Thanks so much!

@@AndrewDotsonvideos did you really just reply an year later

@@KRATOS-dq6sd did you really just reply an eight months later

@@jakwhyld4406 did you really just replay a month later?

@@mevnesldau8408 Did you really just reply a month later?

I agree

Please make video on dirac equation

I concur

yeah.. get on with the dirac eqn, you beardy hipster

Done

I'd like to start at the top left, but my light has a terrible glare around there! That's why I have a black outline in the corner in all of my math videos. So I know where not to write.

It's not too important for videos as we can rewind when needed. As long as you are careful in the classroom you should be okay. :)

i am on a two and a half year journey so far, i am 44...

Carlos C Imho, Klein-Gordon equation actually only work for bosons. Electrons in hidrogen atoms are fermions, so these can be handled by Dirac Equation. May this helps.

Carlos C Yes, it gives the wrong results for those particles because the equation is not applicable for those particles. The Klein-Gorden equation is a boson equation, since, by having a scalar wavefunction, it assumes integer spin.

Hmm. Griffiths uses the squared version. I guess as long as you define what you're saying it means.

Liviu Stoica Box, square... eh, really minor difference

David Tong Quantum Field Theory lecture notes www.damtp.cam.ac.uk/user/tong/qft/one.pdf => box only but both usages are allowed

first time I saw d'Alambertian I thought it was an error in the file lol

@@lPlanetarizado Good one😂😂😂

People there are extremely intelligent. Like sheldon cooper type people. Fine okay, maybe just leonard intelligent. However, I would suspect that to get into that university for physics, you would need an iq of like 160 at least. This guy might be very smart, but he is not a genius. He still probably has an iq from 140 to 150 though.

@@robertmorrison1657 I don't think Caltech judges people by IQ.

@@heavennoes Not exactly. They don't have you take an iq test to see if you will do well in university. What they will do is see your grades, scores on SAT's, and other things that show how smart you are, plus maybe some extracurriculars just to see how driven and committed. So they kind of see what your iq is like, whether it is high enough or not. So they sort of do judge by iq.

It's usually given as \Box which, however, is strange. It is more logical to make the transition from "nabla" to "quabla" (quadri-nabla) and to denote the latter as \Box which for contravariant components is equal to \partial^{\mu}. Then a Minkowski-space scalar product can be written as \Box^2 = \partial^{\mu} \partial_{\mu} which is just the d'Alembertian. I think this is the reason why some authors disagree with the more widespread convention of d'Alembertian = \Box. And I do, too.

@@TheBaluchiterium Thank you for the explanation, as I was only familiar with the \Box notation for the d'Alembertian. Also, I really like how you wrote it with latex code :)

ThePrimebrook same problem here, ive used 200+ pieces of paper to do problems and that is wasteful IMO. I really need a whiteboard, but I don't know which

I feel completely the same

Francisco Russo thx for support :D

Global but not local. You always need to replace derivatives with gauge covariant derivatives if you require local gauge invariance. But yes it does still have a global phase invariance and a corresponding conserved charge

@@AndrewDotsonvideos thanks for your answer!! And such a nice video 💯

It tells you how not-flat the spacetime you're working in is. If it is flat, you get a diagonal of 1's (with some minus signs) which basically just tells you a^2+b^2 = c^2 still holds.

Mohannad Islaieh u are right

The Ultimate Reductionist You are wrong in several aspects. Firstly, the Klein-Gordon equation and the Dirac equation are incorporations of *special* relativity into quantum physics. There is no *general* relativity involved. Secondly, quantum field theories DO *mostly* combine general relativity with quantum physics. In fact, the only real conflict between both theories is that the Einstein field equations are classical field equations, not quantum field equations, but unlike all other field equations that we have worked with, Einstein's field equations in particular cannot be quantized by any methods currently known. The problem lies with quantizing the metric tensor, which yields singularities upon singularities. These are non-regularizable and non-renormalizable.

@@angelmendez-rivera351 "Firstly, the Klein-Gordon equation and the Dirac equation are incorporations of special relativity into quantum physics. " Ah. Oops. Rookie mistake.

@@angelmendez-rivera351 why need to quantize spacetime curvature? what if just let it operate on wave-function continuously?

@@drbeanut You cannot have wavefunctions without quantization.

@@angelmendez-rivera351 How so? A wave-function merely assigns a continuous probability amplitude distribution to some arbitrary field.

Its not an algebraic product, it's an operator manipulation. The linearity of derivatives allows you to get away with notation tricks like that for some things even though its not strictly correct.

x17 Exeventien yes, i saw this in papa flammy videos, it is just using the operator twice

(dy/dx)^2 and d^2y/dx^2 are not the same. What he did was the squaring before applying it to wavefunction. E=d/dt i.e E as an operator which you apply to a function. an operator applied to an operator is an operator. i.e (d/dt)(d/dt)y = d/dt(dy/dt) = d^2y/dt^2 BUT (dy/dt)(dy/dt) is a function on a function ie d/dt = operator dy/dt = function. So grad^2[y]+m(y)+E^2[y] represents d2y/dx2 + my + d2t/dE2 square brackets indicate operator acting on y and regular brackets are functional, or even numeric. The important part is that the operator (the derivative operator in this case is squared not the function). The way I remember it is: if Im a function and you give me a number I can tell you the resulting number. If Im an operator and you give me a function I can give you the resulting function.

Francisco Russo He never wrote (dx/dt)^2. He wrote d^2(φ)/dx^2. They are different.

Yup you’re right, sorry about that

helped*

And for non-SM, there’s the axion pseudoscalar for charge parity violation, all I can think of off the top of my head

Qft is all about that. But you automatically get all integer particle numbers and even their superpositions.

I do not believe it does that if you work with bosons, only with fermions. That is because the equation only works with bosons.

For the schrodinger equation you have the wave function whichs depends on space and on time and moves through space and time seperately, here you get a field that propagates in space-time. So its like the wave function, except now its a field in space-time.

Sebastiaan van Rijk Yes, but he is asking what does the field stand for. The answer is that it is a probability amplitude.

nope you can use - + + + instead if youd like but keep it consistent

@@oni8337 didn't know 2 years ago. I know now XD

It's just a definition. You could define a smiley face operator to equal the right hand side so long as it's consistent with what you're using it for. But, Griffiths also uses the box squared for what its worth.

Andrew Dotson Yeah, I get that. But it’s always convenient to have some kind of a universal notation. It’s like the metric, you can use + - or - + notation. There should be some SI equivalent of such stuff. Anyway, keep it up, Andrew.

Caterina Cevallos I'm pretty sure in this case covariant refers to the form of the equation. Indeed there are no indices in the final form of the KG equation. I don't think "contravariant equation" is a thing

Sorry I'm just seeing this now. When I say Lorentz-covariant, I mean it transforms properly under a Lorentz transformation. The equation will still change, but under the prescription given by the L.T.

@@thephysicistcuber175 maxwells equations in vacuum are covariant, and they have unpaired indices... Covariant in this context means the equation is tensorial with respect to the Poincaré group. In other context covariant and contravariant just refers to tangent vector and cotangent vector components.

@@zoltankurti I wasn't implying that there must not be unpaired indices, I was just saying that in this case the equation is obviously covariant because of the absence of such indices

@@thephysicistcuber175 many not invariant objects have no indices. Pressure, energy density, and I don't need to go on because I'm sure you can list a bunch of these too.

Phi usually represents a complex function that is not a scalar field.

Zoltán Kürti What is your definition of quantum mechanics? Because my understanding of the term is that there is no obvious reason this should not be called quantum mechanics. It can be derived from the axiomatic postulates of a quantum theory, so it is quantum. And, there is an operator called the d'Alembert involved.

d'Alembert's operator is the 4-wavevector operator squared and multiplied by a constant.

Ritesh Bachhar The wavefunction is a scalar field.

Smit Shilpatul Yes, but the Klein Gordon equation is only valid for spin 0 particles, so your point is moot.

Basically, it is just measuring velocities in fractions of c and measuring action (or angular momentum) in hbars.

In what sense is physics simple? It is extremely complex... Even compact equations, which are certainly not the rule, are notation heavy and many are notoriously long even with compact notation. And to be honest, although great mathematical prowess is absolutely necessary, I think the greatest challenge - and also beauty - of physics is gathering a sensible intuitive interpretation of said mathematical jargon to explain the world around you. I really don't understand where the triviality begins.

DeltaX Gamer PT physics is the most rudimentary and least complex of all the scientific disciplines. It’s the pre-school of science. Basic idealism and simple assumptions. Nice geometric forms and convenient system behaviour. Even when you examine Quantum Mechanics, it’s mathematical foundations are simpler than Newtonian Mechanics. I am not trivialising they field of Physics just stating an obvious fact about its nature and the functional limits. The predictions made by theories in physics are orders of magnitude easier to validate than all other scientific fields Ever wondered why? Physicists often take it personally which is the wrong attitude. Their field is a basal fundamental footing for other scientific disciplines to build upon.

@@PetraKann It is that way because it is our own interpretation of the language of our universe. I understand where you are coming with this and agree with most of it. I just don't like how you say it is not a complex field. Its assumptions and implications are far from the whole truth of the theories, which is why it so rare to find someone who can communicate them diligently and in a simple manner. It is simpler in the sense that it is the science where you skip the least logical steps apart from mathematics. It is still, however, one of the hardest sciences to be able to grasp.

Yeah, because it's more difficult to learn a bunch of long and complicated names than learning how to reason.