Babe wake up, new spinnors for beginnors video dropped!
@MattHudsonAtx
3 ай бұрын
If you liked Spaniards for Beganiards or Skynyrd for Begynyrds...
@SpenceReam
3 ай бұрын
Shhhhhhh…. I’m putting it on to sleeeeeeeeeep
@jamesondasilva6502
3 ай бұрын
It's like christmas morning
@aburaihan5512
2 ай бұрын
Dear @eigenchris, I covered your all lectures..and really your teaching capability is outstanding.. You are more than a professor for me. Firstly tensor for beginners,then tensor calculus and recently i finished the general relativity series.. I don't know where are you from,who you are I will say.. you are awesome. A true genius. Now my next plan is to cover your new series "spinors ".. A little request for you.. If you can then please make a detailed series on " kerr black hole".. Since I am doing a thesis on "Spacetime Singularly and black holes" , your whole lectures gave me an extra power.. I am from Bangladesh. Always love and respect for you my professor ❤
@eigenchris
2 ай бұрын
Thanks for the positive comments. I did start looking at the Kerr black hole solution, but it was too complicated for me to understand. I believe other channels have covered it, though.
@aburaihan5512
2 ай бұрын
@@eigenchris yes there is available some lecture series on kerr metric and kerr geodesics.. But I am waiting for you. I love your explanation.. I hope like the other series you will also make a great series on Kerr black hole for us. Best wishes for you. Thank you so much professor for replying me.. ❤️
@star_lings
3 ай бұрын
this is pure gold. Thank You!
@georgesas7090
3 ай бұрын
@44:00 There is one detail that in the past has haunted me for some time, namely that this is true only if both fields are expressed in the frequency domain, in which case they can take up complex values. If they are expressed in the time domain, then they are only real valued, and the phase is not expressed as a complex number of magnitude one, but is only visible in the argument of the sinusoidal function (not shown here). You have indirectly said it in previous videos, but I thought it would be good to explicitly mention it here again.
@eigenchris
3 ай бұрын
I think you can still express the E and B fields as complex numbers in the time domain. You just "ignore" the imaginary part when it comes to getting real results. Multiplication by "i" still works as a way of changing the phase in this case.
@georgesas7090
3 ай бұрын
When mathematically expressing the fields as exponentials e^{i/omega t} you always imply +c.c. (plus complex conjugate). In fact this is usually explicitly written in the formula, otherwise stated in the text. If you multiply with i in the formula then you must multiply the c.c. with -i. Some people say to take the real part, this can be done as long as a factor of 2 is also considered. In both cases you multiply before you take the real part, so the \pi/2 difference in the exponent translates into difference into the arguments of the sine and the cosine (Euler identity). This method is sort of cheating because what you actually do by this is to Fourier transform back into the time domain and works like this only with plain waves. If e.g. you have a laser envelope, then this is not correct anymore. (I am a condensed matter physicist). I forgot to mention that I consider your videos amazing work, and I hope all professors and lecturers would research their stuff as much as you do.
@descheleschilder401
3 ай бұрын
Isn't the frequency the time domain? And the wave-vector the space domain?
@georgesas7090
3 ай бұрын
@@descheleschilder401 In the time domain the fields are functions of time, E(t) and B(t), and are purely real. In the frequency domain the fields are functions of omega, E(omega) and B(omega), and they are complex. You go from the one domain to the other through Fourier transform (back-transform), E(omega) = S E(t) e^{-i omega t} dt. Here S stands for integral. I use here the physics convention. The Fourier transform of cos(at) is pi [delta(omega-a) + delta (omega+a)], while the transform of sin(at) is the same but multiplied with -i namely with a phase difference of -pi/2. In real space you would have to rewrite sin through cos to see it, namely sin(at) = cos (at - pi/2). If your fields are not expressed as sums of plane waves (that is cos and sin functions), then this trig identity cannot be used anymore and you will have to perform the integration. You can also Fourier transform the spatial coordinates x, y, z and go from the real space to the so-called reciprocal space where your fields or other functions are expressed as functions of the wavevector (with components k_x, k_y, k_z). This is often done in quantum mechanics. You can transform time, f(omega,x,y,z), space, f(t,k_x,k_y,k_z) or both f(omega,k_x,k_y,k_z).
@gpilnstar8957
2 ай бұрын
Hi, informative videos. I'm still trying to study and fully understand spinors. I wonder if you could maybe do some videos and tutorials about "twistors for beginners"?
@eigenchris
2 ай бұрын
I don't know anything about twistors unfortunately. I don't plan on making videos on them.
@slaxer570
2 ай бұрын
Hey Chris could you do a video on the wigner-eckart theorem at some point?
@eigenchris
2 ай бұрын
I'm not familiar with that theorem and I'm kind of burned out on videos right now. I just plan to finish this series amd that's it.
@slaxer570
2 ай бұрын
@@eigenchris Thanks anyways!
@lizzie0196
3 ай бұрын
Can you update your github repository that you provide a link to in the description to include the powerpoint presentations SfB15 through SfB20. It looks like the repository was last updated 8 months ago.
@eigenchris
3 ай бұрын
I just uploaded SfB 16-20. Unfortunately SfB15 is stuck on an old hard drive from a dead laptop. I might be able to track it down, but I won't be dealing with that today.
@boffo25
3 ай бұрын
I was just studying this on my quantum field theory book with 800 pages.😊
@temp8420
2 ай бұрын
Brilliant
@arcturusgd
3 ай бұрын
Hi chris Would you like to see my eventual series about pure math and physics? If not, i dont mind.
@it6647
3 ай бұрын
12:30 whole section 13:47
@vulpetite
3 ай бұрын
Based
@jasimmathsandphysics
3 ай бұрын
wow
@Achrononmaster
3 ай бұрын
West coast propaganda.🤣
@moneteezee
3 ай бұрын
?, this is just covering math
@cademosley4886
3 ай бұрын
More like left-chiral propaganda. 🧐
@nice3294
3 ай бұрын
Stunning work, you really laid out all the details for the different representations in a way that made it easy to follow and understand
@GeodesicBruh
3 ай бұрын
Oh yeah baby new Eigencrhis video
@diribigal
3 ай бұрын
This video referenced a lot of previous videos. It was nice to pull everything together like this.
@kevinmchugh2302
3 ай бұрын
Sir, thank you for this series of videos. And also, thank you for your series on general a d special relativity. I've seen them all, and they make Peskin and Schroeder, and Wheeler, Thorne (Gravitation) so much clearer. I will now re-read these texts in light of the clarity you provided in these videos. Bless you sir.
@BakedAlaska187
2 ай бұрын
Finally, a clear Faraday tensor presentation I wept. 🔥
@MegaManki
3 ай бұрын
But do spinors have rizz?
@eigenchris
3 ай бұрын
About as much rizz as my monotone voice.
@peacekeeper9687
3 ай бұрын
Thanks a lot 👍👍👍
@Wibrr
3 ай бұрын
New banger just dropped
@BakedAlaska187
3 ай бұрын
A lot of material here to work with. Very much appreciated. Thanks so much …🔥🔥
@dilanlakmal6622
3 ай бұрын
Thank you very much..... We hope your video ❤
@changethiswhenyouareok
3 ай бұрын
Next topology for beginners please
@g3452sgp
3 ай бұрын
Astonishing! Is this what is called the spinor algebra?
@eigenchris
3 ай бұрын
I've never heard the term "spinor algebra" before, so I don't know what that is. The Js/Ks or As/Bs are typically referred to as making up the "Lorentz Algebra", which is denoted so+(1,3) or sl(2,C).
@g3452sgp
3 ай бұрын
@@eigenchris Well, you can name it.
@rudrapandey6473
2 ай бұрын
A weyl vector is constructed out of minkowski 4-vector [x y z t]. What happens to weyl vector if x, y, z, t are complex numbers? How to construct a weyl vector for complexified spacetime?
@eigenchris
2 ай бұрын
I assune the fornula would be the same.
@georgesas7090
3 ай бұрын
I have two questions which pertain to 7:54 and 22:07, namely As and Bs in sl(2)_c. The first question concerns the definition of Bx, By, and Bz, which if I calculate explicitly for su(2) matrices. I get all of them equal to zero (since the Ks are actually the Js multiplied by i). In fact I always thought that there are only three linearly independent matrices, often denoted X, Y, and H. This brings me to my second question. In the real algebra (i.e., with real coefficients) I can understand the need for 6 basis matrices (2 for the real and imaginary part of each sigma), but in the complexified algebra this is not necessary anymore leaving us once again with only three matrices (since now all Ks are just multiples of Js). Am I doing something wrong here?
@eigenchris
3 ай бұрын
The Bs would go to zero in the left spinor (1/2, 0) representation. In the right spinor (0, 1/2) representation, where the spatial directions are reversed, the As go to zero. The X, Y, and H are math notation what physicists normally call the "raising", "lowering", and "eigenvalue" operators, which I show at 8:30 (although there might be an extra factor of 1/2 in the physicist notation... math notation usually has raising/lowering the eigenvalues by steps of 2 instead of steps of 1). Those are the 3 possible 2x2 matrices with trace zero, up to multiplication by a complex number. When it comes to defining the Lie algebras, they are ultimately defined by their abstract commutation relations. The matrices only pop up *after* you choose a representation. So the Lie algebra I show at 7:54 is 6-dimensional because there are 6 generators. It's possible to invent representations where some of these generators end up having the same matrices (for example, in the trivial representation, all matrices go to [0]; this is called a "non-faithful" or "non-injective" representation). But the Lie algebra rules don't depend on any particular choice of matrices. They exist in terms of abstract symbols.
@georgesas7090
3 ай бұрын
@@eigenchris Clear enough now, thank you. I am not used to working with Lie Algebras, I usually work with groups, and therefore I failed to realize that 0 matrices yield the trivial representation.
@francescodraisci6404
2 ай бұрын
Insane contents, taking notes of everything for my exams. Please continue 🙏
@TheDD1255
3 ай бұрын
Hi, I would love to have the slides of this video as a support to take notes on them is it available ? Thanks a lot. Great content as always. ^^
@thescichan7638
3 ай бұрын
See description
@TheDD1255
3 ай бұрын
@@thescichan7638 I don't think it's updated :/
@eigenchris
3 ай бұрын
I'm a bit behind uploading the slides. I'll try to get them out this week. I'll reply to this comment when theybare up if I remember.
@eigenchris
3 ай бұрын
I just uploaded SfB 16-20. I appear to have misplaced the SfB 15 notes when I migrated to a new laptop. I'll upload them if I manage to find them.
@AMADEOSAM
3 ай бұрын
Very impressive work! Thanks
@UPu-tq8zm
3 ай бұрын
Great timing for this vid ! I got a QFT exam comin up
@soma7891
3 ай бұрын
First
@desvonbladet
3 ай бұрын
Help me, @eigenchris, you're my only hope! When all this started, one of the motivating examples for the series was the characterisation of plane polarized light by what traditional optics calls "Jones vectors", which are (to the untrained eye) complex 2-vectors. I got on board because I have wondered for many years why these vectors and the state vectors of electron spin share the same representation (and live on the very isomorphic Poincaré and Bloch spheres, respectively, even though these are clearly both the same thing). Polarisation has been out of this story for many episodes and in this episode it makes a welcome comeback, but it shows up in the (1,1) representation of SL2 (I think; I follow along, but much of the time I am barely hanging on and sometimes not even that.) I assume it is very obvious to people who understand this stuff well and deeply but it isn't to me: how do you get from these 3D rotate and boost representations back down to the SU(2) form for specifically plane waves? At the most hand-wavingly level I can imagine saying "to make plane waves you have to fix a spatial dimension, and this drops the algebra down a couple of dimensions (one for each spinor) and there you go!" but I would really really like to have the details laid out by someone who knew what they were doing. Obviously you don't owe me anything, and this course is a wonderful thing in its own right, but that was the question I started with on this journey and it's the question I still yearn to have answered, so if you are in the mood to take pity on a hapless viewer it would delight me beyond measure.
@eigenchris
3 ай бұрын
Yeah, back in videos 2-3, we started with a 3D electric field vector, but because the wave was travelling in the z-direction and the wave is transverse, that means only the X and Y components of the field are non-zero, so we just ignored the z-component and got a 2-component Jones vector. I do still find it a little confusing why Jones vectors end up being spinors that transform with SU(2). Photons are spin-1 particles as opposed to spin-1/2 particles. However, since photons are massless, the longitudinal polarization that massive spin-1 particles get is forbidden, so photons only end up having two spin states (you can call them +1 and -1, or "left" and "right", depending on the basis). Maybe this makes them somewhat more similar to spin-1/2 particles which also have 2 states. I admit I'm still a bit confused by this myself, however.
@PK-ru8dg
3 ай бұрын
Are there any particular textbooks that you used for these videos (or that you would recommend in general)? Amazing series btw.
@eigenchris
3 ай бұрын
I read some parts of "Group Theory in Physics" by Wu-Ki Tang, butsometimes it was pretty terse and a bit hard to follow. The wikipedia article on "Representation Theory of the Lorentz Algebra" isn't bad either. For this video I did have to sit down and figure out a lot of the details on my own though. I didn't realize gamma_5 could be used to project out left/right circularly polarized waves until I started playing around with the algebra, so that was a nice surprise. Most other sources I've seen use the hodge start or levi-civita symbol, but those don't work as nicely.
@NakintuLazia-mu1sj
3 ай бұрын
Is it possible to use spontaneous symmetry breaking of mass creation to describe spacetime curvature
@eigenchris
3 ай бұрын
I don't know much abkut spontaneous symmetry breaking, unfortunately.
@crochou8173
3 ай бұрын
2:32 I think those are spacetime bivectors!
@eigenchris
3 ай бұрын
I believe they are, since they are anti-symmetric. It's not always obvious at first because matrix representations can always have their basis changed, so they don't jump out as being exactly equal to what the bivectors look like in the Chiral basis of the gammas. But they should belong to the same representation.
@ghkthILAY
3 ай бұрын
in 17:20 you write rho(e^A* e^B)= rho_L(e^A) X rho_R(e^B) which to me looks like part of the definition of an isomorphism- can those actually be viewed as isomorphism or is this just a coincidence?
@eigenchris
3 ай бұрын
I'm not sure if there's a name for this. An isomorphism is usually a single function. Here, rho, rho_L and rho_R are 3 separate functions. Normally an isomorphism is just a single function. It looks similar though.
@ghkthILAY
3 ай бұрын
@@eigenchris I see, thanks!
@b6kf367
3 ай бұрын
hey chris, which video in the playlist goes through adjoint representation
@eigenchris
3 ай бұрын
I never got around to that topic unfortunately.
@josephmellor7641
3 ай бұрын
@@eigenchris Do you have any resources on it you could recommend?
@b6kf367
3 ай бұрын
@@eigenchris Thought somehow I missed it, anyways keep the great work! your playlist greatly helped me correlate concepts, it was blast to realize my blind calculation of different spin matrices in qm is actually different reps of su(2) algebra (it was hard face palm). Your are doing a great work enlightening, Have a great day!
Пікірлер: 73