A qualitative question about Blackbody Radiation

In summary, the conversation discusses the concept of blackbody radiation and how it relates to the intensity versus wavelength curve. It explains that the intensity is proportional to the number of oscillators emitting radiation, which is in turn proportional to the probability of a mode being occupied, known as the Bose factor. The relation between photon energy, frequency, and wavelength is also discussed, and it is noted that the complete Planck blackbody function can be derived with further details. The conversation also mentions the density of oscillator modes and how it is proportional to frequency.
  • #1
Rishabh Narula
61
5
would this be a correct understanding
of blackckbody radiation phenomena?
in particular the intensity versus
wavelength curve?
"A Blackbody consists of oscillators of
molecular dimensions.

Intensity is proportional to number of oscillators
with sufficient energy hv
emitting radiation,
and that number of oscillators
is proportional to e^-(hv/kT)=1/e^hv/kT.
now if v(frequency) increases,1/e^hv/kT decreases,and
thus number of oscillators decreases.
Hence there are more oscillators emitting
radiation for lesser values of v,i.e larger
values of lambda(wavelength).Hence more
radiations(more intensity) of shorter
wavelengths are observed since more oscillators
emit these radiations.

And thus at a given temperature T,Intensity
increases with Wavelength in the curve.

If we increase temperature T,value of
fraction 1/e^hv/kT increases and thus consequently
number of oscillators increase.thus even if
wavelengths of radiations are low,
at higher temperature their intensities
will still be higher compared to at
lower temperatures.

Hence at higher T even shorter wavelengths correspond
to high Intensities in the curve.

This explains the curve till maxima,
After the maxima I think the Intensity
falls with increasing wavelength since
now the range of radiations are having too
high a wavelength,the energy for these radiations
is too less,there are not many oscillators
vibrating at this less a frequency...hence
the drop in intensity (bit unsure about this
last para do tell me if wrong)."

-Source,mostly me explaining
some university chemistry study material to
myself.
 
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  • #2
You sort of have a handle on it, but not completely. Detailed calculations show that the density of oscillator modes per frequency interval is proportional to ## \nu^2 ##. Meanwhile, the probability of a mode being occupied, [edit:correction=the mean occupancy of a given mode] is the Bose factor ## \frac{1}{(e^{h \nu /(kT)}-1)} ##.
(This is approximately the Boltzmann factor ## e^{-h \nu /(kT)} ## for small ## T ##. For large ## T ##, the Bose factor is approximately ## kT/(h \nu) ##).
In addition, the photon energy is ## h \nu ##. With a couple more details, the complete Planck blackbody function can be derived.
See also https://www.physicsforums.com/threa...-density-of-a-black-body.956343/#post-6063569
 
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  • #3
The most simple way to describe black-body radiation is to think about it in terms as a gas of photons, i.e., the electromagnetic radiation in a finite-volume box described with finite-temperature QED. This leads immediately to the Planck Law of the phase-space distribution:
$$f(\vec{q})=\frac{2}{(2 \pi \hbar)^3} \frac{1}{\exp(\beta c q)-1}, \quad q=|\vec{q}|.$$
The relation to photon energy, frequency and wavelength is [EDIT: corrected factors of ##c##; thanks to @Charles Link for pointing them out to me]
$$E=c q = \hbar \omega = 2 \pi \hbar c/\lambda.$$
Then you have to remember how distribution functions transform one one variable to another, i.e., you must keep ##\mathrm{d}^3 q f(\vec{q})## the same when changing the independent variable. E.g., you get the distribution with respect to energy or frequency using
$$\mathrm{d}^3 q=4 \pi q^2 \mathrm{d} q=\frac{4 \pi E^2}{c^3} \mathrm{d} E=\frac{4 \pi \hbar^3 \omega^2}{c^3} \mathrm{d} \omega,$$
and from this finally to wavelength
$$\mathrm{d} \omega=\frac{2 \pi c}{\lambda^2} \mathrm{d} \lambda \; \Rightarrow \; \mathrm{d}^3 q = \frac{32 \hbar^3 \pi^4}{\lambda^4} \mathrm{d} \lambda.$$
Multiplying these elements with the phase-space distribution factor ##2/(2 \pi \hbar)^3 f_{\text{B}}## gives the photon-number-phase-space density. Here
$$f_{\text{B}}=\frac{1}{\exp(\beta \hbar \omega)-1}=\frac{1}{\exp(\beta h \nu)-1}=\frac{1}{\exp(\beta c/\lambda)-1}=...$$
with ##\beta=1/(k_{\text{B}} T)##.
 
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  • #4
Charles Link said:
You sort of have a handle on it, but not completely. Detailed calculations show that the density of oscillator modes per frequency interval is proportional to ## \nu^2 ##. Meanwhile, the probability of a mode being occupied, [edit:correction=the mean occupancy of a given mode] is the Bose factor ## \frac{1}{(e^{h \nu /(kT)}-1)} ##.
(This is approximately the Boltzmann factor ## e^{-h \nu /(kT)} ## for small ## T ##. For large ## T ##, the Bose factor is approximately ## kT/(h \nu) ##).
In addition, the photon energy is ## h \nu ##. With a couple more details, the complete Planck blackbody function can be derived.
See also https://www.physicsforums.com/threa...-density-of-a-black-body.956343/#post-6063569
thanks for replying,will reply further when I know more. :3
 
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Related to A qualitative question about Blackbody Radiation

1. What is blackbody radiation?

Blackbody radiation is the electromagnetic radiation emitted by a perfect absorber and emitter of energy, known as a blackbody. It is a type of thermal radiation that is emitted by all objects with a temperature above absolute zero.

2. How does blackbody radiation relate to the color of an object?

The color of an object is determined by the wavelengths of light that it reflects or absorbs. A blackbody absorbs all wavelengths of light, so it appears black. As the temperature of a blackbody increases, the peak wavelength of its emitted radiation shifts towards the visible spectrum, causing it to appear red, orange, yellow, and eventually white as the temperature increases.

3. What is the Stefan-Boltzmann law and how is it related to blackbody radiation?

The Stefan-Boltzmann law states that the total radiant heat energy emitted by a blackbody is proportional to the fourth power of its absolute temperature. This means that as the temperature of a blackbody increases, the amount of energy it emits also increases significantly.

4. How is blackbody radiation used in astronomy?

Blackbody radiation is used in astronomy to study the temperatures of stars and other celestial bodies. By analyzing the wavelengths of radiation emitted by these objects, scientists can determine their temperatures and other properties.

5. What is the difference between blackbody radiation and thermal radiation?

Blackbody radiation is a type of thermal radiation, but not all thermal radiation is blackbody radiation. Thermal radiation is any form of electromagnetic radiation emitted by a body due to its temperature, while blackbody radiation specifically refers to the radiation emitted by a perfect absorber and emitter of energy, or a blackbody.

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