studiot Posted September 15, 2017 Posted September 15, 2017 So far you seem to think that there are three possibilities (though not mutually exclusive as with your chlorophyll example) when light approaches something. Transmission, reflection or absorbtion. That is not the case. Do you think that the radio wave is transmitted through the solid copper bar of the aerial? We have just agreed that the wave appears on the downstream side of the bar (it is not blocked by it) Having agreed that an aerial is frequency selective, do you understand how one works? I suggest you forget photons here. I have already suggested that for the purposes of this thread and your on topic question classical wave theory is adequate. I agree that there are many additional observable effects that require alternative theories, but they are not on topic here. Directionality is a not function of the wave alone, it is also dependent upon the source. The issue of a propagation medium is off topic here, but I would just observe that Faraday's notion that the wave generates its own medium as is goes along is adequate here. Finally your response to my observation that waves are larger than the obstruction leaves much to be desired. If the waves are propagating in the z direction then they are enormous in the x and y directions since they are so far from the source they are effectively plane waves. The 'waves' of light from the Sun are vastly bigger than the Earth itself. Hopefully you understand this much.
BanterinBoson Posted September 16, 2017 Author Posted September 16, 2017 1 hour ago, studiot said: So far you seem to think that there are three possibilities (though not mutually exclusive as with your chlorophyll example) when light approaches something. Transmission, reflection or absorbtion. That is not the case. Yes, I think there are 3 possibilities. Why is that not the case? Even if a wave is wide enough to flow around an object, the part that contacts the object still must flow through or bounce off of the face of the object. See 7:08 here: Quote Do you think that the radio wave is transmitted through the solid copper bar of the aerial? We have just agreed that the wave appears on the downstream side of the bar (it is not blocked by it) Having agreed that an aerial is frequency selective, do you understand how one works? I can't answer yes or no, I have to explain: Radio waves are produced by current that flows up and down an aerial at a specific frequency. The radio waves propagate through space inducing current flows on anything that will resonate with the wave. When the wave meets a receiving antenna, the wave induces current in the antenna = to the frequency of the wave. So to answer your question, I don't think the wave is transmitted down the copper bar, but the wave is transmitted through the copper bar in the direction of the wave travel. The current that is induced is not the same as the wave because it is current and not a EM wave. We could say a "signal" is = to both the wave and the current, but I think the wave and the current are separate things. Also, the induced current should steal energy from the wave. Radio waves sufficiently low can propagate through 100s of feet of earth, therefore they must be using the earth as a medium for propagation (ie not flowing around the earth). If they can do that, then a copper bar is no obstacle. As we climb higher and higher in frequency up the EM scale, the less able the waves will be to pass through the copper bar until we reach a point where the waves reflect. As we go further, we should reach frequencies that can penetrate the copper, but I don't think visible light is high enough. Quote I suggest you forget photons here. I have already suggested that for the purposes of this thread and your on topic question classical wave theory is adequate. I agree that there are many additional observable effects that require alternative theories, but they are not on topic here. Do you mean I should forget photons as particles? I can do that. I don't even think of electrons as particles, but super-directional waves where the wave-front has a high density of energy. Quote Directionality is a not function of the wave alone, it is also dependent upon the source. I agree. The frequency of the wave is determined by the source and therefore the directionality is determined by the source. At 12:00 here, they say it can take 150,000 years for light to find its way out of the sun. I believe that if the light were radio waves, that would not be true. It's the directionality of the wave that prevents it from finding its way out sooner. Directionality is dependent upon frequency and the higher the frequency, the more like a particle a wave will act. What do you think? Quote The issue of a propagation medium is off topic here, but I would just observe that Faraday's notion that the wave generates its own medium as is goes along is adequate here. That's fine. If light generates its own medium as it goes, that is fine. Quote Finally your response to my observation that waves are larger than the obstruction leaves much to be desired. If the waves are propagating in the z direction then they are enormous in the x and y directions since they are so far from the source they are effectively plane waves. The 'waves' of light from the Sun are vastly bigger than the Earth itself. Hopefully you understand this much. I don't understand. I don't see the light from the sun as one big wave a million times bigger than earth, but a collection of small waves emitted from tiny particles inside the sun and those waves don't spread just like a laser beam wouldn't. I think of the sun analogous to trillions and trillions of tweeters emitting high-frequency sound that is directional and aimed at earth... not one giant bass woofer emitting sound in all directions. I suspect that you're trying to get me to accept that high-frequency light has the same properties as low-frequency radio waves and I can't do that with my current understanding. I need a reason to believe that gamma rays, for instance, would be omnidirectional. Astrophysicists say that if a quasar is not aimed directly at earth, then we have nothing to fear because the gamma rays do not spread very much, even over astronomical distances. So I have trouble equating gamma rays with radio waves and therefore when a highly-directional (particle-ish-like) wave meets a co2 molecule, I don't see it simply flowing around it like a radio wave would to an antenna. But even if it did, I think it would still reflect part of the wave back. The difference could be a moot point.... whether a co2 molecule reflects an entire small wave back or if it reflects part of a large wave back, the result is probably the same. Therefore the co2 is taking a co2-size bite out of the visible light that would otherwise be reaching earth while only re-emitting a small part of IR back to earth. That action should be cooling. I hope this post doesn't make you mad. I'm trying hard not to offend you, but I can't say I understand when I don't.
swansont Posted September 16, 2017 Posted September 16, 2017 14 hours ago, BanterinBoson said: I agree and I think it is because the photon is a wave, not a particle. Light is both. Quote What do you think about the analogy of the directionality of sound increasing as frequency increases to that of EMWs? Are radio waves more omnidirectional than higher frequencies of light? Whether light is omnidirectional depends on the source. Quote And the electron is so super-directional that we describe it as a particle? I'm trying to equate "particle-ness" to directionality. Electrons have both wave and particle behavior. It's not directionality that governs this. Quote Isn't that also true of sound through a tube? Wave guide cutoff is due to cross-section. The resonance you describe here is based on length. You can get that behavior with light, as well: standing waves in cavities. Quote I think I can see what you're saying. The atom "looks big" if the photon is at resonant frequency because the photon interacts significantly and it "looks small" if the photon is at a frequency higher than resonance because there is little interaction, but then I'm having trouble seeing it "looking small" if the photon is below resonance because lower frequencies will interact. CO2 is transparent to light at several tens of microns, or less than a micron, but it has absorption bands in between. (you included this in your first post) You are clinging to this notion that frequency below resonance is not transmitted and just isn't true. There are multiple resonances. As you move off-resonance, in either direction, absorption decreases. Quote For instance, holding a weight suspended by a rubber band in your hand and then slowly move your hand up and down, the weight should follow the movement of your hand. As you move faster, you'll find the resonant frequency where the weight and your hand move in opposite directions. As you move your hand faster than resonance, the weight will not move. Resonant light absorption by atoms is not classical. Your model doesn't apply. Quote Don Lincoln says here (at 1:30) that electrons and top quarks have no size: As I stated earlier. (point particles) Quote I can see that, but what happens outside those values is what I'm having trouble seeing. And you have to learn about quantum mechanics to do so. Quote I'm not sure light having a constant speed regardless of the speed of the observer necessarily disproves a medium. I can see how it would seem so, but space is actually spacetime in 4 dimensions and having zero velocity with respect to the medium necessitates maximum velocity through time and maximum velocity through space necessitates zero velocity through time, so it seems not so easy to disprove an aether with relativity due to the extra time variable and the fact that, from the point of view of light, there is no time or space. How do we use something that sees no space or time to disprove something that exists within spacetime? Probably a discussion for another thread, but if a physical medium exists, you have to have a velocity with respect to it, and that just doesn't work. Quote If you're saying that light will pass straight through a molecule if there are no resonances, then how does light reflect off of chlorophyll? We know that grass is green because it absorbs all frequencies of visible light except green, therefore green is reflected and that is what we see. All other frequencies are absorbed and converted to sugar and IR radiation which is re-emitted. So if green light reflects off of chlorophyll, then why can't light reflect off of co2? Bulk material behaves differently. If you turned chlorophyll into a gas, its optical behavior would change. Light would reflect off of solid CO2 8 hours ago, BanterinBoson said: So I have trouble equating gamma rays with radio waves and therefore when a highly-directional (particle-ish-like) wave meets a co2 molecule, I don't see it simply flowing around it like a radio wave would to an antenna. But even if it did, I think it would still reflect part of the wave back. The difference could be a moot point.... whether a co2 molecule reflects an entire small wave back or if it reflects part of a large wave back, the result is probably the same. Therefore the co2 is taking a co2-size bite out of the visible light that would otherwise be reaching earth while only re-emitting a small part of IR back to earth. That action should be cooling. Light that never reaches the earth doesn't get counted; it doesn't contribute to warming. Saying that this is cooling would be double-counting this effect.
studiot Posted September 16, 2017 Posted September 16, 2017 (edited) 9 hours ago, BanterinBoson said: I don't understand. I don't see the light from the sun as one big wave a million times bigger than earth, but a collection of small waves emitted from tiny particles inside the sun and those waves don't spread just like a laser beam wouldn't. I think of the sun analogous to trillions and trillions of tweeters emitting high-frequency sound that is directional and aimed at earth... not one giant bass woofer emitting sound in all directions. I suspect that you're trying to get me to accept that high-frequency light has the same properties as low-frequency radio waves and I can't do that with my current understanding. I need a reason to believe that gamma rays, for instance, would be omnidirectional. Your are too suspicious. I am trying to work through the physics of something with you so that that you can see for yourself what the facts are. It is fundamental to wave theory that a point source produces a spherical wave that spreads out in all directions. So a point source on the surface of the Sun (Point P in my fig1) creates an expanding spherical wavefront as I have shown dashed in Fig1. This has an average radius of 150 million kilometres by the time that front reaches Earth. The source at P does not have a choice in this, nor does it have collimators or focusing devices. But yes there are many such sources on the surface of the Sun and they are not coherently phased, like a laser. (Thank the Lord Huygens) or we would not be here if they were. I said that as a result of that distance that wavefront is effectively a plane wave so the (very simple) maths of this is shown in fig2. The deviation of a circular curve from a straight line is given by the formula (distance along the straight line squared) divided by (twice the circle radius) With the figures shown this works out at about one tenth of a millimetre over the linear distance of the radius of the Earth. So think how much straighter the wavefront must be over the size of a carbon dioxide molecule, whose bond lenght is 1.16 x 10-7 millimetres long. This is why we can say that in another model - that of geometrical optics which treats light as a series of 'rays', that the rays from the Sun are parallel. I mentioned bond length of carbon dioxide which is important because this determines the frequencies for the four fundamental modes of vibration of the molecule. The plan is to fully understand how a straight radio aerial works and see if we can use this to understand how the stretching modes of the carbon dioxide bonds comes to be in the IR and microwave bands and not other bands of the EM spectrum. But first we must look at your thoughts on how a radio aerial actually works. 9 hours ago, BanterinBoson said: I can't answer yes or no, I have to explain: Radio waves are produced by current that flows up and down an aerial at a specific frequency. The radio waves propagate through space inducing current flows on anything that will resonate with the wave. When the wave meets a receiving antenna, the wave induces current in the antenna = to the frequency of the wave. So to answer your question, I don't think the wave is transmitted down the copper bar, but the wave is transmitted through the copper bar in the direction of the wave travel. The current that is induced is not the same as the wave because it is current and not a EM wave. We could say a "signal" is = to both the wave and the current, but I think the wave and the current are separate things. Also, the induced current should steal energy from the wave. This is not far off the truth but just need a couple of amendments. I have underlined the relevant bits. Firstly and most importantly. Forget resonance - the word will get you into trouble when we consider charge separation in the carbon dioxide molecule. It has a specific and quite different meaning for chemists and bonding. The radio waves induce current in any conductor of any length. Resonance is not required. The issue is what happens to those currents when they have been induced. We shall see that conductor length then becomes vitally important. I prefer the phrase charge separation, rather than current because that is what happens in both the radio aerial and the carbon dioxide molecule. The effects are potential driven, not current driven. Finally we spoke about reflection, absorbtion and transmission. What about refraction and diffraction of waves? I think we are doing well and making real progress Edited September 16, 2017 by studiot defeat timeout feature of forum 1
BanterinBoson Posted September 21, 2017 Author Posted September 21, 2017 Apologies for the delay, but I have accomplished some studying. On 9/16/2017 at 7:41 AM, swansont said: Light is both. How can it be both? The only way is if light is a wave that is thought of as a particle in some situations. Quote Electrons have both wave and particle behavior. It's not directionality that governs this. I read that protons often pass through each other in particle accelerators. Particles cannot do that because particles cannot occupy the same space. @user104: (1) you cannot. (2) yes indeed, the LHC collides protons with protons and most times the protons just go through each other without scattering. – John Rennie Feb 22 '16 at 12:42 @ConstantineBlack: particles aren't points. They are excitations in a quantum field and don't have a position or a size in the sense that macroscopic objects do. They are pointlike in the sense that any experiment to measure a minimum size will fail. Any two particles, electrons, quarks and photons, can have overlapping probability distributions, and there is a finite probability they can both be detected in any volume element no matter how small that volume element is. However it is meaningless to ask if any two particles of any kind can be at the same point in space. – John Rennie Feb 22 '16 at 17:12 @ConstantineBlack: the probability of finding a particle in volume dVdV is P=ψ∗ψdVP=ψ∗ψdV and this goes to zero as dVdV goes to zero. So the probability of finding any particle in a point of zero volume is zero. – John Rennie Feb 22 '16 at 17:26 https://physics.stackexchange.com/questions/238976/why-doesnt-an-electron-ever-hit-and-stick-on-a-proton?noredirect=1&lq=1 Quote Wave guide cutoff is due to cross-section. The resonance you describe here is based on length. You can get that behavior with light, as well: standing waves in cavities. Aren't electrons considered to be standing waves in their orbitals? Quote You are clinging to this notion that frequency below resonance is not transmitted and just isn't true. Above resonance is what I thought, but you're right. What I failed to see is that EM waves only interact with charges, not matter itself. Charged particles do not have that much mass, so they are accelerated by the EMR. But hypothetically, if a charge could be massive enough or attached strongly to enough mass, what would happen to the EM wave as it passed by that charge? If a force is applied in effort to move a charge, but the charge cannot move, then where does the force go? That is a little analogous to a speaker and signal where the speaker cannot keep up with the signal due to its mass. In that case, the energy goes to heat in burning the voicecoil, but what happens when an EM wave attempts to accelerate a charge in the opposite direction of its momentum? Is heat generated? If so, how? The heat would have to be radiated in IR resonances, so how would those resonances develop? Quote Resonant light absorption by atoms is not classical. Your model doesn't apply. As I stated earlier. (point particles) And you have to learn about quantum mechanics to do so. Probably a discussion for another thread, but if a physical medium exists, you have to have a velocity with respect to it, and that just doesn't work. Let's save that for other discussions Quote Bulk material behaves differently. If you turned chlorophyll into a gas, its optical behavior would change. Light would reflect off of solid CO2 Agree. Quote Light that never reaches the earth doesn't get counted; it doesn't contribute to warming. Saying that this is cooling would be double-counting this effect. What I mean is if something is introduced into the atmosphere that reflects light away, then the effect of that introduction is cooling relative to the non-introduction of the element. I'll state more about this after I reply to Studiot. On 9/16/2017 at 9:09 AM, studiot said: Your are too suspicious. I am trying to work through the physics of something with you so that that you can see for yourself what the facts are. It is fundamental to wave theory that a point source produces a spherical wave that spreads out in all directions. So a point source on the surface of the Sun (Point P in my fig1) creates an expanding spherical wavefront as I have shown dashed in Fig1. This has an average radius of 150 million kilometres by the time that front reaches Earth. The source at P does not have a choice in this, nor does it have collimators or focusing devices. But yes there are many such sources on the surface of the Sun and they are not coherently phased, like a laser. (Thank the Lord Huygens) or we would not be here if they were. I said that as a result of that distance that wavefront is effectively a plane wave so the (very simple) maths of this is shown in fig2. The deviation of a circular curve from a straight line is given by the formula (distance along the straight line squared) divided by (twice the circle radius) With the figures shown this works out at about one tenth of a millimetre over the linear distance of the radius of the Earth. So think how much straighter the wavefront must be over the size of a carbon dioxide molecule, whose bond lenght is 1.16 x 10-7 millimetres long. Good point! I know waves should propagate in all directions, but I know from experience that high-frequency sound does not. I don't know why that is. Quote This is why we can say that in another model - that of geometrical optics which treats light as a series of 'rays', that the rays from the Sun are parallel. That makes sense. Quote I mentioned bond length of carbon dioxide which is important because this determines the frequencies for the four fundamental modes of vibration of the molecule. Not just bond length and strength, but mass as well. Right? What I failed to see before was that EMWs only interact with charges, not matter itself. So the charges will be the "handles" that the EMW "grabs" to then shake the molecule and how the molecule behaves is determined by the bonds and the mass of the atoms. Quote The plan is to fully understand how a straight radio aerial works and see if we can use this to understand how the stretching modes of the carbon dioxide bonds comes to be in the IR and microwave bands and not other bands of the EM spectrum. I'm a little confused how the co2 molecule would interact with the microwave band because: Often molecules contain dipolar groups, but have no overall dipole moment. This occurs if there is symmetry within the molecule that causes the dipoles to cancel each other out. This occurs in molecules such as tetrachloromethane and carbon dioxide. The dipole-dipole interaction between two individual atoms is usually zero, since atoms rarely carry a permanent dipole. https://en.wikipedia.org/wiki/Intermolecular_force The way I understand microwave heating with water is that the MW rotates the water molecule which then interacts kinetically with other molecules that cause resonances in the water molecule in the IR band. Water is a dipole, but co2 is shaped more like - + - so I'm curious if MW could rotate the molecule in the same way. Quote Forget resonance - the word will get you into trouble when we consider charge separation in the carbon dioxide molecule. It has a specific and quite different meaning for chemists and bonding. I can't forget resonances. For instance when an EMW causes co2 to resonate, it absorbs energy which is expressed by the exaggerated movement and an electron being kicked up to the next energy level. Resonance makes too much sense to me. Quote The radio waves induce current in any conductor of any length. Resonance is not required. Yes, any bits of metal lying around within radio wave transmission will have currents induced in them. Resonance is not required, I agree. Quote The issue is what happens to those currents when they have been induced. We shall see that conductor length then becomes vitally important. Hmm.. well the induced currents radiate EMWs as well, right? Quote I prefer the phrase charge separation, rather than current because that is what happens in both the radio aerial and the carbon dioxide molecule. The effects are potential driven, not current driven. I don't understand "phase charge separation". Quote Finally we spoke about reflection, absorbtion and transmission. What about refraction and diffraction of waves? This is a good opportunity to tell you what I have learned since we last spoke. In the case of an EM wave and glass, the wave excites the charges causing an acceleration which cause a re-emission of an EM wave at a later point in time which causes a phase shift from the original wave. The summation of the two waves causes refraction through the glass and the reflection is simply the re-radiated wave traveling in the opposite direction that isn't completely cancelled by the incident wave. In the case of opaque objects, the EM wave causes resonances which absorb the wave and hinder further propagation. The re-emission of EM waves from the excited charged particles are shifted 180 degrees out of phase and cancel the incoming wave completely. If the object is not a black body and it has a color, then the color is from light that was not cancelled by the phase shift and is re-radiated at the frequency of the color (that is why grass is green). I'm not sure diffraction is a property of re-emission. Maybe you could tell me more about that. Here's where I really need some help: Any molecule that does not absorb will reflect and refract due to the acceleration of charges produced by the force from the electric fields interacting. Since we know for sure that co2 does not absorb in the visible light band, then we know for sure that it reflects (re-emits) in that band. What we don't know is the amount of the reflection. So the problem has now been reduced to determining the amount of reflected 10^14 light in comparison to the amount of 10^13 light that is absorbed and re-radiated. 10^14 light has 10x more energy than 10^13 so the breakeven point would be at 1/10 reflection; therefore, if 1/10 of visible light is reflected, then it's a wash. But if we consider that the cone of IR re-emission towards earth is roughly 90 degrees, then only 1/4 of the IR re-radiation from co2 will strike earth (360/90=4). So, does that mean 1/40 of visible light needs to be reflected for it to be even? What is confusing me is energy being solely a function of frequency and dividing light in half does not halve the energy just like cutting a cake in half does not reduce its temperature, but it must have an effect on something. Someone will have to show me what that means. That's the last piece to the puzzle. Quote I think we are doing well and making real progress I'm trying my best to understand this. Thanks for your help! ________________________________________________________ Ah crap... I forgot to wait an hour for the "reply merging" behavior to expire. I can't imagine why someone would like that. If I had wanted it to all be one post, I would have constructed it so.
swansont Posted September 21, 2017 Posted September 21, 2017 On 9/20/2017 at 9:44 PM, BanterinBoson said: How can it be both? The only way is if light is a wave that is thought of as a particle in some situations. Bingo! Quote I read that protons often pass through each other in particle accelerators. Particles cannot do that because particles cannot occupy the same space. Wave behavior, as all things exhibit. Quote Aren't electrons considered to be standing waves in their orbitals? No, that's the Bohr model interpretation. That model has been replaced. The QM solution does not lend itself to that interpretation. Quote But hypothetically, if a charge could be massive enough or attached strongly to enough mass, what would happen to the EM wave as it passed by that charge? If a force is applied in effort to move a charge, but the charge cannot move, then where does the force go? Force doesn't have to "go" anywhere. If You had a charge on a larger mass, the mass would undergo a proportionally smaller acceleration. A proton is not accelerated as much as an electron by a given electric field, and a heavy ion is accelerated even less. Quote That is a little analogous to a speaker and signal where the speaker cannot keep up with the signal due to its mass. In that case, the energy goes to heat in burning the voicecoil, but what happens when an EM wave attempts to accelerate a charge in the opposite direction of its momentum? Is heat generated? If so, how? The heat would have to be radiated in IR resonances, so how would those resonances develop? Accelerating charge is a classical interpretation, and is not consistent with discussion of quantum mechanics. QM is quite different behavior, not seen in the macroscopic world. You have to let go of classical models of behavior. Quote What I mean is if something is introduced into the atmosphere that reflects light away, then the effect of that introduction is cooling relative to the non-introduction of the element. True, but if it's there already that's not the level everyone else is using in these discussions. Further, the reflection is minimal. Air at STP has an index of refraction of 1.000277 (higher as you get colder and the air becomes more dense). reflection for normal incidence is ((n1-n2)/(n1+n2))^2 Meaning if you went directly from a vacuum to air near the ground, you would reflect about 2 x 10^-8 of the light at that interface. But we aren't doing that. The density of the atmosphere increases as you get closer, so the transition is more gradual, and the reflection is even smaller. Reflection from the atmosphere is small. And now you're talking about the effect of adding in a few ppm, or tens of ppm, of additional CO2. A small change in a small effect.
studiot Posted September 21, 2017 Posted September 21, 2017 (edited) I have started a new discussion about wave/ particle duality in the quantum theory section http://www.scienceforums.net/topic/110124-wave-particle-duality-inspired-by-a-thread-in-chemistry/ I will address the (useful) remarks by BanterinBoson in a second reply. Edited September 21, 2017 by studiot
BanterinBoson Posted September 22, 2017 Author Posted September 22, 2017 14 hours ago, swansont said: Bingo! Yay! Quote No, that's the Bohr model interpretation. That model has been replaced. The QM solution does not lend itself to that interpretation. What is the interpretation of the new model? From: https://en.wikipedia.org/wiki/Atomic_orbital#Electron_properties With the development of quantum mechanics and experimental findings (such as the two slits diffraction of electrons), it was found that the orbiting electrons around a nucleus could not be fully described as particles, but needed to be explained by the wave-particle duality. In this sense, the electrons have the following properties: Wave-like properties: - The electrons do not orbit the nucleus in the manner of a planet orbiting the sun, but instead exist as standing waves. Thus the lowest possible energy an electron can take is similar to the fundamental frequency of a wave on a string. Higher energy states are similar to harmonics of that fundamental frequency. - The electrons are never in a single point location, although the probability of interacting with the electron at a single point can be found from the wave function of the electron. Particle-like properties: - There is always an integer number of electrons orbiting the nucleus. - Electrons jump between orbitals in a particle-like fashion. For example, if a single photon strikes the electrons, only a single electron changes states in response to the photon. - The electrons retain particle-like properties such as: each wave state has the same electrical charge as the electron particle. Each wave state has a single discrete spin (spin up or spin down). This can depend upon its superposition. I don't believe an integer number of electrons necessarily implies particles just like a wine glass resonance doesn't imply sound is a particle. Electrons jumping between orbitals doesn't imply particles either and for the same reason. Piano keys may be discrete, but that doesn't mean sound is. Spin might imply a particle, but I don't know much about spin. Quote Force doesn't have to "go" anywhere. If You had a charge on a larger mass, the mass would undergo a proportionally smaller acceleration. A proton is not accelerated as much as an electron by a given electric field, and a heavy ion is accelerated even less. I suppose force goes to acceleration as F=ma and if there is no acceleration, then there is no force or else the mass is infinite and cannot be accelerated. So it seems safe to say that if an EMW causes a force, then there will be an acceleration. Quote Accelerating charge is a classical interpretation, and is not consistent with discussion of quantum mechanics. QM is quite different behavior, not seen in the macroscopic world. You have to let go of classical models of behavior. Accelerating charges do not cause EMWs? What do I let go of? Quote True, but if it's there already that's not the level everyone else is using in these discussions. Idk what others are doing; I'm just interested in the change. They say co2 makes earth hotter. Hotter than what? Hotter than before the co2 was there. Quote Further, the reflection is minimal. Air at STP has an index of refraction of 1.000277 (higher as you get colder and the air becomes more dense). CO2 is 1.00045 https://en.wikipedia.org/wiki/Refractive_index So that means v= 299,792,458/1.00045 But that doesn't tell me the reflection because that number is a composite of the original wave and the re-emitted wave. How do I figure only the re-emitted wave? That is the reflected wave I need to know. My next problem is at what frequency is that refraction number? I need a dispersion index https://en.wikipedia.org/wiki/Dispersion_(optics) This page says spinach chloroplasts have a refractive index of 1.02 - 1.06, which seems quite low compared to glass (1.50) which is more clear than plant leaves. https://watermark.silverchair.com/api/watermark?token=AQECAHi208BE49Ooan9kkhW_Ercy7Dm3ZL_9Cf3qfKAc485ysgAAAekwggHlBgkqhkiG9w0BBwagggHWMIIB0gIBADCCAcsGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQM9k0k4vpbFmtAwcB5AgEQgIIBnCGSV_FzHr0ZM7YrjqnI7sxln05MN9JX0B-CnFOHOvtUuK-GG_mkxZCRRbJhKeX6I-5ckJ3ojxbz9WPPqBYutP2aEV4Mu3L_4cCoAfZoU-Ywncf3RMLV1jWbm2yKtPXD7ZqO5148Xy3dOKNQQOT3Fv682OmGWnB_kwqpj3cN2iaMzjFktP3tIseZre7Vg0VnksDnIEn4oi9hxzEDq5qXYHTdVPoDNA8DnpbL4x3QbaUGZB8_TuLac61hr4tRHsafDqlbwSxxbTz565_3-SU8N2GIElLrJqJSm4sL38b75m3d1nAkrkErTHX6Ia3joM4ZnytHoYAggpyuLP2q8MAqAWzdhzNtElhmw6dy_gJdbEFhi5alDYLVBWx7uuZ0aMiaZu9k1PesthCUYGvfNHNA0u64vY4SZAE6ghK5oJJh4VIpVRf8es5plsTjcrMqQ1RtkY9S8zZAdvh68jf_fwvgJ_B-jpBULQFagsqK0GnUXRE8WWVQcpjXAhKOVymNO9FHK6Skcdr3zGDCExdUCGy4bts2FLJs5xUTZDLKryk But still, without a dispersion index, I'm not sure what I am comparing. Quote reflection for normal incidence is (n1-n2/n1+n2)^2 Meaning if you went directly from a vacuum to air near the ground, you would reflect about 2 x 10^-8 of the light at that interface. 2 x 10^-8 what units? Idk what "of the light" means and this is another area where I need insight. Temperature is degree of hotness while heat is quantity of hotness, but the definition of heat is rigorously defended by those who assert heat is a measure of transferred energy. Heat is the amount of energy flowing from one body of matter to another spontaneously due to their temperature difference, or by any means other than through work or the transfer of matter. https://en.wikipedia.org/wiki/Heat So WTH? How can something contain heat when heat is not anything that can be contained? Major confusion here. What is the measure of energy contained inside a bucket of hot water? And if we put a pane of glass between a light bulb and a thermometer, the meter will read lower, yet the energy (E=hf) of any specific f is the same regardless if it's reflected or transmitted. So what is it that makes the meter read cooler? Are you quite sure physics is not an art? lol Because so many loose definitions and rampant ambiguity makes for an unintuitive science. Quote But we aren't doing that. The density of the atmosphere increases as you get closer, so the transition is more gradual, and the reflection is even smaller. Reflection from the atmosphere is small. Are you sure reflection decreases with increasing density? That doesn't seem right. If we go from .5 atm to 1 atm, why does it matter how gradual it is? I didn't see time in the equation. Quote And now you're talking about the effect of adding in a few ppm, or tens of ppm, of additional CO2. A small change in a small effect. It seems like you're saying co2 is meaningless, and if you are, then I agree. I just want to know the science involved. If I can't understand it, then I have to believe someone on faith.
swansont Posted September 22, 2017 Posted September 22, 2017 7 hours ago, BanterinBoson said: I suppose force goes to acceleration as F=ma and if there is no acceleration, then there is no force or else the mass is infinite and cannot be accelerated. So it seems safe to say that if an EMW causes a force, then there will be an acceleration. Accelerating charges do not cause EMWs? What do I let go of? Not all EMWs are caused by accelerating charge. Quantum transitions in an atom or nucleus cannot be traced back to an accelerating charge. Quote Idk what others are doing; I'm just interested in the change. They say co2 makes earth hotter. Hotter than what? Hotter than before the co2 was there. CO2 is 1.00045 https://en.wikipedia.org/wiki/Refractive_index So that means v= 299,792,458/1.00045 But that doesn't tell me the reflection because that number is a composite of the original wave and the re-emitted wave. How do I figure only the re-emitted wave? That is the reflected wave I need to know. I gave you the equation for reflection at normal incidence (i.e. perpendicular): ((n1-n2)/(n1+n2))^2 Quote My next problem is at what frequency is that refraction number? I need a dispersion index https://en.wikipedia.org/wiki/Dispersion_(optics) This page says spinach chloroplasts have a refractive index of 1.02 - 1.06, which seems quite low compared to glass (1.50) which is more clear than plant leaves. Index and opacity are not necessarily correlated. The dispersion is likely small over the wavelength range we're discussing. Quote 2 x 10^-8 what units? Idk what "of the light" means and this is another area where I need insight. It's a fraction. There are no units. 20 parts per billion. Quote Heat is the amount of energy flowing from one body of matter to another spontaneously due to their temperature difference, or by any means other than through work or the transfer of matter. https://en.wikipedia.org/wiki/Heat So WTH? How can something contain heat when heat is not anything that can be contained? Major confusion here. What is the measure of energy contained inside a bucket of hot water? And if we put a pane of glass between a light bulb and a thermometer, the meter will read lower, yet the energy (E=hf) of any specific f is the same regardless if it's reflected or transmitted. So what is it that makes the meter read cooler? A pane of glass block the IR — glass is a really good absorber at those wavelengths. Quote Are you quite sure physics is not an art? lol Because so many loose definitions and rampant ambiguity makes for an unintuitive science. This isn't a classroom or a textbook. You've jumped into the middle of this without a proper background. People took years of classes to get that background. Quote Are you sure reflection decreases with increasing density? That doesn't seem right. If we go from .5 atm to 1 atm, why does it matter how gradual it is? I didn't see time in the equation. Gradual in a spatial sense. Gradient. Quote It seems like you're saying co2 is meaningless, and if you are, then I agree. I just want to know the science involved. If I can't understand it, then I have to believe someone on faith. Meaningless for reflection, which is not the mechanism involved in the greenhouse effect.
John Cuthber Posted September 22, 2017 Posted September 22, 2017 14 hours ago, BanterinBoson said: Because so many loose definitions and rampant ambiguity makes for an unintuitive science. Just because you don't understand them, doesn't mean the definitions are loose or ambiguous. Reality , and consequently science, are under no obligation to match your intuition. It's quite complicated; you actually need to study it.
studiot Posted September 22, 2017 Posted September 22, 2017 (edited) I am going to be away for a few days so before I go I will respond to some of your comments as promised. I really suggest you take the time to read them more fully as they contain information you are clearly missing. On 21/09/2017 at 2:44 AM, BanterinBoson said: Quote Studiot I prefer the phrase charge separation, rather than current because that is what happens in both the radio aerial and the carbon dioxide molecule. The effects are potential driven, not current driven. I don't understand "phase charge separation". That would be because you misread my text. I didn't say phase charge separation, I said the phrase charge separation. I can only assume you understand the meaning of charge separation. On 21/09/2017 at 2:44 AM, BanterinBoson said: Quote Studiot I mentioned bond length of carbon dioxide which is important because this determines the frequencies for the four fundamental modes of vibration of the molecule. Not just bond length and strength, but mass as well. Right? What I failed to see before was that EMWs only interact with charges, not matter itself. So the charges will be the "handles" that the EMW "grabs" to then shake the molecule and how the molecule behaves is determined by the bonds and the mass of the atoms. Yes mass also plays its part, I was going to go on to present the formula that results for the carbon dioxide molecule along with the expected bond lengths and frequencies, but you 'jumped the gun'. Note that each my explanatory notes is there for a reason so when I recommend forgetting resonance I also offered a reason so I was suprised you did not ask for this rather than your actual response. On 21/09/2017 at 2:44 AM, BanterinBoson said: Quote Studiot - Forget resonance - the word will get you into trouble when we consider charge separation in the carbon dioxide molecule. It has a specific and quite different meaning for chemists and bonding. I can't forget resonances. For instance when an EMW causes co2 to resonate, it absorbs energy which is expressed by the exaggerated movement and an electron being kicked up to the next energy level. Resonance makes too much sense to me. Here is a list of 'normal' bond lengths. Note that the actual measured length in carbon dioxide is 1.16 Angtrom, a significant shortening on the 1.23 shown above. Chemists attribute this to the three resonance structures of carbon dioxide and the resulting so called resonance energy. Again I was going to discuss this in more detail until you rejected it. On 21/09/2017 at 2:44 AM, BanterinBoson said: Quote The plan is to fully understand how a straight radio aerial works and see if we can use this to understand how the stretching modes of the carbon dioxide bonds comes to be in the IR and microwave bands and not other bands of the EM spectrum. I'm a little confused how the co2 molecule would interact with the microwave band because: Often molecules contain dipolar groups, but have no overall dipole moment. This occurs if there is symmetry within the molecule that causes the dipoles to cancel each other out. This occurs in molecules such as tetrachloromethane and carbon dioxide. The dipole-dipole interaction between two individual atoms is usually zero, since atoms rarely carry a permanent dipole. https://en.wikipedia.org/wiki/Intermolecular_force The way I understand microwave heating with water is that the MW rotates the water molecule which then interacts kinetically with other molecules that cause resonances in the water molecule in the IR band. Water is a dipole, but co2 is shaped more like - + - so I'm curious if MW could rotate the molecule in the same way. Right so the plan was to show the link between the charge separation that occurs in a radio aerial and the charge separation in a carbon dioxide molecule and how they are similar. I have already told you that the vibrations concerned are stretching and waggling modes. You are correct that carbon dioxide has no dipole moment on symmetric stretching (but does have a qudrupole moment) so no absorbtion band there, but it does have an asymmetric stretching mode, associated with the resonance structures. It also has two waggling modes that are degenerate. Degenerate means they have the same energies because they are simply the same waggling motion in two planes at right angles. Carbon dioxide is a linear molecule, unlike water. So all the molecular rotational modes appear as fine spectroscopic structure on the bond distortion modes. None of the electron promotion energies correspond to IR or microwave regions. If we continue with the plan we can explore pictures/animations of this next time. Meanwhile I will leave you with an picture of an actual carbon dioxide spectrum readout showing the three main modes and the 'missing' symmetric mode. And also some spectra from various molecules in the atmosphere, that may prove useful later. Edited September 22, 2017 by studiot
BanterinBoson Posted September 23, 2017 Author Posted September 23, 2017 6 hours ago, John Cuthber said: Just because you don't understand them, doesn't mean the definitions are loose or ambiguous. Be nice. The correct answer is "Yeah, lol, I can see what you mean." Not, "Well, you're just stupid." Your not resonating with what I'm talking about makes me wonder just how much you've been studying, because: Gamma Rays Gamma rays typically have frequencies above 10 exahertz (or >1019 Hz), and therefore have energies above 100 keV and wavelengths less than 10 picometers (10−11 m), which is less than the diameter of an atom. However, this is not a strict definition, but rather only a rule-of-thumb description for natural processes. Electromagnetic radiation from radioactive decay of atomic nuclei is referred to as "gamma rays" no matter its energy, so that there is no lower limit to gamma energy derived from radioactive decay. This radiation commonly has energy of a few hundred keV, and almost always less than 10 MeV. In astronomy, gamma rays are defined by their energy, and no production process needs to be specified. The energies of gamma rays from astronomical sources range to over 10 TeV, an energy far too large to result from radioactive decay. So are gamma rays defined by their source or by their energy? Loose and ambiguous. On the subject of reflection and refraction, there are two ways to describe the same process: One is the classical wave propagation where an EMW causes an electron to move which induces another EMW which causes a phase shift in the resultant wave. The other is the QM method where photons travel every possible path between atoms and the superposition of all those paths cause the shift in phase. Which is reality? I can't see how anyone can come away from that video thinking any physicist has a good handle on reality. One of those professors on that sixtysymbols channel likes to challenge himself to explain something in 10 min. 10 min??? Do atoms touch? Well, that's another 12 min video: https://www.youtube.com/watch?v=P0TNJrTlbBQ Then we have absolute zero which seems very much like it should be an absolute limit, but yet we also have negative temperatures, which are unintuitively hotter than any positive temperature. It takes 13 min to explain that https://www.youtube.com/watch?v=yTeBUpR17Rw Light can't travel faster because the speed of light is the speed of causality, yet somehow the universe is expanding faster than the speed of causality. https://www.youtube.com/watch?v=msVuCEs8Ydo The definition of heat is often subjective or is used interchangeably for internal energy and temperature. F=ma only works when velocities are low and doesn't account for the change in mass. For some reason, no one wants to tell the students that until it circulates around the back of the classroom under the topic of "Hey, did ya know F=ma is wrong?" All that off the top of my head. Let me know if I need to google some more; no doubt someone has a list somewhere. Quote Reality , and consequently science, are under no obligation to match your intuition. You're correct about reality, but science does have an obligation to be clear in its definitions and using intuitive nomenclature would help as well. Most arguments I see on forums are simply bickering about the definition of words and the opponents would actually agree if they had their semantics standardized. Quote It's quite complicated; you actually need to study it. So is music. 15 hours ago, swansont said: Not all EMWs are caused by accelerating charge. Quantum transitions in an atom or nucleus cannot be traced back to an accelerating charge. Thanks. I am going to look into that. Quote I gave you the equation for reflection at normal incidence (i.e. perpendicular): ((n1-n2)/(n1+n2))^2 So that equation is just the reflected bit and not the composite refracted wave? Oh ok. https://en.wikipedia.org/wiki/Fresnel_equations The fraction of the incident power that is reflected from the interface is given by the reflectance or reflectivity, R, and the fraction that is refracted is given by the transmittance or transmissivity, T, (unrelated to the transmission through a medium). So n1 = 1 and n2 = 1.00045 at some unknown frequency, then R = 5 x 10^-8 And T = 1 - R = .99999995. Ah, so it's a percent. Got it. Thanks! Quote Index and opacity are not necessarily correlated. The dispersion is likely small over the wavelength range we're discussing. How can that be? I know in the IR band that T goes to zero and therefore R goes to 1. n1 = 1 and n2 = x as x -> infinity, then ((n-x)/(n+x))^2 converges to 1. R and T seem variable across the bands. Are there not equations to describe R and T in terms of frequency? Or at least n in terms of frequency? n = c / phase velocity. How does co2 fit into determining the phase velocity at specific frequencies? Quote A pane of glass block the IR — glass is a really good absorber at those wavelengths. Then use an led light that doesn't emit IR. The thermometer will still read cooler. Quote This isn't a classroom or a textbook. You've jumped into the middle of this without a proper background. People took years of classes to get that background. That's true. Quote Gradual in a spatial sense. Gradient. I still don't see a difference in gradual or abrupt if we go from .5 atm to 1 atm. Quote Meaningless for reflection, which is not the mechanism involved in the greenhouse effect. The mechanism is that co2 is a highpass filter which allows high-energy light to enter but does not allow low-energy light to leave (easily). When I first heard that, I immediately thought "Hey, wait a minute!" Now, here I am. 6 hours ago, studiot said: I am going to be away for a few days so before I go I will respond to some of your comments as promised. I really suggest you take the time to read them more fully as they contain information you are clearly missing. Thanks for the heads-up. I'll study your post before I reply. Thanks for helping me.. I really appreciate it.
BanterinBoson Posted September 23, 2017 Author Posted September 23, 2017 I can add "phase" to the "list". The term phase can refer to several different things: It can refer to a specified reference, such as {\displaystyle \scriptstyle \cos(2\pi ft)\,}, in which case we would say the phase of {\displaystyle \scriptstyle x(t)\,} is {\displaystyle \scriptstyle \varphi \,}, and the phase of {\displaystyle \scriptstyle y(t)\,} is {\displaystyle \scriptstyle \varphi \,-\,{\frac {\pi }{2}}\,}. It can refer to {\displaystyle \scriptstyle \varphi \,}, in which case we would say {\displaystyle \scriptstyle x(t)\,} and {\displaystyle \scriptstyle y(t)\,} have the same phase but are relative to their own specific references. In the context of communication waveforms, the time-variant angle {\displaystyle \scriptstyle 2\pi ft\,+\,\varphi }, or its principal value, is referred to as instantaneous phase, often just phase. https://en.wikipedia.org/wiki/Phase_(waves) I know what phase means, but I don't know what science says it means because the definition is loose and ambiguous. I just sat down to study some science and the first thing I encounter is "the term you would like defined is ambiguous". I couldn't believe it, so I immediately came here to elucidate the frustration I'm experiencing. And the defense was that it takes years of study to learn physics whereas I'm jumping in the middle. I'm not sure about that... it takes years of study because of stuff like that ^. It's rampant! I'm on the learning-end here and I'm telling ya it's as if the muddling-up were deliberate to hinder learning as much as possible. Why can't "science" state clearly its definitions? Without clear definitions, it's impossible to communicate. If I say the phase is x, then you immediately have to ask what I mean. If I throw "gamma rays" in, then you'll really be uninformed. Do I mean sub-xray-energy from a nucleus or high-energy waves from something else? Previously, I thought gamma rays meant high-energy waves, but now I have to stipulate super-xray-energy gamma rays of nucleic or non-nucleic origin to convey what was so much easier before I bothered to read the real definition (or lack of one). Off soapbox, back to studying...
swansont Posted September 23, 2017 Posted September 23, 2017 10 hours ago, BanterinBoson said: Be nice. The correct answer is "Yeah, lol, I can see what you mean." Not, "Well, you're just stupid." You're wrong, so why would anyone reply with the former? And nobody has suggested the latter. 10 hours ago, BanterinBoson said: Your not resonating with what I'm talking about makes me wonder just how much you've been studying, because: Gamma Rays Gamma rays typically have frequencies above 10 exahertz (or >1019 Hz), and therefore have energies above 100 keV and wavelengths less than 10 picometers (10−11 m), which is less than the diameter of an atom. However, this is not a strict definition, but rather only a rule-of-thumb description for natural processes. Electromagnetic radiation from radioactive decay of atomic nuclei is referred to as "gamma rays" no matter its energy, so that there is no lower limit to gamma energy derived from radioactive decay. This radiation commonly has energy of a few hundred keV, and almost always less than 10 MeV. In astronomy, gamma rays are defined by their energy, and no production process needs to be specified. The energies of gamma rays from astronomical sources range to over 10 TeV, an energy far too large to result from radioactive decay. So are gamma rays defined by their source or by their energy? Loose and ambiguous. Physics ≠ Astronomy I'll bet you can find lots of terms whose definition differs from one field to the next. There are definitely definitions in physics that are not what is used in everyday conversation. Quote On the subject of reflection and refraction, there are two ways to describe the same process: One is the classical wave propagation where an EMW causes an electron to move which induces another EMW which causes a phase shift in the resultant wave. The other is the QM method where photons travel every possible path between atoms and the superposition of all those paths cause the shift in phase. Which is reality? Reality is something you might discuss in philosophy. Physics uses models to describe how nature behaves. You use the model best suited to the job. Quote I can't see how anyone can come away from that video thinking any physicist has a good handle on reality. It's not the job of physics to describe reality. It describes how nature behaves.
John Cuthber Posted September 23, 2017 Posted September 23, 2017 4 hours ago, BanterinBoson said: Why can't "science" state clearly its definitions? It usually can. What are you struggling with? Complaining that the definitions are not absolutely concrete is like complaining that you can't use English because it's full of words like "stone" that have more than 1 meaning. Well, if you know what you are talking about the meaning is usually clear from the context, and if it's not, you can ask for clarification.
BanterinBoson Posted September 24, 2017 Author Posted September 24, 2017 14 hours ago, swansont said: You're wrong, so why would anyone reply with the former? Because I am not wrong and have supplied evidence illustrating that I am not. If you need more evidence, then I'd enthusiastically accept the invitation to vent frustration while simultaneously proving just how right I am if I could post here every time I find ambiguity while researching. Here is another: https://en.wikipedia.org/wiki/Wavenumber#Definition 14 hours ago, swansont said: Physics ≠ Astronomy That's a weak defense as astronomy depends on physics. 14 hours ago, swansont said: I'll bet you can find lots of terms whose definition differs from one field to the next. You're probably right and that isn't something to be proud of. 14 hours ago, swansont said: Reality is something you might discuss in philosophy. Physics uses models to describe how nature behaves. You use the model best suited to the job. It's not the job of physics to describe reality. It describes how nature behaves. What's the difference between nature and reality?
BanterinBoson Posted September 24, 2017 Author Posted September 24, 2017 14 hours ago, John Cuthber said: It usually can. What are you struggling with? Complaining that the definitions are not absolutely concrete is like complaining that you can't use English because it's full of words like "stone" that have more than 1 meaning. Well, if you know what you are talking about the meaning is usually clear from the context, and if it's not, you can ask for clarification. Probably why they say English is one of the hardest languages to learn and I'm not such a big fan of it because I find most arguments have semantics for fuel. Quote What are you struggling with? Thanks for asking! I'm struggling with how to describe the amount of... I don't know what to call it... energy, heat, internal energy... in a co2 molecule due to the absorption of IR radiation and then compare that to the amount reflected in the visible band. I don't know how to do that because E=hf and therefore the amount reflected > the amount absorbed because E is purely a function of frequency. Another problem I have is how to differentiate the amount of IR re-radiation traveling towards earth vs the amount travelling to space. If we assume the re-radiation forms a cone of 90 degrees over the earth (which is the total possible paths a ray could take to strike a sphere), then it "seems" like energy should be divided by 4 (360/90 = 4), but it's not because E=hf. I am confused by the fact that heat is energy in transmission and not energy in storage, so the energy contained in a vibrating molecule is not heat, but the energy emitted is. So is the energy of the vibrating molecule called internal energy? Do I quarter the internal energy to find what I need? Or am I looking for heat? And in the case of the visible light reflection, am I looking for heat? I'm totally lost and I'm not even sure how to ask a question properly. I also need a diffusion index for co2 because I need to know how much light is reflected at each frequency. On 9/22/2017 at 4:01 PM, studiot said: That would be because you misread my text. I didn't say phase charge separation, I said the phrase charge separation. I can only assume you understand the meaning of charge separation. Oh goodness... I'm such a goofball! I thought you misspelled "phase" as "phrase" and upon being corrected, I then googled "phrase charge separation" like an idiot lol. I was thinking, "What the heck is 'phrase charge separation? Google has nothing about it." Then it occurred to me that you meant the phrase "charge separation." Next time use quotes when dealing with dummies I knew the meaning of "charge separation", but I did not know it was called that. Wiki says - Photoinduced charge separation is the process of an electron in an atom or molecule, being excited to a higher energy level by the absorption of a photon and then leaving the atom or molecule to a nearby electron acceptor. https://en.wikipedia.org/wiki/Photoinduced_charge_separation I have a question about the last statement on that page: If a photon of light hits the atom it will be absorbed if, and only if, energy of that photon is equal to the difference between the ground state and another energy level in that atom. This raises the electron to a higher energy level. Why is it not better to say that the difference between the ground state and another energy level is the point where a photon is absorbed? Why does the photon depend on the energy level and not the energy level depend on the photon? Do you see what I am asking? It seems like it would be the resonance of the atom that determines the point where a photon would be absorbed and that would define the difference in energy levels. Not the other way around, which is the difference in energy levels determines if a photon is absorbed and that defines the resonance because it doesn't make sense that way. It's putting the cart before the horse. Resonance is a function of mass and bond strength and resonance determines if a photon is absorbed and therefore is it resonance that determines the difference in energy levels. Why is that not correct? ______________________________________________________________ Damn merging! I hope Studiot notices that I addressed him. Can someone please give me a reason why anyone would desire to have their replies all merged together? I can't see how that is a handy feature and it's exceedingly frustrating to have what I wrote to Studiot to be merged with what I wrote to John.
John Cuthber Posted September 24, 2017 Posted September 24, 2017 4 hours ago, BanterinBoson said: I don't know how to do that because E=hf and therefore the amount reflected > the amount absorbed because E is purely a function of frequency. You are muddling several things together there. The amount of light reflected by gases is usually so small you can ignore it. As was calculate earlier the reflection by the Earth's atmosphere is only something like 20 parts in a billion. E=hf is true for a single photon. But it's not the point. A photon will only be absorbed if it's the "right" energy to promote the molecule to an excited state with a higher energy- for IR that's usually a case of setting the molecule vibrating. So for carbon dioxide at wavelengths where it's absorbed light won't be reflected, it gets absorbed and the energy gets redistributed among the gas molecules and the gas gets warmer. For wavelengths where the gas doesn't absorb, the IR will just pass straight through. Very little of it would be reflected.(some of it will be scattered- again that's a small effect) Generally, reflection takes place at interfaces between things- like the surface of a piece of glass. The atmosphere doesn't have a "surface"- it fades gradually as you get higher, so there's very little reflection.
swansont Posted September 24, 2017 Posted September 24, 2017 7 hours ago, BanterinBoson said: Because I am not wrong and have supplied evidence illustrating that I am not. If you need more evidence, then I'd enthusiastically accept the invitation to vent frustration while simultaneously proving just how right I am if I could post here every time I find ambiguity while researching. Here is another: https://en.wikipedia.org/wiki/Wavenumber#Definition Again, different fields. Quote That's a weak defense as astronomy depends on physics. So do a lot of disciplines. All of the sciences, and engineering, depend in some way upon physics. Quote You're probably right and that isn't something to be proud of. Not proud, but that's the way it is. Speciation/evolution, as applied to etymology. (see what I did there? "/" means division in math, but has a different interpretation in language.) Quote What's the difference between nature and reality? We can observe nature, but we never can be sure we are observing reality.
BanterinBoson Posted September 25, 2017 Author Posted September 25, 2017 16 hours ago, John Cuthber said: You are muddling several things together there. The amount of light reflected by gases is usually so small you can ignore it. As was calculate earlier the reflection by the Earth's atmosphere is only something like 20 parts in a billion. E=hf is true for a single photon. But it's not the point. A photon will only be absorbed if it's the "right" energy to promote the molecule to an excited state with a higher energy- for IR that's usually a case of setting the molecule vibrating. So for carbon dioxide at wavelengths where it's absorbed light won't be reflected, it gets absorbed and the energy gets redistributed among the gas molecules and the gas gets warmer. For wavelengths where the gas doesn't absorb, the IR will just pass straight through. Very little of it would be reflected.(some of it will be scattered- again that's a small effect) Generally, reflection takes place at interfaces between things- like the surface of a piece of glass. The atmosphere doesn't have a "surface"- it fades gradually as you get higher, so there's very little reflection. Ok, I understand all that. So how do I figure the amount of energy, heat, whatever-it's-called absorbed by the co2 molecule at the right IR frequency? And after I've done that, then how do I figure the amount radiated back to earth in a 90 degree cone? Also, I need help with nomenclature. Light = heat because heat is energy in travel. Is that right? What is the energy inside a vibrating atom called? Kinetic, I know, but does it have a more specific name? Thanks! 14 hours ago, swansont said: Not proud, but that's the way it is. Speciation/evolution, as applied to etymology. (see what I did there? "/" means division in math, but has a different interpretation in language.) Yeah, but that's hijacking for convenience and not formal definition. The numerator and denominator are opposites and so in language the convention carries over for true/false, on/off, etc. It is also the numerator in terms of the denominator, as you wrote it, so speciation in terms of evolution instead of speciation in terms of creation. Computers have changed our way of thinking and now "/" often means a subset, so it would be Evolution/Speciation/next subset which is a lot like the reciprocal of the "in terms of" definition, so none of it really deviates far from the mathematical meaning of the symbol. 14 hours ago, swansont said: We can observe nature, but we never can be sure we are observing reality. Now that's deep! Good one
swansont Posted September 25, 2017 Posted September 25, 2017 12 hours ago, BanterinBoson said: Ok, I understand all that. So how do I figure the amount of energy, heat, whatever-it's-called absorbed by the co2 molecule at the right IR frequency? And after I've done that, then how do I figure the amount radiated back to earth in a 90 degree cone? It gets complicated. You'd have to integrate the earth's emission over the absorption band of the CO2, the look at the CO2 emission profile (probably a dipole radiation pattern, so it's not isotropic, but that might not matter). And that's just one absorption. You would have to look at whether re-radiated is then absorbed, and then a third absorption, etc. for the entire thickness of the atmosphere. Fortunately, people have done this already. Probably no need to re-invent the wheel. Just find what they've done. edit: such as http://clivebest.com/blog/?p=4265 models calculating this (i.e. search terms to use) will include Beer's law and Kirchoff's law Quote Also, I need help with nomenclature. Light = heat because heat is energy in travel. Is that right? Light is heat when it comes from thermal radiation. Light from non-thermal sources e.g. a laser, or fluorescent light, is not heat. (it would be included in "work" in the energy equation) The energy being discussed in our context is heat; the radiation from the sun, and from the earth, is blackbody radiation.
studiot Posted September 25, 2017 Posted September 25, 2017 On 22/09/2017 at 9:01 PM, studiot said: Note that each my explanatory notes is there for a reason so when I recommend forgetting resonance I also offered a reason so I was suprised you did not ask for this rather than your actual response. On 24/09/2017 at 5:35 AM, BanterinBoson said: Damn merging! I hope Studiot notices that I addressed him. Yes I noticed so hopefully you understand the significance of the separation of charge But sadly I can only conclude the following from this address as the only response to an explanation, already twice given:- On 24/09/2017 at 5:35 AM, BanterinBoson said: Resonance is a function of mass and bond strength and resonance determines if a photon is absorbed and therefore is it resonance that determines the difference in energy levels. Why is that not correct? I take this last bit to mean "My mind is made up so don't confuse me with the facts"
swansont Posted September 25, 2017 Posted September 25, 2017 On 9/24/2017 at 0:35 AM, BanterinBoson said: I have a question about the last statement on that page: If a photon of light hits the atom it will be absorbed if, and only if, energy of that photon is equal to the difference between the ground state and another energy level in that atom. This raises the electron to a higher energy level. Why is it not better to say that the difference between the ground state and another energy level is the point where a photon is absorbed? Why does the photon depend on the energy level and not the energy level depend on the photon? Do you see what I am asking? You can excite an atom in other ways (e.g. collisions with other atoms). The energy levels exist independent of photons. Therefore it would be wrong to say that the energy level depends on the photon. On 9/24/2017 at 0:35 AM, BanterinBoson said: It seems like it would be the resonance of the atom that determines the point where a photon would be absorbed and that would define the difference in energy levels. Not the other way around, which is the difference in energy levels determines if a photon is absorbed and that defines the resonance because it doesn't make sense that way. It's putting the cart before the horse. On 9/24/2017 at 0:35 AM, BanterinBoson said: Resonance is a function of mass and bond strength and resonance determines if a photon is absorbed and therefore is it resonance that determines the difference in energy levels. Why is that not correct? Again, atomic structure can be calculated and/or investigated, and it exists, independent of the proximity of any photons.
John Cuthber Posted September 25, 2017 Posted September 25, 2017 14 hours ago, BanterinBoson said: So how do I figure the amount of energy, heat, whatever-it's-called absorbed by the co2 molecule at the right IR frequency? First, you stop trying to run before you can walk. Then you take classes in physics, maths and quantum mechanics. Then, with luck you can apply time dependent perturbation theory to the question. Alternatively, you can read it off the graph posted in the OP
BanterinBoson Posted September 26, 2017 Author Posted September 26, 2017 14 hours ago, swansont said: It gets complicated. You'd have to integrate the earth's emission over the absorption band of the CO2, the look at the CO2 emission profile (probably a dipole radiation pattern, so it's not isotropic, but that might not matter). And that's just one absorption. You would have to look at whether re-radiated is then absorbed, and then a third absorption, etc. for the entire thickness of the atmosphere. Fortunately, people have done this already. Probably no need to re-invent the wheel. Just find what they've done. I'd be content just knowing how to figure it for one abstract molecule without the complications of the entire atmosphere. Reinventing wheels teaches me how to make wheels 14 hours ago, swansont said: edit: such as http://clivebest.com/blog/?p=4265 models calculating this (i.e. search terms to use) will include Beer's law and Kirchoff's law That's a helpful link and I will have to study that more. The last comment on the page tipped me off to Stefan-Boltzmann which allows me to describe power in terms of area of a blackbody, which is what co2 is at a certain frequency. At least I can say "this many watts goes to space and that many heads to earth", but first I have to decide how much area a co2 molecule has and what T will be. The Beer's law requires an optical depth, so I'm not sure that applies here. I'm still struggling to learn the nomenclature: Roughly, the temperature of a body at rest is a measure of the mean of the energy of the translational, vibrational and rotational motions of matter's particle constituents, such as molecules, atoms, and subatomic particles. The full variety of these kinetic motions, along with potential energies of particles, and also occasionally certain other types of particle energy in equilibrium with these, make up the total internal energy of a substance. Internal energy is loosely called the heat energy or thermal energy in conditions when no work is done upon the substance by its surroundings, or by the substance upon the surroundings. https://en.wikipedia.org/wiki/Thermodynamic_temperature That looseness is what makes this difficult and the last thing I want to start doing is associating heat with stored energy. Maybe the best way to attack this is to make a spreadsheet of every term (heat, internal energy, thermal energy, radiance, spectral radiance, radiant flux, temperature, thermodynamic temperature, yada yada) so I can have it all in front of me to make sense of. The math and the concepts are straightforward, so the language is the only barrier. 14 hours ago, swansont said: Light is heat when it comes from thermal radiation. Light from non-thermal sources e.g. a laser, or fluorescent light, is not heat. (it would be included in "work" in the energy equation) Huh? Why do you want to confuse me like that? lol! How is light from a laser not heat? How is a fluorescent light doing work? I'm getting a variety of answers: 1) In thermodynamics, heat is often contrasted with work: heat applies to individual particles (such as atoms or molecules), work applies to objects (or a system as a whole). Heat involves stochastic (or random) motion equally distributed among all degrees of freedom, while work is directional, confined to one or more specific degrees of freedom. https://en.wikipedia.org/wiki/Heat 2) Work is any energy transfer that does not carry entropy. Heat, on the other hand, is any energy transfer that carries entropy. 3) Work is the mechanical transfer of energy to a system or from a system by an external force on it. Heat is the non-mechanical transfer of energy from the environment to the system or from the system to the environment because of a temperature difference between the two. The conversion of mechanical energy (work) to heat is very efficient, nearly 100%. But the conversion of thermal energy (heat) to work is not so because heat is a low grade energy. https://www.quora.com/In-thermodynamics-what-is-the-difference-between-work-and-heat-I-was-thinking-about-this-when-studying-Carnot-Cycles-and-adiabatic-processes-How-can-a-system-do-work-in-isolation-What-is-it-doing-work-on-Where-is-that-energy-going I like this one: 4) You are right. Microscopically, work and heat are just about the same. Both involve molecular collisions transfer energy from one object to the other. Work involves a kind of "coherent" transfer in a manner of speaking, in which the collisions are predominantly, and to an extreme degree, in one direction. Also, typically the force is applied to one location on the object. And importantly, the boundary of the system is displaced. (E.g. translation or deformation) On the other hand, transfer of energy by heat is "incoherent", many directions, and typically in all directions. And importantly, the boundary of the system is not displaced. Finally, everyday phenomena fall into one or the other category, and they differ in their macroscopic behavior. Loosely speaking, when heat is transfered the temperature rises. When work is done, the boundary of the system changes. Of course adding heat generally also results in the boundary moving [say, expanding], and work generally results in a temperature change [Joule's experiment]. I'm trying to motivate the macroscopic separation between work and heat without a lengthy discussion. To the engineers who first worked all of this out, the very existence of atoms was unknown. To them, the separation between heat and work was very clear. They had little reason to view them as manifestations of the same microscopic process. They thought that heat was a physical fluid. In any event, their remarkable achievements have stood the test of time. https://physics.stackexchange.com/questions/135539/why-work-w-and-heat-q-are-different-concepts What do you think? What's the best way to write distinct definitions for work and heat? 12 hours ago, swansont said: You can excite an atom in other ways (e.g. collisions with other atoms). The energy levels exist independent of photons. Therefore it would be wrong to say that the energy level depends on the photon. Regardless how we excite the atom (photon or collision) it is still resonance that determines the energy level of the electron, right? So it's the mass and bond strengths that determine the energy levels. I don't understand why that would not be correct. Here is what I said before: It seems like it would be the resonance of the atom that determines the point where a photon would be absorbed and that would define the difference in energy levels. Not the other way around, which is the difference in energy levels determines if a photon is absorbed and that defines the resonance because it doesn't make sense that way. It's putting the cart before the horse. 12 hours ago, swansont said: Again, atomic structure can be calculated and/or investigated, and it exists, independent of the proximity of any photons. Right, as a function of mass and bond strengths which determine both resonance and energy levels. 14 hours ago, studiot said: Yes I noticed so hopefully you understand the significance of the separation of charge But sadly I can only conclude the following from this address as the only response to an explanation, already twice given:- Well, I wasn't done replying to your post as I wasn't done studying it in hopes of replying with something that would be to your satisfaction. 14 hours ago, studiot said: I take this last bit to mean "My mind is made up so don't confuse me with the facts" Not at all and I have no idea how you came to that conclusion as I took special care to prevent anyone from coming to that conclusion. Anyway, I've discovered the limits to my patience for walking on eggshells only to still be incapable of making you happy, so whatever. I'll figure it out with or without your help and if having your help means the continual negotiation of derogatory implication then I'd prefer to go-it alone. Thanks for trying. 10 hours ago, John Cuthber said: First, you stop trying to run before you can walk. Then you take classes in physics, maths and quantum mechanics. Then, with luck you can apply time dependent perturbation theory to the question. Alternatively, you can read it off the graph posted in the OP Thanks
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