Light Transmission Through Window Glass

[Swampeast Mike]

Do photons in the form of visible light actually pass through ordinary window glass?

I ask because window glass is assumed to be opaque to photons in the form of infrared energy. The IR photons are accepted on one side of the glass, conduct through the body of the glass in the form of heat and [presumably] new photons are emitted from the other side.

Are the very photons on the "out" side of the glass the exact same ones we see as visible light on the "in" side?

 

[zoobyshoe]

Are the very photons on the "out" side of the glass the exact same ones we see as visible light on the "in" side?

A hard question to answer, depending on what you mean by "the same ones".

Read this thread for a detailed discussion of the same question:

https://www.physicsforums.com/showthread.php?t=76246

 

[Swampeast Mike]

Why does it not surprise me that the "answer" depends on your perspective of the photon. Every other question about photons seems to depend similarly so why not this one as well?

Just when I think I can begin to understand something defined as a "massless bundle of energy", I find myself no closer than in the beginning.

 

[Swampeast Mike]

Now Another Question

In the case of IR through window glass where I suppose it's safe to say that these are not the same photons, is all of the energy conducting through the body of the glass detectable as heat? Is some of the energy purely kinetic?

 

[Claude Bile]

To answer the original post, visible photons (with the exception of a few absorption bands) are indeed transmitted through the glass. The properties of the photon 'going in' (in the case of transmission) are exactly the same as photons 'coming out' (i.e. momentum and energy are conserved).

Claude.

 

[krab]

Are the very photons on the "out" side of the glass the exact same ones we see as visible light on the "in" side?

Photons are elementary particles. They are indistinguishable. If one photon goes in and another comes out, it makes no sense physically to ask whether that is the same photon. You cannot say "Oh I recognize this photon; it has a wart on it." Photons have no features beyond their energy and polarization. It is impossible to design an experiment to tell you whether a photon that is subjected to some experiment is the same one going in as coming out, or OTOH is destroyed when going in and a new one emitted. This is also not because we don't know enough about the photon being scrutinized. Photons cannot be distinguished even in principle; they have no hidden properties unknown to us.

 

[Swampeast Mike]

Fairly easily deal with a visible light photon striking one surface, its' energy passing through the body without being absorbed or altered and an identical photon being liberated from the other surface.

Again, in the context of the sun shining through ordinary window glass into a room, what about the IR photon? It strikes one surface and its' energy is absorbed into the body. Does each striking photon result in the release of a significantly less energetic photon from the other surface or does the energy of multiple photons combine to produce a fewer numer of relatively higher energy but still lower than the original photons? Impossible to know? Depends on the nature of what the inner surface "sees" in the room? Has all of the energy passed through the body in the form of heat?

 

[ZapperZ]

OK, against my better judgement, I'll do this once more. Some of you may have read this before when I responded to a similar thread - so you may skip this.

First of all, let me just say that "glass" is a VERY bad example to use to describe the physics of optical transmission/conductivity in matter. It has a very complex structure, the typical ones are amorphous and tend to be analogous to a liquid phase. With that in mind, I will first describe the GENERIC physics of optical transmission in dielectric solids, and then, at the end, if I still have the energy, try to tackle transmission in glass.

There are two principles to keep in mind when we deal with optical transport: (i) the band structure of the material and (ii) the vibrational states (phonon states).

When light hits a dielectric solid, there is a "skin depth" that is a characteristic penetration of that EM radiation into the material. At the microscopic level, what you have is the E-field of the EM radiation causing a vibration of either the dipole mode of the molecule of the material OR the vibration of the phonon modes of the solid. So this is the vibrational states that is the principle one has to keep in mind. This is why light normally slows down in an ordinary dispersive medium.

HOWEVER, this is where you then have to look at the band structure. If the energy of that light (photon) is SMALLER than the band gap of the solid, than there is NO ABSORPTION of that energy or frequency. The oscillation or vibration cannot be sustained or absorbed by the material and it passes out of the material. This results in the transmission if the penetration depth is larger than the thickness of the material.

If the photon energy is larger than the band gap, then the solid can absorb that energy. What happens later can be complicated and depends on the fine details of that material. The vibrational energy can be absorbed by the entire crystal and turns into heat. Or, it can reradiate, even at different frequencies, etc... The possibilities are endless. I have also only discussed the optical band gap and not the electronic band gap (which is a whole other beast).

Coming back to glass, the problem here is the exact chemical composition in "ordinary glass". Quartz and fused silica, for example, have different properties than "ordinary" amorphous glass that you buy in the store. All of them certainly have a band gap larger than the photons in the visible spectrum. However, ordinary glass is quite absorbant in the UV region (band gap around this value), while not so much for quartz and fused silica (band gap slightly larger).

Now, in the IR region, almost EVERYTHING is absorbant. But here, I think, the mechanism has more to do with vibrations at the molecular level of the material, rather than the bulk material. Someone in chemistry can correct this if I'm wrong, but the IR spectrum range corresponds to "stretching" modes of many molecules, especially organic ones. I would not be surprised if it corresponds to the Si-O or Si-O2, etc. modes of glass. That would explain why it absorbs some, not all, IR spectrum. After all, you do feel heat from the sun going through your glass window.

Zz.

 

[Astronuc]

Now, in the IR region, almost EVERYTHING is absorbant. But here, I think, the mechanism has more to do with vibrations at the molecular level of the material, rather than the bulk material. Someone in chemistry can correct this if I'm wrong, but the IR spectrum range corresponds to "stretching" modes of many molecules, especially organic ones. I would not be surprised if it corresponds to the Si-O or Si-O2, etc. modes of glass. That would explain why it absorbs some, not all, IR spectrum. After all, you do feel heat from the sun going through your glass window.

Zz.

That is correct.

Perhaps a good reference is - Introduction to Solid State Physics, 8th Edition
Charles Kittel - http://www.wiley.com/WileyCDA/WileyTitle/productCd-047141526X,descCd-tableOfContents.html

Infrared spectroscopy is used to charaterize the structure and properties of inorganic oxides.

 

[Dr.Brain]

Refraction can be thought of absorb-and-emit rule ,although there has been no evidence and any reference experiment that I can tell but as per my textual knowledge goes , the explanation for refraction is pretty plain and simple.

When light enters from one medium to another, the photons in it are absorbed by the atoms present on the boundary of the two mediums, now the photons in the light are being obstructed every now and then ,due to the atoms of the medium that are present , so apparently it appears to us that light has slowed that but actually the photons which always travel at the speed of light but seem to be slowed due to frequent obstructions in their path.

 

[Swampeast Mike]

Zapper Z;

I too agree with, "I'd rather do science than talk about science".

The true purpose behind this question has nothing to do with window glass--sorry for the poor choice of material.

The thread, https://www.physicsforums.com/showthread.php?t=73153 has the real problem. It's as scientific as I can muster in a real-world and exceptionally dynamic system.

Have spent many, many hours staring through my office window wondering about the problem. Then, it seemed that I might be looking through the explanation.

If some IR energy is accepted and re-radiated through window glass without being converted to heat, why not via water in this system as well? In that case, the house itself (primarily the outside walls and particularly the window glass) is the missing "heat sink". Some of the radiation from the burner has passed through the system to house without becoming measurable heat. Possible?

 

[ZapperZ]

The thread, https://www.physicsforums.com/showthread.php?t=73153 has the real problem. It's as scientific as I can muster in a real-world and exceptionally dynamic system.

Have spent many, many hours staring through my office window wondering about the problem. Then, it seemed that I might be looking through the explanation.

If some IR energy is accepted and re-radiated through window glass without being converted to heat, why not via water in this system as well? In that case, the house itself (primarily the outside walls and particularly the window glass) is the missing "heat sink". Some of the radiation from the burner has passed through the system to house without becoming measurable heat. Possible?

Unfortunately, I don't understand what you're asking here, nor in this one:

A very small device adds energy to a system primarily via radiation. Temperature differential is high—at least 1,500°F.

The system itself is rather massive—tons.

Intermediate heat transfer is via forced convection of a fluid (water) to iron.

Energy output is again primarily via radiation. Temperature differential is low—say 20°F.

Is it possible for any amount of the transferred energy to pass through the system without appearing as detectable temperature? If not, what law prevents this?

For example, "temperature differential is high" and "low" between WHAT? I don't understand the scenerio of both of the above.

Water can gain heat energy from IR radiation the same way most other organic molecules can. Put a bucket of water out in the sun and see what happens. It doesn't mean it absorbs all of it the same way a pane of glass doesn't.

Zz.

 

[Swampeast Mike]

For example, "temperature differential is high" and "low" between WHAT?

Radiation enters the system with high differential temperature between the radiant burner and the heat exchanger coil.

Radiation leaves the system with low differential temperature between the radiators and their surroundings.

The real problem is compared to the exact same system with a different (but not radiant) burner, I'm getting significantly more heat from the system than should be possible given the temperatures I can measure. No formula I can find for any form of heat transfer can explain getting the same heat output from a lower temperature system...

Well...one thing still has the expected temperature...the water in the boiler itself.

 

[ZapperZ]

Radiation enters the system with high differential temperature between the radiant burner and the heat exchanger coil.

Radiation leaves the system with low differential temperature between the radiators and their surroundings.

Back up a bit here. You may not be aware of the terminology you are using. When you say "radiant heat" as in IR (which is what you asked in this thread, no?), this is NOT the same the heat transfer in conduction/convection in which a thermal gradient between two bodies is required. I can transfer IR radiation from one body to another without regard to the temperature gradient. IR is an EM radiation. A blackbody will absorb as much heat as I'll give it regardless of its temperature.

The real problem is compared to the exact same system with a different (but not radiant) burner, I'm getting significantly more heat from the system than should be possible given the temperatures I can measure. No formula I can find for any form of heat transfer can explain getting the same heat output from a lower temperature system...

Well...one thing still has the expected temperature...the water in the boiler itself.

You need to keep in mind that the temperature of a body depends on how much heat it absorbs versus how much heat it is giving out. If it is a good heat absorber but a bad heat radiator, it'll continue to increase in temperature no matter the temperature difference between it and the source.

Zz.

 

[Swampeast Mike]

No problems with understanding the differences between conduction, convection and radiation.

The principal mode of heat transfer into the system is via radiation of that unique burner to the stainless steel heat exchanger.

Heat is being liberated to the house with a high degree of radiation via freestanding cast iron radiators. Yes, the radiators are definitely convecting some as well, but it's extremely difficult to even estimate the proportion of radiation to convection. Conduction is almost completely absent, being confined to the four small feet of each radiator. If memory serves, my lowest possible estimate of the percentage of radiation is 43% although I suspect it could be somewhat higher.

Both average radiator temperature and average supply (temp available to the radiator) temperature decreased yet average room temperature stayed nearly identical for similar outdoor/indoor conditions. Air temperature stratification in the rooms (either the driving force or result of natural convection depending on perspective) showed no significant change--if anything it decreased slightly.

Most of the radiators are coated in a custom-made "paint" composed mainly of linseed oil and small amounts of various colors of mica. This approximates original bronzing that contained large amounts of powdered bronze and other metals. Emissivity of the "mica bronze" should be at least 0.8 and likely a bit higher--similar to most paints and construction materials.