55:43 - - the soundwave of a certain b*tch. I know about them b*itches. Man your work is phenomenal, it's like a year's course in 2 hours.
@krzysztofciuba27110 күн бұрын
OK not completely: at 17:50nn-the typical textbook nonsense on "time dilation". In the "Moving" system (x',t') these parameters x',t' represents not the values "recorded" by the moving "mythical observer=the set of synchronized clocks) but the ones as "been seen" by the observer in the system at rest! The "moving" observer records the same values of x',t' as the stationary observer,i.e., x,t; otherwise, it would violate the 1st Relativity Principle, then also that the unit time (of a clock) and distance (of a "rigid rod") is not "on2=1 sec,ore else)! Consequently, the case of "muon" (as a clock) is the same textbooks BS: a "muon" is a statistical "being=wave packet"; hence, all these experimental data can only be explained if one treats this "muon" as a "wave"; otherwise, the 1/3 of experiments data cannot be counted for -see the diagram for both radioactive "objects" at rest and "moving": in t>T(1/2-a halftime) the values of both functions are almost the same even graphically! One a better exposition of the Subject but not completely again
@MrFischvogel17 күн бұрын
This is great! I tried to understand Heisenberg's magical paper for so long. This really helps! =)) THANK YOU SO MUCH, SIR
@LightningHelix10121 күн бұрын
That was 🎉
@anantsharma31424 күн бұрын
I am totally in love with your work and content on youtube. You are providing something that many books fail to deliver, developing the subject in a chronological order. I just wanted to ask you that, will you be covering relativistic quantum mechanics and dirac equation anytime soon in future? eagerly waiting for your upcoming videos.
@mohitsinha273224 күн бұрын
You have done great service to the small community of Physics Students who wonder How & where from was so much Linear Algebra & Operator theory was forced upon humanity in the Historical development, esp when we study QMech in the Modern way from J J Sakurai and the like... Thx a tonne!
@vtrandal26 күн бұрын
Your work explaining the historical development of quantum mechanical models if fantastic and greatly appreciated. Thank you!
@maximusideal1525Ай бұрын
This answers so many questions I had about the transition from old quantum mechanics to current QM for so long!
@tilkeshАй бұрын
Thx
@FormalSymmetryАй бұрын
18:56 Is S₃ supposed to have the signs reversed? It looks like it is supposed to be the Pauli Y matrix. EDIT: I should have watched to the end. I see you are already aware of the Pauli matrices.
@bhaskar08Ай бұрын
Finally the wait is over!
@ihmejakki2731Ай бұрын
I just come across your videos again and again and I can't understand how you're not a bigger name on youtube. Super good stuff!
@mre_physvidsАй бұрын
This is outstanding. I have Stat Mech in the fall and when we get to black body radiation, I may take a day off and just run this video.
@felixwaldherr8848Ай бұрын
amazing
@skbhatta1Ай бұрын
I enjoyed the video and thank you very much. I am trying to relearn all the things that I tried to learn in my university days from your great explanations. Thanks
@lepidoptera9337Ай бұрын
Unless you were doing science history in university, the "derivation" won't help you much because it is physically not correct (but Heisenberg would not have been able to know that at the time). The structure of quantum mechanics does not come from this kind of reasoning, even though it dominated the early guesswork. In hindsight we can only say the founders guessed the correct solution but they arrived at it by all the wrong means. Today we know better, but we still teach it wrong. Having said that, certain aspects of Heisenberg's formalism are far closer to the actual facts than e.g. the Schroedinger/von Neumann approach.
@RemyieАй бұрын
Wow finally a video that actually SHOWS how it works, thank you 👍
@declanwk12 ай бұрын
I have always struggled with this topic, thank you for this very clear exposition. One thing that used to confuse me, was I thought that in the Carnot engine, gas had to flow from the hot to the cold reservoir. But from your video (and from considering a toy Stirling engine) it is clear that only heat needs to flow from the hot reservoir into the piston and then from the piston into the cold reservoir (with some of the heat energy converted into work). In many heat engines gas does flow from the hot to the cold reservoir but it is not necessary to consider for this analysis. The same amount of gas can remain in the piston throughout the cycle.
@dag-vidarbauer802 ай бұрын
Great lecture, but why did you remove your cover of anti-hero?😢
@user-kc2iu8df4p2 ай бұрын
Think you so much great work
@Pidrittel2 ай бұрын
Simply fantastic video! Thanky for sharing and well done!
@Pidrittel2 ай бұрын
Thank you so much for uploading all these videos. Some of them contain very unique ways to explain topics optics that you dont find elsewhere (on youtube at least). Your video about abberations is for example one of the cleanest ways to explain and differentiate the different terms I have come across. Very well done and thanks again! I got one question regarding the notation of polarisation (S- or P-Polarization or TE/TM) since you used it in this video in the chapter about fresnel equations: I regularly get confused by the TE / TM notation: It is used in electric engineering when describing modes in waveguides, it is used to describe modes in laser resonators (eg. TEM00).. Could you elaborate when we talk about S/P polarisation and when about TE / TM / TEM, and how to distinguish the terms properly? I have a feeling that TE/TM/TEM mean different things, depending on context. (Edit: in the end its the same question as in physics.stackexchange.com/questions/732802/te-and-tm-modes-vs-polarization )
@brisingreye52092 ай бұрын
@Sander thanks for your great video! I am trying to reproduce your results shown at @30:29 using python. However I am unsure how to do so. I did define a 2D array containing 9 point sources. However I am unsure how to then determine the diffraction patterns as shown in your image. Could you maybe explain your steps taken (or provide a part if your code)?
@brisingreye52092 ай бұрын
Looking a bit closer at the expressions I am a bit confused as how you define the ''h'' in your Delta W. It appears as the magnitude of the vector (from optical axis to the point source). I did use the equation in which we set x=rho*cos(theta) (which then puts the h along the x-axis). So i guess I need to use the polar system instead. However then I still dont understand how you determine the angle theta and the magnitude rho in order to produce your plots.
@SanderKonijnenberg2 ай бұрын
@@brisingreye5209 Phi and rho are the polar coordinates of the pupil, which is in Fourier space. So you'd do the following: - Pick a point source in the object plane with a certain position vector . - Fourier transform the point source (this should give a linear phase ) and truncate it with the pupil/aperture. - Apply the phase error \Delta W in the pupil. rho and phi are the polar coordinates in the pupil (where phi is the angle between and , to ensure rotational symmetry of the system). At the edge of the pupil, rho should equal 1. The strength of this phase error can depend on h, i.e. the farther away the point source from the optical axis, the larger the phase error in the pupil. - (Inverse) Fourier transform the pupil function to obtain the PSF for the point source. - Do this for all point sources in the grid to obtain a figure as shown in the video I explain in more detail how to calculate PSFs with Fourier transforms here: kzread.info/dash/bejne/rGFm1sakepO_qsY.htmlsi=ytQRR-rmCooB1sfO
@brisingreye52092 ай бұрын
@@SanderKonijnenberg First and foremost thanks for your reply! Point 3 isnt clear to me. So if I understand you correctly: 1: define a point source in the object plane: Point = zeros(3000x3000) and then set Point(1000,2000)=1 for example. 2: Next take the FFT: Four_Point = fftshift(fft2(Point)) 3: truncate Four_point with the aperature function -> What exactly do you mean if you say ‘’truncate it with’’? Do you mean to say that I should set all values in the aperture matrix to zero whenever the Four_Point is nonzero? Should the aperture function (say P0) be defined in real space or should it also be the fourier transform? 4: Apply \Delta W: So the result from point 3 (say Matrix ‘’R”): R*exp(^i*k*\Delta W) 5: Inverse Fourier: this is straightforward (and if not, I guess the previous points are the problem for the time being anyways). On a more general note: Am I correct to say that the wavefront error is the additional optical path difference we need to add for rays to pass through the aperture to converge to the ideal image in the image plane? When calculating the PSF using the wavefront error, do we then determine it at the hight of the aperture or the image plane (in other words: If I wish to determine the image on my detector, do I still need to propagate the result. If yes: I guess I can use the FFT(PSF)*h, where h is the propagation function). Thanks a lot!
@SanderKonijnenberg2 ай бұрын
@@brisingreye5209 About point 3: the aperture function is defined in Fourier space. Your quantity 'Four_Point' is a plane wave, so it is non-zero everywhere. It should be truncated by the aperture, i.e. the field should be set to zero outside the aperture. If you (inverse) Fourier transform this truncated plane wave (without introducing the wafevront error \Delta W), you should get an Airy disk whose position depends on the location of the point source in the object plane (see 17:53). The smaller you make your aperture, the larger the Airy disk. The wavefront error is the additional optical pathlength compared to the ideal spherical wavefront. If the spherical wavefront is approximated as quadratic (Fresnel approximation), then it cancels out with the Fresnel propagator when we propagate the field from the pupil plane to the image plane (see 18:27). Therefore, the wavefront error is determined at the height of the aperture plane, and the PSF is determined at the height of the image plane. The propagation from aperture plane to image plane is fully incorporated in the Fourier transform, because the Fresnel propagator cancels with the ideal quadratic wavefront, onto which the wavefront error is added.
@brisingreye52092 ай бұрын
@@SanderKonijnenberg Thanks alot! I got 3 more questions I would like to ask: Question 1: I got it to work somewhat properly, in the sense that it now forms a PSF based on the distance with respect to the optical axis. However it does not correct for the angle it makes with regards to the x-axis. (so the PSF of the upper left corner would be identical to that of the right upper corner and (besides the somewhat smaller h) to those on the x-axis). Question 2: I am somewhat confused by the notion that the aperture is defined in fourier space, although it depends on spatial coordinates. As an experimentalist (trying out python calculations) I know that you can build an 4f system and place an (physical) aperture to block part of the spatial frequencies within the fourier plane. So I wrongly assumed the radius of the aperture used in the code (rho) to correspond to a spatial distance. In actually fact it thus corresponds to the spatial frequencies that are allowed to pass through the aperture. How then are the physical dimensions of the aperture related to those in fourier space (at what hight rho, would you expect a certain frequency k_y)? I am also wondering how it relates to the impulse response of an optical system (as I wrongly assumed them to be one and the same thing). Question 3: Did you make a video (or are you planning to make one) relating the aperture and the wavefront errors to an optical system (say some simple lens system)? I would be really helpful to how they relate and thus how one can effect the image quality by changing/adjusting the optical system? If the answer is no, do you happen to know a great source of information that can help me on my way (once I have successfully reproduced the results as shown in your great video!)
@KarlFredrik2 ай бұрын
Brilliant video! Thanks!
@swag_designs54703 ай бұрын
This is so great I love it
@eamonnsiocain64543 ай бұрын
Well written and produced
@davidhand97213 ай бұрын
Sure, but compatibleism. My judgment here has nothing to do with morality, either. The brain is a physical object which evolves according to all relevant natural laws, so it is deterministic if physics is deterministic. But you _are_ your brain, and your brain is you; this is an observational fact. Your brain determines what you do, and you are your brain, therefore you determine what you do. End of story. That being said, my preference for determinism in physics is based instead on information and the principles of science. If the evolution of the universe dU/dt is not wholly a function of U then it must have additional parameters beyond U. There is literally not enough information in U(t) to calculate U(t+dt). But U is the state of the universe, which can only be defined as everything that exists, so if there's information outside of U, then we clearly aren't really talking about U. Furthermore, the basis of scientific reasoning is that we can infer functions of U by meticulously controlling and observing U itself, i.e. discover natural laws that apply at all t. If U does not include all of its own parameters, then it's no longer reasonable to use the scientific method to draw any general conclusions. The results of our experiments are contaminated with extra information that can never be interrogated out of U. We used the scientific method to discover quantum mechanics, so if this particular non-deterministic aspect of QM is true, then there is no reason to believe that QM is true. I choose to believe that truth can be known and science is valuable. Finally, determinism is restored by the Everett interpretation. You can have a deterministic view of QM so long as you acknowledge that you are a part of the wavefunction yourself, not some kind of magical god-like observer on the outside, beyond the influence of physics. The difficulty there is in fact the same mess that arises in free will when you assume dualism. Yes, of course the laws of physics apply to you, but that doesn't mean you are powerless, and it doesn't mean you have to twist your brain in knots with non-determinism.
@michael-varney-music3 ай бұрын
Interestingly, in matlab2023b if you create the aperture with even number of pixels you do not have a linear phase, but you do with odd number of pixels. Opposite of what this video shows.
@SanderKonijnenberg3 ай бұрын
Yes, I noticed the same thing some time after I had uploaded this video. Oh well...
@paoloberra51414 ай бұрын
Great video!! Really complete. I've seen you have uploaded the pdf slides only for Heisenberg matrixes. Could you also upload the slides of the other videos? It could be useful to study them better. Anyway thanks for your effort! Regards Paolo Berra
@sammyapsel14434 ай бұрын
At 13:09 , why did you ignore the 1/lambda*z term when you calculated the fresnel propagated field? And why didn't you take the abs^2 when presenting it?
@SanderKonijnenberg4 ай бұрын
I ignored the factor 1/lambda*z because it's a global factor that is typically uninteresting since it doesn't change the 'shape' of the field. We typically plot the field amplitude/intensity in arbitrary or normalized units anyway. I didn't calculate the intensity by taking the squared modulus, because then lots of subtle features tend to become much less visible in the plot. So it's basically just a plotting trick (just like you can plot the log to make weak features visible), but you're right that the squared modulus would be the physical quantity that a camera would measure.
@replyasapreplyasap4 ай бұрын
This is great. Thank you!
@jacobvandijk65254 ай бұрын
@ 03:02 The comparison is a bit unfortunate. A classical wave function describes something physical, while the quantum mechanical wave function has no physical meaning. It's just a mathematical tool. @ 12:40 This picture (on the right) is NOT correct. No photon travels back in time. You should have used the picture on the left and changed the z-axis into a time-axis. Then, we're doing physics again. Read about "time-slicing", Sander.
@jacobvandijk65254 ай бұрын
@ 17:42 This is NOT a source-term. A real source-term is like the driving-force for a harmonic oscillator. But this "source-term" is the potential energy of that oscillator. There is no driving-force in this explanation; like there is no driving-force in QFT's. Quantum fluctuations have no physical explanation, only a mathematical one (the HUP). The famous Big Bang Theory has no starting point either; it is only an evolution theory. We have no clue where the initial impuls for The Big Bang or for quantum fluctuations comes from; IF these theories make any sense at all.
@jacobvandijk65254 ай бұрын
The name Snell (with double L) is an anglicisation of the Dutch name Snel (with single L). Anglicisation is like colonisation. You want to impose your own culture on everything :-(
@jacobvandijk65254 ай бұрын
@ 0:00 Here, he leaves out the QUANTUM MODEL of light. In video 9. he corrects this.
@pavelrozsypal89564 ай бұрын
Systematic, beautiful and historically correct exposition ! Such a difference from many other "trivialized" explanations appearing even in respected textbooks crudely distorting actual historical development of physical ideas of great physical science founders. All your videos are deep, illuminating and very interesting. Great work !
@jacobvandijk65255 ай бұрын
@ 0:55 Here, in H's Principle, the source of light is called a field too. One could call this starting field U(0,0). @ 1:40 If you also put in U(0,0), then you have an expression for "the propagator" from QFT. Here, U(0,0) = 1.
@malikjavadov3665 ай бұрын
great work!
@marsyeticjj5 ай бұрын
Great work! Helps me a lot in understanding aberrations. I have one question though, that you mentioned multiple times that in an ideal optical system, rays from a point converges to a point. But in wavefront aberration calculation, how does this starting point determined? will the starting point'l location affect the wavefront aberration we get? And how do we determine the shape of the idea optical system, when we don't have one design, say in corneal wavefront aberration analysis? Looking forward to your reply!
@SanderKonijnenberg4 ай бұрын
Indeed, in general the wavefront error can be different for different object points. The Seidel aberrations for example express a dependence on the object point location. However, there are many cases where the wavefront error can be approximated to be independent of the object point location (i.e. the aberrations are isoplanatic), for example in imaging systems with a low numerical aperture. Unfortunately, I have no particular expertise in corneal wavefront analysis. But if I had to guess, I would try to parametrize the shape of the cornea, measure the wavefront error for various wavelengths, and then try to fit the parameters to the measured data?
@JX-ow8vq5 ай бұрын
excellent works!!!
@user-jq5ut4jg5d5 ай бұрын
Can you share the MATLAB code?
@SanderKonijnenberg5 ай бұрын
The Google drive link in the video description also contains a zip-file with the Matlab codes. Hopefully those include what you're looking for.
@McRingil5 ай бұрын
Wow, incredible work, never seen put eberything in such an order.
@joeboxter36355 ай бұрын
@27:28 the simplified form of clausis inequality may have an error. You have used Th and Tc and so that is the same as the original equation where you got a positive result. However, you get a negative result the second time and say that you are not considering the Temperature of the reservoirs as before. But the equation says otherwise. You say you are considering the temperature of the surrounding. And indeed that is what you say in the integral equation. I believe what you mean is Qin/Ts -+ Qout/Ts < 0, not what you have Qin/Th + Qout/Tc. Because the Work is 0. Qin/Ts is > 0. But Qout/Ts is < 0 and Qout < Qin which means it has to be negative. As oppose to the original Qin/Th + Qout/Tc is > 0 because Th > Tc and Qin and Qout are opposite. And by convention Qin is positive and Qout is negative. It's the denominator Tc > Th that makes the whole thing positive. Excellent video, though. I don't think I've ever seen such a comprehensive treatment. You almost sound more like a chemist or mathematician and not a physicist. For whatever reason, I find physicists tend to be more sloppy with the book keeping than are chemists and mathematicians.
@SanderKonijnenberg5 ай бұрын
Thank you very much for this comment. I hope to learn from it, but I'm afraid I don't fully understand it yet at the present moment. - '[You] say that you are not considering the Temperature of the reservoirs as before. But the equation says otherwise.' Is this referring to what I say at 27:03? My point there was that in the equation, we don't consider the *entropy* change of the reservoirs, because the signs of Q_in and Q_out are defined with respect to the engine. I.e. a positive Q_in would imply a *negative* change in entropy in the reservoir, but the term in the equation would be *positive* ,so the equation does not express the entropy change in the reservoir. And since the equation does not express the change in total entropy (i.e. of both the engine and the reservoirs), it does not contradict the Second Law of Thermodynamics. - 'It's the denominator Tc > Th that makes the whole thing positive.' I'm not sure I follow. Should it not be Tc<Th, since the temperature of the cold reservoir is lower than that of the hot reservoir? The way I see it is the following: 1) |Q_in|=|Q_out| because W=0. 2) Q_in is positive, Q_out is negative (because they are defined with respect to the engine). 3) Because Th>Tc, the negative term Q_out/Tc dominates the positive term Q_in/Th, making their sum negative. Could you perhaps indicate in which of these three claims you find something objectionable?
@asiaasia44125 ай бұрын
Thats amazing! Thank you, great lecture, helped me to better understand holography princples
@ribamarsantarosa44655 ай бұрын
Oh no I'm afraid I run through all the playlist :( Gonna miss these explanations!!
@ribamarsantarosa44656 ай бұрын
Hele fijne structure, snel, zonder muziek etc bedankt.
@PaulLebow6 ай бұрын
Lectures are so helpful. Would be great to see treatment of phase-only holograms such as produced by a spatial light modulator - "kinoform", I believe. Would like to projected a "resolved" intensity pattern at intermediate ranges. Can one back propagate to determine the need phase on the SLM or is too much information missing?
@SanderKonijnenberg6 ай бұрын
When performing Fourier transforms, phase information is typically much more important than amplitude information. A common demonstration is the following: take some image or piece of music, Fourier transform it, discard amplitude information (so only keep phase), and inverse Fourier transform it. The resulting image or piece of music is still well recognizable, though somewhat degraded in quality. Since the propagation of optical fields is described by Fourier transforms (i.e. Angular Spectrum Method, Fresnel integral, Fraunhofer integral; I have another video on that kzread.info/dash/bejne/i2l4w8mGkrOrd84.html ) , the same result holds: if you back-propagate a field, only keep its phase, and forward-propagate it, then the original field should be recognizable, though there will likely be additional speckle noise present. To minimize the noise, one could use iterative algorithms to optimize the phase-only hologram, see e.g. doi.org/10.1016/j.optcom.2018.09.076 (used search key words: beam shaping Gerchberg-Saxton).
@PaulLebow6 ай бұрын
@@SanderKonijnenbergThanks. Yes seems like an iterative method would be needed. Will have to ping on a colleague who was Feinig's grad student. Currently we project a pattern in the far field by imposing the FFT of a random pattern on our SLM. It does a pretty good job, except for the resulting random speckle in the far field.
@intuitivelyrigorous6 ай бұрын
27:26 is little confusing to me. What i understand is first you defined entropy as [ heat given(+) or heat taken(-) to system] / T . Which always increases as you showed from 2nd law assumption. Then you said "in clausius inequality we are calculating entropy of engine not reservoir moreover we are dividing it by temperature of reservoir, hence it does not contradicts that entropy increases " In this statement their are lot of holes in my mind, the way you said it ... I feel like it means that in clausius inequality we are not measuring the entropy at all since dividing given heat to engine by temperature of reservoir is not how you defined entropy and hence it does not contradicts 2nd law. But another teacher actually defined entropy and derived 2nd law using clausius inequality infact what your 2nd law states is what actually he referred to as the clausius inequality. I know different teacher call things with different names. But the confusion from 27:26 remains in my mind and it's unclear what you actually meant. Thanks. Very insightful video.
@SanderKonijnenberg6 ай бұрын
Perhaps the following quote helps (from hyperphysics.phy-astr.gsu.edu/hbase/thermo/clausius.html ): 'The Clausius Inequality applies to any real engine cycle and implies a negative change in entropy on the cycle. That is, the entropy given to the environment during the cycle is larger than the entropy transferred to the engine by heat from the hot reservoir.' In other words, Clausius' inequality does not state that the total entropy (i.e. of both engine and the two reservoirs combined) decreases, because that would contradict the Second Law of Thermodynamics. Evaluating the *total* change in entropy would take into account the change in entropies of both the engines and the two reservoirs. But what is calculated here is the negative of the entropy transferred from the engine to the reservoirs (so the fact that it's equal to or less than 0 indicates that the entropy of the two reservoirs combined remains constant or increases).
@intuitivelyrigorous6 ай бұрын
@@SanderKonijnenberg Okay now i think it makes sense with what I was taught previously is similar to following integrals in worded form : (entropy given to engine) + (entropy taken out of engine) < 0 <=> (the entropy given to engine)-(entropy given to reservoir) < 0 <=> (the entropy given to engine) < (entropy given to reservoir) <=> ( Considering reservoir as surrounding, engine as system and processes in surrounding to be reversible. Equality will hold true if Q/Tsurr is also reversible) We can write Q/Tsurr < ∆S Which is our second law.
@tengstone17836 ай бұрын
Great work! Learn a lot from this video. Thanks!!
@arturocatalanogonzaga78746 ай бұрын
Hi! Thanks a lot for the extremely helpful videos! Could you please clarify how and why, at minute 21:16, you went from having an Object in the (xo,yo) object plane to the (x/M,y/M) plane? why is it the lens plane that gets demagnified? shouldn't it be (xo/M,yo/M)?
@SanderKonijnenberg6 ай бұрын
My apologies for the confusion: in this slide, (x,y) denotes the image plane coordinates (not the lens plane). If we ignore the effects of the PSF (i.e. assume that it's a delta peak), then the image I(x,y) should be a magnified copy of |O(x_o,y_o)|^2. That is, I(x,y)=|O(x/M,y/M)|^2, so that the object field at object coordinates (x_o,y_o) is mapped to image coordinates (x,y)=(M*x_o,M*y_o).
@arturocatalanogonzaga78745 ай бұрын
@@SanderKonijnenberg thank you so much for the reply!
@MarcoBarbato6 ай бұрын
Fantastic lecture❤ I was trying to decode original Heisenberg paper, but now I understand that I miss some other element.
@SanderKonijnenberg6 ай бұрын
Thanks for the comment, I've made the slides and transcript available here: drive.google.com/drive/folders/1TFzYibPY8t9y4jH84Ifq-l-3NJjB_cHt?usp=drive_link
Пікірлер
55:43 - - the soundwave of a certain b*tch. I know about them b*itches. Man your work is phenomenal, it's like a year's course in 2 hours.
OK not completely: at 17:50nn-the typical textbook nonsense on "time dilation". In the "Moving" system (x',t') these parameters x',t' represents not the values "recorded" by the moving "mythical observer=the set of synchronized clocks) but the ones as "been seen" by the observer in the system at rest! The "moving" observer records the same values of x',t' as the stationary observer,i.e., x,t; otherwise, it would violate the 1st Relativity Principle, then also that the unit time (of a clock) and distance (of a "rigid rod") is not "on2=1 sec,ore else)! Consequently, the case of "muon" (as a clock) is the same textbooks BS: a "muon" is a statistical "being=wave packet"; hence, all these experimental data can only be explained if one treats this "muon" as a "wave"; otherwise, the 1/3 of experiments data cannot be counted for -see the diagram for both radioactive "objects" at rest and "moving": in t>T(1/2-a halftime) the values of both functions are almost the same even graphically! One a better exposition of the Subject but not completely again
This is great! I tried to understand Heisenberg's magical paper for so long. This really helps! =)) THANK YOU SO MUCH, SIR
That was 🎉
I am totally in love with your work and content on youtube. You are providing something that many books fail to deliver, developing the subject in a chronological order. I just wanted to ask you that, will you be covering relativistic quantum mechanics and dirac equation anytime soon in future? eagerly waiting for your upcoming videos.
You have done great service to the small community of Physics Students who wonder How & where from was so much Linear Algebra & Operator theory was forced upon humanity in the Historical development, esp when we study QMech in the Modern way from J J Sakurai and the like... Thx a tonne!
Your work explaining the historical development of quantum mechanical models if fantastic and greatly appreciated. Thank you!
This answers so many questions I had about the transition from old quantum mechanics to current QM for so long!
Thx
18:56 Is S₃ supposed to have the signs reversed? It looks like it is supposed to be the Pauli Y matrix. EDIT: I should have watched to the end. I see you are already aware of the Pauli matrices.
Finally the wait is over!
I just come across your videos again and again and I can't understand how you're not a bigger name on youtube. Super good stuff!
This is outstanding. I have Stat Mech in the fall and when we get to black body radiation, I may take a day off and just run this video.
amazing
I enjoyed the video and thank you very much. I am trying to relearn all the things that I tried to learn in my university days from your great explanations. Thanks
Unless you were doing science history in university, the "derivation" won't help you much because it is physically not correct (but Heisenberg would not have been able to know that at the time). The structure of quantum mechanics does not come from this kind of reasoning, even though it dominated the early guesswork. In hindsight we can only say the founders guessed the correct solution but they arrived at it by all the wrong means. Today we know better, but we still teach it wrong. Having said that, certain aspects of Heisenberg's formalism are far closer to the actual facts than e.g. the Schroedinger/von Neumann approach.
Wow finally a video that actually SHOWS how it works, thank you 👍
I have always struggled with this topic, thank you for this very clear exposition. One thing that used to confuse me, was I thought that in the Carnot engine, gas had to flow from the hot to the cold reservoir. But from your video (and from considering a toy Stirling engine) it is clear that only heat needs to flow from the hot reservoir into the piston and then from the piston into the cold reservoir (with some of the heat energy converted into work). In many heat engines gas does flow from the hot to the cold reservoir but it is not necessary to consider for this analysis. The same amount of gas can remain in the piston throughout the cycle.
Great lecture, but why did you remove your cover of anti-hero?😢
Think you so much great work
Simply fantastic video! Thanky for sharing and well done!
Thank you so much for uploading all these videos. Some of them contain very unique ways to explain topics optics that you dont find elsewhere (on youtube at least). Your video about abberations is for example one of the cleanest ways to explain and differentiate the different terms I have come across. Very well done and thanks again! I got one question regarding the notation of polarisation (S- or P-Polarization or TE/TM) since you used it in this video in the chapter about fresnel equations: I regularly get confused by the TE / TM notation: It is used in electric engineering when describing modes in waveguides, it is used to describe modes in laser resonators (eg. TEM00).. Could you elaborate when we talk about S/P polarisation and when about TE / TM / TEM, and how to distinguish the terms properly? I have a feeling that TE/TM/TEM mean different things, depending on context. (Edit: in the end its the same question as in physics.stackexchange.com/questions/732802/te-and-tm-modes-vs-polarization )
@Sander thanks for your great video! I am trying to reproduce your results shown at @30:29 using python. However I am unsure how to do so. I did define a 2D array containing 9 point sources. However I am unsure how to then determine the diffraction patterns as shown in your image. Could you maybe explain your steps taken (or provide a part if your code)?
Looking a bit closer at the expressions I am a bit confused as how you define the ''h'' in your Delta W. It appears as the magnitude of the vector (from optical axis to the point source). I did use the equation in which we set x=rho*cos(theta) (which then puts the h along the x-axis). So i guess I need to use the polar system instead. However then I still dont understand how you determine the angle theta and the magnitude rho in order to produce your plots.
@@brisingreye5209 Phi and rho are the polar coordinates of the pupil, which is in Fourier space. So you'd do the following: - Pick a point source in the object plane with a certain position vector . - Fourier transform the point source (this should give a linear phase ) and truncate it with the pupil/aperture. - Apply the phase error \Delta W in the pupil. rho and phi are the polar coordinates in the pupil (where phi is the angle between and , to ensure rotational symmetry of the system). At the edge of the pupil, rho should equal 1. The strength of this phase error can depend on h, i.e. the farther away the point source from the optical axis, the larger the phase error in the pupil. - (Inverse) Fourier transform the pupil function to obtain the PSF for the point source. - Do this for all point sources in the grid to obtain a figure as shown in the video I explain in more detail how to calculate PSFs with Fourier transforms here: kzread.info/dash/bejne/rGFm1sakepO_qsY.htmlsi=ytQRR-rmCooB1sfO
@@SanderKonijnenberg First and foremost thanks for your reply! Point 3 isnt clear to me. So if I understand you correctly: 1: define a point source in the object plane: Point = zeros(3000x3000) and then set Point(1000,2000)=1 for example. 2: Next take the FFT: Four_Point = fftshift(fft2(Point)) 3: truncate Four_point with the aperature function -> What exactly do you mean if you say ‘’truncate it with’’? Do you mean to say that I should set all values in the aperture matrix to zero whenever the Four_Point is nonzero? Should the aperture function (say P0) be defined in real space or should it also be the fourier transform? 4: Apply \Delta W: So the result from point 3 (say Matrix ‘’R”): R*exp(^i*k*\Delta W) 5: Inverse Fourier: this is straightforward (and if not, I guess the previous points are the problem for the time being anyways). On a more general note: Am I correct to say that the wavefront error is the additional optical path difference we need to add for rays to pass through the aperture to converge to the ideal image in the image plane? When calculating the PSF using the wavefront error, do we then determine it at the hight of the aperture or the image plane (in other words: If I wish to determine the image on my detector, do I still need to propagate the result. If yes: I guess I can use the FFT(PSF)*h, where h is the propagation function). Thanks a lot!
@@brisingreye5209 About point 3: the aperture function is defined in Fourier space. Your quantity 'Four_Point' is a plane wave, so it is non-zero everywhere. It should be truncated by the aperture, i.e. the field should be set to zero outside the aperture. If you (inverse) Fourier transform this truncated plane wave (without introducing the wafevront error \Delta W), you should get an Airy disk whose position depends on the location of the point source in the object plane (see 17:53). The smaller you make your aperture, the larger the Airy disk. The wavefront error is the additional optical pathlength compared to the ideal spherical wavefront. If the spherical wavefront is approximated as quadratic (Fresnel approximation), then it cancels out with the Fresnel propagator when we propagate the field from the pupil plane to the image plane (see 18:27). Therefore, the wavefront error is determined at the height of the aperture plane, and the PSF is determined at the height of the image plane. The propagation from aperture plane to image plane is fully incorporated in the Fourier transform, because the Fresnel propagator cancels with the ideal quadratic wavefront, onto which the wavefront error is added.
@@SanderKonijnenberg Thanks alot! I got 3 more questions I would like to ask: Question 1: I got it to work somewhat properly, in the sense that it now forms a PSF based on the distance with respect to the optical axis. However it does not correct for the angle it makes with regards to the x-axis. (so the PSF of the upper left corner would be identical to that of the right upper corner and (besides the somewhat smaller h) to those on the x-axis). Question 2: I am somewhat confused by the notion that the aperture is defined in fourier space, although it depends on spatial coordinates. As an experimentalist (trying out python calculations) I know that you can build an 4f system and place an (physical) aperture to block part of the spatial frequencies within the fourier plane. So I wrongly assumed the radius of the aperture used in the code (rho) to correspond to a spatial distance. In actually fact it thus corresponds to the spatial frequencies that are allowed to pass through the aperture. How then are the physical dimensions of the aperture related to those in fourier space (at what hight rho, would you expect a certain frequency k_y)? I am also wondering how it relates to the impulse response of an optical system (as I wrongly assumed them to be one and the same thing). Question 3: Did you make a video (or are you planning to make one) relating the aperture and the wavefront errors to an optical system (say some simple lens system)? I would be really helpful to how they relate and thus how one can effect the image quality by changing/adjusting the optical system? If the answer is no, do you happen to know a great source of information that can help me on my way (once I have successfully reproduced the results as shown in your great video!)
Brilliant video! Thanks!
This is so great I love it
Well written and produced
Sure, but compatibleism. My judgment here has nothing to do with morality, either. The brain is a physical object which evolves according to all relevant natural laws, so it is deterministic if physics is deterministic. But you _are_ your brain, and your brain is you; this is an observational fact. Your brain determines what you do, and you are your brain, therefore you determine what you do. End of story. That being said, my preference for determinism in physics is based instead on information and the principles of science. If the evolution of the universe dU/dt is not wholly a function of U then it must have additional parameters beyond U. There is literally not enough information in U(t) to calculate U(t+dt). But U is the state of the universe, which can only be defined as everything that exists, so if there's information outside of U, then we clearly aren't really talking about U. Furthermore, the basis of scientific reasoning is that we can infer functions of U by meticulously controlling and observing U itself, i.e. discover natural laws that apply at all t. If U does not include all of its own parameters, then it's no longer reasonable to use the scientific method to draw any general conclusions. The results of our experiments are contaminated with extra information that can never be interrogated out of U. We used the scientific method to discover quantum mechanics, so if this particular non-deterministic aspect of QM is true, then there is no reason to believe that QM is true. I choose to believe that truth can be known and science is valuable. Finally, determinism is restored by the Everett interpretation. You can have a deterministic view of QM so long as you acknowledge that you are a part of the wavefunction yourself, not some kind of magical god-like observer on the outside, beyond the influence of physics. The difficulty there is in fact the same mess that arises in free will when you assume dualism. Yes, of course the laws of physics apply to you, but that doesn't mean you are powerless, and it doesn't mean you have to twist your brain in knots with non-determinism.
Interestingly, in matlab2023b if you create the aperture with even number of pixels you do not have a linear phase, but you do with odd number of pixels. Opposite of what this video shows.
Yes, I noticed the same thing some time after I had uploaded this video. Oh well...
Great video!! Really complete. I've seen you have uploaded the pdf slides only for Heisenberg matrixes. Could you also upload the slides of the other videos? It could be useful to study them better. Anyway thanks for your effort! Regards Paolo Berra
At 13:09 , why did you ignore the 1/lambda*z term when you calculated the fresnel propagated field? And why didn't you take the abs^2 when presenting it?
I ignored the factor 1/lambda*z because it's a global factor that is typically uninteresting since it doesn't change the 'shape' of the field. We typically plot the field amplitude/intensity in arbitrary or normalized units anyway. I didn't calculate the intensity by taking the squared modulus, because then lots of subtle features tend to become much less visible in the plot. So it's basically just a plotting trick (just like you can plot the log to make weak features visible), but you're right that the squared modulus would be the physical quantity that a camera would measure.
This is great. Thank you!
@ 03:02 The comparison is a bit unfortunate. A classical wave function describes something physical, while the quantum mechanical wave function has no physical meaning. It's just a mathematical tool. @ 12:40 This picture (on the right) is NOT correct. No photon travels back in time. You should have used the picture on the left and changed the z-axis into a time-axis. Then, we're doing physics again. Read about "time-slicing", Sander.
@ 17:42 This is NOT a source-term. A real source-term is like the driving-force for a harmonic oscillator. But this "source-term" is the potential energy of that oscillator. There is no driving-force in this explanation; like there is no driving-force in QFT's. Quantum fluctuations have no physical explanation, only a mathematical one (the HUP). The famous Big Bang Theory has no starting point either; it is only an evolution theory. We have no clue where the initial impuls for The Big Bang or for quantum fluctuations comes from; IF these theories make any sense at all.
The name Snell (with double L) is an anglicisation of the Dutch name Snel (with single L). Anglicisation is like colonisation. You want to impose your own culture on everything :-(
@ 0:00 Here, he leaves out the QUANTUM MODEL of light. In video 9. he corrects this.
Systematic, beautiful and historically correct exposition ! Such a difference from many other "trivialized" explanations appearing even in respected textbooks crudely distorting actual historical development of physical ideas of great physical science founders. All your videos are deep, illuminating and very interesting. Great work !
@ 0:55 Here, in H's Principle, the source of light is called a field too. One could call this starting field U(0,0). @ 1:40 If you also put in U(0,0), then you have an expression for "the propagator" from QFT. Here, U(0,0) = 1.
great work!
Great work! Helps me a lot in understanding aberrations. I have one question though, that you mentioned multiple times that in an ideal optical system, rays from a point converges to a point. But in wavefront aberration calculation, how does this starting point determined? will the starting point'l location affect the wavefront aberration we get? And how do we determine the shape of the idea optical system, when we don't have one design, say in corneal wavefront aberration analysis? Looking forward to your reply!
Indeed, in general the wavefront error can be different for different object points. The Seidel aberrations for example express a dependence on the object point location. However, there are many cases where the wavefront error can be approximated to be independent of the object point location (i.e. the aberrations are isoplanatic), for example in imaging systems with a low numerical aperture. Unfortunately, I have no particular expertise in corneal wavefront analysis. But if I had to guess, I would try to parametrize the shape of the cornea, measure the wavefront error for various wavelengths, and then try to fit the parameters to the measured data?
excellent works!!!
Can you share the MATLAB code?
The Google drive link in the video description also contains a zip-file with the Matlab codes. Hopefully those include what you're looking for.
Wow, incredible work, never seen put eberything in such an order.
@27:28 the simplified form of clausis inequality may have an error. You have used Th and Tc and so that is the same as the original equation where you got a positive result. However, you get a negative result the second time and say that you are not considering the Temperature of the reservoirs as before. But the equation says otherwise. You say you are considering the temperature of the surrounding. And indeed that is what you say in the integral equation. I believe what you mean is Qin/Ts -+ Qout/Ts < 0, not what you have Qin/Th + Qout/Tc. Because the Work is 0. Qin/Ts is > 0. But Qout/Ts is < 0 and Qout < Qin which means it has to be negative. As oppose to the original Qin/Th + Qout/Tc is > 0 because Th > Tc and Qin and Qout are opposite. And by convention Qin is positive and Qout is negative. It's the denominator Tc > Th that makes the whole thing positive. Excellent video, though. I don't think I've ever seen such a comprehensive treatment. You almost sound more like a chemist or mathematician and not a physicist. For whatever reason, I find physicists tend to be more sloppy with the book keeping than are chemists and mathematicians.
Thank you very much for this comment. I hope to learn from it, but I'm afraid I don't fully understand it yet at the present moment. - '[You] say that you are not considering the Temperature of the reservoirs as before. But the equation says otherwise.' Is this referring to what I say at 27:03? My point there was that in the equation, we don't consider the *entropy* change of the reservoirs, because the signs of Q_in and Q_out are defined with respect to the engine. I.e. a positive Q_in would imply a *negative* change in entropy in the reservoir, but the term in the equation would be *positive* ,so the equation does not express the entropy change in the reservoir. And since the equation does not express the change in total entropy (i.e. of both the engine and the reservoirs), it does not contradict the Second Law of Thermodynamics. - 'It's the denominator Tc > Th that makes the whole thing positive.' I'm not sure I follow. Should it not be Tc<Th, since the temperature of the cold reservoir is lower than that of the hot reservoir? The way I see it is the following: 1) |Q_in|=|Q_out| because W=0. 2) Q_in is positive, Q_out is negative (because they are defined with respect to the engine). 3) Because Th>Tc, the negative term Q_out/Tc dominates the positive term Q_in/Th, making their sum negative. Could you perhaps indicate in which of these three claims you find something objectionable?
Thats amazing! Thank you, great lecture, helped me to better understand holography princples
Oh no I'm afraid I run through all the playlist :( Gonna miss these explanations!!
Hele fijne structure, snel, zonder muziek etc bedankt.
Lectures are so helpful. Would be great to see treatment of phase-only holograms such as produced by a spatial light modulator - "kinoform", I believe. Would like to projected a "resolved" intensity pattern at intermediate ranges. Can one back propagate to determine the need phase on the SLM or is too much information missing?
When performing Fourier transforms, phase information is typically much more important than amplitude information. A common demonstration is the following: take some image or piece of music, Fourier transform it, discard amplitude information (so only keep phase), and inverse Fourier transform it. The resulting image or piece of music is still well recognizable, though somewhat degraded in quality. Since the propagation of optical fields is described by Fourier transforms (i.e. Angular Spectrum Method, Fresnel integral, Fraunhofer integral; I have another video on that kzread.info/dash/bejne/i2l4w8mGkrOrd84.html ) , the same result holds: if you back-propagate a field, only keep its phase, and forward-propagate it, then the original field should be recognizable, though there will likely be additional speckle noise present. To minimize the noise, one could use iterative algorithms to optimize the phase-only hologram, see e.g. doi.org/10.1016/j.optcom.2018.09.076 (used search key words: beam shaping Gerchberg-Saxton).
@@SanderKonijnenbergThanks. Yes seems like an iterative method would be needed. Will have to ping on a colleague who was Feinig's grad student. Currently we project a pattern in the far field by imposing the FFT of a random pattern on our SLM. It does a pretty good job, except for the resulting random speckle in the far field.
27:26 is little confusing to me. What i understand is first you defined entropy as [ heat given(+) or heat taken(-) to system] / T . Which always increases as you showed from 2nd law assumption. Then you said "in clausius inequality we are calculating entropy of engine not reservoir moreover we are dividing it by temperature of reservoir, hence it does not contradicts that entropy increases " In this statement their are lot of holes in my mind, the way you said it ... I feel like it means that in clausius inequality we are not measuring the entropy at all since dividing given heat to engine by temperature of reservoir is not how you defined entropy and hence it does not contradicts 2nd law. But another teacher actually defined entropy and derived 2nd law using clausius inequality infact what your 2nd law states is what actually he referred to as the clausius inequality. I know different teacher call things with different names. But the confusion from 27:26 remains in my mind and it's unclear what you actually meant. Thanks. Very insightful video.
Perhaps the following quote helps (from hyperphysics.phy-astr.gsu.edu/hbase/thermo/clausius.html ): 'The Clausius Inequality applies to any real engine cycle and implies a negative change in entropy on the cycle. That is, the entropy given to the environment during the cycle is larger than the entropy transferred to the engine by heat from the hot reservoir.' In other words, Clausius' inequality does not state that the total entropy (i.e. of both engine and the two reservoirs combined) decreases, because that would contradict the Second Law of Thermodynamics. Evaluating the *total* change in entropy would take into account the change in entropies of both the engines and the two reservoirs. But what is calculated here is the negative of the entropy transferred from the engine to the reservoirs (so the fact that it's equal to or less than 0 indicates that the entropy of the two reservoirs combined remains constant or increases).
@@SanderKonijnenberg Okay now i think it makes sense with what I was taught previously is similar to following integrals in worded form : (entropy given to engine) + (entropy taken out of engine) < 0 <=> (the entropy given to engine)-(entropy given to reservoir) < 0 <=> (the entropy given to engine) < (entropy given to reservoir) <=> ( Considering reservoir as surrounding, engine as system and processes in surrounding to be reversible. Equality will hold true if Q/Tsurr is also reversible) We can write Q/Tsurr < ∆S Which is our second law.
Great work! Learn a lot from this video. Thanks!!
Hi! Thanks a lot for the extremely helpful videos! Could you please clarify how and why, at minute 21:16, you went from having an Object in the (xo,yo) object plane to the (x/M,y/M) plane? why is it the lens plane that gets demagnified? shouldn't it be (xo/M,yo/M)?
My apologies for the confusion: in this slide, (x,y) denotes the image plane coordinates (not the lens plane). If we ignore the effects of the PSF (i.e. assume that it's a delta peak), then the image I(x,y) should be a magnified copy of |O(x_o,y_o)|^2. That is, I(x,y)=|O(x/M,y/M)|^2, so that the object field at object coordinates (x_o,y_o) is mapped to image coordinates (x,y)=(M*x_o,M*y_o).
@@SanderKonijnenberg thank you so much for the reply!
Fantastic lecture❤ I was trying to decode original Heisenberg paper, but now I understand that I miss some other element.
Thanks for the comment, I've made the slides and transcript available here: drive.google.com/drive/folders/1TFzYibPY8t9y4jH84Ifq-l-3NJjB_cHt?usp=drive_link