Spectral radiancy curve of a black body

In summary, In response to a general question, a black body has all wavelengths absorbed, while a gray body does not. For a gray body, the power absorbed is equal to the power emitted over all wavelengths. However, for a black body, the power absorbed is not equal to the power emitted over all wavelengths.
  • #1
Ajit Agrawal
10
0
I was wondering about the spectral radiancy curve of a black body. How does the spectral radiancy of any other body looks like.
I have seen one among many possibilities:
http://www.giss.nasa.gov/research/briefs/schmidt_05/

Q 1. But i was thinking that for a general body (not a black body); are there some wavelengths missing?

Q 2. Can anyone explain in simple language, the reflection of light by photon theory.

I believe that both the questions are one and same. Can anyone remove my confusion?
 
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  • #2
Welcome to PF;

Q1. Yes - in general, some wavelengths get absorbed so you get the lines in the spectra photographs. You should be able to google for examples.

Q2. photon interacts with electrons in material - many possible things can happen, one of them is that a photon gets reflected. The details depend on the model being used ... i.e. Richard Feynman has QED model for refection where the "reflected" photon leaves the surface at a random angle.

In another thread you asked:
The way i understood this is:
1) When a electromagnetic spectrum falls on a body which is NOT a black body, then some photons are reflected back completely.
2) in cavity radiation, when light enters the cavity, the photons those are reflected by the walls of the cavity are sooner or later absorbed by it.
Basically, a blackbody is an idealization of a body that absorbs all light incident upon it, and radiates energy at the same rate (so it is in thermal equilibrium with it's surroundings). Cavity radiation looks a great deal like blackbody radiation - the idea is that there is a very small chance that radiation entering the cavity will exit before it is absorbed.
 
  • #3
thank you for reply

A very nice and clear answer. I understood that, but i have a different question.

I have confusion regarding use of formula P = σεAT4.

Suppose there is a general body; and radiations of particular band of spectrum are falling on it.
Then the power absorbed by it is P = σεAT4. (Where, T is the temperature of the body).

Suppose the temperature of the very same body is kept constant (T) by some means. But this time, entire electromagnetic spectrum is falling on it.
Then will it be proper to say that power absorbed is same as P = σεAT4.
 
  • #4
That would be Stephan's law.
Have a look at the situations in which Stephan's law applies and how it was derived and you'll have your answer ;)

But you have not said what confuses you?
 
  • #5
thank you for replying

I have seen the derivation; the integration was done from 0 to ∞. Thus, it means that the sum of intensity of all wavelengths comes out to be P = σAT4 for a black body.

Now, when i take a body which is not a black body (let me call it a grey body), then the result will be P = σεAT4 when the integration is done for all wavelengths from 0 to ∞.

But, my confusion is, if all the wavelengths are not falling on the grey body, then then the summation should come out to be less than σεAT4 <-- this is what i understand. Am i right on this?
 
  • #6
I'll take a shot...

The Stefan-Boltzmann constant (σ) relates the energy (which can be related to λ via E=hf and c=fλ) to the Temperature (T) of the body.

It's not that the T of the body predicts what wavelengths are falling on it, but rather the summation of the wavelengths that are. Further, if the temperature would be a certain value for a black body, it would be less than this value for a gray body, and this difference would be reflected (no pun intended) as such in the Power (P) calculated via Stefan-Boltzmann law.
 
  • #7
@10me: that's pretty good.

@Ajit: Have you tried reworking the derivation for a limited range of energies?

Consider: If you heat something to a temperature and thermal equilibrium with just blue light, would it radiate any differently than if it were heated to the same temperature with red light? It's just as hot and heat is heat is heat right?
 
  • #8
Excellent point, but so sorry I never did that. But I understand that the if the temperature is same, then be it be red light or blue light, heat radiated should be same. Am I right on this?
 
  • #9
You should get used to doing the math instead of trying to work stuff out qualitatively like this.

Well you have an example of the calculation you need to do, which has been performed over all wavelengths.
You should be able to do the required calculation for a limited range of wavelengths for each of your colors.

Generally, an object thermally radiating in blue is hotter than one radiating in red though.
It is possible for an object radiating in blue to lose energy at the same rate as an object radiating in red - it just has to be a weaker emitter. So it depends on the circumstances... what are the circumstances you have in mind?

The answer to that should guide your calculation.
 
  • #10
I do not know this much maths :) but I can understand it somewhat. That is why I was trying to do qualitatively.
 
  • #11
Ajit Agrawal said:
I do not know this much maths :) but I can understand it somewhat. That is why I was trying to do qualitatively.
Then you need the practice. ;)

Mathematics is the language of physics - to understand physics (especially at the level you are asking questions about it) you must get proficient at the maths. Anything else will be like trying to relate the poetry of Imru' al-Qais (or Shakespear or whatever) using pidgin English.
 
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  • #12
Fully agreed, can you suggest the best book on thermal radiation.
 
  • #13
There is no "best book", start with your level and work up.
Blackbody radiation tends to be introduced briefly at freshman college level and more in-depth the following year, across several papers ... so you want calculus to senior secondary level at least.
 
  • #14
i have read calculus till that level, but in my profession i never needed it, therefore over years i forgot them all. Ofcourse there is no best book, but can you guide me to one book that you prefer; or all the books. I have capacity to read, so that is not a problem.
 
  • #15
Used to like stuff by Tipler ... but I hardly used a book since my profs gave good lecture notes.
Since this is "stage II" college, it is hard to know where to advise you. It's normally studied as part of a broader course on statistical mechanics or quantum physics - so those would be the broader areas to look through.
There are usually lectures online covering specific sub-topics like "blackbody radiation".

There is a textbook discussion forum - but I would seriously google for freshman college courses and get a HS calc primer from a bookshop: any will do.

Meantime: the comments bottom of post #9 should give you an idea what to expect.
Enjoy :)
 
  • #16
I want lecture notes of your prof :) if you want any educational material, try me, i might have that too original stuff.
 

What is a spectral radiancy curve?

A spectral radiancy curve is a graph that shows the amount of energy emitted by a black body at different wavelengths. It is also known as a Planck curve or black body radiation curve.

What is a black body?

A black body is an idealized object that absorbs all radiation that falls on it and emits radiation at all wavelengths. It is often used as a standard reference for comparing the radiation emitted by other objects.

What does the shape of a spectral radiancy curve indicate?

The shape of a spectral radiancy curve indicates the distribution of energy emitted by a black body at different wavelengths. It follows a characteristic shape known as the Planck distribution, which peaks at a specific wavelength depending on the temperature of the black body.

How does the temperature of a black body affect its spectral radiancy curve?

The temperature of a black body directly affects its spectral radiancy curve. As the temperature increases, the peak of the curve shifts to shorter wavelengths and the amount of energy emitted at all wavelengths increases.

Why is the spectral radiancy curve of a black body important?

The spectral radiancy curve of a black body is important because it provides insights into the behavior of radiation and the properties of matter. It is also used in various fields of science, such as astronomy and thermodynamics, to understand and analyze the emission of energy from objects.

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