Derivation of blackbody radiation equations for stars

In summary: And there are two things I don't understand in this expression based on what you just explained me. According to equation 43 the intensity is the energy density (uv) over the solid angle and then multiplied by c. The first thing I don't understand is why is it divided by the solid angle, I think it's related to the fact we are calculating the intensity of all the star? And I also don't know where c comes from. I'm sorry if I'm asking too many questions but I can't find this expression anywhere and so it's almost impossible to understand...The energy density (uv) over the solid angle is given by equation 43. The c in equation 43 is a constant that relates the energy density (
  • #1
JulioHC
15
1
Good evening,

As part of my course, I had this week two lectures about the blackbody radiation and its relation to the stars. While I do understand how to use results such as the Stefan-Boltzmann law and Wien's Law I'm lost in other parts. I think the only parts that I don't understand yet are the intensity, the flux and the luminosity, especially how the expressions for these properties are derived and what's the role of the solid angle in the derivations.

So far I have reread my notes, read the lecture notes from the lecturer and I have been searching for an explanation on different pages without results. I will take "An introduction to modern astrophysics" from the library and I hope I will find what I need there but in case I don't I wanted to ask you if there's any book, web, video... or other sources that can help me.
 
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  • #3
phyzguy said:
Well, there's always Wikipedia:

https://en.wikipedia.org/wiki/Planck's_law
https://en.wikipedia.org/wiki/Black-body_radiation

It's probably best for you to ask specific questions.
Thank you, I tried Wikipedia but it's not really oriented to the stars and doesn't include all the derivations. Also I thought that because they are all related quantities it would be possible to find an explanation of them all together. In any case, I think the strangest thing for me right now is how you use the solid angle to integrate the intensity, that is, what's the relationship between the solid angle and the intensity?
 
  • #4
JulioHC said:
In any case, I think the strangest thing for me right now is how you use the solid angle to integrate the intensity, that is, what's the relationship between the solid angle and the intensity?

Well, the star emits a total amount of radiated power P, as given by the Stefan-Boltzmann law. This power would have units energy/time. Imagine a spherical surface surrounding the star. Since the star is spherically symmetric, this power is spread evenly over the surface of that sphere. The sphere has a total solid angle of 4π. So a part of the sphere with solid angle Ω would intercept a power of P*Ω/(4π). This still has units of energy/time. To calculate the intensity of a given solid angle, you need to know how far away you are, or the radius R of your imaginary spherical surface. If your detector has an area A, the solid angle it covers is given by Ω=A/(R^2), so it will collect a power of P*A/(4πR^2). Intensity has units of energy/area/time., so to get this, you have to divide the intercepted power by the area. The intensity measured by a detector of area A at a distance R would be given by I = P*Ω/(4πA) = P/(4πR^2). I think I have all of this right. Let me know if this helps and if you have more questions.
 
  • #5
phyzguy said:
Well, the star emits a total amount of radiated power P, as given by the Stefan-Boltzmann law. This power would have units energy/time. Imagine a spherical surface surrounding the star. Since the star is spherically symmetric, this power is spread evenly over the surface of that sphere. The sphere has a total solid angle of 4π. So a part of the sphere with solid angle Ω would intercept a power of P*Ω/(4π). This still has units of energy/time. To calculate the intensity of a given solid angle, you need to know how far away you are, or the radius R of your imaginary spherical surface. If your detector has an area A, the solid angle it covers is given by Ω=A/(R^2), so it will collect a power of P*A/(4πR^2). Intensity has units of energy/area/time., so to get this, you have to divide the intercepted power by the area. The intensity measured by a detector of area A at a distance R would be given by I = P*Ω/(4πA) = P/(4πR^2). I think I have all of this right. Let me know if this helps and if you have more questions.

Thank you! It definitely helps. However, I have this expression in the lecture notes
upload_2019-2-3_19-46-18.png

And there are two things I don't understand in this expression based on what you just explained me. According to equation 43 the intensity is the energy density (uv) over the solid angle and then multiplied by c. The first thing I don't understand is why is it divided by the solid angle, I think it's related to the fact we are calculating the intensity of all the star? And I also don't know where c comes from. I'm sorry if I'm asking too many questions but I can't find this expression anywhere and so it's almost impossible to understand it.
 

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  • #6
The quantity u here is the energy density in the black body. It is an energy density, so has units of energy/volume. Think of a volume of the black body as being filled with a large number of photons of different energy traveling in many different directions. The quantity [itex] u_\nu [/itex] is the energy per unit volume with a frequency ν. The quantity [itex] \frac{du_\nu}{d\Omega}[/itex] is the energy per unit volume with frequency ν traveling in the direction specified by [itex] d \Omega[/itex]. Now you ask how many of these photons cross a unit area per second. Since they are moving with velocity c, this is just [itex] c \frac{du_\nu }{d \Omega}[/itex]. This may not be that easy to see, but if you think about a volume having a density of some quantity ρ, I think you will see that the rate at which this quantity crosses a boundary is just given by ρ*v, where v is the velocity. Dimensionally, if you multiply an energy density with units energy/volume times a velocity with units length/time, you get an intensity in energy /area/time. So at the boundary of the black body (or the star), this will give the intensity that escapes the black body into space. Hope this helps.

The classic textbook for these topics is Rybicki and Lightman, "Radiative Processes in Astrophysics".
 
  • #7
I'm uploading the relevant text from Rybicki and Lightman on this topic. Maybe their explanation will make more sense to you than mine.
 

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  • #8
Sorry for taking so long to answer. Your explanation makes sense and with the pdf that you sent with the figures, I think I understand it now. If you don't mind I will try to explain what I understand about all this so that maybe someone will read it and point out my errors.

The energy density is the energy per volume per frequency so that $$dE=u_{f}*dV*df$$ But ##dV=c*dA*dt##. Therefore $$dE=u_{f}*c*dA*dt*df$$ Then, the intensity is the energy per area per solid angle per frequency and per time(What I understand here is that the solid angle in a way defines the portion of the area where the rays are allowed to go through, though I'm not very sure about this). $$dE=I_{v}*dA*d\Omega*df*dt$$ Equating the two previous equations $$u_{f}*c*dA*dt*df=I_{v}*dA*d\Omega*df*dt$$ $$u_{f}*c=I_{v}*d\Omega$$ $$I_{v}=\frac{u_{f}}{d\Omega}*c$$ Which is what I was trying to understand. Now, I think what the flux means is that we limit the intensity to the component of the rays that are parallel to the normal vector of the area so that $$dF_{v}=I_{v}cos(\theta)d\theta$$

I'm sure I must have made many mistakes but I'm happy because I'm starting to understand at least the basics of all these concepts and I have to thank you for having helped me to achieve that.
 

1. What is blackbody radiation and why is it important for stars?

Blackbody radiation is the electromagnetic radiation emitted by a perfect absorber and emitter of radiation, known as a blackbody. It is important for stars because it is the primary mechanism by which they emit energy into space.

2. How were the blackbody radiation equations derived for stars?

The blackbody radiation equations for stars were derived using Planck's law, which describes the spectral energy density of electromagnetic radiation emitted by a blackbody at a given temperature. This law was then applied to the Stefan-Boltzmann law, which relates the total energy emitted by a blackbody to its surface temperature.

3. What factors affect the blackbody radiation emitted by stars?

The amount and type of radiation emitted by a star is affected by its surface temperature, size, and composition. Higher temperatures and larger sizes result in a higher amount of radiation being emitted, while the composition of the star's atmosphere can affect the type of radiation emitted.

4. How do the blackbody radiation equations for stars differ from those for other objects?

The blackbody radiation equations for stars are similar to those for other objects, but they take into account the unique properties of stars, such as their high temperatures and complex compositions. They also consider the effects of factors like stellar atmospheres and nuclear reactions, which can impact the type and amount of radiation emitted.

5. Can the blackbody radiation equations be used to accurately predict the radiation emitted by stars?

Yes, the blackbody radiation equations are widely used in astrophysics to accurately predict the radiation emitted by stars. These equations have been tested and validated through observations of various types of stars, and they continue to be a valuable tool for studying the properties and behavior of stars.

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