Just a couple corrections: the fact that Rayleigh scattering is polarized has nothing to do with the shape of the nitrogen and oxygen molecules. Even for spherical molecules and atoms such as methane or argon one observes the same polarization properties of the scattered light which is due to the angular distribution of an electric dipole emitting light and the fact that light is a transverse electromagnetic wave. The electric field of the light shakes the electrons of the molecules which then radiate light with an angular distribution of an electric dipole. Also, the atoms in the molecules don’t vibrate during Rayleigh scattering, it’s just the electrons which are moving in response to the electric field of the light.
A great teacher teaching a complex subject in English WITHOUT a thick accent. This is where I would put a lot of value this day in the age of information.
I am sure you are a polymath, you can teach many topic from totally different industry with huge understand about the course and taught in the way that easily understand by beginner like us and only use a marker pen without any fancy editing ,massive respect to you by using a single marker pen and white board to change the world. A massive respect of applied physic student form Malaysia.
@@MichelvanBiezen because i knew you since 3 years ago about assignment of L'hopital rule, (that time not even know how to pronounce L'hopital rule). then , I am really surprise and astound that you also teach particle physic , a totally different field ,really respect to you
+Michel van Biezen I get that the purple is 400nm where as red is 700nm lenght. And so blue has longer wave length than purple but less that red. My question is, and what I don't understand for some reason at all, is why doesn't that make the sky purple, since if I understand this correctly, purple should scatter even more than blue?
+Ellis D In astronomy we don't talk about purple light. We lump the blue and purple together and just call it blue. That stems from the fact that we don't see purple stars and as you mentioned we don't see a purple sky. The wavelength of purple is a bit shorter than the wavelength of blue but again we lump those colors together.
Dear Biezen, I have a question around 3:17. Why will the molecules absorb the energy and then re-emit the energy "without loss" ? The point I couldn't understand is the "without loss" statement. May you please explain about it? Thank you very much!
In the macro world, energy is lost during almost any physical interaction. In the micro world, everything happens in quantum steps and can only happen in quantum steps and consequently energy is transferred in quantum step.
Why then do we see much more orange/red light during sunrise and sunset time periods? It cannot all be due to the N factor increasing the number of molecules along the light's path, can it? If so, then that would be a very big coincidence that our atmosphere is exactly enough to scatter mostly blue when sunlight shines through it vertically but exactly enough to scatter mostly red when sunlight shines through it largely horizontally.
Michel van Biezen But then why do the red(dish) colors overwhelm the blue(ish) ones, which presumably are also still being scattered? And why do they not overwhelm them when shining through a lesser distance?
+stellarfirefly It is a matter of numbers. When the rays come from a high angle, many more of the blue photons scatter compared to the red ones. But when traveling from a smaller angle (greater distance through the atmosphere) there are more of the yellow, orange, and red photons scattered.
This video explains the right answer. ua-cam.com/video/0b1fqodmZJ0/v-deo.htmlm5s Basically most of the blue light has scattered because it scatters more! None reaches us. Only the red light which has scattered less reaches us.
When a molecule vibrates, is not the resultant wave a spherical wave front regardless of the orientation of the molecule? Oh. I see. The resultant wave front has an amplitude proportional to the angle of incidence. A resultant wave front is weaker in one orientation versus another. Thanks. I’ll look around, but I’ve been trying to find an explanation for why reflected light is polarized parallel to the reflective surface while refracted light is polarized perpendicular to the surface. I am surmising that it has something to do with the freedom of movement for valence electrons as well as the freedom of movement for the nuclei of a crystal’s constituent atoms. Metals having a valence band that is loosely bound to individual atoms creates a kind of sea that has more freedom of movement across the metal’s surface as opposed to perpendicular to it. Whereas a transparent crystal is rigidly locked into its lattice restricting vibrations to along those bonds? Or is the valence band electrons playing more of a role for wave propagation through a crystal more so than the nuclei? Thanks for your time.
Keep in mind that the emissions of an atom or a molecule, moving by itself in the air, is very different than the emission of an atom or molecule locked in a solid (which is more like black body radiation).
@@MichelvanBiezen I did. That’s why I specifically mentioned the oscillations induced in valence band electrons which are weakly coupled to the material they are bound to. For metals, it is a very weak coupling allowing for far greater amplitudes of induced oscillations. I am just a bit fuzzy on what is oscillating in a crystal with more strongly bound valence electrons. Is it still valence electrons that are oscillating within a transparent crystalline lattice, or are individual atoms oscillating? Or is it both depending on the angle of incidence and wavelength of the incident light which would account for reflection at a particular angle instead of refraction as utilized by fiber optics?
Wait. Maybe I missed a lecture or a lab at some point, but I believe that nitrogen is a linear molecule with 2 atoms so it has 3(n) - 5 or 3(2) - 5 = 1 normal vibrational mode. For me, vibrating in that other axis just doesn't compute. Water (vapor) is a non-linear molecule with 3 atoms so it follows 3(n) - 6 or 3(3) - 6 or 3 normal vibrational modes. Like nitrogen, carbon dioxide is also a linear molecule but with 3 atoms so it has 3(n) - 5 or 3(3) - 5 = 4 normal vibrational modes, one of which is doubly degenerate. Correct me if I'm wrong.
+Michel Van Biezen, I get it why oxygen and nitrogen absorb the energy. But why do oxygen and nitrogen molecules radiate back the energy? What is the trigger?
So does a single ray of light get split into multiple rays of light after it is absorbed and re-released from the electron? If so, does that mean subsequent ray of light is weaker or less energetic than the parent ray of light? Your illustration seems to imply this.
+SogMosee If the photon is scattered it is not absorbed. It is as if the photon is deflected. The illustration gives you the probability of which direction it will be scattered. It will be more likely scattered in one direction over another direction. That is the nature of Rayleigh scattering.
thanks for a perfect explanation of this topic one question left, so in principle, if the molucule just absorbs the photons energy and then emmits it back, the scattering angle is absolute random? or I understand it completely wrong?
@@MichelvanBiezen then one more question concerning scattering. We have a rubidium atom cloud which is cooled down (Kinetic energy close to zero) by a laser and caught in a magnetic-optical trap. Now the Idea is to detect the movement of the cold atom cloud along the trap based on the scattering of laser, if I understand it right, the Rayleigh scattering should be dominant in this case, still the laser light has to interact not only with atoms, but with electrons, meaning that Compton scattering will take place too, but can't it's influence be negliable?
The wavelengths that are scattered less are absorbed by the molecule ? can we explain rayleigh scattering with quantum mechanics without em waves but photons ?
@@MichelvanBiezen Evidently, Martian sunsets are blue b/c the atmosphere is thin (letting the blue through) and fine dust scatters the red (according to a Bing science quiz)
But scattering means changes in frequency lenght or amplitude of the wave or it has something to do with photon particle i. e. it is phisical collision? Best greatings and thank You.
You may be referring to the Compton Effect. But changes in the frequency of the photon due to a collisions causing it to scatter is only detectable when the original frequency of the photon is very high. With light photons that is not the case.
It is just a little bit confusing because we have light as a photon and wave, what is more the electron is a cloud of probability or part of matter and the interaction is based on molecule movement and jupms of electrons between shels. Which combination is right for light scattering?
Michel van Biezen sir but in my text book its given that " the angle between incident light and direction in which the intensity of scattered light is observed is called angle of scattering it is experimentally observed that intensity of scattered light varies with angle of scattering. The intensity is maximum at 90 degrees angle of scattering" is it correct
Note the equation of Rayleigh scattering and the dependency of the scattered intensity and the angle. There are different scattering principles. This equation is specifically for Rayleigh scattering.
@@MichelvanBiezen yeah! The higher the frequency the more they're affected right! Ive been philosophising about to what extent Rayleigh Scattering could cause red-shifting of em waves in the intergalactic medium. The Density and distance are of the same power. Density is very low but distances are also very large. With the discovery of necessary dark matter maybe the intergalactic medium is more dense than earlier presumed and perhaps could explain some portion of red-shifting in distant galaxies. If my math is correct if scattering is a factor it will also have an Indirect relationship with the expansion of the universe, seemingly multiplicative which could be misinterpreted as an accelerating expansion rate, if the model is incomplete. I'm writing up a proper hypothesis, but I'm just an electrical engineer no cosmologist.
I imagined the interaction as if a paper boat is sitting on a quiet lake then it start vibrating due to an incoming water wave approaches. is this imaginary view acceptable?
That is what happens during sunrise and sunset. When the sunlight travels a greater distance through the atmosphere (at a lower angle), more of the red light will scatter out and the sky will look red.
@@MichelvanBiezen I dont think that is correct. Dust would reflect light but not scatter it. I did some research yesterday and found the following: "The result is that the emitted radiative power is proportional to the square of the dipole moment and to the fourth power of the oscillation frequency. This is a consequence of the smallness of the dipole compared with the wavelength of the radiation. For a transmitting aerial (antenna) the optimal size is of the order of about half the wavelength, for maximum transmitting power. The fraction “dipole-size over wavelength” is more favourable for short wavelengths. Thus, more light with short waves is scattered than with long waves, inversely proportional to the fourth power of the wavelength. " www.itp.uni-hannover.de/fileadmin/arbeitsgruppen/zawischa/static_html/scattering.html
The scattering is caused by a combination of dust particles, water vapor droplets, and the gas molecules of the air. The smaller the wavelength of the radiation, the more likely they will scatter.
+Duham Zanotto (SwimmingStrings) The scattered particles will be moving in a different direction, the remainder of the sunlight will continue in the same direction.
If we compare blue wavelength with indigo and violet the scattering is more likely in the violet than in blue color and sky should look violet instead of blue. Is that not because alpha The interaction factor of a nitrogen molecule is higher for blue instead of violet? or it is because violet color has lesser intensity than blue in incident light ?
In astronomy, the two colors, purple and green, are ignored. The reason is that we don't see purple stars, nor green stars. These colors are overwhelmed by the other colors, (even though they are there), and thus are not considered.
So the light waves causes polarization in the molecule making it a dipole? So does the molecule go to a virtual energy state and then deexcite? Or have I understood something incorrectly?
Nice video! It seems intuitive that N2 absorbs the vibrations of the electromagnetic field more readily in the longitudinal direction. However, you did not explain why this happen. Could you help me out with this, please? It is quite confusing because the cross section is smaller in this direction....
The molecule does not absorb the radiation, because it is not a polar molecule. (That is why N2 cannot participate in the greenhouse effect). Instead the molecule will scatter the radiation, and that is why it is called the Rayleigh scattering.
Thank you!! you are really talented to explain hard theory easily. I'm so impressed. But I am wondering it is correct that photon is absorbed to molecules and the electron in the molecules make vibrations and it changes electric field inside the photon and the changed photon is emitted?
you said electron makes vibrations and it changes electric field inside the photon but some report i see is saying separation of charge within the molecule makes electromagnetic radiation(not change), thereby resulting in scattered light. I'm so confusing...
@@MichelvanBiezen What do you mean “typically.” Wireless communications operates solely on the oscillations of valence electrons in a non-quantum manner. Quantum jumps release a photon of a particular wavelength determines by the difference between the two energy levels the electron jumped. Oscillating electrons can create any wavelength of light depending upon the wavelength of the electron’s oscillations. Oscillation is not a quantum phenomenon driven by elemental electrons. I thought most, if not all, of the photons coming from the Sun are due to heat induced oscillations of both electrons and nuclei, not energy level transitions of electrons.
@@김희영-l3i The Sun is a large ball of plasma. Plasma is a fluid of charged particles. These particles are charged because the high levels of energy do not allow electrons to settle into a stable orbit around a nucleus. Electrons and bare nuclei are electric charges that oscillate at high frequencies, and it is those oscillations that create the photons; not electrons jumping between shells of an atom. Moreover, the photons that strike the Moon do not “bounce off” of the Moon. They cause oscillations in the surface atoms of the Moon and those particles create new photons that radiate toward the Earth. Technically, the Moon radiates its own light.
The wavelength of blue light is small enough to be scattered by the particles in the atmosphere as compared to the other light colors, which have larger wavelengths. (Like a small tire has trouble getting over a boulder, but a bigger tire can more easily get across a boulder).
I think the MOST Important thing I saw in this Entire video was .... (wait for it ...) you remembered to put a Cap on your Little Stick Figure guy..!! LOL!!.. just kidding ofcourse.. Nicely Done Michel!!.. You are the One Stop Shop for all things Physics, Engineering and Mathematical... oh and Astronomical!! Thank you!
Think of molecules as little objects connected with springs. If you push one of them it will begin to vibrate (like a mass on a spring) and that will push against the other atoms which will begin to vibrate as well.
Well yeah, where did he say something else? He said that there will be four (in terms of lewis dots) electrons between two oxygen atoms, as the covalent bond has formed. Not that oxygen has 4 valence electrons. Listen carefully.
These are the real heroes. People who make science interesting and easy to understand. Keep doing what you’re doing!
Just a couple corrections: the fact that Rayleigh scattering is polarized has nothing to do with the shape of the nitrogen and oxygen molecules. Even for spherical molecules and atoms such as methane or argon one observes the same polarization properties of the scattered light which is due to the angular distribution of an electric dipole emitting light and the fact that light is a transverse electromagnetic wave. The electric field of the light shakes the electrons of the molecules which then radiate light with an angular distribution of an electric dipole. Also, the atoms in the molecules don’t vibrate during Rayleigh scattering, it’s just the electrons which are moving in response to the electric field of the light.
Excellent input. The shape of the molecule does affect the charge distribution, but your input is spot on.
A great teacher teaching a complex subject in English WITHOUT a thick accent. This is where I would put a lot of value this day in the age of information.
I am sure you are a polymath, you can teach many topic from totally different industry with huge understand about the course and taught in the way that easily understand by beginner like us and only use a marker pen without any fancy editing ,massive respect to you by using a single marker pen and white board to change the world. A massive respect of applied physic student form Malaysia.
Thank you for your kind words and welcome to the channel!
@@MichelvanBiezen because i knew you since 3 years ago about assignment of L'hopital rule, (that time not even know how to pronounce L'hopital rule). then , I am really surprise and astound that you also teach particle physic , a totally different field ,really respect to you
Great job in turning complexity into simplicity. It was very insightful. Thank you for sharing the knowledge.
Glad you like it. 🙂
Great explanation. I have been wondering how scattering works. Thank you.
Finally I understand why the sky is blue. Love learning this shit. Thank you Michel.
good teaching skills, got understanding easily :)
+Michel van Biezen I get that the purple is 400nm where as red is 700nm lenght. And so blue has longer wave length than purple but less that red. My question is, and what I don't understand for some reason at all, is why doesn't that make the sky purple, since if I understand this correctly, purple should scatter even more than blue?
+Ellis D In astronomy we don't talk about purple light. We lump the blue and purple together and just call it blue. That stems from the fact that we don't see purple stars and as you mentioned we don't see a purple sky. The wavelength of purple is a bit shorter than the wavelength of blue but again we lump those colors together.
Dear Biezen, I have a question around 3:17.
Why will the molecules absorb the energy and then re-emit the energy "without loss" ?
The point I couldn't understand is the "without loss" statement.
May you please explain about it?
Thank you very much!
In the macro world, energy is lost during almost any physical interaction. In the micro world, everything happens in quantum steps and can only happen in quantum steps and consequently energy is transferred in quantum step.
Why then do we see much more orange/red light during sunrise and sunset time periods? It cannot all be due to the N factor increasing the number of molecules along the light's path, can it? If so, then that would be a very big coincidence that our atmosphere is exactly enough to scatter mostly blue when sunlight shines through it vertically but exactly enough to scatter mostly red when sunlight shines through it largely horizontally.
+stellarfirefly
Longer path equates to the other colors being scattered as well.
Michel van Biezen
But then why do the red(dish) colors overwhelm the blue(ish) ones, which presumably are also still being scattered? And why do they not overwhelm them when shining through a lesser distance?
+stellarfirefly It is a matter of numbers. When the rays come from a high angle, many more of the blue photons scatter compared to the red ones. But when traveling from a smaller angle (greater distance through the atmosphere) there are more of the yellow, orange, and red photons scattered.
stellarfirefly my question exactly. im still not clear on this...driving me nuts.
This video explains the right answer. ua-cam.com/video/0b1fqodmZJ0/v-deo.htmlm5s
Basically most of the blue light has scattered because it scatters more! None reaches us. Only the red light which has scattered less reaches us.
When a molecule vibrates, is not the resultant wave a spherical wave front regardless of the orientation of the molecule?
Oh. I see. The resultant wave front has an amplitude proportional to the angle of incidence.
A resultant wave front is weaker in one orientation versus another. Thanks.
I’ll look around, but I’ve been trying to find an explanation for why reflected light is polarized parallel to the reflective surface while refracted light is polarized perpendicular to the surface.
I am surmising that it has something to do with the freedom of movement for valence electrons as well as the freedom of movement for the nuclei of a crystal’s constituent atoms. Metals having a valence band that is loosely bound to individual atoms creates a kind of sea that has more freedom of movement across the metal’s surface as opposed to perpendicular to it. Whereas a transparent crystal is rigidly locked into its lattice restricting vibrations to along those bonds? Or is the valence band electrons playing more of a role for wave propagation through a crystal more so than the nuclei?
Thanks for your time.
Keep in mind that the emissions of an atom or a molecule, moving by itself in the air, is very different than the emission of an atom or molecule locked in a solid (which is more like black body radiation).
@@MichelvanBiezen I did. That’s why I specifically mentioned the oscillations induced in valence band electrons which are weakly coupled to the material they are bound to. For metals, it is a very weak coupling allowing for far greater amplitudes of induced oscillations.
I am just a bit fuzzy on what is oscillating in a crystal with more strongly bound valence electrons. Is it still valence electrons that are oscillating within a transparent crystalline lattice, or are individual atoms oscillating? Or is it both depending on the angle of incidence and wavelength of the incident light which would account for reflection at a particular angle instead of refraction as utilized by fiber optics?
Wait. Maybe I missed a lecture or a lab at some point, but I believe that nitrogen is a linear molecule with 2 atoms so it has 3(n) - 5 or 3(2) - 5 = 1 normal vibrational mode. For me, vibrating in that other axis just doesn't compute.
Water (vapor) is a non-linear molecule with 3 atoms so it follows 3(n) - 6 or 3(3) - 6 or 3 normal vibrational modes. Like nitrogen, carbon dioxide is also a linear molecule but with 3 atoms so it has 3(n) - 5 or 3(3) - 5 = 4 normal vibrational modes, one of which is doubly degenerate.
Correct me if I'm wrong.
Thank you for awasome lecture
Where to come (derivation) that formula of scattered intercity??
+Michel Van Biezen, I get it why oxygen and nitrogen absorb the energy. But why do oxygen and nitrogen molecules radiate back the energy? What is the trigger?
So does a single ray of light get split into multiple rays of light after it is absorbed and re-released from the electron? If so, does that mean subsequent ray of light is weaker or less energetic than the parent ray of light? Your illustration seems to imply this.
+SogMosee
If the photon is scattered it is not absorbed. It is as if the photon is deflected. The illustration gives you the probability of which direction it will be scattered. It will be more likely scattered in one direction over another direction. That is the nature of Rayleigh scattering.
Michel van Biezen Thank you for the response that clarifies things greatly.
thanks for a perfect explanation of this topic
one question left, so in principle, if the molucule just absorbs the photons energy and then emmits it back, the scattering angle is absolute random? or I understand it completely wrong?
For typical scattering, yes. But for Rayleigh scattering some directions are more frequent than others.
@@MichelvanBiezen then one more question concerning scattering. We have a rubidium atom cloud which is cooled down (Kinetic energy close to zero) by a laser and caught in a magnetic-optical trap. Now the Idea is to detect the movement of the cold atom cloud along the trap based on the scattering of laser, if I understand it right, the Rayleigh scattering should be dominant in this case, still the laser light has to interact not only with atoms, but with electrons, meaning that Compton scattering will take place too, but can't it's influence be negliable?
Beautiful presentation, thank you so much!
Our pleasure!
The wavelengths that are scattered less are absorbed by the molecule ? can we explain rayleigh scattering with quantum mechanics without em waves but photons ?
Might exoplanets orbiting Red Dwarfs lack significant Rayleigh Scattering and so have less daytime airglow and fainter daytime skies?
That is an interesting thought!
@@MichelvanBiezen Evidently, Martian sunsets are blue b/c the atmosphere is thin (letting the blue through) and fine dust scatters the red (according to a Bing science quiz)
But scattering means changes in frequency lenght or amplitude of the wave or it has something to do with photon particle i. e. it is phisical collision? Best greatings and thank You.
You may be referring to the Compton Effect. But changes in the frequency of the photon due to a collisions causing it to scatter is only detectable when the original frequency of the photon is very high. With light photons that is not the case.
It is just a little bit confusing because we have light as a photon and wave, what is more the electron is a cloud of probability or part of matter and the interaction is based on molecule movement and jupms of electrons between shels. Which combination is right for light scattering?
There are a lot of different pieces of information and theories wrapped up in your questions. Better to take it one step at at time.
I will try. Thank You.
Great explanation !!!
Glad you liked it!
Sir, in which direction is the intensity of scattered light more. In other words at what angle the intensity is more
The intensity of the scattered light is higher along the direction parallel to the length of the molecule.
Michel van Biezen sir but in my text book its given that " the angle between incident light and direction in which the intensity of scattered light is observed is called angle of scattering it is experimentally observed that intensity of scattered light varies with angle of scattering. The intensity is maximum at 90 degrees angle of scattering" is it correct
Note the equation of Rayleigh scattering and the dependency of the scattered intensity and the angle. There are different scattering principles. This equation is specifically for Rayleigh scattering.
@@MichelvanBiezen p
Lovely. However isn't all Electromagnetic Radiation subject to Rayleigh Scattering, rather than just light?
Yes, but some wavelengths are much more affected than others.
@@MichelvanBiezen yeah! The higher the frequency the more they're affected right!
Ive been philosophising about to what extent Rayleigh Scattering could cause red-shifting of em waves in the intergalactic medium. The Density and distance are of the same power. Density is very low but distances are also very large. With the discovery of necessary dark matter maybe the intergalactic medium is more dense than earlier presumed and perhaps could explain some portion of red-shifting in distant galaxies. If my math is correct if scattering is a factor it will also have an Indirect relationship with the expansion of the universe, seemingly multiplicative which could be misinterpreted as an accelerating expansion rate, if the model is incomplete. I'm writing up a proper hypothesis, but I'm just an electrical engineer no cosmologist.
That was wonderful teaching! Thankyou 😃
Thanks a lot sir. This is one of 5he best video on this topic..
Glad you liked it
I imagined the interaction as if a paper boat is sitting on a quiet lake then it start vibrating due to an incoming water wave approaches. is this imaginary view acceptable?
There are some similarities. Although in this video the topic is the scattering of the incoming E&M radiation.
Hello. You mentioned Nitrogen & Oxygen molecules. Can you help elaborate about interactions with aerosols? And how that would be different or similar?
2:23. Do they/can they only vibrate like this because they are symmetric charged? Or is it also seen in non-symmetric molecules?
Makes better sense. Thank you!
what would it take for our sky to be Red (scatter Red)? which gas molecule should be common in our atmosphere?
That is what happens during sunrise and sunset. When the sunlight travels a greater distance through the atmosphere (at a lower angle), more of the red light will scatter out and the sky will look red.
But why do electromagnetic waves get scattered more, if they have shorter wavelength?
When they have shorter wavelengths, it is more difficult for the light waves to make it through the dust particles in the air.
@@MichelvanBiezen I dont think that is correct. Dust would reflect light but not scatter it. I did some research yesterday and found the following:
"The result is that the emitted radiative power is proportional to the square of the dipole moment and to the fourth power of the oscillation frequency. This is a consequence of the smallness of the dipole compared with the wavelength of the radiation. For a transmitting aerial (antenna) the optimal size is of the order of about half the wavelength, for maximum transmitting power. The fraction “dipole-size over wavelength” is more favourable for short wavelengths. Thus, more light with short waves is scattered than with long waves, inversely proportional to the fourth power of the wavelength. "
www.itp.uni-hannover.de/fileadmin/arbeitsgruppen/zawischa/static_html/scattering.html
The scattering is caused by a combination of dust particles, water vapor droplets, and the gas molecules of the air. The smaller the wavelength of the radiation, the more likely they will scatter.
I don't understand, does the scattering particles change the direction of propagation of sunlight or not?
+Duham Zanotto (SwimmingStrings) The scattered particles will be moving in a different direction, the remainder of the sunlight will continue in the same direction.
What happens to the ones that don't get scattered?
More of the long wavelength radiation makes it through the atmosphere, less of the short wavelengths make it through
If we compare blue wavelength with indigo and violet the scattering is more likely in the violet than in blue color and sky should look violet instead of blue. Is that not because alpha The interaction factor of a nitrogen molecule is higher for blue instead of violet? or it is because violet color has lesser intensity than blue in incident light ?
In astronomy, the two colors, purple and green, are ignored. The reason is that we don't see purple stars, nor green stars. These colors are overwhelmed by the other colors, (even though they are there), and thus are not considered.
such a good teacher!
So the light waves causes polarization in the molecule making it a dipole? So does the molecule go to a virtual energy state and then deexcite? Or have I understood something incorrectly?
It is more a matter of preferential scattering direction
Hello, can anyone help me, what does I_0 stands for in the rayleigh scattering equation?
Io is the unscatterred intensity. (The intensity of the incoming E&M before scattering).
@@MichelvanBiezen thank you so much for the answer sir
@@MichelvanBiezen Sorry sir, can I ask another question?
Is the unit of intensity of scattering watt/m^2 ?
Thank you, that was really helpful!
why does the molecule absorb the energy of the photon?
Molecules have multiple ways of absorbing photons. (increased kinetic energy and multiple vibrational and rotational modes)
Michel van Biezen okey, thank you very much ! I am doing my annual task about this topic and your video really helps a lot.
Nice video! It seems intuitive that N2 absorbs the vibrations of the electromagnetic field more readily in the longitudinal direction. However, you did not explain why this happen. Could you help me out with this, please? It is quite confusing because the cross section is smaller in this direction....
The molecule does not absorb the radiation, because it is not a polar molecule. (That is why N2 cannot participate in the greenhouse effect). Instead the molecule will scatter the radiation, and that is why it is called the Rayleigh scattering.
@@MichelvanBiezen
nice explanation sir
My thoughts get saturation with this video 🙏🙏
hello, sir, where can i get the slides please? thanks
The slides (of the thumbnails) are not available.
You're a life saver! Thank you!!
Happy to help!
Thank you!! you are really talented to explain hard theory easily. I'm so impressed. But I am wondering it is correct that photon is absorbed to molecules and the electron in the molecules make vibrations and it changes electric field inside the photon and the changed photon is emitted?
Yes, that is correct. Although the emission of photons is typically caused by the jumping of electrons between energy levels.
you said electron makes vibrations and it changes electric field inside the photon but some report i see is saying separation of charge within the molecule makes electromagnetic radiation(not change), thereby resulting in scattered light. I'm so confusing...
@@MichelvanBiezen What do you mean “typically.”
Wireless communications operates solely on the oscillations of valence electrons in a non-quantum manner.
Quantum jumps release a photon of a particular wavelength determines by the difference between the two energy levels the electron jumped.
Oscillating electrons can create any wavelength of light depending upon the wavelength of the electron’s oscillations.
Oscillation is not a quantum phenomenon driven by elemental electrons. I thought most, if not all, of the photons coming from the Sun are due to heat induced oscillations of both electrons and nuclei, not energy level transitions of electrons.
@@김희영-l3i The Sun is a large ball of plasma. Plasma is a fluid of charged particles. These particles are charged because the high levels of energy do not allow electrons to settle into a stable orbit around a nucleus. Electrons and bare nuclei are electric charges that oscillate at high frequencies, and it is those oscillations that create the photons; not electrons jumping between shells of an atom.
Moreover, the photons that strike the Moon do not “bounce off” of the Moon. They cause oscillations in the surface atoms of the Moon and those particles create new photons that radiate toward the Earth.
Technically, the Moon radiates its own light.
The best explanation!!
Glad it was helpful!
That was amazing 😭 Thanks a lot ❤❤
Is electron configuration the biggest factor in determining whether scattering is elastic or inelastic?
You didn't confirm why only blue light got scattered?
The wavelength of blue light is small enough to be scattered by the particles in the atmosphere as compared to the other light colors, which have larger wavelengths. (Like a small tire has trouble getting over a boulder, but a bigger tire can more easily get across a boulder).
Than you Michel👍😍
I think the MOST Important thing I saw in this Entire video was .... (wait for it ...) you remembered to put a Cap on your Little Stick Figure guy..!! LOL!!.. just kidding ofcourse.. Nicely Done Michel!!.. You are the One Stop Shop for all things Physics, Engineering and Mathematical... oh and Astronomical!! Thank you!
Did I forget that cap again? Have to be more careful or I'll get a sun burn on that bald spot.
@@MichelvanBiezen actually you did NOT forget... it's just one of your Trademark Contributions to your Videos.. LOL..
Ah good.
why do molecules vibrate?
Think of molecules as little objects connected with springs. If you push one of them it will begin to vibrate (like a mass on a spring) and that will push against the other atoms which will begin to vibrate as well.
very educative.thank you.regards
Thanks for the explanation
This is a great video, it helps my dumb chemistry brain understand these physics concepts very easily!
thanks a lot 😁
Nice sir
Thanks and welcome
Appreciate lot🤗 thanks...
Our pleasure 😊
Thx alot 🌸
oxygen has 6 valence electrons, other than that good lecture.
Well yeah, where did he say something else? He said that there will be four (in terms of lewis dots) electrons between two oxygen atoms, as the covalent bond has formed. Not that oxygen has 4 valence electrons. Listen carefully.
So the nitrogen is responsible for the blue color, not the ozone ?!!!
The scattering effect of primarily the nitrogen.
Tks!
na bro na na man you cant be all evil! you telling me we are not learning?