Chris @chrism@social.cleverlibre.org

RT @fermatslibrary

Euler was the first person to prove the interchangeability of partial derivatives https://t.co/C4H1sxk0CQ

Install

Just found this article, an excellent read!

"Over the past few years, Mastodon has become the model for a friendlier kind of social network, promising to keep out the hateful or ugly content that proliferates on larger and more centralized networks. Journalists hailed it as “Twitter without Nazis” and for years, it’s generally lived up to that promise. But last week, the social network Gab migrated to Mastodon — and Mastodon’s admins have been forced to deal with the internet’s Nazi problem head-on."

https://www.theverge.com/2019/7/12/20691957/mastodon-decentralized-social-network-gab-migration-fediverse-app-blocking

If you are a lay person and dont understand physics but want to in a fun easy way without math, check this out. Feynman was a wonderful mind and great at explanations:

https://youtu.be/P1ww1IXRfTA

@zeccano

NOTE: for those who cant read the below equations (they aren't rendered) you can follow this link to see: http://mathb.in/38381

Here you go a full derivation from the math equations you started with to the one I provided you several times...

axioms (given by you, agreed by physicist):

\[E = {m}_{rel} \cdot {C}^{2}\]

\[{E}^{2} = {p}^{2} \cdot {C}^{2} + {{m}_{rest}}^{2} \cdot {C}^{4}\]

De Broglie's Equation for the momentum of a photon:

\[p = \frac{h}{\lambda}\]

Formula to convert wavelength \(\lambda\) into a frequency:

\[\lambda = \frac{C}{f}\]

Simplify given axioms:

\[{({m}_{rel} \cdot {C}^{2})}^{2} = {p}^{2} \cdot {C}^{2} + {0}^{2} \cdot {C}^{4}\]

\[{{m}_{rel}}^{2} \cdot {C}^{4} = {p}^{2} \cdot {C}^{2}\]

\[{{m}_{rel}}^{2} = \frac{{p}^{2} \cdot {C}^{2}}{{C}^{4}}\]

\[{{m}_{rel}}^{2} = \frac{{p}^{2}}{{C}^{2}}\]

\[{m}_{rel} = \sqrt{\frac{{p}^{2}}{{C}^{2}}}\]

then use de broglie's equation for the momentum...

\[{m}_{rel} = \sqrt{\frac{{(\frac{h}{\lambda})}^{2}}{{C}^{2}}}\]

\[{m}_{rel} = \sqrt{\frac{\frac{{h}^{2}}{{\lambda}^{2}}}{{C}^{2}}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot {\lambda}^{2}}}\]

Now convert wavelength to frequency...

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot {(\frac{C}{f})}^{2}}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot \frac{C^2}{f^2}}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{\frac{C^4}{f^2}}}\]

\[{m}_{rel} = \sqrt{{h}^{2} \cdot \frac{f^2}{C^4}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2} \cdot f^2}{C^4}}\]

and bam you arrive at the equation for the relativistic mass of a photon.

\[{m}_{rel} = \frac{h \cdot f}{c^2} \]

#physics #relativity #einstein #math #maths #mathematics

@CCoinTradingIdeas