Black body radiation

  • Thread starter oheaveno
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Main Question or Discussion Point

can i ask ..... how could planck's idea of quantised energy explain the ultraviolet catastrophe?
WHAT is being quantised? the oscillator atoms of the blackbody cavity? or???

thanks
 

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  • #2
dextercioby
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The energy of those oscillators.

It yielded a spectral distribution which didn't "blow up" for any portion of the em. spectrum.

Daniel.
 
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thanks for replying:D but what do you mean by "didnt blow up for any portion of the em spectrum"??
you know that graph of wavelength VS intensity.... i understand that the higher the temperature the peak of the graph shifts towards the shorter wavelength because there is more energy (tell me if im wrong) but why is the distribution of wavelength the way it is? why cant there be more longer wavelengths with higher intensity?

one website explains this idea by saying this:: say now you have a certain amount of money. you can spend it on one expensive stuff, or several middle priced stuff, or a lot of cheap stuff. you are still spending that definite amount of money but it is just HOW you decide to spend it. so in the case of BBR, there is a lot of middle wavelengths, a few short wavelengths and a few long wavelengths.

that is what i don't understand. why are there a lot of middle wavelengths but not more shorter wavelengths? i can still distribute the same amount of energy but just in a different way.
i hope you get what i mean because i think my understanding of the concept is very bad
 
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dextercioby
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oheaveno said:
thanks for replying:D but what do you mean by "didnt blow up for any portion of the em spectrum"??
Planck's distribution is bounded (for any temperature) and moreover the area under its graph is finite.

oheaveno said:
you know that graph of wavelength VS intensity.... i understand that the higher the temperature the peak of the graph shifts towards the shorter wavelength because there is more energy (tell me if im wrong) but why is the distribution of wavelength the way it is? why cant there be more longer wavelengths with higher intensity?
Thta's tipically for Bose-Einstein statistics. If you study both mathematical statistics and quantum statistical physics, everything will be clear.

Daniel.
 
  • #5
As the example you posted, the main point of it is that if you are going to buy cheap stuffs, you'll have more choices(more combinations) to distribute your finite money, while for the expensive ones, you'll have much less choices. More choices means larger probability. Thus higher intensity.
 
  • #6
reilly
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Since this is a very standard part of physics, details can be found in virtually any freshman physics book.
Regards,
Reilly Atkinson
 

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