Blackbody Curve and Wien's Law Explained (UPDATED)

This is the UPDATED version of my video on Wien's Law. I discuss blackbodies in the context of stars and discuss Wien's Law. Apart form seeing me on the screen, I give much more detail on the actual formula.
(Original here - • Wien's Law (old version) )
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[@douglasstrother6584]

Planck's approach was to analyze the entropy of blackbody radiation as a function of energy. To make both high-frequency and low-frequency data consistent with the Second Law of Thermodynamics, he included an additional "guess" term proportional to the frequency (hf); this results in Planck's Law. Planck's application of Boltzmann's Statistical Mechanics to justify his guess then led to his revolutionary conclusion that the material of the walls emit and absorb radiation in discrete quanta. A paper titled "Planck’s Route to the Black Body Radiation Formula and Quantization" by Michael Fowler (7/25/08) gives a nice discussion.

[@CharlesLeTower]

I’m 36 yrs old, came here out of curiosity and I can’t believe they teach this in high school nowadays. Nice vid.

[@PhysicsHigh]

Thanks. Sure do. Modern physics and the lead up to, is an important topic nowadays.

[@577niccy]

i agree

[@rayirthsivakumar8202]

this was one of the best explanations i have ever heard. Thank you sir

[@tristanchristensen5963]

This was the best explanation of Wien's Law I've seen. Whatever that simulation was is really helpful

[@caklnl]

At the 5:24 mark there’s a conflation between spectral intensity (radiance) and the physicist’s definition of intensity being the power arriving at a unit area some distance from the source. The simulator is labeled “Intensity” which leads to the confusion. When you set the simulator to the Sun’s temperature of 5750 K (it’s actually 5770 but you have to go with the closest setting available) you get 6.20 x 10^7 W/m2. That’s exactly what you get if you divide the Sun’s luminosity by the area of the photosphere. So this “Intensity” is actually the Radiance or Spectral Intensity. Shouldn’t be referring to the Inverse Square Law because that applies to the physicist’s intensity. The blackbody curve should be the same for a 3000 K red dwarf and a 3000 K red giant. It is if we’re considering radiance (how much power it’s emitting through each unit area of its surface.). It would not be the same curve if we were measuring power arriving at a unit area some distance from the star. They would peak at same wavelength but the red giant’s curve would have much more area since it is far more luminous.

[@yugzed2233]

This is what I am confused with, too.

[@samh-l5711]

Thank you! Very well made and easy to understand. :)

[@PhysicsHigh]

Thankyou.

[@johnjeremias9437]

thanks, very cleay explanation, enjoyed your presentation

[@juliagli1973]

Thank you, very helpful for university

[@enochbrown8178]

Simply wonderful. Thank you!!!

[@alarialessard-northrup8834]

this was so helpful thank you

[@arghojitsaha3568]

Thank you sir...im a school student of class12,from India...this video was helpful!

[@4pharaoh]

Since the wavelength is the inverse of frequency times the speed of light isn’t another relationship simply: The peak temp T=6.6 *F where F is the Peak frequency in THz (THz =1e12 Hz)

[@manuellorena1322]

Excelent video! Thanks

[@ZiadHelou]

Finally understood..Thank you

[@kniridgaming1005]

loved the video

[@sollinw]

Very good!

[@soweitwegvonperfection7150]

Thank you Paul :)

[@PhysicsHigh]

You’re welcome. Glad you like it.

[@lilcookie7118]

I really wish you help me in this ! 2:03 I watched Planck's video but can't quite understand the concept. Why is the ultra violet intensity ( radiation ) very low and another question why is their a peak in the graph. Thank U a lot for all the hard work you put in !

[@shoutitallloud]

If you have to carry away some amount of energy, you need more low capacity carriers, but less high capacity carriers. UV waves are like high capacity, and IR are low capacity. And there's one rule that you can't use any type of carrier until it's fully filled. (that relates to quantum nature) So if you total energy is small (low temperature), you can't use half-filled high capacity UV waves. You would emmit energy in small portions with low capacity IR waves. But if temperature is high you would rather prefer to use high capacity UV waves.

[@lilcookie7118]

@@shoutitallloud So, the ultraviolet radiation has high capacity carriers that has more energy thus less intensity but the IR has low capacity carriers thus more carriers - radiation -, high intensity and less energy with each individual photon. Am I correct?

[@shoutitallloud]

@@lilcookie7118 If your reasoning comes to conlusion that I understand as "IR photons have low energy, and UV photons have high energy" - you are correct.

[@djtailik3761]

Why are you sitting in front of the Y-Axis?

[@z4c_7]

Y 0

[@billygraham5589]

A question: Is the total energy represented by the area under the intensity curve to the LEFT of the peak intensity point EQUAL TO the total energy to the right of it?

[@hollywoodbanayad7755]

Wein's constant is more roughly equal to 2.898e-3, but you still got all of your points across to me. Thank you!!

[@eatyourvegetables226]

What program are you using?

[@richardoberhammer1730]

In the first minute you say that a black body absorbs all radiation that falls on it and then re-emits electromagnetic radiation, and a star is a generator of light with no reflection going on. I don’t get the connection between the two. Why are you talking about a star as if it has a relation to a black body?

[@PhysicsHigh]

A star is the closest thing to a black body. True, it’s not absorbing energy externally but it is internally through fusion. It’s emission curve is a black body curve. Maybe if I say that a black body only radiates energy through emission and there’s no reflection.

[@fanR6580]

Ojalá alguien hiciera un video así en español :c

[@ajaykumar-im9hb]

Nobody explains why intensity is less in UV area. The lecture could be made more simple and to the point.

[@shoutitallloud]

If you have to carry away some amount of energy, you need more low capacity carriers, but less high capacity carriers. UV waves are like high capacity, and IR are low capacity. And there's one rule that you can't use any type of carrier until it's fully filled. (that relates to quantum nature) So if you total energy is small (low temperature), you can't use half-filled high capacity UV waves. You would emmit energy in small portions with low capacity IR waves. But if temperature is high you would rather prefer to use high capacity UV waves.

[@SUCKYPICK]

What program is this you are using?

[@SUCKYPICK]

Never mind, thanks google! The interactive scale he is using is here: phet.colorado.edu/sims/html/blackbody-spectrum/latest/blackbody-spectrum_en.html