How to Derive the Most Famous Equation (Unconventionally) | E=mc^2
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- čas přidán 28. 08. 2024
- Tried to make this a short but it was too long. Anyway enjoy!
Sources: cs.stanford.ed...
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You can’t just start with a formula that came from nowhere. You have to derive that equation first.
Yes I know, but I wanted to keep this video short. A longer form video covering all the concepts will be coming soon
Sorry, but you got that wrong. Light definitely has no mass, also not relativistic mass! E=mc^2 is the equation for particles with mass. It is not applied for photons, for photons you have p*c where p depends on the wave length. It is wrong to use m*c for the momentum and it also doesn't help conceptually in my opinion
Sorry if this video didn’t help, I tried my best! If you want to learn more about my derivation, please look at the source for this video in the description. That should explain it better if I didn’t. Thank you for bringing up your concern and I will try to do better in my next videos!
34 years after graduation I still know there is a process (using integration) to get E=mcc in Eisberg & Resnick Appendix A.
It is almost trivial using 4-vectors.
Never thought that i would learn that much from a short, very nice keep it up!
this some great content for my left ear
😂
Amazing short unc I really learned a lot
Nice, this is very much the conventional way. And it is very incomplete. The quantity for the energy squared you show at first should be the one we are deriving. And that requires a course in SR. It is pretty simple though. (really believe me, if you know algebra and perhaps a little bit of calculus and it seems like you do, you can easily understand it.)
Would you not need to use the relativistic momentum formula p = γmv?
The γ in front represents the Lorentz factor, basically how the mass changes because of relativistic effects. I know I said we can just use p=mv for light even though yes you are correct because we are considering extremely high velocities we would need to use the Lorentz factor, but I ignored it because the objects we are considering generally for E=mc^2 are low velocity objects. Hope this answered your question, stay tuned for the longer derivation tho as I will answer your question better.
keen for the in detail one
What do you mean?
@@squarerootofscience1847 you mentioned you were making another video that goes into further detail :)
@@callle ah gotcha I think I read that wrong :)
No!
LOL. Your argument is circular. You have assumed a completely equivalent result to prove the result. That is not how Einstein did it. Neither result was known before Einstein derived them.
Yes I know that’s not how Einstein did it exactly, which I do mention in the video, and I mention that an exact derivation will be uploaded soon. Thanks for the feedback though!
First