Black Body Radiation - Understanding the black body spectra using classical and quantum physics

[@heatszy2069]

Maybe your channel doesn't get a lot of views, but please know that your explanation is very useful for us.I'm glad I found yoyr videos in youtube and I'll be looking forward for your future videos.

[@PhysicsMadeEasy]

Thank you for your encouragement! very much appreciated! I am not really looking to become a KZbin superstar lol. My objective is to help students succeed, and inspire teachers. It's true that a little more exposure from KZbin would allow me to reach more students and teachers, but I refuse to have ads in my videos that disrupts the viewers focus. I think that non-monetized channels are not put forward because YT doesn't make money with these. Still I received an email recently from YT: They can force ads on non-monetized channels. If so, I will be forced to enter their program, so that I can have a little control over where the ads show up. If you see any ads on my videos, please let me know!

[@m.suriyanarayanan8157]

For 20 years I was learning Quantum mechanics from Dirac to Griffiths. In none of the books a so clear relationship between the variables was given. May the Lord shower light and prosperity on you and this Channel.

[@PhysicsMadeEasy]

wow, thank you very much Suriyana. Your words really make my day! And I am glad I was able to clarify these relationships for you.

[@ckgan8130]

Excellent explanation! I am preparing to teach this topic and I remember to come back to your channel as you always give the down to earth explanation! Keep up the outstanding works.

[@user-rp2yg7nc2x]

You are the best teacher in the world

[@prabhatsinha5462]

A very crisp and clear explanation by the Professor - even a school student can comprehend it easily. Thanks a ton professor - from India.

[@PhysicsMadeEasy]

Hello Prabbat. Thank you for your kind words. These video are for high school students, this is why I vulgarize a bit, but make sure to keep the information provided rigorous. I try also to make these videos so that they can easily understand the concepts presented, and enjoy the process of doing so! Final goal: when they review the notes they took during their formal school lesson, they see everything clearer!

[@yme5398]

I was so confused with this topic! You made it so much simpler! Thank you so muchhh and the insights about how the sky gets its color really broadened the spectrum of my thinking

[@PhysicsMadeEasy]

Broadening the spectrum of the viewers thinking, that's exactly what my channel tries to do (in regards to physics of course). Thank you so much for your feedback, it made my day :-)

[@douglasstrother6584]

"Planck’s Route to the Black Body Radiation Formula and Quantization" by Michael Fowler details Planck's thermodynamic analysis of the entropy of Blackbody Radiation, which motivated his hypothesis to satisfy Wien's Law at high frequencies. Planck's application of Boltzmann's Statistical Mechanics led to his conclusion that the material of the walls emit and absorb radiation in discrete quanta. It's a great read.

[@PhysicsMadeEasy]

Thank you for the suggestion Douglas.

[@douglasstrother6584]

@@PhysicsMadeEasy I found it captivating, and it's at a level one would see in an Upper Division Thermodynamics/Statistical Mechanics course.

[@alinemoreira5518]

Certainly, this is the best video I've watched about quantum physics. I simply cannot understand what all these equations mean and you've made it so easy. Thanks a lot!!

[@PhysicsMadeEasy]

Hello Aline, classical and quantum physics are models of reality that are not difficult to understand. It is just that it is often presented in a complicated way, which is in my opinion not necessary when learning the basics (as long as the presentation remains rigorous). Just check wikipedia, it looks so elitist sometimes! My goal is that once a student views one of my video, he gains an understanding of the fundamentals, that allows him to dig deeper and actually understand more detailed and complex presentations of the subject.

[@doc491]

@@PhysicsMadeEasy prof plss continue uploading the vdos they are very helpful

[@JetStream_RPL]

Thank you. i recently started to study quantum physics using a book called quantum physics for dummies where it teaches the basics of quantum physics. but without videos it's really hard to understand the concept. again thank you for creating this video. if you reading this have a great day

[@PhysicsMadeEasy]

Hi, well, you are on the right path for understanding the main aspects of QM. Having read a basic reference book and then watching videos within the context of your new acquired knowledge is a good strategy to improve. When you are ready, you should consider exploring other books too. Thank you for your kind words, and good luck with your rocket launching!

[@JetStream_RPL]

@@PhysicsMadeEasy thank you. Good luck on your Goals . Have a nice day

[@ekeys6897]

That was a very instructive little video, thanks for making it and sharing it with us!

[@LuLu-zx2hk]

one of my college courses is quantum mechanics and the professor explains it very poorly. thank you for your great explanations!

[@PhysicsMadeEasy]

Thank you Lulu. Yes, as Feynman says himself, nobody really understands Quantum Mechanics. So it can be difficult for teachers to explain it. I am glad my work here helped a little. PME, is more of a generalist physics channel for high school / early Uni students. There are other great youtube channel that go beyond High School lessons and dive more profoundly in QM. if you really want to dive in Quantum Mechanics, you should explore what's out there also!

[@busrayigit487]

I am studying my astrophysics exam right now with your videos thank you from Turkey 😊

[@PhysicsMadeEasy]

Hi, I am glad my work was able to help you with exam :-) Thanks your comment.

[@zhaadd]

omg this just made my life way easier dude thanks for making this

[@PhysicsMadeEasy]

Cheers Boinehelo, I am glad my work helped you!

[@War_mania]

There is always a huge difference between people who want to teach for passion and people who are teaching for money.... Sir you radiate energy like black body 😅

[@PhysicsMadeEasy]

Thank you so much for your words. It truly made my day! Yes, it is not with KZbin that I'll pay my bills, that's for sure haha. But as you said, I don't care about that: I consider my channel like my contribution and my gratitude for what the world is giving me. I try to give some back. And I am so glad that all the beautiful comments I receive show that I am succeeding. Be well !

[@sidhudafan785]

I got this topic for my presentation I didn't knew anything about this topic because in my previous class it was deleted today is my presentation I made my presentation yesterday and after 3 hours I'll be having my presentation..... You know best part after watching this video I've prepared notes with help of google also and I think I'll be having best presentation.... Literally these 2 videos are very much helpful thanks you sir for your easy and simple explanation 🖤

[@PhysicsMadeEasy]

Hello Sidhu, I am glad my videos are able to help you with your presentation. How did it go?

[@sidhudafan785]

@@PhysicsMadeEasy it was good but i forgot "thermal imaging" But I got 12 out of 15........thanku sir 🖤

[@alicehyeinyoo1837]

Excellent video! I hated physics at high school, because it was so complicated and difficult but you made it easy now! Thanks a bunch!!! Thank you for your work!

[@PhysicsMadeEasy]

Thank you Alice. You know , it's never to late to get interested! Dive a little, learn the basics, and the little maths that goes with it and that you can use as a language or a tool. Then, suddenly, exploring Physics becomes a little like reading a science fiction novel, just better, because weirder yet real!

[@adithyabharath6453]

you are awesome, please make more

[@PhysicsMadeEasy]

Thank you Adithya. This year I have been in a slower period in terms of video production: so many projects on my plate right now! But I will come back! In the meantime, check out the rest of the channel, there are tons of videos like this one.

[@ElLenadorLA]

Oh my god this is the first time I finally understood this thank you!!

[@PhysicsMadeEasy]

Thank you for your encouraging feedback Ellena!

[@fancynancy4560]

This really helped me to understand this topic more crystal clear ....thank u so much ..ur explanation is too good

[@PhysicsMadeEasy]

Hello Nancy, thank you very much for your feedback. I am glad my work helped your understanding of that subject.

[@odyssey9013]

Sir, you have explained such kind of topic in so easy language, thank you

[@PhysicsMadeEasy]

You are welcome Odyssey!

[@kathirvelp2319]

I just hit the like button within 52 sec. Cause you inspired me

[@PhysicsMadeEasy]

Thank you Kathirvel!

[@jima4799]

Amazing video you deserve more views!

[@PhysicsMadeEasy]

Thank you for your encouragement Ji!

[@adityabiswal5506]

Congrats! You got another subscriber. Edit: You EARNED another subscriber.

[@PhysicsMadeEasy]

Great, welcome to the channel! I hope you will enjoy your visit!

[@adityabiswal5506]

@@PhysicsMadeEasy I'm sure I will.

[@lorebloodlast5712]

please dont stop making us understand about how nature works!

[@PhysicsMadeEasy]

Thank you Lore!

[@ShopperPlug]

The sample formula you solving at the end helps explain the content. I was really confused of such complicated black body radiation. I also read that paint used in church glass also have some quantum physics since the pigments are super tiny.

[@PhysicsMadeEasy]

I am happy my work was able to lift your confusion about that topic. The pigments in church glass involve compounds with energy levels that give their colorations. So yes, pigments (not only those used in churches), but all pigments, including the natural ones required for life to exist like chlorophyll, are related to quantum mechanical processes :-). Let's feel gratitude for a universe rules by QM. Without it, we wouldn't be there!

[@pakhtoonyar8402]

I just got shocked when I saw it Amazing amazing

[@PhysicsMadeEasy]

Shocked? It's the harmonious beauty of nature that does that :-)! It did this for me, that's why now I love physics, and enjoy teaching it.

[@francheska6212]

I am so glad to have stumbled onto your channel! I am taking an atmospheric course and just simply wanted to find more information about Wein's Law and Stefan-Boltzman Law (we are currently on the topic of radiation). I found Wikipedia to be inaccessible as someone who is not familiar with Physics. Thank you for breaking this down so well!

[@PhysicsMadeEasy]

Yes Wikipedia often dives straight into the maths, while just skimming over the conceptual part. Scientific pages on wikipedia, while usually rigorous, are not always very appealing, especially for those not too fond of higher level mathematics. I believe that when a student understands conceptually the physical phenomena, the maths become much easier to grasp. Hence, my channel :-)

[@imenziani]

Thank you sooooooo much i am studying quantum chemistry and your explanation is 🌌AMAZING🌃

[@PhysicsMadeEasy]

Hello Imen, I am glad my video helped you understand one of the basis of this difficult topic!

[@user-iz2et4kj9y]

Your videos are really good and it helps me grasp some things even better. Thank you a lot !

[@PhysicsMadeEasy]

Thank you for your nice words. I am glad my videos are useful to you :-)

[@kgaugelolebelo1299]

This was really helpful!

[@PhysicsMadeEasy]

Thank you for your feedback. I am happy my work helps!

[@sharon69121]

Your videos are really amazing it literally makes physics easily understandable..

[@PhysicsMadeEasy]

Thank you so much for your encouragement Parmanand!

[@SaidAnwar-kp2lr]

This channel is very useful for us.

[@anantj18]

" I'm not Picasso, I'm a physics teacher. " A very cool physics teacher I must say.

[@finallyy3959]

as a physics teacher I know hot blooded animals will emits some black body radiation

[@comedy_with_aashka]

Your videos really help us for understanding in physics Thank you so much Sir For teaching

[@PhysicsMadeEasy]

Hi, you are welcome... but you know, I 🧡 teaching Physics!

[@comedy_with_aashka]

@@PhysicsMadeEasy That's why we love your physics teaching

[@fadhilshiddiq2231]

wow, at first I didn't see the amount of views quite surprised this amazing video have only 600 views

[@PhysicsMadeEasy]

Yes Fadhil, I always refused to put ads in my videos (I do not produce these videos for money, but by ideology, and I do not want students to get distracted). I learned recently that's why KZbin never really promoted my content. On the other hand, I realised a week ago, that they were putting ads on my videos anyway, without my consent! So I suppose I am forced to enter their Program...

[@saveearth9816]

So I think that to start with explanation of how they construct the curve will help of better understanding

[@fractalnomics]

With respect to the UV catastrophe, wasn't it that not all the energy from modes were not shared throughout the atom via equipartition . Didn't Bohr's and later Schrodinger quantum model of the atom have us rethink blackbody distribution completely? That atom absorb and emit at their emission spectral mode/lines. That these lines are predicted and observed via Frank-Hertz and Raman spectroscopy.

[@PhysicsMadeEasy]

Hello Blair, The oscillation modes that are solutions of the wave equation of an electron in a potential well such as in an atom, are not related to the radiation of a black body… Black body radiation does not originate from an emission/absorption process like in a Bohr model. It comes from the generation of EM waves based on the motion of charged particles (motion generated via heat). In an atom, the quantisation of atomic energies was needed to avoid everything ending up as neutrons (and explain emission/absorption spectras 😉). The UV catastrophe is related to the fact that a black body spectra calculated with classical physics would imply that everything with a temperature would fry everything around it. To avoid this, we use a similar idea but consider the full body instead of just an atom: I see it a little like like "statistical quantum physics". The charged particles of the black body can have only certain energies which are quantized by the plank constant. The number of levels is huge and the energy gap between them is so small that the spectra does appear continuous. But we know better… because we are not toast!!

[@fractalnomics]

@@PhysicsMadeEasy Thank you.

[@fractalnomics]

@@PhysicsMadeEasy But didn't the quantum revolution make BB curves redundant? We now explain radiation by spectra lines predicted by QM, this is Bohr. These lines are observed (in the infrared) by either Raman or "IR" spectrometers. Why are we still using this mid 19th century theory and technology? This was a path to a revolution: the old should die. It is totally redundant, and it is not true anymore. The likes of N2 has a mode in the infrared. It is predicted by the Schrodinger equation, observed by Raman laser spectrometers and with same instrument, so is the temperature N2 (via the Boltzmann constraint) measured. This same mode of N2 is central to the operation of the N2-O2 laser where it (the N2) is excited by electron discharge or IR photons (heat) at its single 2338cm-1 mode and from this CO2s 2349cm-1 mode is excited.

[@jenniferrussell5135]

This is so interesting and informative - using it for a college chemistry class!

[@PhysicsMadeEasy]

I am glad it helps Jennifer! Good luck with your class.

[@gowrissshanker9109]

Respected sir, silver is good reflector at room temperature (almost an white body), but when it is melted to liquid state it glows like an black body...what happens to it....does emissivity of body changes with respect to temperature? Can white body change into black body with change in temperature?

[@PhysicsMadeEasy]

Hi Gowri. This is quite a complex question and in order to answer it with precision, one must be a real specialist ion materials science, but I will try to give you a direction of reflection. When you heat a metal you increase the magnitude of the vibrational motion of its components (the lattice of ions and the surrounding electrons). As the reflection and black body emission of light is related to this motion, it is reasonable to assume that temperature will have an effect on emissivity. Now, in a liquid , it is the whole structural configuration of the metal that change, one can easily suspect that this will have a significant effect on optical properties! So, yes the emissivity will change too… Think about water. Shiny ice, then liquid water, than vapor… They do not have the same properties optical properties! I hope this helps!

[@gowrissshanker9109]

@@PhysicsMadeEasyIts helpful and thank you,sir

[@christophergame7977]

The critical feature of Planck's law is that the temperature appears in the exponent, multiplied by a constant. The exponent cannot be made to go to a zero limit for a finite temperature. That is the reason for Planck's quantal hypothesis.

[@fundacjawega]

A wonderful channel, a wonderful physics teacher!

[@PhysicsMadeEasy]

Thank you Fundacja ❤

[@adben]

Brilliant ❤

[@Sara_qanei]

amazing explanation i ever seen❤❤❤thanksss

[@duane56]

Looks like a typo at around the 7:30 mark. The argument of the exponent should be positive. Imagine what would happen to the Wien limit if it were negative. Another catastrophe! Also, "Plank" is usually spelled "Planck." Nice work overall. Much thanks for putting this together.

[@PhysicsMadeEasy]

Well spotted on both fronts! Yes, the exponent is of course positive (Typo), and Planck is indeed written with "ck" at the end. Thank you Duane .

[@saveearth9816]

... I had watched many videos about this phenomenon (BLACK BODY RADIATION) what was confusing me is how they construct the curve for each certain temperature.... Then I understand that they heat the matter for e. g to 3000 K the body will glow (emit visible light +IR&UV ray) then they will pass this light through glassy prism which will analyse it into the spectrum including visible colors +IR&UV... Then they measure the intensity of each part of the spectrum (starting from IR...then visible light colors ended by UV) Plot wave length (horizontal) & intensity longitudinal THANKS & SORRY FOR THE BAD ENGLISH I USED

[@PhysicsMadeEasy]

That's it, you perfectly understood the concept of black body spectra!

[@kumaramit1622]

Excellent explanation

[@viac381]

Totally underrated......

[@PhysicsMadeEasy]

Thank you Viac, The most important for me is not the fame haha, it is the fact that my work helps those in need find an understanding of basic concepts, thus helping them to progress with their studies!

[@AxelScheithauer]

You make the point that according to classical physics we should be careful not to heat a pan, because then we would get a large amount of ultraviolett radiation. Actually, wouldn't we get this ultraviolett radiation with every temperature greater than 0? Of course, we would get a little more when the pan is hot, since it is proportional to the temperature, but would that make such a difference in the temperature range of a pan?

[@PhysicsMadeEasy]

Hello Axel, you are correct, The UV catastrophe implies that the BB radiation diverges towards the shorter wavelength. Any temperature, even very low ones, would lead to the emission of high values of energetic EM radiation. But luckily for us, quantum physics saved the game :-).

[@DustinJamesCrawford]

very clear and helpful! Thank you, sir!

[@PhysicsMadeEasy]

Hi Dustin and thanks for the kind words :-)

[@MaheshNM-jh7yh]

Excellent explanation sir Please make video on Schrodinger wave equation it will be very helpful Your videos are very informative, always excites me to move forward

[@PhysicsMadeEasy]

Thank you Mahesh, it would be of a little higher level than suited for thsi channel, but good suggestion. I'll keep it in mind, in case I find a way to make it easy ;-)

[@MaheshNM-jh7yh]

@@PhysicsMadeEasy thank you so much sir I will wait for your video

[@swarupkumarsatpathy]

Hello Sir..I m from India...I request you to make a video on How does actually electron or any body absorb s or emits energy... Actually what happens when we say electron absorbs photon... Please

[@PhysicsMadeEasy]

When energy is provided to a particle, the energy of the particle will increase. Seems quite trivial haha. There are actually three ways this provided energy will increase the particle's energy: _ It can speed it up (increase of Kinetic Energy) _ It can change it's position relatively to other particles it is interacting with (increase of Potential Energy) _ if the particle is travelling at relativistic speeds when it absorbed the extra energy, the energy can convert to mass (the particle becomes heavier) What the extra energy becomes depends on the specific situation of the particle, and of course, it can be a combination of two or three ways.

[@Littleprinceleon]

@@PhysicsMadeEasy Rather concise, informative video, thanks. I have read (I think it was S. Weinberg's book for laypeople) a somewhat intuitive way (using logic and the definition of length) to describe how quantisation saves from "UV catastrophe": I only remember (maybe incorrectly) that for some reason into a given volume (explained via a cubical box) we could "pack" an infinite amount of waves if they can have an arbitrary wavelength. It's a shame that 4-5years ago I've spent some time (more than two hours) to digest the concept of why does the distribution of waves matter and now knowing at least the basics about quantum waves I can hardly recall any of that logic. Since then I also checked some stuff about the nature of infinity/ies: so it seems to me now that this UV catastrophe is more about the "absurdity" of infinity than about actual physics. That the Newtonian view handles space (and objects) as if composed of infinitely many point-like "stuff" but even a freely moving electron occupies some volume of the macroscopic space, isn't it? Also, I have heard that it can exist in 1D space (confined via some potential well...) Well that's way above my understanding, but gathering info like this may help at least to limit the misconceptions, hopefully. Can there be quantum waves which do not occupy any dimension of our regular space until they interact with something? How do they "move"? (And what is motion anyway 😉🙂?) That's the problem with persons like me, too much curiosity and not enough scientific intuition, not to talk about my mathematical "visualisation"....

[@georgemugenyi6284]

I love my professor

[@annitmul1614]

U so so simplified ❤❤❤

[@paulaborges7726]

AWESOME video

[@user-yn2yy9mq8t]

Keep up the good work! I like all your videos!

[@PhysicsMadeEasy]

Thank you Назар!

[@playityourself]

Great video - thank you. One small mistake is that you tell the black body changes through all the colors of the rainbow. It doesn't. It starts out dark red, then red, orange, yellow, white, blue.

[@PhysicsMadeEasy]

Actually, what does go through all the colors of the rainbow is the wavelength at which the maximum power is emitted. So it does go though all the colors of the rainbow. It's just that our eyes do not detect the same intensity for all colors. so around yellow-green, it appears white to us (like the sun seen from space looks white).

[@arunms8696]

Thank you ❤ SIR

[@dimit73]

very nice video. How the Planck ypothesis explains the Ultraviolet Catastrophe ? I mean why the fact that emitted radiation has energy E= nhf, leads us to predict the actual distribution of the black body radiation at all wavelengths? why most of the emitted energy is located at a very small area of the spectrum? I understand that his equation agrees with the experimental results but how would you explain to a student that without too much mathematics?

[@PhysicsMadeEasy]

Hi Jim, very difficult question, and I am afraid I have to answer: I don’t know… Without using the abstraction of Maths, I am not sure I can explain it in a easy to understand, yet rigorous, physical sense (like I try to do on most of my videos). Originally, Rayleigh’s formula is based on a Newtonian model where particles are bound like little springs, for which the oscillations can be continuous in frequency. If one starts to develop models from that starting point, you end up with the UV catastrophy. The fact that light has to be considered like quantas of energy changes the starting point of the model (the "little springs" can only vibrate according to certain modes now). Developping this allows this exponential to show up and cool things down at shorter wavelengths… Why is that so physically? and why does it lead to the distribution as it is? well, that question could be analogue to why the universe is the way it is… ;-)

[@dimit73]

@@PhysicsMadeEasy Ι understand that when Rayleigh - Jeans use integral, due to his ypothesis Planck uses sum and this leads him to the right equation. But why does UV catastrophe happens?

[@melaniefranklin7607]

@@PhysicsMadeEasy This was a good question by the person, because I had the same question, as when one delves into physics, it seems there is alot to take for granted previously OR the people who originally made these equations had a line of reasoning which we have forgotten, making it difficult to explain. I am inspired to answer this question now, because the explanation does not feel complete about spectrography as a whole unless you understand what Newton and others had in mind for the basis of these equations... sounds like a hypothesis based on some metaphysical or more abstract/ideal version, which was then turned into the equations. I am a bigger picture thinker, who sees most of the scientists doing technical work with tools handed down, but without the comprehensive picture beneath it. Thankyou for the great channel, excellent starting point.

[@gossedejong9248]

brilliant, thank you!!!

[@PhysicsMadeEasy]

You are welcome, I am glad you enjoyed the video :-)

[@ryanthegooch5952]

Amazing

[@mariashahzadi6897]

Best explanation ever♥️🤩

[@PhysicsMadeEasy]

Thank you Maria!

[@dr.klausschustereder5975]

Excellent. Thank you.

[@kottapallipalashanjan2370]

You the best bro

[@AAAli-nv5tv]

Thank you for excellent videos. Informative and very well explained. This helps a lot.

[@PhysicsMadeEasy]

Thank you Ali :-)

[@simongross3122]

Thanks for the video. You're the first I've found to explain that those distributions are not continuous. Does this mean that there are some frequencies of light that are actually impossible? Or does it simply mean that there are frequencies that are impossible for a given black body?

[@PhysicsMadeEasy]

Hi Simon, in absolute, by approaching its description quantum theory, a black body spectra is discrete. Otherwise we arrive to unreasonable results (UV catastrophe) But a black body is an object with trillions of trillions of particles, and thus the possible energy states that can take these particles becomes an exponential of that number. That leads to a quantization that is ridiculously small, so small that it is reasonable to consider the spectra continuous Analogy: it is reasonable to consider space continuous in general relativity and Quantum Mechanics in typical conditions. But in extreme cases of very high energy density, space needs to become discrete too to avoid the occurrence of infinite values (that approach is called Quantum Loop Gravity. FYI, QLG attempts to derive a theory of quantum gravity by quantizing space).

[@simongross3122]

@@PhysicsMadeEasy Thank you again for a great clarification. I think that I need to go and read up about energy states and how they are attained before I can fully understand this. I had thought that in the context of an atom, electrons could only have certain energy states, and since there are only a finite number of elements each with a finite number of possible energy states for electrons, that it then follows that the number of possible energy states in total is not trillions, but something much smaller. But of course, perhaps energy states can pertain to situations other than electrons in the context of atoms. I need to learn more about that. Thank you again. I really appreciate you taking the time.

[@layla.2261]

حبيت شرحك و طاقتك😍😍.

[@PhysicsMadeEasy]

شكرا جزيلا :-)

[@layla.2261]

@@PhysicsMadeEasy على الرُحب.

[@RomanXIl]

Sir,black body does not related to the colour of human being? Means he or she is fair or not. Is body is something which is touched no na? This is what the beauty of physics which doesn't means to be physical. Thank you sir.

[@PhysicsMadeEasy]

What is the color of a human being seen by a snake: InfraRed! Yes, you and I we glow in Infrared because we have a body temperature of 37degrees and are good black bodies. You and I cannot see infrared, so the color we see are just the reflection of whatever light is shining on us... This is not our real color. All humans and animals are Infrared. Fun fact: cats are of a sligthly brighter infrared than us, because they body temperature is at 38.5C

[@LittleWaffle]

Thanks !!!

[@billygraham5589]

I thought that the sun only emitted (mostly) just a few discreet bands of light? Those of the hydrogen spectrum. So why do all the colors of the rainbow show up when we filter sunlight through a prism?

[@PhysicsMadeEasy]

Hi Billy, The sun is a black body, so it emits a black body spectra that covers fully a range of wavelength, not just discrete bands. What you are mentioning is the atomic spectra of hydrogen. The Sun's atmosphere, has a similar composition as that of the sun itself. It is mainly made of hydrogen. When the black body light of the sun's surface passes through its atmosphere, some of it is absorbed by the hydrogen of the atmosphere. So if you take a spectra of the sun from space, you will obtain a black body spectra with absorption lines corresponding to hydrogen (and whatever other elements present in the sun's atmosphere). This is how we are able to determine the composition of other stars (of their atmospheres to be more precise). Check this spectra of the sun: i.ytimg.com/vi/PKk3sSZOcH8/maxresdefault.jpg I hope this helps!

[@hamdaniyusuf_dani]

Black bodies never look green. If temperature is increased, after yellow they will look white, and then bluish white. That's because the intensity of red color doesn't decrease by heating.

[@PhysicsMadeEasy]

The intensity of the red color does decrease (relatively to the others), because of Wien's law... The sun should look greeny yellow, but you are right, we see it white. That's because at its temperature, all visible light frequencies are very intense, and their combination lead to white light for the human eye's perception... With stars at very high surface temperatures, the relative intensity of the blue becomes significantly higher than the others (Again WIen's Law), that is why we see these tend towards the blue.

[@AxelScheithauer]

@@PhysicsMadeEasy But it never becomes green or purple, as Yusuf has pointed out. The green and blue parts of the spectrum might become relatively stronger, as you say, but mixed with the still existing and also increasing red the emitted light is perceived as bluish white. If the temperature rises to move the maximum well into the ultraviolett it will still emit the visible spectrum and therefore you ball will never start looking black. There is an excellent video that explains why there are no purple stars: kzbin.info/www/bejne/bYOzdmCVmqt1hdk

[@asmaakadry6535]

Thank you so much💖💖💖

[@PhysicsMadeEasy]

You are welcome Asmaa

[@franckschwartz1945]

Many thanks for the quality of your videos. I have a question related to the gas. If we look through an infrared camera an hair dryer, we can see the hair dryer but not the hot air coming out from it. This mean that a gas (heaven hot) is not behaving like a blackbody and therefore does not emit infrared radiation. Am I right ?

[@PhysicsMadeEasy]

Hi Frank, I had to check this out on youtube, and yes, the IR camera doesn’t pick up the hot air coming out of the airdryer. I have to admit that at first, this confused me… because gas, like any matter has black body properties (for example, we detect cold and hot gas in space by its black body radiation). After reflecting on this, I realised the answer to why is pretty straightforward. It is all a question of intensity. The density of the air is way less than that of a solid object, therefore the signal emitted will just be much fainter, too faint for the camera to pick up. You would need a kind of coronograph to mask the denser objects and the airdryer itself, and a very sensitive camera, and you should see a light haze of IR where the air is hotter…

[@87711jimmy]

Thanks for explanation, it's better than wikipedia lol.

[@PhysicsMadeEasy]

Better than Wikipedia! Wow ! :-)

[@christophegermain115]

Thanks for your videos, very clear to understand; do you know how to understand our planet’s emissivity or is it correct to assume earth as a black body?

[@PhysicsMadeEasy]

Hi Christophe, the emissivity of a planet depends on many factors as for example its atmosphere or the materials at its surface. For the Earth it is around 0.8-0.9, so not a grey body, but not too far from a black body either. For the moon it's around 0.95-0.98. For info, deserts are around 0.6-0.88, and ice, water or vegetation around 0.95. You can see this on the map of emissivity by NASA: www.jpl.nasa.gov/images/pia18833-nasa-spacecraft-maps-earths-global-emissivity

[@user-dx2ih3dc4q]

Nice video. I'm not sure why you chose to put shiny spots on the 'blackbody', but overall, nicely done.

[@PhysicsMadeEasy]

haha, it's the esthetics of the cool thumbnail that imposed it. I actually don't even remember how I made this image! Note that if there are shiny spots, it just means that these are hotter (the body is maybe not homogeneous - thermal conductivity is not constant everywhere on its surface)

[@maitland1007]

@@PhysicsMadeEasy would hotter areas make shiny spots, or would they make spots that are reddish?

[@PhysicsMadeEasy]

@@maitland1007 Yes hotter area would make the spots brighter (P = Sigma x A x T^4). The spots would also see their wavelength decrease (Wiens Law LambdaMax = 0.002898/T). In other words the color would shift towards the blue side of a rainbow, for example, a spot already red would become orange or yellow when T increases.

[@elenileivaditi3771]

it is very helpful! Thank you!!

[@ts37924]

Thanks a ton sir ! But could you explain why emmision of light waves would result in continuous decrease energy where as emmision of photons is resulting in discrete decrease in energy. (I know the explanation for photons and discrete energy but can't understand why if em waves were used the energy decrease would be continuous)

[@PhysicsMadeEasy]

Simply because EM waves are a classical concept. In classical physics, quantities likes energy, momentum etc... can take any value, which is not the case in quantum mechanics.

[@zendude]

Nice chanel , subscribed already.

[@PhysicsMadeEasy]

Thank you Farouk, I am glad you find it useful!

[@meetpatil4240]

Thanku.

[@LaxmiNarayan-st1nd]

One of best acting in the world

[@PhysicsMadeEasy]

?

[@tribudylestari6995]

May i know which book or journal source you used to say that the color of the sun is actually white? Because im very interested in this discussion and i will make a thesis on this matter

[@PhysicsMadeEasy]

No source, just pure reflection: Our eyesight has developed so to have a maximum sensitivity for a range of wavelength where the sun emits most power. This range is the visible range. The peak of the black body diagram encompasses all visible colors, greeny-yellow should be the max, but just by a tiny tiny anount. Overall, we perceive all colors with approximately the same intensity, thus white.

[@karnati.vijayalakshmi7828]

Sir waiting for more videos

[@PhysicsMadeEasy]

I know, many are... but I have got so much on my plate these days, and all the steps of producing these videos is very time consuming for one man... On the other hand, it gives me time for great ideas of new videos! So when, time allows me a little freedom, you can expect some cool stuff being posted!

[@NaveenKumar-vx7yt]

Very interesting

[@aashishlearningbeastkarki9552]

You are the best.

[@fireflywaterfall4532]

God bless you!

[@magdishan8729]

thank you sir , really helped me a lot

[@PhysicsMadeEasy]

You are welcome Magdishan, I am glad it did :-)

[@zimufan8550]

GOAT

[@user-rx2jt2bv5m]

Hi thanks for the video! Please let us know if there is a mistake in the Planks law formula - a minus sign in the power of (e) as it is in the denominator already

[@PhysicsMadeEasy]

Absolutely Сергей, there is a typo in the video. There shouldn't be a minus sign in the exponential term when on the denominator... I added a "Typo Alert" in the description of the video.

[@thapelogerry6980]

really made easy the ultraviolet catastrophe

[@PhysicsMadeEasy]

Well, I do not mention the UVC in the video... but it is indeed interesting to mention. Information for viewers, the UVC is when you try to describe the behavior of a BB using only classical physics. Any BB with temperature would be emitting deadly UV, X and gamma rays. Obviously (and thanksfully) this is not the case haha! Plank in the early 1900 modified the classical description using a quantum mechanical idea (light is made of quanta of energy, i.e. photons). The prediction this implies matchs reality: no more UVC.

[@kritanadhikari503]

awesome

[@physicsaddiction9784]

hello, sir please clear my doubt sir i have an question that why the black body/object absorbs all the wavelength of the light. another question when the light fall on an apple they emit the red colour of wavelength and absorbs all other wavelength of light what does it happen in the quantum level please sir clear my doubt

[@PhysicsMadeEasy]

Hi (nice nickname haha!) "why the black body/object absorbs all the wavelength of the light.": A black body is an idea... No object is a perfect black body. The closest we have a balls of plasma (stars) which have an emissivity close to 1. In that latter case, where charged particles are free to move around, it is easy for light energy (an oscillating electric field) to be converted to kinetic energy (motion of particles). Still, this classical representation leads to problems which requires a quantum physics approach to get solved. " when the light fall on an apple...": in the skin of the apple, there are pigments. These pigments are molecules that absorb certain colors. In the case of the apple, these colors are Blue and Green which correspond to the absorption by the pigments of photons of a certain energy. The remaining light is reflected (Red). I hope this helps!

[@sumitbhadola]

I understand thank you india loves you

[@PhysicsMadeEasy]

Hey Sumit. And I love India :-)!

[@mcwulf25]

How come the blue sunlight travels further than the yellow at sunset?

[@PhysicsMadeEasy]

At sunset, light has to travel more to reach the ground omapred to midday, (has more distance to cover through the atmosphere). Shorter the wavelength of the visible light, more it interacts with the air and gets scatterered. So after a shorter distance, blue gets scattered, then at a little longer distance, green, then Yellow then Orange etc... So in the end blue travels less (in a straight line) than yellow...

[@mcwulf25]

@@PhysicsMadeEasy I see what you mean. Thanks.

[@samahashah9826]

How do we see a black body if it doesn't reflect?

[@PhysicsMadeEasy]

Because it also emits light... It absorbs all light thrown at it, and therefore gains internal energy: it's temperature increases. And then, because it has temperature, it emits energy (under the form of light). That's why you can see it.

[@jossey-yn3zo]

Wow

[@danieldanieldadada]

4:22 why are shorter wavelengths scattered?

[@PhysicsMadeEasy]

Diffraction is the bending of light due to an obstacle. when light is diffracted, it results in the of the light rays. This phenomena is more effective for wavelengths in the range of the obstacle or slit. . Visible light has wavelength in the range of 400-700 nanometer. molecules in the air are in the range of the of the half a nanometers. roughly a factor 1000. So the scattering effect is quite weak, but not negligible. example: if you look through a pinhole of 0.1mm (=100000nm), with a good eyesight, you can see on the sides of the hole some deformation due to the diffraction of visible light (500 nm). This effect will be more pronounced for shorter wavelength because these are closer to the size of the molecules in the air. It occurs also for red and orange but less than for blue (you can see the orange/red scattering at sunset)

[@danieldanieldadada]

@@PhysicsMadeEasy Awesome! Thank you for your answer! I remember learning about diffraction in my second semester of Physics in high school! We picked Optics as the area of focus back then.

[@namratasrinivasa7568]

How does the weins distribution law come into picture? (E*d(lambda) = A*(lambda)^-5*e^(-B/(lambda*T))d(lambda))

[@PhysicsMadeEasy]

If you take Plank's law and consider only shorter wavelengths, you realize that they end up being the same... In other words, Plank's law for short wavelength (higher energy) reduces to Wiens Law. . In the video I go very fast on this just to give a starting point (around 7:40). if you want a more detailed proof, you can find it by searching a little: for example: edurev.in/t/98345/Wein%E2%80%99s-Distribution-Law---Rayleigh-Jeans-Law-and-P

[@namratasrinivasa7568]

@@PhysicsMadeEasy thank you so much!!

[@mohamedAyman_2003]

genius

[@adammontgomery7980]

Does anyone think that perhaps the sun's spectral peak at green has anything to do with the evolution of chlorophyll. My mind is blown sitting here thinking nature built a tuned resonator to maximize power transfer from the sun.

[@PhysicsMadeEasy]

Absolutely Adam! The chlorophyll pigment was chosen by nature (probably by natural selection when life started) because it must have been the most effective in converting our sun's light energy in regards to the sun's spectra. Around another star, with a black body spectra shifted compared to the sun, the pigment chosen by nature would probably be different. Note though that Green is not the color absorbed by chlorophyll... It looks green because it is reflected. Chlorophyll absorbs Red and Blue. These two combined must harvest more energy than just green, even if green is the wavelength of largest intensity.

[@VarunSaiHere]

GITAM❤

[@PhysicsMadeEasy]

That was attempts to speaking French haha! Thank you for your affection :-)!