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.

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 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

Thank you. Very clear and elegant presentation.

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

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

grateful for you.

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)?

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.

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.

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....

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