Blackbody radiation
Posted: 15 April 2007 03:22 PM   [ Ignore ]
Member
Avatar
RankRankRankRankRankRank
Total Posts:  138
Joined  2007-03-15

I’m looking to have a dialogue with someone on this subject (largely to help prepare me for an upcoming exam)
It was because I tried understanding blackbody radiation via this site that I suddenly felt I was lacking in some very fundamental knowledge about the nature of light, color, vision, temperature, god knows what else.

Perhaps one of you might feel warranted to help me understand the following excerpt from the blackbody site linked above:

[quote:29b9b2f464]
An object is a "blackbody" if the radiation it emits into space originates completely from its temperature. This means the radiation produced by the object comes from light waves [/quote:29b9b2f464]

First of all, what sort of examples do we have of objects whose radiation is a mix between temperature and…._____?

Radiation originating [i:29b9b2f464]completely[/i:29b9b2f464] from temperature?  Well, what causes the activity that causes temperature?

And what is the difference between radiation and lightwaves?

Later in the article, they say:

[quote:29b9b2f464] The important idea is that a blackbody pumps radiation into space in a very special way. Anything which has heat and is dense enough will emit as a blackbody.  That means you, the chair you╠re sitting on and the Earth on which the chair rests are all blackbodies. [/quote:29b9b2f464]

Ok, so I’m confused here.  Earlier they stated that an object is a "blackbody" [i:29b9b2f464]if[/i:29b9b2f464] the radiation it emits into space originates completely from its temperature.

However, in this later excerpt, they are saying: any old old thing, a chair, a tree, a rock is a blackbody that emits radiation (apparently an imperceptibly dim radiation).

Is there ever an instance when light is not emited completely because of temperature? 

I’ve never given much thought to why objects emit light(radiation)

Oh yeah, and is there really any significant difference between light and radiation?

Anyway, I’d appreciate it if anyone could point out any deluded assumptions that I might be exhibiting in my wonderings above.

Profile
 
 
Posted: 16 April 2007 09:33 AM   [ Ignore ]   [ # 1 ]
Moderator
RankRankRankRankRankRankRankRankRankRank
Total Posts:  5508
Joined  2006-10-22

One example of non-temperature related radiation is a luminescent material such as, say, activated calcium sulfide which glows blue in the dark.  Another would be a radioactive material.  I’m not sure but it would also seem that the light given off when, say, a green leaf is in sunlight is another example of radiation that’s not temperature mediated.

Lightwaves are a subset of the set that ranges from cosmic rays to radio waves and lower.

The idea is that blackbody radiation is a special type or source that is related to temperature.  In other words, a lump of coal, a diamond, a block of iron, and your body will all radiate the exact same spectrum of energy if they are all at, say, 98.6 degrees F.  They may ALSO be radiating additional energy from some other cause, such as listed above.

Hope these comments help.

Occam

Profile
 
 
Posted: 16 April 2007 12:25 PM   [ Ignore ]   [ # 2 ]
Moderator
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  4071
Joined  2006-11-28

As I understand it, blackbodies (true blackbodies, anyway) do not reflect or transmit any kind of electromagnetic energy, so all they emit is light generated from their temperature (thermal radiation). Temperature is, essentially, a measure of the movement (kinetic energy) of atoms, and due to this movement photons are emitted at a wavelength related to the temperature. If an object reflects light or emits electromagnetic radiation from radioactive decay or other sources, it is not a true blackbody. And of course most of what comes off ordinary objects in our daily lives is reflected light, so these things are not true blackbodies. Any object will emit radiation in some relationship to its temperature, but there are other factors that determine how effective and emitter it is, and most objects are not true blackbodies. As for why the blackbody has a temperature in the first place, that’s a good question and way beyond my rudimentary undergrad physics.

Light is one kind of electromagnetic radiation, but there are other kinds (radio waves, x-rays, etc)

 Signature 

The SkeptVet
The SkeptVet Blog
Militant Agnostic: I don’t know, and neither do you!

Profile
 
 
Posted: 17 April 2007 09:43 AM   [ Ignore ]   [ # 3 ]
Moderator
RankRankRankRankRankRankRankRankRankRank
Total Posts:  5508
Joined  2006-10-22

Brennen, my response is, well, yes and no.  A true blackbody is as you describe it.  However, it can accept radiated energy if from a warmer source, that is, a hotter black body or any other radiating item.

However, everything does radiate energy dependent on its temperature.  This is what he was referring to as blackbody radiation.  That, however, does not make the thing a blackbody. 

I recall in high school, making a pseudo-black body by putting together a stack of razor blades.  Light went in (between the blade edges), but couldn’t be reflected so the only radiation it put out was based on the temperature of the blades. 

Unfortunately, since it’s been more than fifty years since I’ve thought of black bodies, I can’t recall anything additional.

Occam

Profile
 
 
Posted: 17 April 2007 09:46 AM   [ Ignore ]   [ # 4 ]
Moderator
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  4071
Joined  2006-11-28

Occam,
Thanks for the clarification. Can you tell I’m a b iologist, not a physicist?  :wink:

 Signature 

The SkeptVet
The SkeptVet Blog
Militant Agnostic: I don’t know, and neither do you!

Profile
 
 
Posted: 18 April 2007 01:30 AM   [ Ignore ]   [ # 5 ]
Member
Avatar
RankRankRankRankRankRank
Total Posts:  138
Joined  2007-03-15

Occam,

[quote author=“Occam”]One example of non-temperature related radiation is a luminescent material such as, say, activated calcium sulfide which glows blue in the dark.  Another would be a radioactive material. 

But what I’m confused about is the relationship each part of the EM spectrum has with temperature.  For instance, I know that infrared wave lengths cannot be produced unless a particular threshold in temperature is reached.  That being said, when calcium sulfide glows blue, what part of the EM spectrum is being produced?  I would think it would be a few notches higher than infrared, in other words, shorter-wave lenghths and greater frequencies.    I guess I’m just a bit confused about the relationship between the EM spectrum and heat.  It seems like in some circumstances, light is produced when a certain threshold of temperature is reached - - - whereas in other circumstances it’s not.    Is this generally a bit puzzeling to most people who think on these things?  Or is it just me?

Another would be a radioactive material.  I’m not sure but it would also seem that the light given off when, say, a green leaf is in sunlight is another example of radiation that’s not temperature mediated.

Well, in the case of the leaf, we can say that the light produced by the leaves is largely temperature mediated, because the leaves are not really producing the light but only reflecting it.  The source of the light is the sun.  But is sunlight purely the result of temperature?  And what is temperature?

Lightwaves are a subset of the set that ranges from cosmic rays to radio waves and lower.

The idea is that blackbody radiation is a special type or source that is related to temperature.  In other words, a lump of coal, a diamond, a block of iron, and your body will all radiate the exact same spectrum of energy if they are all at, say, 98.6 degrees F. 

The EM energy waves won’t be visible to the eye of course - infrared EM waves are what I presume you are getting at.

Question: I know that wave-lengths longer than infrared waves are microwaves and radiowaves - - but are they generated only when a certain threshold of temperature is reached?


Thank’s for your time Occam.

Profile
 
 
Posted: 18 April 2007 01:49 AM   [ Ignore ]   [ # 6 ]
Member
Avatar
RankRankRankRankRankRank
Total Posts:  138
Joined  2007-03-15

[quote author=“mckenzievmd”]As I understand it, blackbodies (true blackbodies, anyway) do not reflect or transmit any kind of electromagnetic energy

A star doesn’t transmit light?  I thought that a blackbody only transmits light, while failing to reflect.

But I guess it all depends on what you mean by transmit.  By transmit, do you mean, recieve and the pass along, kind of like a radio recieves a signal from another source and communicates it?

If that’s the case, then you’re right, a star doesnt transmit light in the sense that it recieved energy from an external source.

So all they emit is light generated from their temperature (thermal radiation).

So you see a significant difference between transmit and emit?  I’m not trying to be picky - - I’m just curious.

Temperature is, essentially, a measure of the movement (kinetic energy) of atoms, and due to this movement photons are emitted at a wavelength related to the temperature.

I guess what I’m confused about is the relationship photons have to the entire EM spectrum.  Are there difference species of photons?  Or do all photons contain the entire EM spectrum? 

If an object reflects light or emits electromagnetic radiation from radioactive decay or other sources, it is not a true blackbody.

Fair enough.

And of course most of what comes off ordinary objects in our daily lives is reflected light, so these things are not true blackbodies. Any object will emit radiation in some relationship to its temperature, but there are other factors that determine how effective and emitter it is, and most objects are not true blackbodies. As for why the blackbody has a temperature in the first place, that’s a good question and way beyond my rudimentary undergrad physics.

Light is one kind of electromagnetic radiation, but there are other kinds (radio waves, x-rays, etc)

Thanks Mckenzie.

Profile
 
 
Posted: 18 April 2007 03:02 AM   [ Ignore ]   [ # 7 ]
Administrator
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  15355
Joined  2006-02-14

I’m not sure if you’ve looked at the Wikipedia site on blackbody radiation , but it gives a pretty clear description of what it amounts to:

[quote author=“Wikipedia”]In physics, a black body is an object that absorbs all electromagnetic radiation that falls onto it. No radiation passes through it and none is reflected. It is this lack of both transmission and reflection to which the name refers. These properties make black bodies ideal sources of purely thermal radiation. That is, the amount and wavelength (color) of electromagnetic radiation they emit is directly related to their temperature. Black bodies below around 700 K (430 ∞C) produce very little radiation at visible wavelengths and appear black (hence the name). Black bodies above this temperature, however, begin to produce radiation at visible wavelengths starting at red, going through orange, yellow, and white before ending up at blue as the temperature increases.

(I think by “transmitted light” they mean light that passes through the object, as glass transmits light).

All color is due to the energy (= wavelength) of the related photons. Reflected light is just as much as is transmitted or emitted light. What happens with reflected light is that the atoms in the object (a leaf, as it might be) are excited by incoming light, absorb some of the energies and emit others. The leaf’s atoms emit light at the energy (= wavelength) necessary to produce an experience of green in us. So there is no real “difference in kind” between reflected and emitted light.

Not sure if that helps. (I’m no physicist, hope this is correct—this is how I understand the situation).

NB: if you want to learn a lot about this process I highly recommend Richard Feynman’s book QED: The Strange Theory of Light and Matter .

 Signature 

Doug

-:- -:—:- -:—:- -:—:- -:—:- -:—:-

El sue├▒o de la raz├│n produce monstruos

Profile
 
 
Posted: 18 April 2007 05:17 AM   [ Ignore ]   [ # 8 ]
Moderator
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  4071
Joined  2006-11-28

Cory,

Yes, I do mean to distinguish emit (light generated by an object and released) from transmit (light passing through an object but originating elsewhere.

I believe the specific wavelength (color) of light emitted has to do with the amount of energy of the photons generated by electrons. Electrons drop from one “orbit” to another and the “distance” they drop is proportional to the energy lost as a photon. The greater the temperature, presumably, the “higher” the electron is to begin with which affects the energy of emitted photons and resultant wavelength of emitted light. (The scare quotes are used because I recognize this is a greatly simplified Newtonian model of the atom, and a real physicist would undoubtedly shudder at all the misconceptions embodied in the description, but it works for a mere biologist like me).

Doug,

I understand reflection differently. I don’t think a leaf is green because incident light excites photons in the leaf which then emit green light. I think some incident light is absorbed and some is reflected (meaning simply bounced off without interaction with electrons in the leaf atoms). The color we see is a conssequence of which wavelengths are least absorbed. Green leaves have specieal pigments (primarily chlorophyl) which absorbs certain wavelengths of light to generate energy for the plant, and the rest of the light is reflected. If the leaf’s electrons were excited by incident light and then changed orbitals and emitted light, this would be fluorescence rather than reflection.

I did find one refereence that suggested that at a quantum level reflected light actually is absorbed but the photon is immediately re-emitted with exactly the same energy. I’ve never heard this idea before. All the mateials I can find on reflection discuss it entirely in wave terms, so maybe they just ignore the behavior of actual photons at the quantum level for convenience, but it seems to me the two explanations are fundamentally different. Am I wrong, or is this just a particle and wave don’t try to pin it down to one or the other issue?

 Signature 

The SkeptVet
The SkeptVet Blog
Militant Agnostic: I don’t know, and neither do you!

Profile
 
 
Posted: 18 April 2007 05:32 AM   [ Ignore ]   [ # 9 ]
Administrator
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  15355
Joined  2006-02-14

[quote author=“mckenzievmd”]Doug,

I understand reflection differently. I don’t think a leaf is green because incident light excites photons in the leaf which then emit green light. I think some incident light is absorbed and some is reflected (meaning simply bounced off without interaction with electrons in the leaf atoms). The color we see is a conssequence of which wavelengths are least absorbed. Green leaves have specieal pigments (primarily chlorophyl) which absorbs certain wavelengths of light to generate energy for the plant, and the rest of the light is reflected. If the leaf’s electrons were excited by incident light and then changed orbitals and emitted light, this would be fluorescence rather than reflection.

I did find one refereence that suggested that at a quantum level reflected light actually is absorbed but the photon is immediately re-emitted with exactly the same energy. I’ve never heard this idea before. All the mateials I can find on reflection discuss it entirely in wave terms, so maybe they just ignore the behavior of actual photons at the quantum level for convenience, but it seems to me the two explanations are fundamentally different. Am I wrong, or is this just a particle and wave don’t try to pin it down to one or the other issue?

Hmmm, well this is really pretty far over my head. I’m going with the quantum mechanical interpretation of light reflection that Feynman discusses in his book. It is, as you note, an example of absorption and re-emission. Indeed, IIRC transmission is the same phenomenon of absorption and re-transmission by the (apparently transparent) material.

(FWIW, this is the subject of quantum electrodynamics—the interaction of photons and electrons at the quantum level—for which Feynman won his Nobel).

I think part of the concern here is at what level of abstraction we’re talking. Presently the QM interpretation is the “deepest” level we have. Although of course you are right that the more conventional interpretation of reflection does not involve absorption. It’s just that I don’t believe this is literally accurate, at the deepest level.

 Signature 

Doug

-:- -:—:- -:—:- -:—:- -:—:- -:—:-

El sue├▒o de la raz├│n produce monstruos

Profile
 
 
Posted: 18 April 2007 07:17 AM   [ Ignore ]   [ # 10 ]
Member
Avatar
RankRankRankRankRankRank
Total Posts:  138
Joined  2007-03-15

Hey again Doug,


[quote author=“Wikipedia”]In physics, a black body is an object that absorbs all electromagnetic radiation that falls onto it. No radiation passes through it and none is reflected. It is this lack of both transmission and reflection to which the name refers.

I just wish they would give an actual example from the real world.  Many people have said the best example of a blackbody is a star.  Others have said that a perfect blackbody is a blackhole.  I can apprehend easily enough how a black hole absorbs all radiation that falls onto it. 

But when I think of something like our Sun, it’s a bit weird to think of it as something that absorbs all of the em radiation that falls onto it. 

Are there some stars like Red Dwarfs who fail to absorb all of the em radiation that falls onto it?

These properties make black bodies ideal sources of purely thermal radiation. That is, the amount and wavelength (color) of electromagnetic radiation they emit is directly related to their temperature. Black bodies below around 700 K (430 ∞C) produce very little radiation at visible wavelengths and appear black (hence the name). Black bodies above this temperature, however, begin to produce radiation at visible wavelengths starting at red, going through orange, yellow, and white before ending up at blue as the temperature increases.

If these blackbodies don’t produce light, then as objects, why don’t they reflect? 

(I think by “transmitted light” they mean light that passes through the object, as glass transmits light).

Yes, well, I can’t imagine a body in outerspace that allows light to pass through it like glass allows light to pass through.  Anything that can survive in outer space, I would think, would have to have the sort of density that light can’t transmit through.

All color is due to the energy (= wavelength) of the related photons. Reflected light is just as much as is transmitted or emitted light. What happens with reflected light is that the atoms in the object (a leaf, as it might be) are excited by incoming light, absorb some of the energies and emit others. The leaf’s atoms emit light at the energy (= wavelength) necessary to produce an experience of green in us. So there is no real “difference in kind” between reflected and emitted light.

Interesting.

if you want to learn a lot about this process I highly recommend Richard Feynman’s book QED: The Strange Theory of Light and Matter .

Yes, after having him recomended to me so much over the past year, I look foward to finally going into Feynman’s work for the first time this summer.

Profile
 
 
Posted: 18 April 2007 07:27 AM   [ Ignore ]   [ # 11 ]
Administrator
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  15355
Joined  2006-02-14

Hi Cory,

[quote author=“CoryDuchesne”]I just wish they would give an actual example from the real world.

Well, I think it’s basically an idealization or ‘thought experiment’, not that there are ever any really, absolutely perfect blackbodies ... even black holes emit Hawking radiation, so wouldn’t appear perfectly black.

For more on ‘real world’ examples, I’ll return to Wikipedia:

[quote author=“Wikipedia”]In the laboratory, the closest thing to black-body radiation is the radiation from a small hole entrance to a larger cavity. Any light entering the hole would have to reflect off the walls of the cavity multiple times before it escaped, in which process it is nearly certain to be absorbed. This occurs regardless of the wavelength of the radiation entering (as long as it is small compared to the hole). The hole, then, is a close approximation of a theoretical black body and, if the cavity is heated, the spectrum of the hole’s radiation (i.e., the amount of light emitted from the hole at each wavelength) will be continuous, and will not depend on the material in the cavity (compare with emission spectrum). By a theorem proved by Kirchhoff, this curve depends only on the temperature of the cavity walls.

So this should give some idea as to what a near-perfect blackbody would have to be like.

[quote author=“CoryDuchesne”]Yes, after having him recomended to me so much over the past year, I look foward to finally going into Feynman’s work for the first time this summer.

His stuff is great. Six Easy Pieces is also an excellent intro to physics and QM for a layperson.

 Signature 

Doug

-:- -:—:- -:—:- -:—:- -:—:- -:—:-

El sue├▒o de la raz├│n produce monstruos

Profile
 
 
Posted: 18 April 2007 08:22 AM   [ Ignore ]   [ # 12 ]
Member
Avatar
RankRankRankRankRankRank
Total Posts:  138
Joined  2007-03-15

[quote author=“mckenzievmd”]Cory,

Yes, I do mean to distinguish emit (light generated by an object and released) from transmit (light passing through an object but originating elsewhere.

I believe the specific wavelength (color) of light emitted has to do with the amount of energy of the photons generated by electrons. Electrons drop from one “orbit” to another and the “distance” they drop is proportional to the energy lost as a photon.

How do you suggest I visualize this ‘drop’?  Do you mean the electrons drop in regards to the speed at which they orbit?  And do electrons orbit as linearly as planets orbit?  Or are they all squiggly and random?

The greater the temperature, presumably, the “higher” the electron is to begin with which affects the energy of emitted photons and resultant wavelength of emitted light.

When you say higher - again are you refering to the speed at which they move around the nucleus?  I can’t visualize electrons as doing anything other than that - - so when we distinguish lower electrons with higher, I can only guess that you are refering to speed.


——————————————————————————————-

I think the crux of my confusion is about what photons are before they are photons.  I’ve read explanations about how accelerating electrons produce photons, but I need to visualize the transformation.  So far, the animations i’ve watched of photon production and the explanations I’ve read havent really given me a sense that I’m coming close to seeing the actuality.  I’m not sure if you know what I mean when I make a distinction between merely accepting an explanation/believing in mere concepts - and actually seeing or beholding the truth of the matter in a way that seems real.

For instance, to use an analogy, what was steam before it transformed into steam?

‘Boiling water’ transforms into steam.  You can’t really understand steam until you understand how and why water as a substance not only is warmed, but how and why it gets pushed to its threshold, boils and escapes into the form of steam. 

I need a similar analogy to help me understand the origins of a photon. 

Surely a photon is the result of a transformation. 

What was a photon before it became a photon?

Profile
 
 
Posted: 18 April 2007 08:40 AM   [ Ignore ]   [ # 13 ]
Administrator
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  15355
Joined  2006-02-14

Cory, you definitely have to read about quantum mechanics ... and you definitely have to read some Feynman. What you are asking is how to understand or visualize subatomic behavior with a brain that is used to macroscopic phenomena.

The short answer is that you can’t. Subatomic, QM behavior is just weird. You will confuse yourself if you try to understand it using familiar objects.

[quote author=“CoryDuchesne”]How do you suggest I visualize this ‘drop’?  Do you mean the electrons drop in regards to the speed at which they orbit?  And do electrons orbit as linearly as planets orbit?  Or are they all squiggly and random?

Electrons are weird ... they can only orbit the nucleus at certain radii, and not in between. This means that when an electron changes its energy state by jumping from an outer to an inner orbit it does this without passing through any intermediate orbits.

I think you will agree that this is very unsettling.

They also do not orbit linearly, as you say. They are “all squiggly and random”. IIRC their orbit basically amounts to a probabilistic cloud of locations.

[quote author=“CoryDuchesne”]When you say higher - again are you refering to the speed at which they move around the nucleus?  I can’t visualize electrons as doing anything other than that - - so when we distinguish lower electrons with higher, I can only guess that you are refering to speed.

“Lower” and “higher” refer to the distance that the electrons orbit the nucleus. But the great discovery of the founders of QM was to realize that the atom is not like a small solar system. So insofar as you are understanding the electrons like planets and the nucleus like the sun, you are fundamentally misunderstanding subatomic physics.

[quote author=“CoryDuchesne”]I need a similar analogy to help me understand the origins of a photon. 

Surely a photon is the result of a transformation. 

What was a photon before it became a photon?

One of the greatest things about Feynman’s discussion of QM is that he does not shy away from the larger questions, and he knows how to answer them in a way that at least makes clear what QM says. IIRC he gives no explanation of the origins of the photon. All we need to know is that the photon is emitted as a packet of energy whenever an electron jumps from a higher (outer-ring) energy state into a lower (inner-ring) energy state. This preserves mass-energy; no energy is lost, since the energy is now in the photon. The farther in that the electron jumps, the higher energy will be the photon.

But so far as I know, there is no further, deeper explanation of the transformation or creation of a photon. Perhaps string theorists might have some deeper answer, but then we’d be up against the further question as to whether string theory is correct or not.

 Signature 

Doug

-:- -:—:- -:—:- -:—:- -:—:- -:—:-

El sue├▒o de la raz├│n produce monstruos

Profile
 
 
Posted: 18 April 2007 02:56 PM   [ Ignore ]   [ # 14 ]
Moderator
RankRankRankRankRankRankRankRankRankRank
Total Posts:  5508
Joined  2006-10-22

Cory, I’m guessing you’re in a highschool or first year college physics course.  All of the questions you’re asking are good ones, but beyond your understanding with the physics knowledge at the introductory course level.  I really hate to ever recommend faith, but I suggest that you accept what the books and lecturers say (at least provisionally) so you get the background to move on to advanced courses or the books Doug is recommending. 

From the kind of mind you seem to have, I don’t think it will take you long before you suddenly see the answers to your questions.

The problem here is that we don’t deal with these things on a daily basis so we’ve learned, understood, accepted, then forgot the arduous track that got us there.  We can’t give you the organized information a good physics book and instructor can give.

Occam

Profile