Blackbody Radiation and the Inverse Square Law

In summary, the conversation discusses the concept of blackbody radiation and the inverse square law. The Planck's function for the radiation of a blackbody is in ##W sr^{-1} m^{-3} ## and is independent of distance. The total intensity may decrease with distance, but the temperature fingerprint remains the same. The term "luminosity" is used to describe the total amount of energy emitted by an astronomical object like a star or planet, while "intensity" is used in conjunction with another term, such as "radiant intensity". The spectral radiance is independent of distance or viewing angle, but the energy received can be distance or orientation dependent. The detection range of a radiation with a MJ energy can be determined using
  • #1
ecastro
254
8
I am currently confused with the concept of the blackbody radiation and the inverse square law.

Planck's function for the radiation of a blackbody is in ##W sr^{-1} m^{-3} ##, is this somehow a form of intensity (because of the watts per square meter unit)? If it does, doesn't intensity decreases with distance, and that would mean that the radiation from a blackbody differs depending on the distance of the subject? If temperature will be calculated using Planck's function, does that mean that the temperature of the object is dependent on the distance to the object?

Please help understand these concepts. Thank you in advance.
 
Astronomy news on Phys.org
  • #2
It is also defined per unit solid angle, and this removes the dependence on the distance (see wikipedia for instance). I think that although the total intensity will decrease with the distance from the source, the temperature fingerprint is the shape of spectra not the intensity.
 
  • #3
soarce said:
It is also defined per unit solid angle, and this removes the dependence on the distance (see wikipedia for instance). I think that although the total intensity will decrease with the distance from the source, the temperature fingerprint is the shape of spectra not the intensity.

How about this equation?

##I = \sigma T^4##
 
  • #4
Ok, I misused the term "total intensity". I was thinking of the energy received by a detector calculated as the integral over the detector area, I wasn't referring to the total intensity of the source.
 
  • #5
I found out that the equation I just showed is luminosity. So, is the total intensity of the source is luminosity, or otherwise?
 
  • #6
I think it is luminosity, ##j(T) = \int d\nu d\Omega I(\nu,T)## and you are left with the energy radiated per unit time (power). Further you will probably need to integrate over the source area.
 
  • #7
Alright. Going back to my original question, does this mean that the value of the radiation from a blackbody is independent of distance? The spectra of the Sun is the same whether here on Earth or on Mars?
 
  • #8
ecastro said:
Planck's function for the radiation of a blackbody is in Wsr−1m−3W sr^{-1} m^{-3} , is this somehow a form of intensity (because of the watts per square meter unit)? If it does, doesn't intensity decreases with distance, and that would mean that the radiation from a blackbody differs depending on the distance of the subject?

That equation describes the spectral radiance of a surface. That is, the power emitted in watts per solid angle per wavelength per square meter.

ecastro said:
If temperature will be calculated using Planck's function, does that mean that the temperature of the object is dependent on the distance to the object?

It does not.

ecastro said:
I found out that the equation I just showed is luminosity. So, is the total intensity of the source is luminosity, or otherwise?

As best I can tell, luminosity is a term used in astronomy to refer to the total amount of energy emitted by an astronomical object like a star or planet. The equivalent SI unit is radiant flux. Intensity doesn't seem to be used by itself, but only in conjunction with another term, such as in radiant intensity.

See the full list of units here: https://en.wikipedia.org/wiki/Radiometry

Note the explanation of irradiance in the table: Radiant flux received by a surface per unit area. This is sometimes also confusingly called "intensity".
 
  • Like
Likes ecastro
  • #9
Thanks for the information. I will not be using the term intensity for awhile. Is there a relationship between the spectral radiance and the absolute square of an electric field (this is also sometimes referred as an intensity)?
 
  • #10
ecastro said:
Is there a relationship between the spectral radiance and the absolute square of an electric field (this is also sometimes referred as an intensity)?

That I can't answer. If you're referring to the amplitude of the EM waves emitted from the surface of the object, I'm sure there's a relationship, I just don't know what it is.
 
  • #11
A perfect blackbody radiator follows Lamberts cosine law so that the surface brightness is independent on distance or viewing angle.
In general the angular extent of the emissor is distance dependent so that the energy received becomes distance or orientation dependent
https://en.wikipedia.org/wiki/Lambertian_reflectance
 
  • #12
http://www.stsci.edu/jwst/instruments/nirspec/sensitivity/

Found that one about James Webb sensitivity, but i don't fully understand it.
What would be the detection range of a MJ radiation based on this data for example? What is R=100 or 1000? S/N?
 
Last edited:
  • #13
GTOM said:
What would be the detection range of a MJ radiation based on this data for example?

What's 'a MJ radiation'?

GTOM said:
What is R=100 or 1000?

It refers to the resolution of the prism or grating. See this link: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/gratres.html

GTOM said:
S/N?

I believe that refers to the signal to noise ratio. For example, the NIRSpec instrument page says this: R=100 mode shall reach a limiting continuum flux of 132 nJy at 3.0 µm at S/N=10 in t=10,000 s.

I think this means that in R=100 mode, which has less resolution but concentrates light better than the R=1000 mode, a flux of 132 nJy (nano Janskys) at a wavelength of 3 microns will generate a S/N ratio of 10 in 10,000 seconds worth of exposure time.
 
  • #14
Thanks. MJ radiation, i meant, a body that radiates a MJ energy in less than 10.000 second, so overall radiated energy is MJ.

Drakkith said:
What's 'a MJ radiation'?
It refers to the resolution of the prism or grating. See this link: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/gratres.html
I believe that refers to the signal to noise ratio. For example, the NIRSpec instrument page says this: R=100 mode shall reach a limiting continuum flux of 132 nJy at 3.0 µm at S/N=10 in t=10,000 s.

I think this means that in R=100 mode, which has less resolution but concentrates light better than the R=1000 mode, a flux of 132 nJy (nano Janskys) at a wavelength of 3 microns will generate a S/N ratio of 10 in 10,000 seconds worth of exposure time.
 
  • #15
GTOM said:
Thanks. MJ radiation, i meant, a body that radiates a MJ energy in less than 10.000 second, so overall radiated energy is MJ.

Oh, okay. The correct phrase should be "A MJ of energy", since a megajoule is a quantity. Like a "gallon of milk" or "meter of fabric".

GTOM said:
What would be the detection range of a MJ radiation based on this data for example?

I don't know, honestly. Probably not very far in astronomical distances. Certainly not outside of the solar system.
 

Related to Blackbody Radiation and the Inverse Square Law

1. What is blackbody radiation?

Blackbody radiation refers to the electromagnetic radiation emitted by an idealized object that absorbs all radiation incident upon it. This type of radiation is characterized by a continuous spectrum of wavelengths and is dependent on the object's temperature.

2. How does the inverse square law apply to blackbody radiation?

The inverse square law states that the intensity of radiation is inversely proportional to the square of the distance from the source. This means that the further away an object is from a blackbody radiator, the less intense the radiation will be.

3. What is the significance of the inverse square law in understanding blackbody radiation?

The inverse square law helps us understand how the intensity of blackbody radiation changes as distance from the source increases. This law is important in many areas of science, including astronomy, where it helps us understand the behavior of stars and other celestial bodies.

4. How is blackbody radiation related to the temperature of an object?

The temperature of an object is directly related to the intensity and spectrum of the blackbody radiation it emits. As the temperature increases, the peak wavelength of the radiation shifts to shorter wavelengths, and the intensity of the radiation also increases.

5. Can the inverse square law be applied to all types of radiation?

Yes, the inverse square law applies to all types of radiation, including light, heat, and sound. It is a fundamental law of physics that describes how the intensity of radiation decreases with distance from the source.

Similar threads

Replies
8
Views
3K
Replies
29
Views
2K
  • Astronomy and Astrophysics
Replies
3
Views
2K
  • Astronomy and Astrophysics
Replies
1
Views
1K
Replies
2
Views
1K
  • Mechanics
Replies
5
Views
1K
  • Astronomy and Astrophysics
Replies
6
Views
2K
Replies
1
Views
64
Replies
3
Views
860
  • Other Physics Topics
Replies
5
Views
2K
Back
Top