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
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.
Indeed. The two postulates (or foundations, as you call them) are that the laws of physics should be the same in all inertial frames, and that the speed of light is the same in all inertial frames. The former feels 'natural', the latter feels weird. But if you step over this and work out the consequences, the Lorentz transformation is the result of it. And in a second wave -- as is shown in this video -- you arrive at E=mc^2. Both the Lorentz transformation and E=mc^2 have observational consequences which turn out to be as predicted, and this justifies in retrospect the adoption of the weird light postulate. (It does not explain, of course, why someone thought about trying this postulate in the first place.)
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)?
Good observation! Well, in the way the reasoning is presented here, the photon emission is not explicitly considered. It is just stated that an observer at rest sees after the decay zero kinetic energy of the system in its ground state. That is possible only by back-to-back emission, so it is implicitly assumed that this is what has happened. As it doesn't show up elsewhere in the reasoning, it has no further consequences.
Yes, correct, thanks. Maybe it is still worthwhile mentioning that usually the decaying body does not stay at rest and does not keep the kinetic energy at 0, because it typically radiates in a certain direction and therefore picks up momentum from the recoil force.
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.
I agree. But isn't that just what the other two 'more complicated' ways of reasoning try to express? The reasoning in the video does what you say, but avoids the 'technical jargon' in order to be more intuitive for a broader audience (which makes it perhaps a bit less clear for someone who is familiar to this jargon).
@@stefaancottenierlecturesan4803 Thanks for your feedback! I think, the concept of conservation of energy should be easy enough for everyone to follow. Furthermore, if anyone does not know or not understand this concept, then it should be the needed homework to be completed before tackling the theory of relativity. By the way, here is another video that more or less also used your way of reasoning: ua-cam.com/video/hW7DW9NIO9M/v-deo.html
@@aydin74 Exactly, that other video was the one that was mentioned in the description box on UA-cam 😉. Same reasoning, yet so fast it requires watching it 10x and in slow motion for the arguments to sink in. I tried to make a version that is longer, but that can - maybe, hopefully -- be understood by watching it only once.
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.
I don't think I agree with this. The step you describe is when the moving observer wants to determine the energy difference between the excited system and the de-excited system. That can be done only by measuring the frequency of the emitted photon. And because frequency is a number of events per unit of time, it depends on how time proceeds in the frame of the moving observer. Einstein had proven in a previous paper how the Lorentz transformations relate the time in two inertial systems, and it is this result that is used at 18:02 to determine the observed frequency and the energy difference that results from it. As said in the video, we accept this step without deriving it explicitly. But this step can be derived independently, based on the Lorentz transformations only, not using E=mc^2 itself.
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....
??? (Excitations in atoms are relevant both for physicists and chemists. Excitations in the nucleus - another case used in this video - is definitely physics. And excitations in an elementary particle is 100% physics too.)
Thanks for the constructive feedback. Accent is a matter of taste and habit. For those having issues with it, AI-generated (Whisper) subtitles and translations have now been added. The level of detail in which math is treated, depends on the purpose. Emphasis is here on the flow of logic, less on the math. I'm sorry if you started watching this with different expectations.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
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
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.
Indeed. The two postulates (or foundations, as you call them) are that the laws of physics should be the same in all inertial frames, and that the speed of light is the same in all inertial frames. The former feels 'natural', the latter feels weird. But if you step over this and work out the consequences, the Lorentz transformation is the result of it. And in a second wave -- as is shown in this video -- you arrive at E=mc^2. Both the Lorentz transformation and E=mc^2 have observational consequences which turn out to be as predicted, and this justifies in retrospect the adoption of the weird light postulate. (It does not explain, of course, why someone thought about trying this postulate in the first place.)
Thank you. Very clear and elegant presentation.
Stefaan, bedankt voor deze mooi uitleg... stefaan desmedt
4:00 What about _Ampere's force law_ ? 🤔
(Not the same as Biot-Savart, btw)
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)?
Good observation! Well, in the way the reasoning is presented here, the photon emission is not explicitly considered. It is just stated that an observer at rest sees after the decay zero kinetic energy of the system in its ground state. That is possible only by back-to-back emission, so it is implicitly assumed that this is what has happened. As it doesn't show up elsewhere in the reasoning, it has no further consequences.
Yes, correct, thanks. Maybe it is still worthwhile mentioning that usually the decaying body does not stay at rest and does not keep the kinetic energy at 0, because it typically radiates in a certain direction and therefore picks up momentum from the recoil force.
Admirable presentation, great intro to the background needee. 10!
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.
I agree. But isn't that just what the other two 'more complicated' ways of reasoning try to express? The reasoning in the video does what you say, but avoids the 'technical jargon' in order to be more intuitive for a broader audience (which makes it perhaps a bit less clear for someone who is familiar to this jargon).
@@stefaancottenierlecturesan4803 Thanks for your feedback! I think, the concept of conservation of energy should be easy enough for everyone to follow. Furthermore, if anyone does not know or not understand this concept, then it should be the needed homework to be completed before tackling the theory of relativity. By the way, here is another video that more or less also used your way of reasoning: ua-cam.com/video/hW7DW9NIO9M/v-deo.html
@@aydin74 Exactly, that other video was the one that was mentioned in the description box on UA-cam 😉. Same reasoning, yet so fast it requires watching it 10x and in slow motion for the arguments to sink in. I tried to make a version that is longer, but that can - maybe, hopefully -- be understood by watching it only once.
@@stefaancottenierlecturesan4803 oh yes, thanks, I overlooked that you had mentioned the video in your description box, that explains 😊
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.
I don't think I agree with this. The step you describe is when the moving observer wants to determine the energy difference between the excited system and the de-excited system. That can be done only by measuring the frequency of the emitted photon. And because frequency is a number of events per unit of time, it depends on how time proceeds in the frame of the moving observer. Einstein had proven in a previous paper how the Lorentz transformations relate the time in two inertial systems, and it is this result that is used at 18:02 to determine the observed frequency and the energy difference that results from it. As said in the video, we accept this step without deriving it explicitly. But this step can be derived independently, based on the Lorentz transformations only, not using E=mc^2 itself.
He got it from E=hf' and f'=γf, so E=γ·hf
where γ=1/√(1-β²), β=v(rel)/c.
Thank you very much
grateful for you.
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....
??? (Excitations in atoms are relevant both for physicists and chemists. Excitations in the nucleus - another case used in this video - is definitely physics. And excitations in an elementary particle is 100% physics too.)
Accent is terrible and he does not present the equations well so too painful to follow.
Thanks for the constructive feedback. Accent is a matter of taste and habit. For those having issues with it, AI-generated (Whisper) subtitles and translations have now been added. The level of detail in which math is treated, depends on the purpose. Emphasis is here on the flow of logic, less on the math. I'm sorry if you started watching this with different expectations.
rrrrrrr