List of H20 breakdown substances

[mjc123]

First, to be pedantic: "Sorta" is not a scientific concept. Second, you seem to be making a number of confusions (which are quite common with beginners): one, between a quantity (e.g. polarisability) and the change in that quantity. All bonds are polarisable to some extent; the point is that to be Raman active a vibration must result in a change of polarisability. It is this change, not polarisability itself, that is cancelled. Two, between molecules and bonds (or vibrational modes, which may involve more than one bond). A symmetrical molecule may have asymmetrical vibrations which have a change of dipole moment and are IR active (like the one illustrated).

A more precise answer to your question involves a more detailed consideration of symmetry, which I'm guessing you haven't studied yet. The polarisability has certain symmetry properties which mean that Raman scattering of a photon can induce a change in vibrational state only for modes of certain symmetry types. In particular, for a centrosymmetric molecule, only centrosymmetric modes can be Raman active.

A more qualitative argument may go like this: Bear in mind that the relative change in bond length during a vibration is much, much smaller than the above diagram suggests (for ease of illustration). We may reasonably consider that, for small changes in length, the change in polarisability with bond length will, to a first approximation, vary linearly with the deformation. If we write dP/dx as P', the change in polarisability in an asymmetric stretching mode like that shown, where one bond lengthens by dx and the other shortens by dx, will be P'dx - P'dx = 0. For a symmetric stretch, where both bonds lengthen, it will be P'dx + P'dx ≠ 0. (For the dipole moment, we must remember that the moments of the two bonds are pointing in opposite directions, so the change in dipole moment for the asymmetric stretch is μ'dx - (-μ')dx ≠ 0, while for the symmetric stretch it is μ'dx + (-μ')dx = 0, so the asymmetric stretch is IR active but the symmetric stretch is not.)

 

[jake jot]

First, to be pedantic: "Sorta" is not a scientific concept. Second, you seem to be making a number of confusions (which are quite common with beginners): one, between a quantity (e.g. polarisability) and the change in that quantity. All bonds are polarisable to some extent; the point is that to be Raman active a vibration must result in a change of polarisability. It is this change, not polarisability itself, that is cancelled. Two, between molecules and bonds (or vibrational modes, which may involve more than one bond). A symmetrical molecule may have asymmetrical vibrations which have a change of dipole moment and are IR active (like the one illustrated).

A more precise answer to your question involves a more detailed consideration of symmetry, which I'm guessing you haven't studied yet. The polarisability has certain symmetry properties which mean that Raman scattering of a photon can induce a change in vibrational state only for modes of certain symmetry types. In particular, for a centrosymmetric molecule, only centrosymmetric modes can be Raman active.

A more qualitative argument may go like this: Bear in mind that the relative change in bond length during a vibration is much, much smaller than the above diagram suggests (for ease of illustration). We may reasonably consider that, for small changes in length, the change in polarisability with bond length will, to a first approximation, vary linearly with the deformation. If we write dP/dx as P', the change in polarisability in an asymmetric stretching mode like that shown, where one bond lengthens by dx and the other shortens by dx, will be P'dx - P'dx = 0. For a symmetric stretch, where both bonds lengthen, it will be P'dx + P'dx ≠ 0. (For the dipole moment, we must remember that the moments of the two bonds are pointing in opposite directions, so the change in dipole moment for the asymmetric stretch is μ'dx - (-μ')dx ≠ 0, while for the symmetric stretch it is μ'dx + (-μ')dx = 0, so the asymmetric stretch is IR active but the symmetric stretch is not.)

Thanks. Let's illustrate this.

We know that during the inelastic scattering process the molecules excite to higher vibrational energy levels and make transitions to the ground level by emitting photons with a frequency different from the incident photons.

What is faster, the transition from higher vibration energy level to ground level or the one asymmetric stretch from left to right?

If the asymmetric stretch is faster, then there is no time for inelastic scattering. But if inelastic scattering is faster, it should release the say strokes photon and this should get out of the molecules to reach the detectors. So how can this change of polarizability be canceled when the photons were already released? Unless the one instance of asymmetric stretch from say left to right is faster so no time for inelastic scattering to occur? (I know the stretching occurs many billions of time a second or so. I'm describing 1/1000000000000000 millisec or less.)

 

[sysprog]

(I know the stretching occurs many billions of time a second or so. I'm describing 1/1000000000000000 millisec or less.)

I'm not here denying your main point, but that number of miiliseconds is a few orders of magnitude too small $--$ I counted 15 instances of the zero numeral in your fractional number of milliseconds $-$ a billionth is 0.000000001 $-$ your number is a quadrillionth $-$ that's a millionth of a billionth $-$ of a millisecond.

 

[jake jot]

I'm not here denying your main point, but that number of miiliseconds is a few orders of magnitude too small $--$ I counted 15 instances of the zero numeral in your fractional number of milliseconds $-$ a billionth is 0.000000001 $-$ your number is a quadrillionth $-$ that's a millionth of a billionth $-$ of a millisecond.

I just typed 000 without counting them.. just to emphasize I knew the stretching didn't occur in seconds but in billions of billions of milliseconds. I don't know exactly how fast, so how fast does one stretching occurs compared to the transition from virtual energy state to ground state, any ideas?

 

[mjc123]

Your reference says "Raman scattering has a lifetime of 10^-14 s", which is comparable to the period of molecular vibrations - e.g. the antisymmetric stretch of CO₂ at 2350 cm^-1 has a period of 1.4 x 10^-14 s. But in any case I don't get your point. There is no change of polarisability during the vibration, so there can be no Raman transition. It is forbidden by symmetry.

 

[jake jot]

Your reference says "Raman scattering has a lifetime of 10^-14 s", which is comparable to the period of molecular vibrations - e.g. the antisymmetric stretch of CO₂ at 2350 cm^-1 has a period of 1.4 x 10^-14 s. But in any case I don't get your point. There is no change of polarisability during the vibration, so there can be no Raman transition. It is forbidden by symmetry.

Change in polarizability is the same as transition from virtual energy state to ground state, right?

So I want to understand in picture what exactly happens that prevents the change of polarizability or transition during the vibration. I know when we hit water with a Raman spectrometer, billions of atoms were inside the laser spot. I have numerous experiences with Raman, so I need to visualize what happens when it the laser hits the molecules that is asymmetric and no Raman inelastic scattering occurs. This means the laser is not inelastically scattered back. Any references of the selection rule with great illustrations to make it logical and intuitive?

 

[mjc123]

Change in polarizability is the same as transition from virtual energy state to ground state, right?

Wrong. It is a property of the molecule (more particularly, of a vibrational mode of the molecule), whether or not there are any photons around to get scattered. But that property is necessary for Raman scattering to occur. BTW, water has different symmetry from CO₂, and all its vibrational modes are Raman active.

 

[jake jot]

Wrong. It is a property of the molecule (more particularly, of a vibrational mode of the molecule), whether or not there are any photons around to get scattered. But that property is necessary for Raman scattering to occur. BTW, water has different symmetry from CO₂, and all its vibrational modes are Raman active.

Ok. I've been googling with these words "group theory molecule symmetry Raman" and reading articles about it. If you know of one with good illustrations and most intuitive even with interactive java or animation, just share it anytime. Thanks a lot.

 

[jake jot]

We know that in order for a molecule to be Raman active, there must be a change in the polarizability, meaning that there must be change in the size, shape or orientation of the electron cloud that surrounds the molecule. And this change said to occur in symmetric stretching, but not asymmetric stretching.

Without using any rules or tables, can anyone describe using plain words why in asymmetric stretching, there is no change in the size, shape or orientation of the electron cloud that surrounds the molecule? (maybe a PF challenge?)

 

[snorkack]

We know that in order for a molecule to be Raman active, there must be a change in the polarizability, meaning that there must be change in the size, shape or orientation of the electron cloud that surrounds the molecule. And this change said to occur in symmetric stretching, but not asymmetric stretching.

Without using any rules or tables, can anyone describe using plain words why in asymmetric stretching, there is no change in the size, shape or orientation of the electron cloud that surrounds the molecule? (maybe a PF challenge?)

Simple reason: "size, shape or orientation" is not enough.
In asymmetric stretching, one bond stretches but the other bond shrinks, and does so by pretty much exactly the same amount as the first bond stretches. Because the polarizability is not a directed vector, it does not matter for polarizability that the two bonds are in opposite directions - the sum is still the same because changes of the halves are equal and opposite. Shape changes but "size" does not, and "size" staying constant as shape changes is what makes Raman scattering impossible.

 

[mjc123]

So I want to understand in picture what exactly happens that prevents the change of polarizability or transition during the vibration.

I guess we belong to different tribes; I am of those who are content to say "if that's what the maths says, that's how it is"; you want to be able to visualise what is happening physically. I suspect what you are asking is impossible, or at least very difficult (I would be happy to be proved wrong). I can visualise in a crude way what is going on with IR and dipole moments; I find it much harder to visualise what's going on with polarisability and Raman scattering. But the maths is the only accurate and scientific description we have; mental models and "visualisations" may help our understanding to some extent, but they are limited, and possibly misleading, especially if they are confined to analogies of classical effects we are familiar with in daily life, whereas there are some quantum effects that have no classical analogue. Ultimately, if you want a proper understanding, you have to grapple with the theory and the maths; purely qualitative descriptions, whether in words or pictures, can only take you so far.

 

[jake jot]

I guess we belong to different tribes; I am of those who are content to say "if that's what the maths says, that's how it is"; you want to be able to visualise what is happening physically. I suspect what you are asking is impossible, or at least very difficult (I would be happy to be proved wrong). I can visualise in a crude way what is going on with IR and dipole moments; I find it much harder to visualise what's going on with polarisability and Raman scattering. But the maths is the only accurate and scientific description we have; mental models and "visualisations" may help our understanding to some extent, but they are limited, and possibly misleading, especially if they are confined to analogies of classical effects we are familiar with in daily life, whereas there are some quantum effects that have no classical analogue. Ultimately, if you want a proper understanding, you have to grapple with the theory and the maths; purely qualitative descriptions, whether in words or pictures, can only take you so far.

I think what you are saying is we must use the lessons in quantum mechanics whose punchline is "In the absence of measurement to determine it's properties, there are no properties". For example, in Stern Gerlach experiment, there is no definite spin without measurement, and in double slit experiiment, there is no position before measurement. So instead of visualizing the asymmetrical molecules with atoms on either side. We must instead think of it like this?

Before I begin to think of solid objects as like that. Please others try to explain it in plain english, if you still can.

 

[mjc123]

I'm not saying "in the absence of measurement there are no properties". To use your example of spin, I'm not saying the spin has no value before it's measured (I leave that question tot he philosophers of QM), but that spin itself is a property that has no anlogue in classical mechanics (nothing that we can "visualise" from experience). To model the electron as a little particle orbiting the nucleus and spinning on its own axis may be helpful to an extent, but it is not "what is really happening", and will let us down at some point if we take it too literally. You ask for plain English, but can plain English describe what is outside our experience, other than by using such only-partly-adequate analogies?

 

[jake jot]

Simple reason: "size, shape or orientation" is not enough.
In asymmetric stretching, one bond stretches but the other bond shrinks, and does so by pretty much exactly the same amount as the first bond stretches. Because the polarizability is not a directed vector, it does not matter for polarizability that the two bonds are in opposite directions - the sum is still the same because changes of the halves are equal and opposite. Shape changes but "size" does not, and "size" staying constant as shape changes is what makes Raman scattering impossible.

Someone told me that in the asymmetric stretch the molecule retains the same shape because as one extends the other contracts and vice versa. But you said it was the size staying constant. Is it not the shape that stays constant?

Whatever. So we must not imagine the oxygen atoms as either on the left and right side of the carbon in the case of carbon dioxide, but smeared in the molecules like in 2 places at once, like in the quantum double slit?

Look. I own a Raman spectrometer and putting up a molecular scanning laboratory to monitor changes of molecular bonds to unknown field and hamiltonians. And everyday whenever the laser hits the sample, I need to visualize or have ideas how it can cause transitions only in symmetric molecules and not in asymmetric. Also need to look at this from the view of an electromagnetic wavelenght that is a thousand times as large

Also the laser beam is light which has electric and magnetic field, it has directed vector, so I'm trying to reconcile it with the polarization that has no directed vector as you described.

 

[jake jot]

First, to be pedantic: "Sorta" is not a scientific concept. Second, you seem to be making a number of confusions (which are quite common with beginners): one, between a quantity (e.g. polarisability) and the change in that quantity. All bonds are polarisable to some extent; the point is that to be Raman active a vibration must result in a change of polarisability. It is this change, not polarisability itself, that is cancelled. Two, between molecules and bonds (or vibrational modes, which may involve more than one bond). A symmetrical molecule may have asymmetrical vibrations which have a change of dipole moment and are IR active (like the one illustrated).

A more precise answer to your question involves a more detailed consideration of symmetry, which I'm guessing you haven't studied yet. The polarisability has certain symmetry properties which mean that Raman scattering of a photon can induce a change in vibrational state only for modes of certain symmetry types. In particular, for a centrosymmetric molecule, only centrosymmetric modes can be Raman active.

A more qualitative argument may go like this: Bear in mind that the relative change in bond length during a vibration is much, much smaller than the above diagram suggests (for ease of illustration). We may reasonably consider that, for small changes in length, the change in polarisability with bond length will, to a first approximation, vary linearly with the deformation. If we write dP/dx as P', the change in polarisability in an asymmetric stretching mode like that shown, where one bond lengthens by dx and the other shortens by dx, will be P'dx - P'dx = 0.

I'm still pondering what you said. Why did you use one single "dx" for either side where one bond lengthens and the other shortens? Wouldn't it be "dx_l" and "dx_s"?

For a symmetric stretch, where both bonds lengthen, it will be P'dx + P'dx ≠ 0. (For the dipole moment, we must remember that the moments of the two bonds are pointing in opposite directions, so the change in dipole moment for the asymmetric stretch is μ'dx - (-μ')dx ≠ 0, while for the symmetric stretch it is μ'dx + (-μ')dx = 0, so the asymmetric stretch is IR active but the symmetric stretch is not.)

 

[DrClaude]

Someone told me that in the asymmetric stretch the molecule retains the same shape because as one extends the other contracts and vice versa. But you said it was the size staying constant. Is it not the shape that stays constant?

Shape is too vague a term. What does it mean exactly?

Whatever. So we must not imagine the oxygen atoms as either on the left and right side of the carbon in the case of carbon dioxide, but smeared in the molecules like in 2 places at once, like in the quantum double slit?

No, the classical picture with a localized oxygen nucleus on both sides is a very good one here.

Look. I own a Raman spectrometer and putting up a molecular scanning laboratory to monitor changes of molecular bonds to unknown field and hamiltonians. And everyday whenever the laser hits the sample, I need to visualize or have ideas how it can cause transitions only in symmetric molecules and not in asymmetric.

There is no way to avoid the math here, meaning quantum mechanics and group theory for symmetry.

Also need to look at this from the view of an electromagnetic wavelenght that is a thousand times as large

This is mostly irrelevant here, except for the fact that it means that you can consider that the electromagnetic field at any given instant is uniform over the molecule (long-wavelength approximation).

Also the laser beam is light which has electric and magnetic field, it has directed vector, so I'm trying to reconcile it with the polarization that has no directed vector as you described.

Polarizability is a tensor, not a vector.

I'm still pondering what you said. Why did you use one single "dx" for either side where one bond lengthens and the other shortens? Wouldn't it be "dx_l" and "dx_s"?

There is a single x coordinate here. You could but labels on P if you want, but I think that what @mjc123 wrote was clear enough for a forum discussion.

 

[jake jot]

In our discussions. We jumped from change in polarizability to transitions from virtual energy state to ground state. We omitted the topic of induced dipole. IR activeness has to do with permanent dipole moment, and Raman has to do with induced dipole.

Can you give an example where induced dipole can cause changes in vibrational states? In plain QM, when you give electrons energy. It got excited to higher electronic energy level. When you cause an induced dipole, how come there is more tendency for the molecule to get to higher energy state? Can induced dipole itself bestow energy to the electrons, or cause distinctions in energy states enough to make transitions possible? But it's not intuitive how induced dipole can cause ease of electronic transition. Note in electronic transitions, it's not moving from negative to positive.

I'm trying to understand the connection between the two. I actually watch many videos at youtube and this is not so clear. An intuitive answer would make it become part of common sense understanding and memorizing formulas or equations.

 

[DrClaude]

Can you give an example where induced dipole can cause changes in vibrational states? In plain QM, when you give electrons energy. It got excited to higher electronic energy level. When you cause an induced dipole, how come there is more tendency for the molecule to get to higher energy state? Can induced dipole itself bestow energy to the electrons, or cause distinctions in energy states enough to make transitions possible? But it's not intuitive how induced dipole can cause ease of electronic transition. Note in electronic transitions, it's not moving from negative to positive.

Transitions between states in an atom or a molecule take place because of the coupling of the atom/molecule with the electromagnetic field. Using perturbation theory, one finds, to first order, that the coupling comes from the interaction of the EM field with the dipole moment of the atom/molecule. Polarizability comes in as a second-order term, where a dipole can be induced by the field, and this dipole then interacts again with the field, causing the transition.

In all cases, what the EM field can do is move charges around, hence its interaction with a permanent or induced dipole moment.

Note that the virtual transitions you mentioned above are simply a visualization of perturbation theory.

I'm trying to understand the connection between the two. I actually watch many videos at youtube and this is not so clear. An intuitive answer would make it become part of common sense understanding and memorizing formulas or equations.

Again, the answers can only be found by diving into the theory.

 

[jake jot]

Transitions between states in an atom or a molecule take place because of the coupling of the atom/molecule with the electromagnetic field. Using perturbation theory, one finds, to first order, that the coupling comes from the interaction of the EM field with the dipole moment of the atom/molecule. Polarizability comes in as a second-order term, where a dipole can be induced by the field, and this dipole then interacts again with the field, causing the transition.

It makes more sense now.

For permanent dipole moments that is IR active where IR spectroscopy is useful. Do all heat and energy increase the dipole moments so you will have immediate effects on absorbance? What kinds of molecules where even when you heat them or introduce energy, there will be no changes in the permanent dipole moments or asymmetric stretching?

When I got the Raman scanner, my priority was studying the bonding of water to see influence to external fields and energy.. Lately I realized the water electrons were so close to the nucleus so the change in polarizability is poorer, hence water is a bad target.

 

[mjc123]

As has been repeatedly stated, a permanent dipole moment is not necessary for IR activity. What is required is a dipole moment that changes during the vibration (even if the time-average is zero). Heating will not affect the dipole moment significantly, but it may affect the spectrum by populating higher rotational and vibrational states (that's one way of measuring the temperature of a gas).

Did you not do some homework before buying a Raman scanner, and realize that water is a bad Raman scatterer? That's one of the advantages of Raman - you can look at species in aqueous solution , which you can't with IR because of the massive absorbance of water.

 

[jake jot]

As has been repeatedly stated, a permanent dipole moment is not necessary for IR activity. What is required is a dipole moment that changes during the vibration (even if the time-average is zero). Heating will not affect the dipole moment significantly, but it may affect the spectrum by populating higher rotational and vibrational states (that's one way of measuring the temperature of a gas).

Did you not do some homework before buying a Raman scanner, and realize that water is a bad Raman scatterer? That's one of the advantages of Raman - you can look at species in aqueous solution , which you can't with IR because of the massive absorbance of water.

I'm assembling a molecular laboratory. I need to acquire an IR Spectrometer. But the cheapest ones available (in fact mostly available) are the NIR spectrometer with range of wavelength of 650 to 2500 nm. The fundamental IR active modes are all above 2500nm. The ones below are just the overtones. I want to ask about overtones and how any external energy introduced into the molecules can show up in the spectrums from 650 to 2500nm.

2,700 nm corresponds to about 3700 cm-1

15,000 nm corresponds to about 650 cm-1

In details. NIR spectrometers have primarily a range of 900nm to 2400nm (or 11,111 cm-1 to 4166 cm-1). In the following table or list of the molecules functional groups, the range is between 2700nm to 15,000 nm (3700 cm-1 and 650 cm-1). Outside of the range of NIR spectrometers.

https://www.sigmaaldrich.com/technical-

the rest the list is in the url above.

So if I get an IR spectrometers that can only detect the overtones of the fundamental IR dipole modes. And external energy is introduced into the molecules. How much will it show up in the 650 to 2500 nm wavenength range? How much can external energy affect the overtones? Or not visible at all? I won't get NIR Spectrometers to study functional groups but just to see if the overtones can be affected by any external energy. What is the wavelengh of higher rotational and vibrational states?

 

[jake jot]

As has been repeatedly stated, a permanent dipole moment is not necessary for IR activity. What is required is a dipole moment that changes during the vibration (even if the time-average is zero).

To clarify something. You were describing polarizability in the second sentence, right? If you were, then you didn't get my second to last message. I wasnt asking about polarizability anymore which we have discussed in many messages already.

What I was inquiring in second to last message was about the equipartition of energy where the energy can be distributed in rotational, vibrational or electronic modes of vibrations.

The fundamental modes of molecular permanent dipole moments are in the range of 3 to 15 micron. Most IR spectrometers I saw are in the NIR range which is 0.78 to 2.5 micron. And these can only detect the overtones of the fundamental modes.

Chemistry: Notes about IR spectroscopy (openchemistryhelp.blogspot.com)

So in my question to you. I wasn't asking about polarizability anymore but how rotational, vibrational, electron degree of freedom in the equipartition of energy is related to the permanent dipole moments. You said not significant. So what wavelength are these rotational, vibrational electronic degree of freedom?

Also I was googling about "thermal imaging and dipole moments". I couldn't find any hits. Thermal imaging seems to involve blackbody radiation. But what molecular modes contribute to blackbody radiation? If you can see the wavelength range above. The thermal modes overlap with the permanent dipole moments modes.

I already owned a thermal imaging system that can probe 7.5 to 13micron wavelength. So I need to understand everything about the infrared wavelength to understand it all. I indeed tried to google and read a lot. So I'm not seeking a tutorial here but just clarifications. That's all. Thanks.

Heating will not affect the dipole moment significantly, but it may affect the spectrum by populating higher rotational and vibrational states (that's one way of measuring the temperature of a gas).

Did you not do some homework before buying a Raman scanner, and realize that water is a bad Raman scatterer? That's one of the advantages of Raman - you can look at species in aqueous solution , which you can't with IR because of the massive absorbance of water.

 

[mjc123]

To clarify something. You were describing polarizability in the second sentence, right?

No. I was describing a dipole moment that changes during vibration due to the changing geometry of the molecule. It is this (and I repeat, NOT a "permanent dipole moment") that is necessary for a vibration to be IR active.
Polarisability refers to the production of an induced dipole moment by the application of an external electric field. Raman activity requires a change in polarisability during the vibration.

 

[jake jot]

No. I was describing a dipole moment that changes during vibration due to the changing geometry of the molecule. It is this (and I repeat, NOT a "permanent dipole moment") that is necessary for a vibration to be IR active.
Polarisability refers to the production of an induced dipole moment by the application of an external electric field. Raman activity requires a change in polarisability during the vibration.

Thanks for the clarifications. So the IR wavelength being seen by thermal cameras/imagers is the same dipole moment that changes due to the changing geometry of the molecule? Note the wavelength of these thermal images are similar to the IR spectroscopy fundamental functional groups wavelength, both about 3 to 15 microns.

But you said "Heating will not affect the dipole moment significantly, but it may affect the spectrum by populating higher rotational and vibrational states (that's one way of measuring the temperature of a gas).".

I still don't know how to relate the blackbody radiation and dipole moments of molecules. Not a single site out of nearly half a hundred sites I read mentioned the connections.

Is it that they are not exactly related because the temperature seen by the thermal imager is the peak of the following Blackbody and not the molecules dipole moment wavelengths? What other vibrational states other than dipole moments and rotations that contribute to thermal vibrations?

 

[Lord Jestocost]

I still don't know how to relate the blackbody radiation and dipole moments of molecules.

There is no relation. Have a look at https://www.itp.uni-hannover.de/fileadmin/arbeitsgruppen/zawischa/static_html/blackbody.html:

"The situation is similar in a glowing solid or liquid. Instead of the single atoms or ions, in this case one has to consider the whole bulk system, as the particles are in permanent interaction with their neighbours. Thermal motion of electrons and ions is again irregular enough so that the above arguments are applicable...
... It is remarkable how little the intensity and its distribution over the different wavelengths depend on the glowing substance. Radiation emerging from a small opening in a cavity depends only on the temperature and not at all on the material of the cavity's walls. The opening of a cavity appears black at low temperatures and represents an almost ideal black body; the radiation emitted is called black-body radiation or cavity radiation. Given the temperature, the black-body radiation is completely specified. Light of the sun is similar to blackbody radiation."

 

[jake jot]

There is no relation. Have a look at https://www.itp.uni-hannover.de/fileadmin/arbeitsgruppen/zawischa/static_html/blackbody.html:

"The situation is similar in a glowing solid or liquid. Instead of the single atoms or ions, in this case one has to consider the whole bulk system, as the particles are in permanent interaction with their neighbours. Thermal motion of electrons and ions is again irregular enough so that the above arguments are applicable...
... It is remarkable how little the intensity and its distribution over the different wavelengths depend on the glowing substance. Radiation emerging from a small opening in a cavity depends only on the temperature and not at all on the material of the cavity's walls. The opening of a cavity appears black at low temperatures and represents an almost ideal black body; the radiation emitted is called black-body radiation or cavity radiation. Given the temperature, the black-body radiation is completely specified. Light of the sun is similar to blackbody radiation."

Ok thanks. But why do the wavelengths of dipole moments and thermal imaging coincide at 3 to `15 micron?

an example of IR spectrum

and reflect (pun unintended) this

 

[DrClaude]

Ok thanks. But why do the wavelengths of dipole moments and thermal imaging coincide at 3 to `15 micron?

The blackbody spectrum is derived from a gas of photons at thermal equilibrium. It turns out that a perfect blackbody, an idealized material that absorbs all EM radiation that falls onto it, has the same emission spectrum as the gas of photons at the same temperature.

Many materials are good approximations to blackbodies. When you look at their spectrum, their emission intensity as a function of wavelength follows more or less what is expected of a blackbody, so that emission can be used to measure their temperature, and this is the basis for thermal imaging.

It turns out that, at temperatures we humans consider normal, the photon energy of the blackbody spectrum peaks in the IR part of the spectrum. This is why thermal imaging is done using IR cameras. It also turns out that from a molecular and solid state point of view, this range of photon energy corresponds to vibrations of bonded atoms. In molecules, the vibrational modes are discrete, which leads to the kind of spectra you just showed, while for the solid state the vibrational modes (phonons) are much closer in energy and are basically continuous.

Note that since homonuclear diatomic molecules are not active in the IR, as written above, the main components of air, N2 and O2, are transparent in the IR, which is why thermal imaging cameras can work.

So there is ultimately a link between polar bonds and blackbody radiation, but there are many layers of understanding that go between the two.

 

[snorkack]

Ok thanks. But why do the wavelengths of dipole moments and thermal imaging coincide at 3 to `15 micron?

an example of IR spectrum

View attachment 274241

Now reflect on this spectrum and imagine a much thicker layer of the same substance.
The top line around 80 % represents reflection of apparatus. The rest is the spectrum of sample.
It is a thin layer. Look at, for example, the double peak at 1440 and 1400 cm^-1.
About 25 % transmission at 1440/cm, 35 % at 1400/cm and 45 % in between, at 1420/cm
Normalized for the 80 % reflection, it is roughly 31 % at 1440/cm, 44% at 1400/cm and 56 % at 1420/cm
Double the thickness and the numbers are 10 %, 19 % and 31 %. Quadruple the thickness and the numbers are just 1%, 4 % and 10 %.

In thick layers, the IR lines disappear because the weaker absorption between the peaks is already nearly 100 %. There will be little visible difference between light where 99 % is absorbed in the first 1 mm, between peaks, and light where 99,9 % is absorbed in the first 0,1 mm, at peak. Likewise, the emitted light will be almost as strong as black body radiation both at spectral peaks and non-peaks.