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geldklopperij

Stefaan, bedankt voor deze mooi uitleg... stefaan desmedt

rrrrrrr

Accent is terrible and he does not present the equations well so too painful to follow.

That clickbait bought me here

So Einstein was chemist and not a physicist then why devote your whole life to physics or is his chemistry not published.. This probably why we think Newton was so much brighter....

i do like to expres my gratitude to the effort that have been made for explain the dft such this level of easier thanks for everybody who make this project becomes real and the entousiasm to keep the fire of dft community for the next generation

I think the first foundation is that c is constant and the conclusion is that time must be variable in a relativistic world. In our daily real (normal) world (non relativistic) we experience time as "constant". THIS i think - is big thing to understand....and made them search for the Lorentz transformation. Stefaan - Great video and great presentation.

How can we know the error when can't determine the true value of a parameter?

Zeer leuk om te zien

14:00 I think that should read as: ∆Eₜₒₜ(S¹→M¹)=½mv²: time {0} ∆Eₜₒₜ(S¹→Sº)=-E=∆Eₜₒₜ(M¹→Mº): times [0→t] ∆Eₜₒₜ(Sº→Mº)=½mv²: time {t} i.e. The first and last line ∆Eₜₒₜ eqns represent the difference between frames S & M of the total energy (=internal+kinetic), and are for an instant of time, either 0 or t, but middle line eqns are for an interval of time 0 _to_ t. So, the middle line eqns are the only eqns that represent work done by a system, from different reference frames. 16:47 S¹→Sº and M¹→Mº have different time intervals according to Relativity, which leads to: ∆Eₜₒₜ(M¹→Mº)= -γE ≈ -E(1+½β²), Where γ=1/√(1-β²) and β=v/c. According to Relativity, on the rhs: Eₜₒₜ(M¹)= E+∆Eₜₒₜ(S¹→M¹) Eₜₒₜ(Mº)= ∆Eₜₒₜ(Sº→Mº) Thus: ∆Eₜₒₜ(M¹→Mº) =∆Eₜₒₜ(Sº→Mº)-(E+∆Eₜₒₜ(S¹→M¹)) = ∆(∆Eₜₒₜ(S→M))-E & from above: ∆Eₜₒₜ(M¹→Mº)= -γE So, must have: ∆(∆Eₜₒₜ(S→M))=(1-γ)E ∆(½m·v²(S→M))=(1-γ)E ½∆(m)·v²(S→M)=(1-γ)E. …Note(†) ≈(1-(1+½β²+…))E ∆K=∆(∆Eₜₒₜ(S→M))≈ -(½β²+…)E ≈ -½(v²/c²)E ~~~~~~~~~~~~~ Notes: (†) v doesn't change for different time values as it's just a relative velocity => As the energy kinetic energy is changing, the only thing left is the mass, so there must've been a mass drop: ∆m<0. => Internal Energy:E= ½∆m·v²/(1-γ). ~~~~~~~~~~~~~~~~~~~~ This '∆K' above represents the loss of "kinetic energy difference in the system between frames", due to the transition change of energy states. If there is a loss of kinetic energy difference, but the relative velocities are the same, there must be a transfer of mass ∆m<0, such that: ∆K≈ -½(v²/c²)E & ∆K= ½∆m·v² Thus, the mass transfer (away from the system) is: ∆m≈ -E/c². Thus: m(Sº→Mº)-m(S¹→M¹)≈ E/c².

4:00 What about _Ampere's force law_ ? 🤔 (Not the same as Biot-Savart, btw)

at 18:02 he got this one -E( 1 v^2/2c^2) by assuming m1=E/c^2 and this was what we wanted to prove and that the way around to use the result to get the result. he didnt do anything about the doppler shift, this is not a proof.

FINALLY. I have been looking for the true proof of this for almost 2 hours. There are so many dumb videos "proving" E=mc^2 😂 Wanted to know the real proof/motivation. Thanks so much!

Admirable presentation, great intro to the background needee. 10!

grateful for you.

fantastic

The electron-electron repulsion term in the Hamiltonian seems to have the wrong sign. @5:50

And here is just an idea to make the reasoning a bit simpler: It is a bit difficult to follow the logic that both possible ways from top left to bottom right should give the same result. In the same way, in the original paper from Einstein it is a bit difficult to follow the logic of subtracting two equations from one another. Instead here is a more intuitive and therefore simpler concept, I think: The real core of both ways to explain it is that the energy in both frames of references is conserved and therefore the difference of the total energy between the two systems is the same before and after the decay.

Very good presentation, thanks! Did you not forget about the change of momentum when the atom emitted the photon (I think that was the reason why Einstein in his paper made the atom emit 2 photons in the opposite direction)?

Thank you. Very clear and elegant presentation.

Splendid Talk........

It is a GREATE lecture, Thanks a lot prof

Great lecture!

really excellent non-technical insight! Of DFT.

Great Lecture. I have a simple doubt @28:19, when we have considered that electrons are not interacting with each other, why the fourth term in the expression isn't dropped?

Amazing!

Do we have any link for another part of this video?

Simply Amazing!!

I always skipped this because of the thumbnail. RUclips recommended me this again. So, I began to read comment section that brought positive vibes in me... Gonna to watch it 👍

Hello, I want to buy a laptop wich will help me (hope) in my doctorate thesis, could you advise me, is gpu (graphical processor unit) important in the ab initio codes such as Wien2k, quantum esspresso, vasp, castep,.... thanks for any help

Opnieuw erg interessante video, Stefaan. Misschien een idee om de links van 15:50 in de beschrijving te zetten als aanklikbare links? Ik denk dat er nog wel plaats is na "extravagante technologie" :-)

Great video, Prof. Stefaan! The automatic English subtitles from RUclips were of great help!

Note: it takes some time to 'switch on' your streaming capabilities on youtube (24 hours). So if you want to stream today, please activate your youtube streaming yesterday.

Dear professor greetings, nice to see you after long time, am learning a lot from you.

Thank you very much! That was a nice way to explain it :))

Excellent explanation!!!

respect

Explaining a concept from various points of view is always the most helpful yet the most labour intensive for the lecturer. Big Thanks for your efforts, this is an excellent lecture which i guess targets at the exact level of knowledge you identified in the beginning of the video

i need a trnslation of this video help me please

May I link to your video for a grant project that I am doing for Texas A&M University, please? If so, how would you prefer it to be cited?

THIS IS EXACTLY WHAT I NEEEED THANKSSSSSS

This is what I called: DFT in a nutshell. Very well done!

thank you

This was awesome, tnx!

Great explanation, thank you for uploading this online.

Thanks!

Thank You So much for this excellent lecture..!!