Compton Scattering

[@marialiyubman]

Words can’t express my gratitude to whoever made these courses available to the public for free. Thank you. ❤️

[@jacobvandijk6525]

Well said. In my studying years (70s/80s) there was nothing like this (of course). We had to make a reservation to make use of a computer, haha! Btw, I think this man is just a brilliant lecturer. Good luck!

[@CARLOSSAA-y6f]

absolutely flawless professor Zwiebach!

[@mjackstewart]

I think it’s adorable how he says: Photons: phot’ns Electrons: electr’ns

[@aravindgundakaram1830]

The way they ended this video in mystery makes the viewer want to continue studying the next video 😂

[@nth2tell]

Wow he made equations look simple by explaining everything simply like it's natural things to happen.

[@user-pt-au-hg]

A mark of a good lecture and professor. :)

[@isatousarr7044]

Compton scattering is a fascinating phenomenon that illustrates the dual nature of light, behaving both as a wave and as a particle. It occurs when a photon collides with a charged particle, typically an electron, resulting in a transfer of energy and momentum. This interaction leads to a longer wavelength of the scattered photon, which is a clear demonstration of the particle-like behavior of light. Compton scattering not only plays a vital role in our understanding of quantum mechanics but also has practical applications in various fields, such as astrophysics, medical imaging, and radiation therapy. The study of this effect has significantly contributed to the development of quantum theory and our understanding of electromagnetic radiation.

[@brendastephanie1403]

I can't believe it, MIT is cliffhanger us with PHYSICS CLASSES

[@skm0916]

hello from India, very helpful video lecture, made the topic crystal clear this what a perfect professor is!! thanks a lot MIT for this open course

[@davehses755]

Fantastic lectures. No wonder MIT is world's number 1.

[@oscaraguilar6906]

Llevo viendo estos videos toda la semana. joder que guapo es tener plata para pagarte un buen maestro

[@jimshilleto5655]

By diagonalizing the matrix M = (cos alpha, sin alpha; cos beta, sin beta)^T, in the new basis B ={v1, v2}, we see that if the beam splitter's lines are set to v1 and v2, the eigenvectors of M with two distinct eigenvalues a1 and a2, then the phase factors become simply a1*v1 and a2* v2. So the calculations become very simple when the light rays travel along lines v1 and v2 instead of 45 degrees up and 45 degrees down.

[@jimshilleto5655]

Small edit: Change 'the phase factors become" to "the new phases are".

[@jimshilleto5655]

Sorry. This comment should be in Lecture 2.3 of the professor's 8.04 course, not in this course. He has so many I got confused.

[@jacobvandijk6525]

Compton Scattering reminds me of the way Rutherford detected the nucleus of an atom!

[@stijndhondt9611]

Boy if there ever was a cliffhanger... It does seem L3.4 doesn't continue where this one stops. Anyone have an explanation to why the first bump is there?

[@stijndhondt9611]

Nvm, I found the correct answer. The first bump is present because some of the electrons which scatter the light are very closely bound to the nucleus of the atom. This means that the m in the Compton equation is not the mass of the electron, but the mass of electron+nucleus which is much larger. This makes Δλ very small.

[@ingenieur_sans_emploi]

@stijn D'hondt thanks

[@sonstephan57]

Thank you for the answer, that makes sense. So, for the Carbon nucleus is accelerated by a photon?

[@nicktohzyu]

can't a photon scatter off the nucleus directly?

[@TIENTI0000]

because of the photon is reflecting from nuclei instead of the electron.

[@johnbruhling8018]

My best guess for the reading of λi is that they are the reflected x-ray particles that did not interact with a virtually free electron. The prepared crystalline structure of the metal might provide places where the x-ray particle can oscillate on the surface without resistance and also without an electron in affecting proximity and then could be reflected by a following x-ray particle with trivial energy loss.

[@oggythebug]

Hello very interesting lecture it is exactly what i studied here in my country in france, in 1980 on my last school year before going for my first year at the university, we studied everything shown here, photoelectric effect and compton scatering plus the special relativity including all the equations writen on the blackboard, we even had to learn the demonstration of the final compton scatering formula i still have this in my book that i kept all these years, i think the level was high in those years , we were only 17 years old. best regards

[@dlbattle100]

Ermahgerd, the squeaky eraser.

[@mjackstewart]

David Battle Right? You’d think MIT could afford the non-squeaky kind.

[@Bai_Su_Zhen]

Now I'm curious. Why is there the other peak? :D

[@yatharthbakshi]

It is due to interaction of photon with the nucleus of the atom.

[@ryanyi8900]

According to the notes from MIT open courseware, the other peak is caused by the collision between the photon and the whole atom. Because of the negligible Compton wavelength of the atom, the wavelength shift is very small.

[@পদার্থবিজ্ঞান-চ৯হ]

The peak at λi represents a process in which an electron receives some momentum from the photon but still remains bound. This is not very unlikely: the typical momentum of a bound electron is actually comparable to the momentum of the photon. In this case the photon scatters at 90◦ and the recoil momentum is carried by the whole atom. The relevant Compton wavelength is therefore that of the atom. Since the mass of the carbon atom is several thousands of times larger than the mass of the electron, the Compton wavelength of the atom is much smaller than the electron Compton wavelength and there should be no detectable change in the wavelength of the photon

[@Jk-r5c]

@@ryanyi8900 Thx a lot😀

[@পথিক-ভ্রমণ]

The Compton Effect KZbin video link: kzbin.info/www/bejne/oIqYaYx-icetjtU

[@nibussss]

I also think photons are just particles that look like a wave when it bounces off other particles. I though it was my idea....last time I checked it was considered wave/particle duality...

[@ryanyi8900]

In the photoelectric effect, we know that the material absorbs light by quanta. But in Compton scattering, the electrons partially *absorbs*(collide) the energy of high energy photon. So are the two (the photoelectric effect and Compton scattering) contradictory?

[@jacobvandijk6525]

I would say NO. Here the photon is a particle with some momentum. In the photoelectric effect the photon gives off ALL of it's momentum, while in Compton scattering the photon only gives off A PART of it's momentum.

[@ryanyi8900]

Jacob van Dijk Thank you!

[@jacobvandijk6525]

@@ryanyi8900 You're (always) welcome.

[@tanfeexulhaqq4616]

​@@jacobvandijk6525 isn't that the question? If a photon carries a certain quantum of energy, then how is it possible for it to lose only a part of it? Consider the photoelectric effect as an example, the photoelectrons don't eject for as long as we provide photons with a frequency lower than the threshold frequency. If it were possible for photons to lose part of their energy, then it would require a higher intensity of light for photocurrent to flow; which is totally against the experimental data. So, what's going on here?

[@jacobvandijk6525]

@@tanfeexulhaqq4616 Nothing wrong with the picture you describe. In the photo-electric effect the emphasis is indeed on photon-absorption. But when you shine light an a material photons can be reflected (or scattered) too. They don't enter the material; think of a mirror. And if the scattering is inelastic, the photon gives up some of its energy to the material (which is heated a bit). Absorption always kills the photon, in collisions it can survive to some degree.

[@صلاحالدينالشرعبي-ش5د]

why is the intensity of the second beak is bigger?

[@FrankPappacoda]

Thank you for sharing this lecture

[@kaushaljain5999]

10:08 I think you have to correct the formula of cross-section. And after correction unit becomes "per unit area per unit solid angle" rather than "area per solid angle".

[@michaelbondin8342]

No he's right

[@nibussss]

Photon must have repelling with some other X particle which makes high energy photons bounce off a lot....

[@pulkitojha6438]

guys, someone please explain to me why is that extra peak there in the experiment described at the end of the video? next video doesn't cover this. as the angle is 90, it can't be that some of the photon are detected without being scattered at all.

[@changcheng73]

Photons are scattered not only by electrons but also by Carbon nuclei. And there is a negligible Compton shift for those nuclear part of scattered photons wavelength

[@adriangheorghe2327]

Si atunci care este mecanismul prin care din fotonul gama incident de mare energie, de mare frecventa, ar aparea fotonul gama imprastiat, de frecventa si energie mai mica decat a fotonului gama incident.? And then what is the mechanism by which the incident gamma photon of high energy, of high frequency, would appear the scattered gamma photon, of lower frequency and energy than the incident gamma photon.?

[@abhijithrambo]

Where can I find the first homework he mentioned?

[@tanfeexulhaqq4616]

If a photon carries a certain quantum of energy, then how is it possible for it to lose only a part of it? Consider the photoelectric effect as an example, the photoelectrons don't eject for as long as we provide photons with a frequency lower than the threshold frequency. If it were possible for photons to lose part of their energy, then it would require a higher intensity of light for photocurrent to flow; which is totally against the experimental data. So, what's going on here?

[@jessereiner5401]

Does anyone know why the peaks aren't sharper? What causes the smaller variation in wavelength of the scattered photons that makes the peaks broad rather than pointy? Is it a detector issue? Some other quantum effect?

[@saikatmaji2917]

yeah good question. collision should be instanteneous

[@AnshulSharma1997]

This is due to Doppler broadening

[@PeterBaumgart1a]

​@@AnshulSharma1997This, plus all detectors have some finite resolution, in angle and in energy, etc.

[@stumbling]

What a cliffhanger!

[@criticalthinking575]

Is linear algebra essential for this course

[@1eV]

no

[@bigymara]

What a cliffhanger

[@shivamkakkar4027]

how can a photon loses its energy? As it has packets of energy how can it be lost after hitting to electron.There is something more which is till Now unnoticable which changes some of photon's property.

[@sleepy314]

it changes wavelength

[@summadayze733]

It is just an inelastic collision and conservation of energy

[@shivamkakkar4027]

@@summadayze733 Yes you are right it is inelastic collision but in what terms it is losing energy?

[@summadayze733]

@@shivamkakkar4027 the lost energy is the kinetic energy of the recoiled electron which was before almost motionless

[@RaihanSergi]

did the Professor say "collision"? So it wasnt like the electron absorbed the entire photon and then re-emit it?

[@iyadabutaimeh5545]

if electron absorb photon.photoelectric happen

[@shahnazgul990]

can we use gamma ray photon for compton and photoelectric effect

[@JohnVKaravitis]

"Thomson," not "Thompson."

[@rezokobaidze8501]

how does electron and photon decide in wich direction (angle) to scater if photon hits electron directly? I understand formula and experime so we take fact final result but if i want to predict direction?

[@PeterBaumgart1a]

In aggregate, all possibilities happen, with the x-section formula governing the probability distribution. In a single event, a single random one.

[@maalls]

The other pick, why is it there??

[@sonstephan57]

On guy up says it is because it is not only an electron detected, but also the entire nucleus for the Carbon, so, if you use the equation for initial lenghtwave, it makes sense

[@magnfiyerlmoro3301]

why when theta=0 there is no collision ? can someone explain?

[@dhruvabhat3261]

@Thejas CS I know you 🤔

[@hywelgriffiths5747]

Theta is the scattering angle, so theta=0 means there was no scattering, hence no collision

[@سأكونيومامااريد-ف1ن]

I am from in Iraq عراقي... #طالب سادس

[@saraesmael545]

منور يستااااا 😂😂

[@atikabrarsourov7500]

Sir, homework done- If electron at rest absorbs photon, the velocity of photon had to decrease i.e change which is against relativity or electron has to move with the speed of light, that makes its mass infinite which is against the law of conservation of momentum.

[@dinaawaell8038]

great!

[@martindorrance8133]

It is counterintuitive to have no mass but still have momentum. Can this be explained in layman’s terms?

[@sleepy314]

E ²=(mc ²) ²+(pc) ² is the full equation. So when m=0, then E ²=(pc) ² or p=E/C

[@PeterBaumgart1a]

The idea that mass is required for momentum is a simplification (oversimplification) from highschool physics. At low speeds p = mv. But at high speeds (relativistic regime) it's more complicated, see the formula given in the other response.

[@yuriboldt5320]

Great!

[@الفيزيائياحمدمشالي]

Thanks very good ❤❤

[@sonstephan57]

But why was the first peak there?

[@TIENTI0000]

because of the photon is reflecting from nuclei instead of the electron.

[@JG-yh2ut]

wave particle duality

[@user-gu7se5iw7b]

I,’m from KMU

[@surojpaul14]

mind blowing 😘

[@nibussss]

Static/uniform particle distributed throughout universe??? Matrix...lol...this is crazy

[@pk3059]

mass of photon=0 ??

[@DebeshDey322]

Rest mass of photon is 0

[@sai7401]

Could be better a little bit,feels something is missing

[@sridharchandramouli9615]

It's unfortunate that the teaching of this subject hasn't changed much since I was a post-grad at a lower ranked university 20+ years ago. Even back then I thought to myself that I wasn't there to take a class in history along with some interesting qualitative ideas with not enough detail in the reference to experiments. Back then I put in down to the professor/university I attended but if MIT is teaching it pretty much the same way 15+ years later, there's a wider problem. Almost no other topic in physics or any other science is taught in this way.

[@javierhernandez1390]

MFTI or MGU are much better!

[@valentinsarmagal]

What is this?

[@javierhernandez1390]

@@valentinsarmagal Moscow PhysTech and Moscow State University

[@valentinsarmagal]

@@javierhernandez1390 do they have open courses in English?

[@javierhernandez1390]

@@valentinsarmagal nope. Learn Russian!

[@tjk9694]

They've got a different style. Whatever suits you best

[@irmaostuga]

.