Assumptions for blackbody spectra vs. emission spectra vs. absorption spectra

  • #1
Hi there,

I am a physical oceanographer teaching an introductory undergraduate earth science class that has a unit on astronomy. I have a physics undergraduate background, took a few astronomy classes at the undergraduate level back in the day, and did a bit of undergraduate research in astronomy, but this was over a decade ago. As I am (re)teaching myself some of this stuff I'm running into questions.

As I understand, objects that can be approximated as blackbodies are objects that are more or less in thermal equilibrium (they are absorbing the same amount of light as they emit) and they are also absorbing and emitting much more light than they are reflecting. (It has been a while since I thought about blackbody radiation too, so please correct me.) Stars fall in this category, and their temperature can be estimated from the wavelength where the intensity of the spectrum peaks.

I understand that emission line spectra are created when the electrons in an atom have been excited and then fall back to their original energy state. When the electrons move back to the lower energy states they emit photons at particular wavelengths. These signals at particular wavelengths can be used to identify the composition of the emitter. Would it be correct to say that objects that express emission line spectra are NOT in thermal equilibrium? Is this the reason they do not have blackbody spectra? And/or does it have to do with their "optical depth"? (I keep coming back to optical depth in my googling but haven't been able to figure out what it means.)

It seems that absorption spectra are created when light from an object that can be approximated as a blackbody runs into an object that absorbs light at certain frequencies due to its composition. Then the resulting spectra is a blackbody spectra but is missing energy at those certain wavelengths. Those wavelengths are at the same wavelengths that you would see in an emission spectra from an object composed of the same type of "stuff". Why does this "stuff" absorb and emit in one case and only absorb in another? Perhaps it is doing both in both cases but it is just different due to the perspective of how we measure it?

I guess my overall question is what are the basic assumptions about the conditions that create a blackbody spectra vs. an emission spectra vs. an absorption spectra?

The textbook we are using (not chosen by me) doesn't answer this question but it seems an important fundamental.

Thank you for your help!
 
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Answers and Replies

  • #2
A few more questions/follow ups:

-Would it be true to say that emission spectra still show a background blackbody curve but the emission lines are just much more energetic?

-Is "continuous spectra" synonymous with "blackbody spectra", or does this just mean a spectra without distinct lines?

-This is how it is explained in our book. Types of emitters are described (although I find this very vague), and the mechanism of how the spectra are created are described in the text, but the reason why these mechanisms are specific to these types of emitters is not clear to me.
1630522258030.png

Thanks again for your help!
 
  • #3
PSS - I prefer to think of spectral lines in terms of the bottom panel, as that is most analagous to my work with ocean wave spectra. When I see a graphic like the one in the top panel, my brain tries to convert it to the one at the bottom.
1630523922429.png
 
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  • #4
256bits
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As I understand, objects that can be approximated as blackbodies are objects that are more or less in thermal equilibrium (they are absorbing the same amount of light as they emit) and they are also absorbing and emitting much more light than they are reflecting. (It has been a while since I thought about blackbody radiation too, so please correct me.) Stars fall in this category, and their temperature can be estimated from the wavelength where the intensity of the spectrum peaks.
Not sure what you mean by 'more or less in thermal equilibrium'. Thermal equilibrium with what.

A blackbody absorbs all radiation impinging upon its surface. None is reflected. Its emissivity, ε, 1.
If there is some reflection, at all wavelengths, it has an emissivity less than 1 ( an emissivity of 0 would mean a perfect mirror ). These objects with emissivity less than one are called a greybody.

For radiation emitted from an object, to be called a blackbody ( or greybody ), the object emits at all wavelengths, following an intensity curve reminisant of the black body spectrum - a blackbody emission curve is the most of any radiative power that any object can emit - all other emission spectrums are less due to the emissivity being less than 1.

a continuous spectrum is necessary for the blackbody spectrum, but the reverse is not true.

Optical depth is how far you can see into an object. A foggy day has less optical depth than a clear day.
It relates to the scatterings a photon produced in the interior of the object has to do before it exits. What you see is actually the very last scattering before the photon exits the surface.
 
  • #5
vela
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As I understand, objects that can be approximated as blackbodies are objects that are more or less in thermal equilibrium (they are absorbing the same amount of light as they emit) and they are also absorbing and emitting much more light than they are reflecting. (It has been a while since I thought about blackbody radiation too, so please correct me.)
A blackbody absorbs all light incident on it. It emits radiation with a characteristic spectrum that depends only on its temperature. It doesn't have to be in thermal equilibrium with its surroundings.
I understand that emission line spectra are created when the electrons in an atom have been excited and then fall back to their original energy state. When the electrons move back to the lower energy states they emit photons at particular wavelengths. These signals at particular wavelengths can be used to identify the composition of the emitter. Would it be correct to say that objects that express emission line spectra are NOT in thermal equilibrium? Is this the reason they do not have blackbody spectra?
Emission line spectra are produced when you have individual atoms and molecules, like in a gas. The reason the filament of a light bulb doesn't is because the atoms are close together so the energy levels combine into continuous bands. Nothing really to do with thermal equilibrium.
It seems that absorption spectra are created when light from an object that can be approximated as a blackbody runs into an object that absorbs light at certain frequencies due to its composition. Then the resulting spectra is a blackbody spectra but is missing energy at those certain wavelengths. Those wavelengths are at the same wavelengths that you would see in an emission spectra from an object composed of the same type of "stuff". Why does this "stuff" absorb and emit in one case and only absorb in another? Perhaps it is doing both in both cases but it is just different due to the perspective of how we measure it?
To be a little more precise, you're referring to absorption line spectra. In astronomy, you typically have a star producing a nice continuous spectrum, and then the gas the light has to pass through to get to us absorbs some wavelengths of light. The gas atoms/molecules will eventually emit the same wavelengths, but they will do so in a random direction. If the direction isn't toward us, we won't see the light so those wavelengths appear to be missing to us.
 
  • #6
vela
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Here's a diagram from Comin's Discovering the Universe. It may give you a better picture of what's going on.
spectra.jpg
 
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The gas atoms/molecules will eventually emit the same wavelengths, but they will do so in a random direction. If the direction isn't toward us, we won't see the light so those wavelengths appear to be missing to us.
Not necessarily - they can relax the excitation by multiple emissions, or by collisions.

Optical depth is specific to wavelength.

Consider water overlying a reflective white sand bottom.
A thin layer of water on reflective white bottom looks clear and colourless because almost all visible light of any colour passes through the shallow water and after reflection back again.
A moderate depth of water on reflective white bottom looks bright blue because the water absorbs red light but blue light is passed through and reflected from bottom.
A deep body of water looks dark blue, because water also absorbs blue light, although less than red light.
A deep body of water does not look black because water also possesses Rayleigh scattering which for blue light is comparable to absorption. Some blue light is scattered back by water before it is absorbed.

One part of a star behaving as a black body is a big optical depth. Weak absorptions in regions away from spectral lines or bands still absorb most of the incident light because of great depth, like water absorbs most blue light and is dark blue not light blue.
 
  • #8
Thanks for your responses everyone. I think I got the equilibrium idea from the wikipedia page on blackbody radiation:
https://en.wikipedia.org/wiki/Black-body_radiation
"Black-body radiation is the thermal electromagnetic radiation within or surrounding a body in thermodynamic equilibrium with its environment, emitted by a black body (an idealized opaque, non-reflective body)."
That said, I know wikipedia is not necessarily a legitimate resource, so thanks for getting the equilibrium idea out of my head. Perhaps there is some nuance here I'm missing.

@256bits
"a blackbody emission curve is the most of any radiative power that any object can emit - all other emission spectrums are less due to the emissivity being less than 1.
a continuous spectrum is necessary for the blackbody spectrum, but the reverse is not true."
Thank you for clarifying these things - very helpful:

@vela - Thank you for the diagram. That was helpful
This is a nice PBS special that talks about the absorption line spectrum of the sun:
This helped me understand that the cloud of "cooler" gas is sometimes the outer layer of a star

The optical depth idea is sort of starting to sink in, but I have more learning to do there. Thanks to all of you for trying to explain it to me. Same for blackbodies- I think I need to find my old thermodynamics textbook...

@snorkack In some of my research I use the light absorption properties of the ocean to estimate coastal bathymetry using multi-spectral satellites. I'm still trying to make the connections between this sort of phenomena and optical depth.
 
  • #9
vela
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Thanks for your responses everyone. I think I got the equilibrium idea from the wikipedia page on blackbody radiation:
https://en.wikipedia.org/wiki/Black-body_radiation
"Black-body radiation is the thermal electromagnetic radiation within or surrounding a body in thermodynamic equilibrium with its environment, emitted by a black body (an idealized opaque, non-reflective body)."
That said, I know wikipedia is not necessarily a legitimate resource, so thanks for getting the equilibrium idea out of my head. Perhaps there is some nuance here I'm missing.
I looked at the Talk page for that entry, and there's been some discussion about that sentence although no one ever fixed it. The sentence is at best misleading, though I'd say it's wrong. Apparently, what was meant is that the internals of a black body have to be in thermal equilibrium so that the black body actually has a temperature.
 
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