the proof of this that I want to establish by experiment.
I am sure James never said that. In several years I have caught James in only 4 errors and that is a totally false statement James would never say.
The mass of the absorbing system, assuming the electon was only excited to a higher energy level, not removed, is increased, WHILE THE SYSTEM IS IN THE EXCITED STATE, but all electrons all have exactly the same rest mass - it is a universal constant, just like their charge is.
Thanks for reminding about that experiment. Were the results accepted and peer reviewed. I like it.
As you said "ONLY to try to get you to stop speaking of IR absorption by a molecule as being by ONE electron of the molecule - at best localization it is by one bonding cloud between two atoms of the molecule. " I could live with this too, "the bonding cloud", yet there are other available electrons as well, and they are possibly more exposed , less shielded by the atoms.
Take a break for a couple of days to recap and research.
It was recently in the Gravity never zero thread. Well that is what I understand he said. We'll check it.
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