Properties of Light: Blackbody Radiation
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- Опубликовано: 20 май 2012
- TRIGGER WARNING: In my youth and foolishness, I committed the grave and unforgivable sin of saying (shudder) "degrees Kelvin" rather than just "Kelvin", as a morally upright person would. Several people, paragons of trivial minutiae, have correctly pointed this out in the comments and reached the obvious conclusion that this error invalidates the other nine minutes and fifty seven seconds of this ten minute video. I, humbly and with a penitent heart, prostrate myself before these sages of technicalities, and, in wailing lamentations, beg forgiveness for my folly, which clearly made the entire rest of the video unintelligible. To others who would read this, be warned, it's just "Kelvin". To add "degrees" in front of that is madness.
A look at the property where, at high temperatures, objects will begin to glow and the spectrum of the light they emit will depend on temperature. This can be used to measure the temperature of distant objects. Part of an introductory series on astronomy
Dude you are a godsend man I'm in an astronomy class right now to get one last science credit before I graduate college and it has me stressing hard! But you explain this stuff very well and it has helped me put the pieces together for the next test. We're studying how stars can be studied and since they are so far away from us we have to use light to divulge every other piece of information and I couldn't really grasp that as well until I saw this vid man. Thank you again keep it up!!!
"Girl... you are so hot, that your radiation spectrum has its peak on a very short wavelength....::"
These videos are excellent. You explain concepts very clearly. Thank you!
I was having a difficult time understanding black body spectra until I watched this video. Thank you so much for the super lucid explanation!
This is the best video by far about black body radiation.
They are very excellent videos.But unfortunately the sound is so extremely low, that it is difficult to hear it.
Turn up the volume. Problem solved.
unfortunately my volume stops at 10
The sound's fine for me
awesome explanation, thank you.
Brilliant stuff, thank you soooo much!
Very clear explanation.
Great lesson
Great video. You explain very well sir.
Thank you for making this understandable.
Thank you !! I read a lot about blackbody but did not get it until I saw this video (y) :)
thank you, very nice explanation
It's not degrees Kelvin , it's just Kelvin ...
+Hypersonic Wrong, in Black bodies the unit Kelvin is not absolute, therefore Degrees Kelvin. Remember, whenever there is degrees involved it is because it is arbitrary (not of a proper system, one based of personal choice - randomness / because it is simpler to use).
GreenDye that's not true it's always just Kelvin , no such thing as degrees Kelvin.
@@SuperZombiekillar Yes, that's why it is just Kelvin because Kelvin is not arbitrary. What is your source of this statement? --> in Black bodies the unit Kelvin is not absolute
I don't understand why everyone must point out he is wrong in saying degrees kelvin. We know it's a measure of temperature it doesn't matter when your calculating things.
because people want to sound smart when commenting ;)
because anyone with any semblance of a scientific education knows that "degrees Kelvin" is an amateur mistake; it makes me wonder what else he is incorrect about in this video.
Thanks for your great work! It is said by some that bright stars in when viewed in space display no color except at the far ends of the spectrum.(no yellow or orange stars) Is this true?
Thank you! 🤘🏻
There is a new complete explanation of black body radiation on the QuantumBaking channel. It is organized as nine videos of five minutes each and actually derives Planck's formula. It assumes calculus, but computer animations let everyone follow the explanation.
Thank you
Most stars are emit fairly accurate black-body spectra based on their temp. There can be stronger absorption or emission lines based on their composition and the temp of their outer layers but for the most part I think that's it. I'm not sure about the "viewed in space" condition... from the ground you have to take into account absorption from the atmosphere which can change the spectra, and the interstellar medium can "redden" the star but it sounds like you're talking about something else
PhysicistMichael, it's not correct to say degrees Kelvin. ... Kelvin scale is just Kelvin.
That was supposed to be "the atoms will absorb the energy and they HEAT up ..." the light energy gets converted to thermal energy of the object "and eventually the object will start to GLOW" like a heated stove-top element. Sorry if I need to enunciate more.
How do infrared goggles help us see the color ? for instance, as you mentioned the temperature our body releases is very less which means a bigger wavelength ( outside the visible light spectrum) so on the basis of what do the goggles show the color red.
The images that you see through infrared goggles are false color images. Basically, the camera itself is sensitive to infrared light coming from different directions. The camera records that infrared image as a digital signal and will assign different colors to different parts of the infrared spectrum. So the camera detects longer wavelength infrared light, that might show up in the display as a darker blue or purple color. More of the near infrared light might show up in the display as a more yellow color. The brightness that is displayed usually corresponds to the intensity of the infrared light that the camera detected. So the colors that you see in the display don't really correspond to any real colors (again our eyes can't detect infrared light); instead these false colors simply represent parts of the spectrum that we can't see.
Thanks
I love you! Thank you
Great Video and great explanation! So I was wondering: If we heat up any given object to high enough temperatures, will it eventually start to glow green/blue or will it reach something like plasma status before it even gets there? (I hope this isn't a dumb question)
Very interesting!
thank u very much
Oh and maybe you can also explain why the sun is a blackbody (given it consists of mainly hydrogen and helium which give off light)? I don't quite see how these two elements can form a "black body".
+Sven Beckmann black body is an ideal case of an object that is a perfect emitter and absorber of radiation. It's like the Carnot cycle for heat engines. No heat engine can exceed the performance of the Carnot cycle since it's an ideal case with not losses to friction and such.
Sounds like melanin to me.
Black Holes? How close to ideal Absorber?
Pretty much perfect. Although in classical relativity black holes absorb everything that crosses their event horizon, introducing quantum mechanical properties into the theory seem to suggest that black holes do have a temperature and would emit black body radiation. However, for a black hole the mass of the sun, this temperature would be about 60 nanokelvin, so low that we have no chance of observing this from distant black holes. Another strange effect is that the lower the mass of a black hole, the higher it's effective temperature, so if a microscopic black hole could be produced in a particle accelerator, we may be able to see the higher radiation, but this is still way beyond current accelerator capabilities (and don't worry, this radiation would cause that tiny black hole to evaporate very quickly so it wouldn't eat up the Earth or anything like that)
Thanks, I appreciate that information :)
what is intensity on the Y axis?
You can basically think about it as how bright the light is at each of the different wavelengths. For example, if you were looking at pure red light then the wavelengths that correspond to red light would have a high brightness and all of the other wavelengths would have zero brightness. For white light all of the different wavelengths have a high brightness (white light is what you see when all of the colors of light mix together).
We can generate these graphs by sending light through a prism, so all of the individual colors are separated from each other, and then measure how bright each of the colors appears. These graphs can be very detailed and contain a huge amount of information about their sources.
but isn't brightness a relative term? are you saying UV light would have more brightness? it's confusing. Intensity isn't temperature?
also, I'm confused about the graph, how are you going to increase the wavelength on the x-axis? shouldn't you first increase the intensity and then the wavelength will increase? and shouldn't that mean the intensity should be on the x-axis? because it's Y(X), so x changes not Y.
First, for each one of these graphs we are looking at individual objects that are heated to a certain temperature, so we aren't really changing intensity or wavelength independently in any of these experiments. We take the light from an object that is heated to a certain temperature, send that light though a prism (or some similar device that separates our the different wavelengths of the light) and then simply measure the intensity (will define better below) of the light at each particular wavelength.
So for more detail, what is plotted on the y-axis in these graphs is usually light power per unit area per unit wavelength. Let's break this up piece by piece. Light power can be though of as light energy released per unit time, most commonly measured in watts (which is Joules of energy per second). This could be in the form of a few high energy UV photons, or a lot of lower energy infrared photons, but astronomers usually compare the overall energy emitted, so it is a fair comparison (and more easily measurable).
As I mentioned, we send the light from the heated object through a prism to break up the different wavelengths of that light. We can look specifically at tiny wavelength ranges and measure how much energy is carried by the photons of light in that wavelength range. So maybe we're looking at red light between 650 and 651 nm. We can then say in this range how much power is carried in this narrow wavelength range, hence, light power per unit wavelength.
But we also want to be careful that we're comparing the light from similar objects. If I was comparing the light from a golf ball sized object heated to a certain temperature, and a beach ball sized object headed to a different temperature, that wouldn't be fair because there is simply more of the second object giving off light. So we scale our light power measurement by the surface area of the object that is emitting the light. So overall what we measure is light power per unit area per unit wavelength, and we make this measurement for all of the different wavelengths of light.
Hope that clarifies things a bit. Sorry for the delay in my response.
What is a spectra?
"Spectra" is the plural of "spectrum". He made a couple of slips.
Very interesting video, however there is no such a thing as Kelvin degrees.
Also, the word “however” isn’t a conjunction.
Kelvin is not a degree
+Martin Kunev Wrong, in Black bodies the unit Kelvin is not absolute, therefore Degrees Kelvin. Remember, whenever there is degrees involved it is because it is arbitrary (not of a proper system, one based of personal choice - randomness / because it is simpler to use).
+SuperZombiekillar I can't find anything confirming your claim. Why wouldn't the unit Kelvin be absolute in black bodies? Do you have any sources explaining this?
Martin Kunev I am on a mobile device, may we speak over skype: daniel.smith6394
Shouldn't there be Green Stars? Brown, Red, Yellow, Blue, White.....No green.....Cosmic Segregration?
There are green stars. However, the peak is in the middle of the visible spectrum (being green), so almost all colours from red to violet are radiated. The combination of all these colours actually makes it appear white to our eyes, even though the peak is at green.
The minute I heard "degrees Kelvin", I was turned off. ;( A physics expert should learn audio normalization in video editing, as well... ;->
My Physics Professor doesn't like "degree" Kelvin. He just writes Kelvin.
My chem professor at UC Berkeley said the same thing. He said degrees should only be used when a measurement is made in respect to a non-zero standard.
That's correct. The standard is to not put that degree symbol (°K).
It looks like everyone noticed it. There are 50 posts about it.
degrees are not used in Kelvin dummy... that just makes this video less credible
Amazing, you used "zero degrees Kelvin." I shouldn't have to tell you that "degrees Kelvin" is not a thing anywhere. The correct unit is simply "Kelvin." This is how I know you're not a serious person. Just what university did you even graduate from?
I've updated the description for you and the others who have pointed this out over the last decade. Thank you for your service.